Feedback Control of Dynamic Systems 7th Franklin Chegg Solutions

Problem 1.01PP Draw a component block diagram for each of the following feedback control systems, (a) The manual steering system of an automobile (b) Drebbel’s incubator (c) The water level controlled by a float and valve (d) Watt’s steam engine with fly-bail governor In each case, indicate the location of the elements listed below and give the units associated with each signal. In each case, indicate the location of the elements listed below and give the units associated with each signal. • The process • The process desired output signal • The sensor • The actuator • The actuator output signal • The controller • The controller output signal • The reference signal • The error signal Notice that in a number of cases the same physical device may perform more than one of these functions. Step-by-step solution step 1 of 4 (A). Manual Steering Systems Step 2 of 4 (B) Step 3 of 4 (c) Set water leveKRef Input) (CorMleror ■ Actuator o/p) Water Ftootvatve Tank Water level Floating (process) Of float (Sensor+contioller+Actualor) "Water Level Controller Step 4 of 4 (D) Fly-Ball Governor Problem 1.02PP Identify the physical principles and describe the operation of the thermostat in your home or office. Step-by-step solution step 1 of 1 Thermostat is used as the sensing element for controlling the room temperature. It performs the task o f automatic reduction o f error to zero, irrespectiTe o f the situation created by disturbance. Working Principle: It contains a fluid which is able to e ^ a n d or contract due to temperature change, which causes the sni^ - action of a switch that makes switching- ON or OFF ofthe heat source. Problem 1.03PP A machine for making paper is diagrammed in Fig. There are two main parameters under feedback control: the density of fibers as controlled by the consistency of the thick stock that flows from the headbox onto the wire, and the moisture content of the final product that comes out of the dryere. Stock from the machine chest is diluted by white water returning from under the wire as controlled by a control valve {CV).A meter supplies a reading of the consistency. At the “dry end” of the machine, there is a moisture sensor. Draw a block diagram and identify the nine components listed in Problem part (d) for the following: (a) Control of consistency (b) Control of moisture Figure A papermaking machine Figure A papermaking machine Watt’s steam engine with fly-ball governor In each case, indicate the location of the elements listed below and give the units associated with each signal. • The process • The process desired output signal • The sensor • The actuator • The actuator output signal • The controller • The controller output signal • The reference signal • The error signal Step-by-step solution step 1 of 13 (a) Step 2 of 13 Figure 1 shows the general block diagram to understand the process of the system. The nine components listed below are identified in figure 1. • Process • Process desired output signal • Sensor • Actuator • Actuator output signal • Controller • Controller output signal • Reference signal • Error signal Actuating Actuator signal ou^ut signal Figure 1 step 3 of 13 Draw the block diagram for the figure 1.12 in the textbook to understand the process of control of consistency. Step 4 of 13 ^ Error Actuator Paper consistency Sensor Figure 2 Step 5 of 13 Compare figure 1 and figure 2. The comparison shows the working of the paper making machine >A/ifh the mainr ^nmnnnonfc in fho nrr»rocc nf nr*ntml nf nrinciotonnw Step 6 of 13 The input reference signal is given to the controller and here the valve acts as an actuator for the system. The valve initiates the mixing process and the consistency of the paper is monitored according to time delay. The consistency checking meter checks the output of the process and sends the signal to error detector. Here, the original required consistency Is checked with the output received and checks for any variation in the output. The corresponding required adjustments may be done in the controller to reduce the error rate, to improve the total output of the process. Step 7 of 13 Thus, the control of consistency is explained and the major components are identified. Step 8 of 13 (b) Step 9 of 13 Draw the block diagram for the figure 1.12 in the textbook to understand the process of control of moisture. Step 10 of 13 Paper moisture Sensor Figures step 11 of 13 Compare figure 1 and figure 3. The comparison shows the working of the paper making machine with the major components in the process of control of moisture. Step 12 of 13 Here, Dryer remains as a main process and the moisture is checked in the moisture checking meter (sensor). The total output is compared in the error detector and required adjustments are done by the controller to improve the accuracy. Step 13 of 13 Thus, the control of moisture is explained and the major components are identified. Problem 1.04PP Many variables in the human body are under feedback control. For each of the following controlled variables, draw a block diagram showing the process being controlled, the sensor that measures the variable, the actuator that causes it to Increase and/or decrease, the information path that completes the feedback path, and the disturbances that upset the variable. You may need to consult an encyclopedia or textbook on human physiology for information on this problem. (a) Blood pressure (b) Blood sugar concentration (c) Heart rate (c) Heart rate (d) Eye-pointing angle (e) Eye-pupil diameter Step-by-step solution step 1 of 2 (A) . (B) . (C) . Block Diagram for Blood Pre ssure. Blood Sugar Concentration and Heart Bate Control Step 2 of 2 (D ). rm * 12^T_*2?2!E!S. Block Diagram o f Eye-Pointing Angle and Eye - P i ^ i l Diameter Problem 1.05PP Draw a block diagram of the components for temperature control in a refrigerator or automobile air-conditioning system. Step-by-step solution step 1 of 2 Figure 1 shows the general block diagram to understand the temperature control of refrigerator and in the automobile air-conditioning system. ActuMor Process and in the automobile air-conditioning system. Figure 1 Step 2 of 2 The required temperature signal is set in the thermostat, and then the controller actuates the compressor for cooling process. The temperature sensor measures the temperature and compares the required reference temperature with the measured temperature in the comparator. When the required temperature is achieved, the controller stops the compressor and maintains the temperature. The temperature sensor checks the temperature of the system periodically and actuates the controller if there is difference in the required reference temperature. This process is continued to maintain the required temperature of the system thereby controlling the ON/OFF input to the compressor. Thus, the temperature control of refrigerator and in the automobile air-conditioning system is explained. Problem 1.06PP Draw a block diagram of the components for an elevator-position control. Indicate how you would measure the position of the elevator car. Consider a combined coarse and fine measurement system. What accuracies do you suggest for each sensor? Your system should be able to correct for the fact that in elevators for tall buildings there is significant cable stretch as a function of cab load. Step-by-step solution step 1 of 2 The comoonents of elevator oosition control are shown in fiouret. Step 1 of 2 The components of elevator position control are shown in figure!. Controller Actuator Process Floor Sensor Figure I step 2 of 2 The input is given to the logic controller through comparator. The logic controller drives the transmission mechanism of the elevator (Motor or hydraulic). The elevator moves up and down and reaches the required floor position. When the corresponding floor button is pressed, the controller reduces the speed of the motor to stop the elevator in the respective floor. The electro switch acts as a sensor for coarse measurement to measure the floor level. Accuracy level can be fixed in the sensor considering the cable stretch due to cab load. This sensor enables to locate the elevator in the respective floor accurately without any deviation in the measurements. This complete mechanism forms a closed loop as in figure 1. The error detector compares the reference input to the output of the closed loop to ensure the exact location of the elevator. Hence, the block diagram of the elevator mechanism considering negligence is explained. Problem 1.07PP Feedback control requires being able to sense the variable being controlled. Because electrical signals can be transmitted, amplified, and processed easily, often we want to have a sensor whose output is a voltage or current proportional to the variable being measured. Describe a sensor that would give an electrical output proportional to the following; (a) Temperature (b) Pressure (c) Liquid level (d) EJpwj^fJinJjiri ainnn a ninp (nr hlnnH alnnn an artprv^ (d) Flow of liquid along a pipe (or blood along an artery) (e) Linear position (t) Rotational position (g) Linear velocity (h) Rotational speed (i) Translational acceleration (j) Torque Step-by-step solution step 1 of 2 (A) . TEMP M A T U R E : Thermocoi^lc. (B) . PRESSURE : Pressure gauge. (C) . U Q UlL? LEVEL : Bourdon tube and LV D T s combination. (Py FLOW OF LIQUID ALONG PIPE FORCE : Any pres sure sensor can be used. LINEAR POSITION : LVDT li n e a r Variable Differential Transformer). Step 2 of 2 ROTATIONAL POSiriON : Potentiometer. (G). U N EA R VELOCITY : Speedometer. 0 ^ . ROTATIONAL SPEED : Tachometer. ( 1 \ TRANSLATIONAL ACCLERATION : LVDT ( J y TO R Q U E: Combination o f Gear and Tachometer. (a) Temperature (b) Pressure (b) Pressure (c) Liquid level (d) Flow of liquid along a pipe (or blood along an artery) (e) Linear position (t) Rotational position (g) Linear velocity (h) Rotational speed (i) Translational acceleration (j) Torque Step-by-step solution step 1 of 1 ^ An actuator amplifies the signal taken from the sensor. Problem 1.- ' OPAMP OP AMP Units o f the Actuator output signal are current (Anq^ere) or Voltage (Volts) . Give the units of the actuator output signal. The Operational Amplifier is the most commonly used Actuator. Problem Feedback control requires being able to sense the variable being controlled. amplified. e . Describe a sensor that would give an electrical output proportional to the following. Describe an actuator that could accept an electrical input and be used to control the variables listed. Because electrical signals can be transmitted. and processed easily.08PP Each of the variables listed in Problem can be brought under feedback control. Any Electronic A n^lifier can be used to do such. often we want to have a sensor whose output is a voltage or current proportional to the variable being measured. and ‘ACTH’ is the Adrenocorticotropic hormone. Step 5 of 9 Thus the block diagram of closed loop negative feedback In a biological system is drawn and shown in figure 1. hypothalamus (in the brain) secretes a hormone called Corticotropin Releasing Factor (CRF) which binds to a receptor in the pituitary gland stimulating It to produce Adrenocorticotropic hormone (ACTH). ‘CRF’ is the Corticotrophin Releasing Factor. Step 8 of 9 O j^tocin Desired . Once the baby is born. (b) Positive Feedback in Biology. Consider the birth process of a baby. Pressure from the head of the baby going through the birth canal causes (b) Positive Feedback in Biology. This happens in some unique circumstances. This In turn shuts down (turns off the stress response) for both CRF and ACTH production by negative feedback via the bloodstream until GC returns to Its normal level. Step 6 of 9 (b) Step 7 of 9 The block diagram to show an example of closed loop positive feedback in a biological system Is shown in figure 2. Pressure from the head of the baby going through the birth canal causes contractions via secretion of a hormone called oxytocin which causes more pressure which in turn intensifies contractions. . Draw a block diagram of this closed-loop system. When a person is under long-term stress {say.09PP Feedback in Biology (a) Negative Feedback in Biology. Problem 1. Consider the birth process of a baby. which in turn stimulates the adrenal cortex (outer part of the adrenal glands) to release the stress hormone Glucocorticoid (GC).off ^Stress response • off Negatice feedback_______ Figure 1 Step 4 of 9 Here. ‘GD’ is the Glucocorticiods. a couple of weeks before an exam!). Draw a block diagram of this closed-loop system. Step-by-step solution step 1 of 9 (a) Step 2 of 9 The block diagram to show an example of closed loop negative feedback in a biological system Is shown in figure 1. Baby Birth 1 ^f 1 ^ Is Pressure Pressure y ? y head canal \ n (Contractions) Positive feedback Figure 2 Step 9 of 9 Thus the block diagram is drawn to show the closed loop positive feedback in a biological system. This happens in some unique circumstances. the system goes back to normal (negative feedback). Step 3 of 9 Long Nofinal Pituitary Adrenal GCs — ^L^H ypolhala m c«L |^^^ Gluco glands Mi Glands 1 oofticoids(CD) I Stress response . .X. Problem 2.41 (c) in the text book for the block diagram of a mechanical system.x i) + * 2 ( X 2 . |_ x . + 4| (x.) + V i + *1 (*i . (x. the differential equations describing the system are. ( x j. tz Write the differential equation describing the system.-X j)+ A |X .AiXj + ^X| + +F + i | X i+ i | ( x i..x . give a reason for your answer.+ A ^ (x .41 (a) in the text book for the block diagram of a mechanical system. the differential equations describing the system are.X |) = f .x. )= 0 Step 4 of 6 Draw the free body diagram of mass . state whether you think the system will eventually decay so that it has no motion at all. -A^x. iWj s -k^x^ . ( x ..x.x i) = F Thus. Thus.X |) + < : . Draw the free body diagram of mass ■ Write the differential equation describing the system.k ^ x ^ + k ^ Al|X2 4'i^(x2~Xi)4'A^A^ = 0 Thus..k ^ x ^ . - *1*1. and that interaction continues until all motion damps out. . m^x^ ^-k^x^-k^x^-^k^x^-¥ k^y + *3 (*1 . + A ^ ( i. . ^ *a*a ^ *1 * Write the differential equation describing the system. ►tjX j * 2 *1 * Write the differential equation describing the system.^ x .*2) = 0 Step 6 of 6 Draw the free body diagram of mass . . -ik. +i^Xj+i\x^ = 0 m.X j ) * 0 m . )* 0 Although friction affects only the motion of the left mass directly.■»!)+*3 . *2*1 Write the differential equation describing the system. Figure Mechanical systems Nofnctioa Nofrictkn (I) h (I) I— w v — ^ — I Step-by-step solution step 1 of 6 (a) Refer to Figure 2. Draw the free body diagram of mass m . the system decays to zero motion for both the masses.x . Step 3 of 6 (b) Refer to Figure 2.b ^ x ^ . Step 5 of 6 (c) Refer to Figure 2. + it. ^*1 - *1*1 . m^x^ . n\Xy + 6 |(x . continuing motion of the right mass excites the left mass.A^Xj . + (xj .01 PP Write the differential equations for the mechanical systems shown in Fig. *-i^ x . Draw the free body diagram of mass m . given that there are nonzero initial conditions for both masses and there is no input. + i|X . For Fig.x.(a) and (b).jt..y ) = 0 Thus. -k^x^-k^x^ Step 2 of 6 Draw the free body diagram of mass . m.-JC j) s 0 There is friction that affects the motion of both the masses. m. . the differential equations describing the system are...x.41 (b) in the text book for the block diagram of a mechanical system. + /T j)x .xj)+JC . and x^. m2 ■ m .40 in the textbook. decays to zero due to the damper.x . Thus.{ x . Draw the free body diagram of left side mass.Xj * 0 »Hj^+(Ar|+Ar. m |. State whether you think the system wili eventuaily decay so that it has nomotion at ail..~ x .+(AT. + Jirj)x2 + ( i j . . as the relative motion between x. the two masses continue oscillating together without decay. m. since there is no friction opposing that motion and also no flexure of the end springs.JITjX.02PP Write the differential equation for the mechanical system shown in Fig. Problem 2.)+ 6 ^ ( x j . m |. m.^ = 0 i^ ( i. + i . the system w ill finally decay. . However. K ^.ii) = 0 (2) Therefore.x ^ Figure 1 Step 2 of 4 Write the equation of motion for left side mass.-J ir2 .) s 0 step 4 of 4 The relative motion between x. and give a reason for your answer. m^ lU X i. x .)aii-iC. m j.) = 0 (1) Draw the free body diagram of right side mass.X 2 + J C 2 (x j-x . and these are the essentials to maintain oscillation of the two masses. Figure 2 Step 3 of 4 Write the equation of motion for right side mass. + JCj (jcj . given that there are nonzero initial conditions for both masses. and ji^ . the differential equations for mechanical system are. J— . Figure Mechanical system M <1 Noftictioo Step-by-step solution Step-by-step solution step 1 of 4 Refer to Figure 2. decays to zero. * (6 .-6 . the above equation is moment equilibrium about the pivot point of the left pendulum. Jt(sinO 2-sinO .)cos0sm /^d. Step 5 of 6 Assume the angles are small. Therefore. Step-by-step solution step 1 of 6 Consider the circuit diagram. Step 6 of 6 ^ Write the equation for moment equilibrium about the pivot point of the right pendulum.-sine|)cos82 =m/^e. Assume that the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal. let one of them be displayed by 6| and other by 6. r m m Double pendulum system Step 2 of 6 ^ Re-draw the circuit diagram. and the springs are attached three-fourths of the way down. then sin0| s 0 |. Step 4 of 6 Writing about point ‘0’ the moment equation is. c o s 9 |S ]a n d c o s 0 2 s l. of length /.)= m /= 9 .T h e above equations are modified as. Problem 2. * (e . ^ / I 1 ^ / / m Here there are two degrees of freedom Step 3 of 6 ^ At any instant.03PP Write the equations of motion for the double-pendulum system shown in Fig. the above equation is moment equilibrium about the pivot point of the right pendulum.. Similarly.) . . .-e . write the moment equation for other pendulu -mg/sin6. The pendulum rods are taken to be massless.^ ^ / j *(sin6.0 Therefore. (7) Substitute equation (6)in equation (7).81 for g in equation (8). the vaiue of length and period for a swing are [0.... -m g = ^ m /*tf+ mg X^ sin =0 f l + ^ 8 i n f l = 0 ..2 > r / .. 2 kg stick of length / suspended from a pivot.5 meters.31 = 2sec Consider the following equation for the value of length.3727 m Thus.. M o = -m g n ^ s m 0 — .. the pendulum length is 1.? § .3727m]a Hence. The coordinates and forces relationship is shown in Figure 1. How long should the rod be in order for the period to be exactly 1 sec? (The inertia I of a thin stick about an end point is Assume that 0 is small enough that sin Q ^ Q.. the 1 second for a swing from one side to the other. (3) 2i Consider ^ is very small in equation (3).... g+ ^e = o (4) Consider the following general equation.. This pendulum is shorter because the period is faster. ^+24^a>e+ai‘ f f = 0 (5) Compare equation (4)and equation (5).= . I 2x2 r = 2 .... / is the body’s mass moment of inertia about its center of mass. /= . But if the period had been 2 second. (1) Where. M ^ I a .(2 ) Substitute for /p in equation (2). (8) Substitute 2 for / and 9. 3x9.81 /* - 8^* * 0...8I = 2.) Why do you think grandfather clocks are typically about 6 ft high? Step-by-step solution step 1 of 3 Consider the following equation for the sum of all external moments about the center of mass of Consider the following equation for the sum of all external moments about the center of mass of a body... 2/ r .. a is the angular acceleration of the body. r=— ...81 for g in equation (9).. o Figure 1 Step 2 of 3 From Figure 1. Step 3 of 3 ^ Consider the following equation for the period for a swing.rj ’'V 3 x 9. the equation of moment about point O is given below. Problem 2...04PP Write the equations of motion of a pendulum consisting of a thin.. . (9) Substitute 9. 000 N/m. for which there is insufficient damping (velocity-dependent force) to warrant including a dashpot in the model. Example A Two-Mass System: Suspension Model Figure 1 shows an automobile suspension system. The system can be approximated by the simplified system shown in Fig.000. overshoot for lower values. which have a mass of 20 kg each. By placing a known weight (an author) directly over a wheel and measuring the car?s deflection. The wheel mass was measured directly to be m1 = 20 kg. a force will result if either the road surface has a bump (r changes from its equilibrium value of zero) or the wheel bounces (x changes).01:2.1)s“ + (5. we find that ks = 130. r is a unit step) using Matlab. we find that ks = 130.6 shows the free-body diagram of each mass. for i = 1:4 b = Bd(i): A=[0 1 0 0. Assume that m1 = 10 kg. A system consisting of one of the four wheel suspensions is usually referred to as a quarter-car model. (1) to each mass and noting that some forces on each mass are in the negative (down) direction yields the system of equations Some rearranging results in F=m a. a stable system always has the same signs on similar variables.1).C. plot the position of the car and the wheel after the car hits a “unit bump”(that is. 4. y=step(A. this force has been omitted.X(D) + ^ X ( . The shock absorber is represented in the schematic diagram by a dashpot symbol with friction constant b. (a)impulse responses. the smallest overshoot is with b is 5. The method for keeping the signs straight in the preceding development entailed mentally picturing the displacement of the masses and drawing the resulting force in the direction that the displacement would produce. Likewise. 2. By using the step response data in Fig. 0 0 0 1. and the system gets too fast for larger values.000 N/m. Bd = [1000 3000 4000 5000]. m l = 10. Step 3 of 3 Thus. kg is the car deflection.1 1 ) M2 M2 The transfer function is obtained in a similar manner as before for zero initial conditions.B.t): subplot(2. the need to include the gravity forces is eliminated. we subtract the mass of the four wheels from the total car mass of 1580 kg and divide by 4 to find that m2 = 375 kg.x m + — o r(s ) . m2 = 250. ) = ^ « ( I) . y(:. the force = The force of gravity could be included in the free-body diagram. 1 + — (t-W + . From figure 1. we conclude that b = 9800 N?sec/m. The equilibrium positions are offset from the springs? unstretched positions because of the force of gravity.i): plot(t. are the displacements of the masses from their equilibrium conditions. there is a constant force of gravity acting on each mass. Substitutino s for d/dt in the differential eouations vields J^X(I)+ 1— (Xd) . x and y.^ 0 .-( ks/ml + kw/ml ) -b/ml ks/ml b /m l. Kw = 500. Write the equations of motion for the automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car mass above one wheel. 2 where two spring constants and a damping coefficient are defined.1.3) * d ) “ J* + (516. However. and Ks = 10. Therefore. (18) = (19) The most common source of error in writing equations for systems like these are sign emors. 3(b) and qualitatively observing that the car?s response to a step change matches the damping coefficient curve for ^ = 0. and the jump in car position is the fastest with this damping value.1 f.2. which have a mass of 20 kg each. the transfer function with the numerical values is Y(s) 131e06(5 + 13.X(D) + . kw/ml :0 .685«04)j“ + (I J (n < 0 ^ + l.1) 4 (J -i)+ t. a check on the signs for systems that are obviously stable from physical reasoning can be quickly carried out.t ) + ^ ( y . ^ s 3. Note that the forces from the spring on the two masses are equal in magnitude but act in opposite directions.13) We will see in Chapter 3 and later chaptero how this sort of transfer function will allow us to find the response of the car body to inputs resulting from the car motion over a bumpy road. Figure 1 Automobile suspension Figure 2 The quarter-car model? T Figure 3 Responses of second-order systems versus ^ . y(:. D=0.0 ]. (2) shows that the signs on the and X terms are all positive. Ml Dll Jfl] Mi ^ n s )+ s — m ) . k^ is the applied weight.^ d -y ) + ^ i = ^ r . including the four wheels. ks/m2 b/m2 -ks/m2 -b/m2 ].(y -i)-* . end Step 2 of 3 The output for the MATLAB code is given in figure 1.^(X (I) .2)): legend(Wheer. C=[1 0 0 0 . (2 . The motion of the simplified car over a bumpy road will result in a value of r(t) that is not constant. Step-by-step solution step 1 of 3 Consider the transfer function. kw = 500000. Figure 2. which is also the case for the damper. as they must be for stability.x m = a which. By placing a known weight (an author) directly over a wheel and measuring the car?s deflection.t. after some algebra and rearranging to eliminate X(s). Once you have obtained the equations for a system. b is the damping.000 N/m. . a positive displacement x of mass m1 will result in a force from the spring ks on m1 in the opposite direction to that drawn in Fig. ks= 10000. Solution. By defining x to be the distance from equilibrium.(i-r )= « iii. as have been the equal and opposite forces from the springs. however. Measuring the wheel?s deflection for the same applied the model is for a car with a mass of 1580 kg. Eq.D. The coordinates of the two masses.'Car’): title = sprintf('Response with b = % 4 .7 in the figure. Measuring the wheel?s deflection for the same applied weight. t = 0:0. 0 0 1 0 ]. A positive displacement y of mass m2 will result in a force from the spring on m2 in the direction shown and a force from the spring on m1 in the direction shown. Assume that the model is for a car with a mass of 1580 kg.000 N/m. i'( 4 '»!'»! 4j J a l 'U A A 1 m i« j Where. B=[0. yields the transfer function I 'd ) ^ (» + > ) * * + ( s r + ^ ) * " + ( ^ r + ^ + f e ) ‘" + ( ^ ) ' + i ^ (112) To determine numerical values. m2 = 250 kg.733«Or 0. (b) step responses Figure 4 Free-body diagrams for suspension system Ki-i 0 i^-r) ■r) il Applying Eq. (2). (2. Problem 2. the signs on the and /te rm s are all positive in Eq. As previously noted. Gravitational forces can always be omitted from vertical-spring mass systems (1) if the position coordinates are defined from the equilibrium position that results when gravity is acting. and (2) if the spring forces used in the analysis are actually the perturbation in spring forces from those forces acting at equilibrium.05PP For the car suspension discussed in Example. The force from this spring is proportional to the distance the tire is compressed and the nominal equilibrium force would be that required to support m1 and m2 against gravity. The system can be approximated by the simplified system shown in Fig. For this system. By defining x and y to be the distance from the equilibrium position. b ). including the four wheels.000 N/m. Find the value of b that you would prefer if you were a passenger in the car. (1 1 0 ) Ml Ml Mi Mi 9 + . is the mass. we find that Mv ~ 1. As we will see when we study stability in Section 6 of Chapter 3. The lower spring kw represents the tire compressibility. as indicated by the minus X term for the spring force. The force from the car suspension acts on both masses in proportion to their relative displacement with spring constant ks.000. The magnitude of the force from the shock absorber is assumed to be proportional to the rate of change of the relative displacement of the two masses?that is.x > = o . with the reference directions as shown. however.000 is the best compromise. Therefore. Write the MATLAB program. its effect is to produce a constant offset of X and y. Carefully define where the body’s displacement is zero. Then. kx+Mg = 0 I. The displacement be comes Zero.'. ^x&ere S= dt =Acceleration due to gravity. Problem 2. Step-by-step solution step 1 of 3 Displacement is Zero when forces are balanced Hence T i-k x T No displecement (di^koement) Step 2 of 3 ^ e n M g = -la . Step 3 of 3 From the above figure. It is shown in the figure where displacement is zero.06PP Write the equations of motion for a body of mass M suspended from a fixed point by a spring with a constant k. kx = -hK I . x ) .* ( » . the transfer function with the numerical values is r(s) 131e06(s + 13. The simplest change is to make shock absorbers with a changeable damping.( y . in opposite directions on the wheel axle and the car body. which have a mass of 20 kg each. Write the equations of motion for the automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car mass above one wheel.nn X terms are all positive.8) .3) S(j) “ J* + (516. A system consisting of one of the four wheel suspensions is usually referred to as a quarter-car model.685«04)j2 + (Ume06)s + l. Measuring the wheel?s deflection for the same applied weight. By placing a known weight (an author) directly over a wheel and measuring the car?s deflection. A positive displacement y of mass m2 will result in a force from the spring on m2 in the direction shown and a force from the spring on m1 in the direction shown.x ) .000. which is also the case for the damper. 2 where two spring constants and a damping coefficient are defined. and (2) if the spring forces used in the analysis are actually the perturbation in spring forces from those forces acting at equilibrium. however. A ( t . Note that the forces from the spring on the two masses are equal in magnitude but act in opposite directions.6 shows the free-body diagram of each mass. a positive displacement x of mass m1 will result in a force from the spring ks on m1 in the opposite direction to that drawn in Fig.1) 4 ( J . (2. yields the transfer function X M _____________________ ^ ( » + > ) «(») ( 2 .07PP Automobile manufacturers are contemplating building active suspension systems.7 in the figure. (a) Modily the equations of motion in Example to include such control inputs.11) M2 M2 The transfer function is obtained In a similar manner as before for zero initial conditions. for which there is insufficient damping (velocity-dependent force) to warrant including a dashpot in the model. By using the step response data in Fig.J ) .r ) = « > i* . * ) . M . as indicated by the minus X term for the spring force. The magnitude of the force from the shock absorber is assumed to be proportional to the rate of change of the relative displacement of the two masses?that is. are the displacements of the masses from their equilibrium conditions. Figure 2.K . \32 and shows 6 as in the FBD of figure 2. The system can be approximated by the simplified system shown In Fig. (b) step responses 3^ z s m Figure 4 Free-body diagrams for suspension system lO-i IJi-H h^ -ii Applying Eq.10) Ml Mi Mi Mi y + — 0 . Likewise. (1) to each mass and noting that some forces on each mass are in the negative (down) direction yields the system of equations Some rearranging results in F=m a. the force = The force of gravity could be included In the free-body diagram.( i. 3(b) and qualitatively observing that the car?s response to a step change matches the damping coefficient curve for ^ = 0. (2. (2).y . this force has been omitted. we find that Mv ~ 1. The motion of the simplified car over a bumpy road will result in a value of r(t) that is not constant. (b) Is the resulting system linear? (c) Is it possible to use the force u2 to completely replace the springs and shock absorber? Is thi a good idea? Example A Two-Mass System: Suspension Model a good idea? Example A Two-Mass System: Suspension Model Figure 1 shows an automobile suspension system.i ) +U j Step 2 of 3 The system is linear with respect to because it is additive.t ) + ^ (y -x )= O . y = . after some algebra and rearranging to eliminate X(s). The method for keeping the signs straight In the preceding development entailed mentally picturing the displacement of the masses and drawing the resulting force in the direction that the displacement would produce. By defining x and y to be the distance from the equilibrium position. By defining x to be the distance from equilibrium. The equilibrium positions are offset from the springs? unstretched positions because of the force of gravity. 12) To determine numerical values. The force from the car suspension acts on both masses in proportion to their relative displacement with spring constant ks.r m + ^ ( X ( j ) .« (2. Is a function o f the control variable ui. we conclude that b = 9800 N?sec/m. a force will result if either the road surface has a bump (r changes from its equilibrium value of zero) or the wheel bounces (x changes). its effect is to produce a constant offset of X and y. Therefore. a check on the signs for systems that are obviously stable from physical reasoning can be quickly carried out. Once you have obtained the equations for a system. with the reference directions as shown. The system can be approximated by the simplified system shown in Fig. The coordinates of the two masses. Substituting s for d/dt in the differential equations yields j^ x c i) + . 4. as have been the equal and opposite forces from the springs. It is also possible to make a device to be placed in parallel with the springs that has the ability to supply an equal force. Gravitational forces can always be omitted from vertical-spring mass systems (1) if the position coordinates are defined from the equilibrium position that results when gravity is acting. However it would take very high forces and thus a lot of power and is therefore not done.9) The most common source of error in writing equations for systems like these are sign emors. here 6.* ( » ) ) = a which. a stable system always has ? tKa oinne. The wheel mass was measured directly to be m l = 20 kg.f c O '. as they must be for stability. The shock absorber is represented in the schematic diagram by a dashpot symbol with friction constant b. u2. However.t. however. Problem 2.* ( ! ) ) + — ()'(j) . The forces below are drawn in the direction that would result from a po sitive displac ement o f ^ O " . 2. + i. Solution.13) We will see in Chapter 3 and later chaptero how this sort of transfer function will allow us to find the response of the car body to inputs resulting from the car motion over a bumpy road. However.733«07' 0. (a)impulse responses. the signs on the and /te rm s are all positive In Eq.5.000 N/m.1)s“ + (5. Figure 1 Automobile suspension Figure 2 The quarter-car model? U Figure 3 Responses of second-order systems versus ^ . The lower spring kw represents the tire compressibility. including the four wheels. As we will see when we study stability In Section 6 of Chapter 3. ( y .y ) + t x = ^ r . we subtract the mass of the four wheels from the total car mass of 1580 kg and divide by 4 to find that m2 = 375 kg.( x . b(u1). . (2. (2.i) + t. the need to include the gravity forces is eliminated.■K'. we find that ks = 130. Step-by-step solution step 1 of 3 (A) The FBD shows the a<iditioa of the variable force. As previously noted.Y(sn+ ^ x w = Mi Mi Mi Mi + » — ()'(») . x and y. there is a constant force of gravity acting on each mass.i { u ^ ) { y .000 N/m. Step 3 of 3 (C) It is technically possible. The force from this spring is proportional to the distance the tire Is compressed and the nominal equilibrium force would be that required to support m1 and m2 against gravity. Assume that the model is for a car with a mass of 1580 kg. i ( x . Figure 1 Communications satellite Source: Courtesy Space Systems/Loral (SSL) Figure 2 Schematic of a system with flexibility Step-by-step solution step 1 of 7 Sketch the schematic o f a sjrstem with flexibility. + A ( i .(As + k ^ + As + it U mMs + (« + M')bs + (M + m) ks^ .— sX + + — Y + — ysY = — U M M M M M Rewrite in matrix form (m ^ + b s + k -(b s + k ) ^0^ [ -(As + Jt) Ms^ + bs + k ) [ Y . k A.08PP In many mechanical positioning systems there is flexibility between one part of the system and another. 1 — X -------. 1 where there is flexibility of the solar panels.s r = 0 — X . we get mx = .k \ . (b) Find the transfer function between the control input u and the output y.(As + it)^ Step 7 of 7 Simplify further. it ^ A . 1 M M M M M s^x+ . * X H— X H— X---. Figure 2 depicts such a situation..X + .y -------. the equation o f motion governing the system is obtained as . (a) Write the equations of motion governing this system.v + — V = — u M M M M M Thus. (m ^ + b s + k 0^ detl T -(te + t) u] + As + it . Step 2 of 7 ^ Sketch the free body diagram o f the given figure. 1 — X -------.y -------.x + V H-------. where a force u is applied to the mass M and another mass m is connected to it. Problem 2.y = 0 m m m m -k A .j^) = u + ^ (x .J>) Simplify further. iibc-y) U b ( f. .r .v + — V = — u M M M M M Step 5 of 7 ^ (b) Find ^ Limlace tra n s f^ n o f the equations . . X H— X H— X ------.y = 0 m m m m -k A . - X H— X H— X ------.(As + it) det -(A s+ it) A fs^+As+it + As + it 7 = {^ms^ + As + itJ^Afe^ + As + itj . The coupling between the objects is often modeled by a spring constant k with a damping coefficient b. [ u ) Step 6 of 7 Solve using the Cramer’s rule. _ jt A. although the actual situation is usually much more complicated than this.y ) Step 3 of 7 > X K x -y ) Step 4 of 7 ^ From the free body diagrams.J t ( x . An example is shown in Fig. k A . y ms^ + As + it ^ + bs + k ^ (^Ms^ + b s + k'j ..y -----.. ^ A .k ^ b .s x . .y = 0 m m m m -k A .x + V H-------. that is.3.2) ^ b. or vw _ i a7) *+ h' This expresskm of die dUferential equation (2. and we find that V _ -!■ yg m 06) «. the output for th b case can be multiplied by the magnitude of the input mep to derive a step re9 onae o f any ampUtude. The equation of motion b found mung Eq. m m EXAMPLE 2. (2. Equations of motion: For sunpBdty we asanme that the rotmkNinl inertia o f the wheeb b negligible and that there b ftiction retarding the motion of the car that b proportional to the car’s speed with a proportionality constant. calculate and plot the time reaponie o f velocity for an input step with n 500>N magnitude. (2. v ( ^ ) . Problem 2.7).2 r>co body diagram for The c* term C H ic e b out. find a value of K that you think would result in a control system In which the actual speed converges as quickly as possible to the reference speed with no objectionable behavior. we have aubstkuled i fof dfdt in Eq. Rgure2. k = [100 200 1000 5000].1 aasnmiiig that the en^ne imparta a force ■ ae abown.1). so that it has a control law. b .1000. ^ can be multiplied by die magnitude of the input step. 2. Note that. The friction force acts opposite to the direction motiMi. m = 1500. T h eresu k b ( 2. 2. Solution 1. Revise the equations of motion with vr as the input and v as the output and find the transfer function. num =K/m. sys=tf(num. find a value of K that you think would result in a control system In which the actual speed converges as quickly as possible to the reference speed with no objectionable behavior.V m m m . Increasing K also results in the need for higher acceleration is less obvious. b = 70. Stepresponsewtth Hatlab % setsupthem odetod«flnetlN transfer ftincUon 9 S• (1/1000]/(s -f 5<V1000). u is the input V is the output m is the mass b is the friction force Substitute J C (v ^ -v ) foru. The coordinate o f the car’a positioo Xu the distaoce from the refotenoe line sbo«m and b chosen ao that positive isto the r i ^ Note that in this case the meitial acceleration u amply the second derivative o f x (that is. Eq. (2. b K K V +— V+ — V= — V- m m m A block diagram of the scheme is shown in Figure 1. Use Matlab to 6nd the response o f the velocity of the car for the case in which the inpntjumpe from being « = Oat time f = 0 to a constant Hgui«2. there is a response in 5 seconds and the steady state error is 5%. therefore k b drawn rqiposhe to the direction o f positive motion m l eiketcd as a negative fioroe in Eq.2:50. den = [1 b/m+K/m]. plot(t. 03) For the case of the automotive cruise control where the variable o f interest b the speed. (2.t). The car can then be approximated for modeling poipoeea using the ftee-body diagram men m Fig.108) where vr= reference speed. m m . “ 0 . y = step(sys. b K K V + — V * — V . hold on t=0:0.109) /C= constant.3 Response of the car v e lo d ty toastepinti Step-by-step solution step 1 of 7 Consider the equation of motion.+ 1 ' For reasons that will become efetf in Quqker 3.y) grid end hold off Step 6 of 7 The output for the MATLAB code is given in Figure 2. Transfer functions of a system will be used in later chapters to design feedback corXrollers such as a cruise control device found in many modern cars. that b . and find the response for a unit step In vr using Matlab. and find the response for a unit step In vr using Matlab.den). (2. Step 7 of 7 Thus. (Z4X T hb transfer Auh tion serves as a mwh model that relates the c h ’ s velocity to the forces propelling the car. V + — V ® — IT Where.4) b called the tra n ti’e r function and wiD be used extensively m later chapters. a * x) because the car position b measured wkh respect to « inertial reference frame. (2.------. The Mep responae b shown m Fig. Equivaleikly. X Time reaponae: The dynamics o fn system can be prescribed to Matlab in terms of its transfer function as can be seen in the Matlab statemenb below that implemeiits E<^ (2.110) This is a “proportionarcontrol law in which the difference between \^rand the actual speed is used as a signal to speed the engine up or slow it down. y (s ) K (x ) b K S + — + — m m Write the inputs for MATLAB.1 A Simple System. 70 N-sec/m.. let u = K (v r-v ). the larger the K is the better the performance with no objectionable behavior for any of the cases.1 Cruise control model « B 500 N thereafter. Using trial and error. the equbion o f motion becomes V+ —V * B —. Assume that the car a I « is 1000 kg and visoous drag coeflkient. b B 50 N*secAn.000 .200. .1). Write die equedon of motion for the ipeed m l forward motion of lhecaribow nm Fig. in essence.2. % plots th e step response for u • 500. however.09PP Modify the equation of motion for the cruise control in Example.K Step 3 of 7 Figure 1 Step 4 of 7 Consider the transfer function of the closed loop system. M X + —X = — . this b often written using capital letters to signify that k b the‘Itunsforro'’o f the solution.7) with the numbers filled l a step(S00*9 s). b u V H— V = —. for i=1:length(k) K=k(i). the essence b that you aaaume a aolotionoftfae form v s p v e a an input o f the form ■ b Then.1. Using trial and error. Take the Laplace transform o f the reauiting differential eqoadon and find the transfer function between the input« and the output v. Assume that/n = 1500 kg and b = 70 N-sec/m.2. . Eq. Cruise Control Model 1. shows all forces acting on the body (heavy lineaX and imficates the accdetaCion (dashed line).5000 Write the MATLAB program. In Figure 2.4 ) H k sobtion of w ch an equation win be cowered in detail in Q u p te r 3. 2. for K . K m den I m mj Step 5 of 7 Consider K=100. % defines th e transfer function from Eq. The step fimetion in Madab calcn- bles die time response o f a Ikictf system to ■ unit step input Bwuuie the system b finear. Step 2 of 7 1 u m . which defines coordinmes. tince v b the differential equation chi be written aa^ 03) t:: Rgurt2. inputs from the accelerator pedal. .1 OPP Determine the dynamic equations for lateral motion of the robot in Fig. Problem 2. there was no need to invoke equation (1). (4) Take differentiation on both sides in equation (4). y= .. Source: AP Images Figure 2 Model for robot motion Step-by-step solution step 1 of 3 Refer FIGURE 2. v^e.. steerable wheel in the front where the controller has direct control of the rate of change of the steering angle... (3) The lateral motion as a function of y is shown in Figure 2.(2) Where. Assume the robot is going in approximately a straight line and its angular deviation from that straight line is very small. (7) Thus. Consider is small and rearrange the above equation. Step 3 of 3 Consider the following equation for the actual change in the carts lateral position. sin 9... is the control input. y » -2 —* • ... a is the angular acceleration of the body. S. . M ^ I a . The dynamic equations relating the lateral velocity of the center of the robot as a result of commands in Usteer are desired.... .(1) Where.... Vo. Consider the following equation for the sum of all external moments about the center of mass of a body. Assume it has three wheels with a single.(o) Substitute equation (1 )in equation (6).... Usteer.46 in the textbook.... 1. F^is the constant speed.. Take differentiation on both sides.. . / is the body’s mass moment of inertia about its center of mass. The turning rate change with respect to x axis is shown in Figure 1.. Therefore. Also assume that the robot is traveling at a constant speed. 2.. Where.U ^ .. Step 2 of 3 Consider the following equation for the carts turningrate of change with respect to x axis. Figure 1 Robot for delivery of hospital supplies. L is the length of the wheel. Consider the following equation for the time rate of change of the steering wheel angle.(5) Substitute equation (3)in equation (5). . is nonzero. there is no dynamics come into equation (7). with geometry as shown in Fig.. ) + I O ^ R . * R .0 Apply KirchhofTs current law at node V_.+«^)+ioX [R^+R/J *'* » + i + i o ’ f — 5s— 1 Therefore.= o v Substitute O V fo r y and ------^ f o r y in equation (1). Problem 2. "(*+i)(J?. R^ + R . the transfer function of the circuit. K. •“[ (/i„+Ji/)(j+l) J + “ »'-[(*+i)(^+«. + R^ “ s+ lJ^ +Rf “ s+lRj^+R^ s+lR^. r^ = — [r. 10’ */ y <«’ R. « .. Figure Circuit for Problem Find the transfer function of the simple amplification circuit shown using this model. R„+R^ V= V^+ ^ r . . y -V y -V — — — sfi-=0 R. is K. ) iQ t R..) K R.: 10^ i+ = i-= 0 ..-L .43 in the circuit.R^ ) R^R.1 0 ’ i?.11 PP A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig. I. \ R. Step-by-step solution step 1 of 2 > Refer to Figure 2. v^= — (o — “ 5+it^ R^+R/ Rm*Rf J ..-r. The output voltage is. > .ioL_L^-^ j+ + S + lR^+Ry " ^ r (s + I ) ( R .+R^ K^LJ2L_5^ 1=. y s + lRi. . - Step 2 of 2 The voltage at noninverting input terminal is. (2) R^+ R .)+ioX]—io’«A ________ . j + 1 + 10’ .] o) 5+ r * The currents at inverting and noninverting terminals is. The voltage at the non-inverting terminal of the op-amp is. Step-by-step solution step 1 of 3 Refer to the circuit diagram in Figure 2. Figure 2 Circuit Find the transfer function of the simple amplification circuit shown using this model. the transfer function. y ^ + A r^ = A V ^ ( l + A ) V ^ = AVi. 2: A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig. of non-ideal op-amp is A \ +A . 1+ . Step 3 of 3 From Figure 1. Problem 2. = n The inverting node of the op-amp is directly connected to output node. the output voltage is. Give the transfer function if the op-amp has the nonideal transfer function of Problem. it is proved. 2: 10^ i+ = i-= 0 .12PP Show that the op-amp connection shown in Fig.V ^ The voltages at the inverting and non-inverting terminals of the op-amp are same for ideal op- amp.) Substitute for and for v_ in the equation. So. Figure 1 Circuit A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig. = .38 in the textbook. 1 results in Vout = Vin if the op-amp is ideal. y^=yu Hence. v.-v .4 Therefore. v . y« A K. Step 2 of 3 The simplified op-amp circuit is shown in Figure 1.4 (v .. y^=y- Substitute for v_ in the equation. J^s -I.. + I^R^^■ K.i^ And the motor equation Find To V„ = l^R. and the connection is calied a current amplifier. R 5. = K. The idea is to have the motor current follow the input voltage. R R. We know that the current anq^lifier has no feedback from the output voltage. 0 -r. with R^-^ R .b y —0 V —0 V —0 Substitute in the equation -----.. . Also show the transfer function when R f= Figure Op-amp circuit for Problem Step-by-step solution step 1 of 3 A common connection for a motor power an^lifier is given. the transfer fimction is obtained as ^ .13PP A common connection for a motor power amplifier is shown in Fig. R^ Rj R R^ 5 . Problem 2.* ■ • J .s ^ b ) R This e^^ression shows that. in the steady state. &om the equation . Thus. e . — = 0 and sinqjlify. write the voltage equation. when s —> 0 . + b e . in which And thus. + * 4 = A'^. j . we get U _ R y>. write the voltage equation.as + At the ou^ut. Sketch the given figure. 0 -f% 0 -V . . 0 -r. Assume that the sense resistor rs is very smail compared with the feedback resistor R. ^ r + r/ ‘ step 3 of 3 ^ The d3mamics o f the motor is modeled with negligible inductance as J^9^ + We can rewrite as 4 .1. and find the transfer function from Vin to la. V L -0 ^ R At node B. the current is proportional to the ii^u t voltage. Step 2 of 3 At node A.M l U R .1 . + I .Al)10’ P oleisat[(P -N )10’ -l]. 1. 2: 10^ i+ = i-= 0 .14PP An op-amp connection with feedback to both the negative and the positive terminals is shown in Fig. N = for the circuit to remain stable. Figure 2 Circuit for Problem Find the transfer function of the simple amplification circuit shown using this model.+ R . Figure 1 Op-amp circuit A first step toward a realistic model of an op-amp is given by the following equations and is shown in Fig. rt-R r] Step 3 of 4 If the Op . Equating it to Zero for limiting case.+ E „ I. Problem 2.1+N. E . • [r „ r . A(say) V„ = A M l . E .+ e „ J V. Step-by-step solution step 1 of 4 Step 2 of 4 Clearly — r+ R E„ E. R . r + R j j R (+ R k R .^ R „ -AR. ' ’ = 7Tir' in terms of the negative feedback ratio.Amp is non ideal then.J L e . .A rA We get 1+ R. R + rJ Step 4 of 4 In the limiting case we put denominator = 0 R*.lO ’ We have P S - 10' Max value is |N+10'^ . give the maximum value possible for the positive feedback ratio.. If the op-amp has the nonideal transfer function given in Problem.4-Rj^ R+r 1+NA=PA 10’ 1=(P-N) 's + 1 s + U C P . “ Therefore. (3) “ R.c . (b) active lead.48 (c) in the textbook for active lag circuit. From the circuit.— v.^ . “ R. the transfer function.(s )-± y ^ (s ) fk W K HI 4-ife] /i.( » ) Write the KirchhofTs cument law equation at node P. the dynamic equation is — + . Z. 0 0 c r „ + ^ = ..j — -o * U 0 .U ) (. C t lK W . d l[ 1 1 .c v „ ~ v . . U f ) : RCV 2RCS+2 2R +2R‘ . R d t' ' R Apply Laplace transform.. Substitute K i + ' ^ f ^ f o r y in equation (2). Substitute .c ^ K . w ) + ^ + c i ( F . Draw the circuit diagram with node voltages. the dynamic equation is c v ^ ..^ r „ . the transfer function. ± y^ “ It.r .ii. .F „(» )) = 0 = a K „W + c iK .(4 ■ K. and (d) passive notch circuits —W ^ ^ V V V . . + 2 C s K ( i) + — 2 < i_ i = 0 R ' ' R Step 10 Of 11 From the circuit diagram. P ro b le m 2.48 (b) in the textbook for active lead circuit. the dynamic equation is Apply Laplace transform on both sides. (b) active lead. ^C s + til Therefore.L ^ .15PP Write the dynamic equations and find the transfer functions for the circuits shown in Fig.. step 11 of 11 Substitute p. ^ J !|j ^ Therefore.. (c) active lag. Apply Laplace transform. “ -) K. ____ !____ ] 2 ^0 +2 2R+2R‘ C sJ Therefore. w .o K/2> ^ 2 C S-------------. n . U C R.2 ^ +2 2R + 2R^Cs) RCV 2R O + 2 2R +2R‘ s-HlJ f . K S. the transfer function. and (d) passive notch circuits Figure (a) Passive iead. Write the KirchhofTs current law equation at node y .y 0 .L ] U c *r R.^ y ^ . “ Apply Laplace transform on both sides of equation (3). = o c v «• .i—v v \ I t vvnJ .r . R H ^ Figure 1 Write the KirchhofTs current law at node y .(« ) and ^ j( ^ ) in the equation. y . Q -+ (ri) |>. . -VcMt Write the KirchhofTs cument law equation at inverting terminal. Z i £ l i ii( s ) (b) Refer to Figure 2.48 (d) in the textbook for passive notch circuit. U f ) : /J.A A o S te p -b y -s te p s o lu tio n (a) Refer to Figure 2. (c) active lag. Draw the circuit diagram with node voltages. — 0 M ^ vW . Apply Laplace transform.. K . . It.48 (a) in the textbook for passive lead circuit.C Therefore. (a) Passive lead circuit (b) Active iead circuit (c) Active lag circuit (d) Passive notch circuit Figure (a) Passive iead. U f ) : K.c 0 ^ v* . + ^0 = .— ri+ -^ ^ ir„ (4) .C) (C) Refer to Figure 2.^ V ■*/ [ r. f _ « ^ + ____ I____ ] K . R. .= C B + -^ ii (1) Therefore. y ^ (s )= -c s K . K Write the KirchhofTs cument law equation at node y . R^ R. y ^ = ^ ^ v . > { i ) = c » i ( i ) + — » (» ) / X Cl+ — ^ u(s) U Therefore..Z l k for y . Apply Laplace transform on both sides.Cs+ til step 8 of 11 A (< i) Refer to Figure 2. Write the KirchhofTs cument law equation at node P. Draw the circuit diagram with node voltages. the transfer function. >..C ^ s J . .5 1.R L J_± RR . C V + A j+ ' ' n. co for J^.16PP The very flexible circuit shown in Fig.. band-pass.) C V + A i + ^ C ij '+ A ^ + J ^ 1 «‘ J J =. By selecting different values for Ra. ( 11) c v + ^ j+ ^ R^ f i' Step 6 of 12 Substitute for in Equation (4). 1. z = [0. and C y . and 1. ‘ 0 ~SV^ S. Substitute — for in Equation (5).. ■ ^ r .. (4) ^ R^ R. (8) U ) ' s. ( 12 ) y. 2 for and 0._ f i!.i l 1 Rj R t ’"* 0 R Cs . ( 10 ) 1 RR.= . the transfer function of the system is A _ __________ ^ Step 10 Of 12 Compare Equation (12) and 2.5..52 in textbook. oo for J^and oo for R^.^ j+ ...0]: hold on for i = 1:3 num = [ A ]. 1 1 R ~ RRR. I 1 =-R f . (2) R = . . Problem 2.1. Figure 1 Thus. 1 * J step 7 of 12 1 C RR. Figure Op-amp biquad Step-by-step solution step 1 of 12 Refer Figure 2.{2) and (3).. is called a biquad because its transfer function can be made to be the ratio of two second-order or quadratic poiynomiais.” I 1 J?.L - R ^ ^ ^ i-H h -kH "- Substitute Equation (10) and (11). Rc. (7) Step 4 of 12 Determine and in Equation (5) and (6).+ a v .. wn = 2. wn = 2.- [ « 1 1 R — +Cs ^ j "Ji. den = [ 1Avn''2 2*z(i)/wn 1 ] step( num.^ R^ R‘ «. (« C )’ i ’ + ^ i + l Thus. R { R. J ____ l_ RC Step 11 of 12 (b) Substitute 2 for A. .111 in Textbook.C’ R fi (R c y j? _5_ (« C )’ i ’ + ^ j + l _R A. y .. or band-reject (notch) filter. y^ Rfi (Rcy RR. R ^ [ rr.1 0. R ^^~ R Step 3 of 12 Take Laplace transform in Equation (1).* R^R^ 4* + —^ 5 + ---- (RC)^ Step 8 of 12 Thus. the step responses using MATLAB is shown in Figure 1. high-pass. write the MATLAB program. [-L . 1 K . 0. compute and plot on the same graph the step responses fo the biquad of Fig. ' R.. (a) Show that if Ra = R and Rb = R c= R d= the transfer function from Vin to Vout can be written as the low-pass filter A where A= « i’ ' RC* R (b) Using the Matlab command step. and Rd.0. .. c'^ ^ j’ +f ' R Rj R^ R2 ^ ) y^ C' R.= o (9) R Step 5 of 12 From Equation (8) and (9) write the matrix form.a . A = 2. R . R\ ' I f c I RR.C {R C f Step 9 of 12 (a) Substitute for J^. 1 *.1 for g in Equation 2... d e n ) end hold on Step 12 of 12 Figure 1 shows the output for the MATLAB program. the transfer function of the system is. for A = 2. (3) 2 i+ il+ iL + iiL . Step 2 of 12 Consider the transfer function of the circuit ^ = -C K .111 in textbook. and ^ = 0. . ^ = -CsV^ (6) R . “ I r c y 1 V I K 'K Rt K? } ' ' ( c f ‘ 1 f 1 + l j RR. Rb. ’ R K C V + -^ s+ -^ c v + . I C s + — S+-rT R.) ^ R. Consider the value of F.' + ' 1 1 K Jt. ' 1 R]-r ... the circuit can realize a low-pass. + R ’ R is C V ^ + R iR ^ js “ C ? V i= 0 .= R i.+ R R .^ V „ V.R ’ sC+R. at V : Step 3 of 4 Also ^ + ^ + ^ r i + s R j ) = 0 R.= R sC V.R’ R js’ c ' r V „ -R R . = . v ^ R . ’ R ^ j+ R .17PP Find the equations and transfer function for the biquad circuit of Fig.R " s C + R .+ R . R ’I Rs J 1 1 ^ RsC +R R^^C ^’l ^ W Rs r ~ Ri Step 4 of 4 [■r J+R’ sC+R’ R / C ? Y Vfc ' I RRs J R. Using property o f Inverting Amplifier.R’ s^C? R i ^ sV „ + R i ’ R ’ sC V „ + R i ^ ’ s^C ?V „+R R i R jV4 +R “ R jV j.R’ R .R " R js " C ? V J. V.= R . R R / R. R iR j+ R iR ’ sC+R.=oo R .s“C?' Vi R. RsC ' ^ = -1 V._______ R R . and Rb = Rc=-^.R^ sC+R. Problem 2.= RsC V . V . R v= R . Figure Op-amp biquad Step-by-step solution step 1 of 4 Step 2 of 4 Given: R . Ri R V „+ V .R d = R1.R j +R. V_ ^-^2 -R. R R . R. if Ra = R .R . V. Writing Node Equations.. 036 x 10"’ N-m 1 —O u n ce-Inch per Amp =7.277 N s H hch =2.— ) ^ l^lOOOrpm) Using equation (y) 1 V /1000 rpm = 1. (Ounce-inches have dimension force x distance.3 5 8 = |33.18PP The torque constant of a motor is the ratio of torque to current and is often given in ounce-inches per ampere. (a) Show that the units ounce-inches per ampere are proportional to volts per 1000 rpm by reducing both to MKS (SI) units.8 N = 2 . In consistent units.358 Ounce .55x10-® c/s 1 V o ltp e rl0 0 0 fp m = 9 .96 Ounce . Kt = 33.54 cm = 0.0283 kg x 9.8 ^ x lk g = 2. Kj = 25 x l.205 Pounds 9 .036 _ ■M ^ 1 VoltperlOOOrpm 9.inch per Ampere .9^ O unce-inch per Ampere Step 4 of 4 (C) .5 5 x l0 ^ a 1 Ounce-inch/Amp 7.28 Ounces 1 Ounce = 0.) The electric constant of a motor is the ratio of back emf to speed and is often given in volts per 1000 rpm. What is its torque constant in ounce- inches per ampere? (c) What is the torque constant of the motor of part (b) in newton-meters per ampere? (c) What is the torque constant of the motor of part (b) in newton-meters per ampere? Step-by-step solution step 1 of 4 (a) 9 . Problem 2. where an ounce is 1/16 of a pound.55 Step 3 of 4 (B) K .8 n = 35.036xlOr^N-m/A = 239x10-® N-m/A = |0. = 9.96x7.'. = 25 f — .rad Nnn. the two constants are the same for a given motor.s = 9.O unce-inch = 7.'.036 X 10"® N -m /a (-) Step 2 of 4 30 1 volt per 1000 amp = — xlQ*^ V o lt/ra d /s n rad.55 c.0254 m 1 .2 0 5 x 16 Ounces 9.inch per Ampere Using equation (a ) K t =33.239N-m/A| .8 “ = 0. (b )A certain motor has a back emf of 25 V at 1000 rpm. Problem 2. The charge q and the voltage e across the plates are related by » = C(i)e. The system consists in part of a parallel plate capacitor connected into an electric circuit.f Step 4 of 6 T h e equation c fx notion fo r fo e above plate.) (b) Can one get a linear model? (c) What is the output of the system? Figure Simplified model for capacitor microphone Step-by-step solution step 1 of 6 W e h av e liie follow ing electxom edianical system : Step 2 of 6 ^ G iv en th at capacitance C is a fooctioo o f th e distance x betw een d ie plates. € is foe c b d ectiic constant o f d ie m aterial betw een die plates. as follows.^ = / A t ) and le A " ’ v = R q -V L q + ^ Step 5 of 6 ^ (b) W e cannot g e t a linear m odel because c£ d ie term s ^ a n d qx p resen t in th e differential equations. fe = 2eA' (a) Write differential equations that describe the operation of this system. (It is acceptable to leave in nonlinear form.19PP The electromechanical system shown in Fig. A is foe sur& ce a rea o f foe plates T h e charge q a nd foe voltage « across foe plates are r d a te d b y 9 = C (x )« T h e electric field in tu rn produces foe follow ing force o n fo e m ovable p late foat opposes its m otkm : Step 3 of 6 T h e fie e b o d y diagram o f f o e c ^ n c i f o r p late (b )ii b . Sound waves pass through the mouthpiece and exert a force fs (t) on plate b. The charge q and the voltage e across the plates are related by A = surface area of the plates. The capacitance C is a function of the distance x between the plates. where £ = dielectric constant of the material between the plates. A = surface area of the plates. The electric field in turn produces the following force fe on the movable plate that opposes its motion. which has mass M and is connected to the frame by a set of springs and dampers.^ eA T herefore d ie tw o coiqiled n o n linear equations are i & + B i+ K x + . a s follows: C (x )= - H ere. represents a simplified model of a capacitor microphone. Step 6 of 6 (c) T h e output o f foe system is foe current |r ( t ) = g | . Capacitor plate a is rigidly fastened to the microphone frame. 6 is T h e equation f ix foe circuit p art is at ( Since e is fo e ^ ^ o fia c itiw o lta g e J X I S in c e i( /) = ^ j dt d i‘ eA v = J 8 j+ l9 + . . K .k ( e .e . The rotor has an inertia J1 and a viscous friction B. the output torque is directly proportional to the armature current. J A * B 0 . The load has an inertia J2. a torque constant Kt.-e . is the torque constant Therefore.)= T .)+ iL { e .e .e .) . an armature inductance La. . is the electrical constant It is known that. ) (2) Step 2 of 2 Apply KirchhofTs voltage law. ) + T .51 in the textbook. ) .20PP A very typical problem of electromechanical position control is an electric motor driving a load that has one dominant vibration mode. and a resistance Ra. The sum of applied torque is equal to the sum of opposing torques on a rotational system. Step-by-step solution step 1 of 2 Refer to Motor with a flexible load in Figure 2.+ b (i^ -e . T . is the inertia of the rotor ^ is the viscous damping g is the viscous friction is the spring constant is the motor torque Write the equation of motion for a motor. (d Here. Write the equations of motion. Figure Motor with a flexible load R. Here.. and many other applications. According to Newton’s second law of motion.b { A . The motor has an electrical constant Ke. the sum of torques acting on a rotational mechanical system is zero. A schematic diagram is sketched in Fig.e . reel-to- reel tape drives. The problem arises in computer-disk-head control.= K .l..k ( e . The two inertias are connected by a shaft with a spring constant k and an equivalent viscous damping b. the equations of motion are. Problem 2.B e . Write the equation of motion for a motor. j A 2 = .b ( g . Here.. J A = . Step 8 of 13 Neglect the translation inertia of the system and write the equation Step 9 of 13 Consider friction force. = 2T^ Step 13 of 13 Thus. inertia. Assume you have access to whatever you need. friction and are both zero.. y^is the inertia of the drive wheel. is the wheel angular acceleration. Figure Robot for delivery of hospital supplies. Add the rotational inertia of the two other wheels and the inertia due to the translation of the cart plus the center of mass of the 3 wheels. Problem 2. assume you have command of the torque on a servo motor that is connected to the drive wheels with gears that have a 2:1 ratio so that the torque on the wheels is increased by a factor of 2 over that delivered by the servo. Your equations will require certain quantities. Step 7 of 13 Let the angular acceleration is and assume the inertia is same as the drive wheel. T is the commanded torque from the motor.) ^ ^ =27^ . Step 4 of 13 So. Step 10 of 13 Substitute r § for a. f Where. for example.21 PP For the robot in Fig. Step 2 of 13 Assume that robot has no mass. mass of vehicle. is the mass of the cart plus all three wheels. Step 6 of 13 The acceleration of the drive wheel is directly related to the acceleration of the robot and its other wheels. Determine the dynamic equations relating the speed of the robot with respect to the torque command of the servo. Source: AP Images Step-by-step solution step 1 of 13 Refer Figure 2. Let’s also assume there is no damping on the motor shaft. Step 3 of 13 The motor must have a gear that is half the size of the gear attached to the wheel. Step 11 of 13 Consider angular inertia Step 12 of 13 Substitute 2 for n. provided there is no slippage. and radius of the wheels.45 in textbook. the dynamic equation relating the speed of the robot is + 3 7 ^ + 4 7 . Step 5 of 13 Consider the wheel attached to the robot Where. ( m „ r j + 3 7 ^ + 4 7 ^ ) ^ . is the motor inertia. So multiply the torque by a factor of 2. r. Step 6 of 13 The angles will change in proportion to the angular velocities e..73 in textbook and eliminating 7J the two equations (y . Problem 2.. is the torque applied on gear 2 by the gear 1. =nT. That is. + J.ZL i = / Step 4 of 13 Consider the torque multiplication Is proportional to the radius of the gear T.. T Is the reaction torque from gear 2 acting back on gear 1.( » ) ( j j + J . Oi AT. n ’ ) s ^ + ( b . Tm.. derive the transfer function between the applied torque. where Ts = -Ks92. and the output. (1) Where. 62.ff.35 in textbook.. N is the number of teeth. S te p 1 3 o f1 3 Thus.. JV... . €0 is the angular velocity. Where. Step 5 of 13 The velocity of the contact tooth of one gear is the same as the velocity of the tooth on the opposite gear velocity=o»’ Where.n')Si + ( i .T . Ts. there is a torque applied to the output load. o. The inertia of gear 2 and all that is attached to body 2 Is J^. +K. Adding the spring torque in body 2. Figure (a) Geometry definitions and forces on teeth.. n is the gear ratio. Step 8 of 13 The spring is only applied to the second rotational mass so torque only effects. Step 2 of 13 The force transmitted by the teeth of one gear is equal and opposite to the force applied to the other gear as shown In Figure 2. torque-ftwcexdistance Step 3 of 13 Consider the torques applied to and from each shaft by the teeth ZL.. N. The friction ^ and . (3) step 11 of 13 Substitute 7J for 7*... Step 7 of 13 Consider the equation of motion for body 1. the transfer function of the system is ( J".. r is the radius of the gear. + . N.* b . (b) definitions for the dynamic analysis “0 ^ Step-by-step solution step 1 of 13 Refer Figure 2.. r. The Inertia of gear 1 and all that is attached to body 1 is J^.35 (a) In textbook.22PP Using Fig.a ’ '^s + K . Step 10 of 13 Substitute for ^ In Equation (1). ■ f A + W .(4) Step 12 of 13 Use Equation (3) and (4) the relationship in Equation 2.( 2 ) Where. The servo motor output torque is attached to gear 1.T . . r . So the servo’s gear 1 is meshed with gear 2 and the angle ^ Is position (body 2).. for the case when there is a spring attached to the output load. Step 9 of 13 Consider the equation of motion for body 2. in Equation (2).. (b) R f= heat-flow coefficient between the actuator and the air. Tamb = ambient air temperature. . Figure (a) Precision table kept level by actuators. and (3) the motion d is proportional to the difference between Tacf and Tamb due to thermal expansion. relies on thermal expansion of actuators under two corners to level the table by raising or lowering their respective corners.23PP A precision table-leveling scheme shown in Fig. Assume that (1) the actuator acts as a pure electric resistance. Step-by-step solution step 1 of 2 Step 2 of 2 Proportionality d = Power In Put 551 HEAT m l R R And using equation 2. Problem 2. R = resistance of the heater. Find the differential equations relating the height of the actuator d versus the applied voltage vi. (b) side view of one actuator 1 . The parameters are as follows: Tact = actuator temperature. C = thermal capacity of the actuator. (2) the heat flow into the actuator is proportional to the electric power input. Bq.73 We get C 7L + d cr„+ Rtqd . Figure Building air-conditioning.(a). R.(b). (2) there is no heat flow through the floors or ceilings. Problem 2. where To = temperature outside the building. Write a set of differential equations governing the temperature in each room.24PP An air conditioner supplies cold air at the same temperature to each room on the fourth floor of the high-rise building shown in Fig. (a) high-rise building. Ro F o r-r = + -(3 ) RequideEquations are (l)& (2 )& (3 ) . (b) floor plan of the fourth floor Step-by-step solution step 1 of 2 Step 2 of 2 “C" Thermal cs^acity o f Air *'1” Corners are equivalent “2” Arc equivalent Assuming. The cold airflow produces an equal amount of heat flow q out of each room. ( 1) F o. ^ >7J > ^ > 7 J For * 3" .(2) s. Figure Building air-conditioning. Ro = resistance to heat flow through the outer walls. = --------. and (3) the temperature in each room is uniform throughout the room. (b) floor plan of the fourth floor ui syiiiineiiy (OTeuube me iiuinuei ui uiiieteiiiiai equauuiis lu iiiiee. Take advantage of symmetry to reduce the number of differential equations to three. Assume that (1) all rooms are perfect squares. Ri = resistance to heat flow through the inner walls. -2 . The floor plan is shown in Fig. (a) high-rise building. is the area of tank 2. . Figure Two-tank fluid-flow system Step-by-step solution step 1 of 2 From the relation between the height of the water and mass flow rate. p is the density of water. Problem 2..25PP For the two-tank fluid-flow system shown in Fig.. the continuity equations are (3) M Here. ^ Is the area of tank 1. find the differentialequations relating the flow into the first tank to the flow out of the second tank... p ^ W (8 ) Therefore. (5) (6) I I Substitute for w In equation (3). the differential equations relating the flow into the first tank to the flow out of the second tank are. = w -w ^ p A ji^ ■ . (4) PA i Here.. And. p is the density of water. Is the height of water in tank 2.. is the height of water in tank 1. Step 2 of 2 From the relation between the pressure and outgoing mass flow rate. A = -L f4 -i I I I i Substitute for w and — f or in equation (4). l+ . Step 4 of 5 (c) With hole B closed and hole A..0I)(60) e w (s) ° 6 0 o r 1 V r*6 0 o J 0. (b) At h1 = 30 cm and h2= ^0 cm.... g=981cm /sec* slOOOdVsec^ Substitute 2 0 0 g /m iifo r for ^a n d lOOOcn^sec* for g in equation (2).lO .. 600j Convert the inflow unit in grams/min • <?//. the outflow is 200 g/min.W ( 22 ) ’ 30l.{ s ) . (25) Substitute 30 cm for Aj^. 20 30l.— + — S W ..i‘ pit I.. lOOOcm/sec^ for g and 30 for R in equation (14).. (5) (6) Consider the linearized equation..L a A . ^Igram /cm ^xlO O O cm /sec’ xlO cm 200g/60s 100 /g c m “ s> J?s30 g^cm 2 30g ^cm Step 3 of 5 (b) Consider the following nonlinear equation from above information.. .-W ^ = pAh.58 in the textbook. 1 for /?. but none at A..S h j) - (1 )(1 0 0 )(3 0 )' ' ^ ( l)( 1 0 0 0 )( 1 0 + 5 ^ ) (1)(100)(30) 1 (15) From equation (13) and equation (15). S k— — < S h . 200 g/inn = -!-[lg ia m /c c x lOOOcn^sec^ x lO c m p 1 T— Jg = 2oo / — [lg ra m /c m * x l0 0 0 c m /s e c * x l0 c m J 1 J ts .(12) ’ 3 0 l.(19) Substitute equation (7) in equation (18).. compute a linearized model and the transfer function from pump flow (in cubic-centimeters per minute) to h2.001 0. describes flow through the equal-sized holes at points A...K ( 10 ) ’ 30i..26 Pump Step-by-step solution step 1 of 5 (a) Step 2 of 5 Refer FIGURE 2. Problem 2. h i > 20 cm. (11) Consider the value of nomina! Inflow Is '^Iven below 10 =y>^cm V sec Substitute equation (11) in equation (10). So. 1 r ^ { l) ( im ) ( 3 0 + S h . 40 ’ 40 100 " Consider the total inflow equation. lOOOctn/sec^ for g and 30 for R in equation (21).2 ^ fi+ . 1 /— 7Z— 7:— r — i Substitute 30 cm for 10 cm for Aj. 100 cm2 for A. the desired transfer function in cmYsec • 3 H .. 1 r^(l)(1000)(30+5A| -2 0 ) - (1)(100)(30)' ' ^(l)(1000)(10+5^) (1)(100)(30) 30l.-S h . (3) W ^-W g = p A h ^ .W . Consider the following nonlinear equation from above information. ^ _ L ) l.. . B.. 20 J 100 s k .26PP A laboratory experiment in the flow of water through two tanks is sketched in Fig. the values given for the heights ensure that the water will flow is below as. W ^ .. holding the nominal flow rate maintains h i at equilibrium but h2 will not stay at equilibrium.... 1 ..001 Thus. . The area of both tanks is A. = _ J _ | 1 + J _ ^ fc 1+J _ ( ( F (23) ^ 20 V 100^ " ■ ^ Consider the value of nominal inflow is remc 301.L h + J -^ fc l+ J . 100 cm2 for A. 1 for p .( 8 ) Substitute equation (7) and equation (8) in equation (5). m (s) 7 T1 (27) V 60oJ f n . (c) Repeat parts (a) and (b) assuming hole B is closed and hole A is open. the standard transfer function will not result. Figure Two-tank fluid-flow system for Problem 2. Figure Two-tank fluid-flow system for Problem 2. (7) .. (1) ( 2) W . Hence.= -^(1)(I0O0)(30+5*i -1O-5^)h(1)(100) (1)(100)(30)' s k = . Assume that when /?2 = 10 cm. Assume that Eq.)+ — m (13) ’ 120 0' ’ 100 Substitute equation (7) and equation (8) in equation (6). 600j (28) Step 5 of 5 Substitute equation (27) in equation (28).. lOOOctn/sec^ for g and 30 for R in equation (25)... 100 cm2 for A.— — S h .(»)_1 (0. the desired transfer function is ("iJ .(14) Substitute 30 cm for 10 cm for Aj. write the equations of motion for this system in terms o h1 and h2. (18) .. (4) Consider the value of gravity. 20 100 * Substitute equation (11) in equation (22)..26 cm. (26) ’ 600 ’ 600 ’ Take Laplace equation (24) and equation (26). 40 ’ 40 lo o ' ” ’ Substitute y ..L ( » ' + ^ » ') . 1 for p . W ^ = W ^ + S W . 20 cm for A^. W . S k . the values given for the heights ensure that the water will flow is below as. the constant term increasing h2. Sh. Assume that h3 = 20 cm.= - = . ( 21 ) Substitute 30 cm for Aj^. . 60o A 600j S H j( s ) 1 0. 1 for p . 100 cm2 for A..01 •Y l. h i > 20 cm.L tfk — L t f t V J . .. 1 ^6003 f : 7 X j r ..v ^ c m ’ /sec for in equation (12). (24) ’ 600 ’ 100 Substitute equation (7) in equation (19). 20 ’ j S k -— Sh. 20 cm for A^. or C. and h2 < 20 cm. (a) With holes at B and C.= p A h . and h2 < 20 cm.-W ^ = pA h.. IQOQcip/sfC^ for g and 30 for R in equation (9). (c) To is the temperature outside the house.1 h). “C. °F per BTU/h. = 60’ F a=0 = 2 F lh r dt .=32‘ F . 000 BTU/h. (1) R '■ ' Step 3 of 3 (ii) T.T 2 ) .y . “ F. = 90.385x10-® F lB T U jh r C= 3. where C is the thermal capacity. Problem 2. the house temperature falls 2°F in 40 min. 0 0 0 / i 5 = 2. Equation 2 t. with the outside temperature at 32°F and the house at 60°F. with the outside temperature at 32°F and the house at 60°F. (1) and (2). and in a particular case can be written with time in hoi/rs as = jr* — HLZlZe dt R * where (a) C is the thermal capacity of the house. (f) u is the furnace switch. 7J=60"i? u=l ^ = 2 < f F lh r + 2 0 c + — =90.3 c + — = 9000 R fiC = 2 8 / 3 -------. the furnace (f) u is the furnace switch. (d) K is the heat rating of the furnace. It is measured that.000 RTUlhr Step 2 of 3 (0 7^ = 32^ F . where q = heat-energy flow. the furnace raises the temperature 2*F in six min (0. *C/J ? sec T = temperature. What are the values of C and R for the house? Equation 1 q = ^ ( T i. T. *F.27PP The equations for heating a house are given by Eqs. BTU/*F. (b) Th is the temperature in the house. S te p -b y -s te p s o lu tio n step 1 of 3 CdZ dt R 90. (e) R is the thermal resistance.000 -------. With the furnace off. = 1 if the furnace is on and = 0 if the furnace is off. (2) Putting (i) we get i^ 5 ^ + 2 8 = 9 0 . = 1 if the furnace is on and = 0 if the furnace is off.913x10® B T U l-F . It is measured that. joules per second (J/sec) R = thermal resistance. 3 t (d) = (e) m = sinh t c C ontrol O f Dynamic Systems (7th Edition) P ro b le m Find the Laplace transfomi of the following time functions: (a )/(f)= 1 + 7 f (b) fl. s + \ s+2 (5+3)‘ Thus.f) = 4 + 7f + f2 + 6(t). « /’ +2/ + l aiep iu or iz Take Laplace transform. the Laplace transfomi is 5+ 1 5+ 2 (5 + 3 )* Step 8 of 12 (d) Step 9 of 12 Consider the time function. Problem 3.L ( / ') + i( 2 / ) + L ( l) 2! 2 I =-r + —+- j s s* + 2 s + 2 . where 5(t) is the unit impulse function (c) f{f) = e .f) = 4 + 7f + f2 + 6(t).2 t + t e . 1 7 s+7 Thus.f + 2 e . the Laplace transfomi is s+7 Step 3 of 12 (b) Step 4 of 12 Consider the time function / ( r ) = 4 + 7 f + r * + ^ ( r ) step 5 of 12 Take Laplace transform. where 5(t) is the unit impulse function (c) fl.2 t+ te .t + 2 e .3 t (d) = (e) = sinh t (e) f{f) = sinh t S te p -b y -s te p s o lu tio n step 1 of 12 (a) Step 2 of 12 Consider the time function / ( 0 = l+ 7 r Take Laplace transform.0 = e . i[ / ( / ) ] .01 PP Find the Laplace transform of the following time functions: (a )/(f)= 1 + 7 f (b) fl. where 5(t) is the unit impulse function (c) fl. «/’ +2/+l aiep iu or iz Take Laplace transform.+ — + - r j s s * + 2 i+ 2 Thus. the Laplace transfomi is 5+1 5+2 (5 +3 )* Step 8 of 12 (d) Step 9 of 12 Consider the time function.02PP Find the Laplace transform of the following time functions: (a )/(f)= 1 + 7 f (b) fl.2 t + t e . the Laplace transfomi is 5 ^ -1 . the Laplace transfomi is Step 3 of 12 (b) Step 4 of 12 Consider the time function / ( r ) = 4 + 7 f + / * + ^ ( r ) step 5 of 12 Take Laplace transform. 1 7 s+7 s+7 Thus.+ -------+ - s + \ s + 2 (5 + 3 )' Thus. 4 7 2 !. = . Problem 3.L ( / ') + i( 2 / ) + L ( l) 2! 2 I =.+ — + — +1 s s s «’ +45* + 7j + 2 Thus. the Laplace transfomi is 5*+25+2 Step 11 of 12 (e) Step 12 of 12 Consider the time function. 1 2 1 .f) = 4 + 7f + f2 + 6(t). 1 *5 *-l Thus.3 t (d) = (e) f(f) = sinh / (e) f{f) = sinh t Step-by-step solution step 1 of 12 (a) Step 2 of 12 Consider the time function / ( 0 = l+ 7 r Take Laplace transform.0 = e . /(/) * s in h r Take Laplace transform. the Laplace transfomi is +7^+2 Step 6 of 12 (c) Step 7 of 12 Consider the time function f { l ) = e " + 2 e ' ^ +16'” Take Laplace transform.f + 2 e . i[ / ( / ) ] . the Laplace transfomi is s'+9 s'+9 (s+l)"+9 Step 5 of 6 (C ) step 6 of 6 Consider the time function / ( » ) = « ’ +«■“ sin 3» Take Laplace transform. s4- s^+ 3 6 Thus. £[/(/)]■ L(/*)+£(«■*sin3/) 2! 3 *. l(/(r)) =i(sin3/)+i(2cos3r)+i(«‘'sin3/) 3 2s 3 j'+9r + s'+9 -^ + -: (s+l)’+9 2s Thus. the Laplace transfomi is .• S + fl Take Laplace transform.j ---.r + s’ (s+2)’ +9 =A 3 °s’ '^(s+2)’+9 2 Thus.r + s’ (s+2)’ +9 .• S +<r Take Laplace transform. the Laplace transfomi is Step 3 of 6 (b) Step 4 of 6 ^ Consider the time function / ( / ) = sin3f+2cos3/+e*'$in3/ • Consider £(smo/)g . Problem 3.03PP Find the Laplace transform of the following time functions: (a) f{t) = 4 cos 6f (b) ^t) = sin 3f + 2 cos 3f + e -t sin 3f (c) f{f) = t2 + e -2 t sin 3t Step-by-step solution Step-by-step solution step 1 of 6 (a) Step 2 of 6 Consider the time function / ( / ) = 4 c o s 6 f Consider £(cos<tf)=. + 2 tc o B t Appiy the Lapiace transfomi on both the sides of the time function. /s in 3 /. Apply the Laplace transfomi on both the sides of the time function. we solve the given time function by separating them. /:{/(/)} = £ { le " +2fcost} F{s) = £ { le " } +2r {/cos/} 1 ( s ‘ + 1 ) . lf 2s j‘ + 2j ’ +1 Therefore.I).l) ( s '+ 9 y ( s 't l) ^ Step 8 of 9 (e) Consider the time function. 4 f Step 9 of 9 Further simplification yields. ^(^‘ . Step 4 of 9 Now. £ { .2 s ’ = 1 -2 ( s '. find the Laplace transform for the whole function. /cos3/ is Step 5 of 9 (C) Consider the time function. l) ( 0 ) . ^ {« (0 } = — G (>) f ( . » . l( /)+ 2 /c o s 2 / is 5 *+ 2 j ’ + 8 s* .I " j ‘ + 2s’ +3s*+4s’ +3s’ + 2j + 1 Therefore. Step 2 of 9 Now. the Laplace transfomi of the time function.0 ~ { s ^ l f (s'+lf (s’ + 1 ) '+ 2 ( s + 1)’ (s* .l ) (s’ + 9 ) '" ( s '+ 1)' Therefore. the Laplace transfomi of the time function.2 s ' r+2 ** c-^21 6s 2(»’ . we solve the given time function by separating them.t + 2 t c o s t (d) f[t) = ts\n 3 / .± G ( s ) -s ‘ -9 + 2 s‘ ° j ' +18s’ +81 s ^ -9 s'+\Zs'*% \ 4 ^ -9 Therefore..2 s * + 4 s * .4 s .04PP Find the Laplace transform of the following time functions: (a) f{t) = / sin / (b) f{t) = t cos 3f (c) f{f) = t e . 1 -s ’ F (s ) {s ^ iy ____1 . j ( 2 £ L + 2 [ ( £ i 1)z ! E ) 1 j ’ + l. ) l (. In order to do so. First. J 2s ( s . te^ + 2 tc o s t '® 3 s *+ 4 s’ + 2 s ^ . i ( /^ 4 '2 / oob2 / Appiy the Lapiace transfomi on both the sides of the time function. the Laplace transfomi of the time function.1) (s+l)’(s*+l)' (s*+2j’ +1)+(j^+2s+l)(2s*. Problem 3. £ { f { i ) } = £ {l(l)+ 2te os2l} F ( s ) = / '{ l ( / ) } + 2 r { / c o s 2 / } ( s* + 4 ) .4 s + 2 s '.s (2 j ) r-2 (*+ > r __ i_ s’ . In order to do so..g ij) } = .2 f c o s f (e)/(0=1(fJ + 2/cos 2f (e) /(0 = 1(f)+ 2 / cos 2f Step-by-step solution step 1 of 9 (a) Consider the time function.r^= to a B 3 f Apply the Laplace transfomi on both the sides of the time function.2 /c o s / is 6s 2 ( s '. Since there are two variables in the above function we need to use the multiplication by time Laplace transform property. First.2) (s“+2s+l)(i* +2s’ +l) ^ s * + 2 s ^ + l + 2 s * . /■ { /{ /)} = COS/ } F ( j ) = ^ { t s in y ) ~ 2 jC {/c o s /} .2 i ' + I -2 Step 6 of 9 Further simplification gives. consider ^ ^ /^ = s in r and find the Laplace transform of it.( 2 j ) 5 = 1 -2 s '+ 4 . Apply the Laplace transfomi on both the sides of the time function. find the Laplace transform for the whole function. £ [ f ( i ) ) = £ { tc o s 2 i\ Since there are two variables in the above function we need to use the multiplication by time Laplace transfomi property.l Step 7 of 9 (d) Consider the time function.-. /]^.4 s .( l) ( 2 .^ s {s ^ + 4 f s * + 2 s ^ + S s ^ -S s + l6 s (s ^ + 4 )" Therefore.2 s‘ + 2s*+s’ + 2s*+4i’ + 2s+s*+ 2i“ + 1 3s* + 4 s ’ + 2 j " . /s in / is s* + 2s^ + \ Step 3 of 9 (b) Consider the time function. the Laplace transfomi of the time function. the Laplace transform of the time function. consider g ^/^ = cos3/ and find the Laplace transform of it. j( j* + 4 ) * ^ s*+Bs^ + \ 6 + 2 s ^ .8 s +16 s(s^ +A '^ . .r)sin rdr 0 Take Laplace transform. (2) Compare Equation (1) and (2) and find a and b. .05PP Find the Laplace transform of the following time functions (* denotes convolution): (a) ^ t) = sin t sin 7t (b) fl:0 = s in 2 f+ 7 c o s 2 f (c) f{f) = {sin t)A (^ ) / ( O = / c o s(^ — t ) sinr d r 0 (^) f i t ) = J cos(f . 2 ..f ( l) d l Jf (j)* =J je -“ / { l) d l Interchange the order of integration..'M Thus.. / ( 0 = ^ = lan‘'(oo)-tan"'(5) . 1 5 *+ 2 5 *+ ! Thus.. ■ KHA) 7^^+16 s (s “ + 4 ) 75*+ 16 Thus. cos(f-r)sinr</rj i(/(7))=i(cos(/)*sid(/)) (8) L (c o s t)= . (1 0 ) ' ' 5 +1 Substitute Equation (9) and (10) in Equation (8). the Laplace transfomi is + 3 6 )^5 ^+ 6 4 ) (b) Consider the time function.. =4+3cos2/ Take Laplace transform. the Laplace transfomi is 1 5* + 2j * + 1 Step 11 of 11 (e) Consider the time function. T) sinxdx 0 Step-by-step solution (a) Consider the time function / ( r ) = s io / s in 7 r (1) Consider the equation. (7) ' ' 5 +1 Substitute Equation (7) in Equation (6). Problem 3.jL . L ( s m ( ) = ..^ (9) f +1 L (s m ()= -jL . /(r)=sin*/+7cos^/ (3) Consider the trigonometric fomiula. /(/) s $ in /* s in /. = i [ ^ ______ f _ l 2U'+36 i*+64j I4« ( j’ +36)(j*+64) 14f Thus.f . (6) Take Laplace transform. /(0 =^cos(|l-7|»)-icos(|l+7|f) =-cos6f--cos8r 2 2 Take Laplace transform. l-cos2/ (4) 2 cos'f *-l+cos2r (5) Substitute Equation (4) and (5) in Equation (3). the Laplace transfomi is step 10 of 11 (d) Consider the time function. a = hb = 7 Substitute 1 for a and 7 for b in Equation (2). / ( / ) = J cos(/ . s u u > /s in 6 /K -^ c o s (|a -6 |/)— . i * + 2 i’ + l Thus. Consider the time function..t a n . the Laplace transfomi is r(j*+4) (c) Consider that division by time is equivalent to integration in the frequency domain F { s ) = ] e . the Laplace transfomi is .. find the Laplace transfonn of the following: (a) g(t) = f{t) cos t tn (b) gV) = fff(r )d rd ti Step-by-step solution step 1 of 2 Step 1 of 2 (a) g(t)-f(t)cost Usingproperty L[x(t)y(t)]-^X(s)*Y(s). fe)•F (s) Step 2 of 2 ^ Usingtheintegrationpropertyoflaplacetransforms» L \£ K x )d t] cks> .06PP Given that the Laplace transform of /(/) is F(s). Problem 3.^ 8 .! ^ = --------. i. the time function of .( F ( 5 ) ) = r '( ^ ] / ( / ) . i.e 'o o s ( / ) + ^ ( c ' s in (/)+ c o s (/)e ') = .) + 8 i„ ( . F ( i) = i — L S 5+ 1 Take Inverse Laplace Transform. / ( / ) = .« ' c o s ( / ) . fa + e 3 s *+ 6 s + 6 (s + i) ( s *+ 6 s + 10) ( s + 2 ) ( s * + 6 f+ 1 0 ) ^ 6 + | j s * + ^ ^ + c + 6 j s + ( c + 6 ) = 3 s *+ 6 s + 6 Step 6 of 16 Equate powers of s to find ^ and c.r ' _4_ + r ‘ _ L 5+ 1 5+5 Thus.1 2 ___ n ( 5 + 1) (5 -+ 1 6 ) 4 4 E _____ u l 19 4 (s + 1 ) ( j^ + 4 * ) 3 4 ( j^ + 4 * ) Take Inverse Laplace Transform.i) g ( < . 64 -+ c =6 _38 ^ " l7 *+ c = 2 Step 11 of 16 4 4 38 Substitute — for a . step 13 of 16 (h ) Consider the Laplace transform equation.( F ( 5 ) ) = r '( i) . Take partial fraction expansion and rewrite ^ ( 5)- ^ 5+1 . i.' ' '( 4 . 17 bs+c 2 (s + 3 ) (s+l) (j* +16^ (s+l)(j* +16) ^ ^ + * j 4 ’ + ( * + b)4 + [ ^ + ' j = 25 + 6 Equate powers of s to find ^ and c. Problem 3. ( 3 5 ’ + 6 s + 6 )| 3 ‘' ” (5 "+ 6 5 + 1 0 ) ''~ '“ 5 Substitute — for a in equation (3) and equate numerators.2 5 + 2 ' (5 + 1)’ +1 ( 5 . f 4 > r 4 . (5 + 1)(5 + 5)u • 1 __5 "~4 .c o s ( /) s in h ( r ) l( /) + s iii( r ) c o s h ( /) l( /) 5 Thus.l ) '+ l / ( / ) = 5-* c o s ( / ) . ------for A and — for c In equation (4).' 17 . ^and c.~ ( e . the time function o <f ^ is |(1 + /)1 (/)|.? -+ J . the time function of (5 + 1)(5=+65 + 10) Step 8 of 16 (e) Consider the Laplace transform equation.i c ‘ 's in 3 r l( / ) Thus. the time function of 3j + 2 j 3e"'cos 3 /1( / ) .07PP Find the time function corresponding to each of the following Laplace transforms using partial- fraction expansions: (a) F(s) - (b) f W — s(s+j)(s+5) (c) F ( l) = 3i+2 i'+ i+ lO (d) F(s) = 3F+6S+6 (i+ l)(i^ ' W + 6t i+ 1 0 ) (j+ D d ^+ d a + lO ) _1 (e) F(s) ? + l6 2(i+3) (f) F(s) = (J+1)(J^+16) (g) m = ^ (h) F(s) = 4 (i) F(s) = ?44 e -‘ (i) F (4 ) = 7 ^ Step-by-step solution Step 1 of 16 (a) Consider the Laplace transform equation.O . “ y c+ 6 -6 CmO 3 12 Substitute j ^ ^ in equation (3). ^ W = 7 Take Inverse Laplace Transform. the time function of ^ is Step 2 of 16 (b) Consider the Laplace transform equation ^ 5{5 + 1)( j + 5) Take partial fraction expansion and rewrite F ( ^ ) • Use the cover up method and find a. / ( / ) = 3e"'cos3f l ( / ) . the time function of «( j + I)(5 + 5 ) Step 4 of 16 (c) Consider the Laplace transform equation F (5 ).'( F ( 5 ) ) = £T' . r '( F ( 5 ) ) = r '( i) + L . Take partial fraction expansion and rewrite F (.—c"* sin 3 /1( /) s ^ * 2 s * l0 Step 5 of 16 (d) Consider the Laplace transform equation.= .j l — is |-c o s ( /)s in h ( /) l( /) + s in ( /) c o s h ( /) l( /) | Step 16 of 16 G) Consider the Laplace transform equation. the time function of — is 120 1(0 step 14 of 16 (i) Consider the Laplace transform equation. 4 4 F ( f ) s ------. the time function of —sin4/l(r) s*+16 Step 9 of 16 (f) Consider the Laplace transform equation. F (s )-f Take Inverse Laplace Transform.' .r (5 + 1) ^ + ! ( 5 .) .r)- F (s ).e " 's in ( / ) + c o s ( / ) « " ') ..2 _ + _ l _ s s+ 1 f+ 5 Take Inverse Laplace Transform. '1 i.r ' 17^ +C i f 19 4 (5 + 1) (5»+4») [ 3 4 ( 5 '+ 4 = ) J ^ / / W = f | e '''( ') .~ ( e ’ s in (/)) = tf"'c o s (f)—j( . ' ' 5’ +25 + 10 Take partial fraction expansion and rewrite F ( j ) - 3(5 + 1) 1 ~(5+1)’ + 3 ' (5+1)'+3* Take Inverse Laplace Transform..I 4 ' i.i) l .e < K ( / ) |2 l^ jl( . the time function of £ — is l ( » . rH' ( F ( 5 ) ) = r ' [ l ) /( ') = ^ '( 0 120 '' Thus. 17 17 17 4 4 38 -T ^^+ T F f 5 l = .) Therefore. is Heaviside step function. 3 12 F (5 ) = y ' ( 5 + 1) ( 5^ + 65 + 10) 3 < 12 5 ( 5 + 1) 5 ( 5^ + 65 + 10) Step 7 of 16 Take Inverse Laplace Transform 3 > r '( F ( 5 ) ) = r ' ( 5 + 1) [ 5 ( 5 ^+ 6 5 + 1 0 ) ] =[|e'''(0-ye'*s‘»(')l(0+y«'*<»s{')l(0] 3s*+6s+6 Thus.r '( ^ ] Thus.i5 + l 5'+ 2 5 + 2 5'.+ . 5 + 1.1 )*+ ! Step 15 of 16 Take Inverse Laplace Transform (.'( F ( 5 ) ) = i . ' 5 ' f j ' L .'( F ( 5 ) ) = i. ( /. ) |z l^ j|( . 4 Step 10 of 16 Substitute j y ^ in equation (4) and equate numerators.' l 4 ( 5 '+ 4 ‘ ) /(/) =lsin4/1(5) Thus._ 5_ | s (5 + \y ^ I "4 Step 3 of 16 Substitute i for a . ^ ^ (5 + 1 )(5 *+ 1 6 ) Take partial fraction expansion and rewrite F ( s ) - bs+c F (5 ) = (4) (5 + 1) ( 5^ + 16) 2 (5 + 3) .l) Here. —— for b ^nd — for c in equation (2).' s in ( / ) ) .3) s*\ j +6^+10 Find the value of a. F (5 )= ^ Take partial fraction expansion and rewrite ^ ( 5)- F ( 5 ) = l+ i 5 5 Take Inverse Laplace Transform. (1) S J + 1 Use the cover up method and find a and b ■ * 4 u .i Substitute i for a and _ ] for b equation (1). the time function of jy C o s 4 / l( / ) + ^ s in 4 / l( r ) (5 + 1)(5*+16) Step 12 of 16 (g) Consider the Laplace transform equation.! 2 .( l] /(/) = ( i+ /) i( /) Thus.l) 9 ( /. = ___ ' ' (5 + l)(5 ’ +65 + 10) Take partial fraction expansion and rewrite F ( s ) - = .^ « » 4 < l( 0 + ^ s in 4 / l( / ) 2 ( 5 + 3) Thus.'( F ( 5 ) ) = r ' i . Take partial fraction expansion and rewrite F ( s ) - Take Inverse Laplace Transform. Thus. 150(e-^“ + « '“ ) + y/1.8 9 5 )e '” | fl.' -/4.8 9 5 Find the value of c.3cos(2»)+3.i .6 4 + y 2. / 2 ^ ) Step 8 of 12 (e) Consider the Laplace transform equation.28cos(2/)+5. Find the value of o=(j+l)F(j)|. Take Inverse Laplace Transform. -4.1S0(2cos(2/))-/rt.28«^ -/4.895 Substitute 1.^ i j( j+ l) * Step 6 of 12 (d) Consider the Laplace transform equation. .280 Find the value of d ^ ( s . j.95(-y2sin(2/)) “|-0. r '[ ^ ] / ( / ) = (2-2/d-')l(0 2 (5 * + 4 + 1) Thus.'f (.e .-4 .8 9 5 e -F '-0 .( 4 .1 5 0 + y i. ^ / ) + lc o s h ( .3/c<»(2()-1.895(«-''“ -e^) r 1. step 5 of 12 Substitute 2 for 0 < —2 for ^ and 0 for c in equation (2). the time function of j * + 5 + 2 i_ s *-A ^ s in h ( .28e-'-»8.895(2sin(2/)) ={l.s) = taa ' ( i ) Step-by-step solution step 1 of 12 (a) Consider the Laplace transform equation.28e-'+3. 2S >1.1 5 0 + y l. . Step 11 of 12 (f) Consider the Laplace transform equation.8 9 5 for 6 .9 5 )« -^ -(4 .') i( 0 1 Thus.= -2 2 ( j* + j+ l) 1 2(2 j + 1 ) 5 .28cos(2/)]l(l) 2(i +2)(j+5)' Thus.6 4 « ^ + J 2 M S e ‘ ^ ) J _ |l. /(/) =[l.7 U . the time function of tan"' - ■ 0 ' is >(0 . Take partial fraction expansion and rewrite ^ ( 4)- As -¥B Cs + D F (s )^ j* .28*'' -/4.r)- Ff$)=— fl±£±l— 1 ^“ 1 Take Inverse Laplace Transform. ^ ( ) + i c o s ( V 2 f ) Thus.150(2cos(2/))+yrt.O S e'” ! ~ |.8 9 5 e '” J Rearrange the equation.150+>1.2 j" + 2 3 1 1 1 —j + — —s — .150+>1. [l.79sin(2<)-8. 4 ___ 2 ■ 5^-2 +2 Step 7 of 12 Take Inverse Laplace Transform (3-5 +-n i-'(F(j))=r' 4 2 + r‘ 4 2 *'-2 i’ +2 ^ J ) = ^ s ‘* * ( ' / 2 ' ) + ^ c o s h ( .79siii(2/)-8.3 ( . ' ( .64+>2.28g-*+3.9<sm(2/)+5.r)- r^/ \ O b C Use the cover up method and find a ./ \ 2 0 -2 Take Inverse Laplace Transform. the time function of .0 .950 ^-4. 1.^ + y 2 . Take partial fraction expansion and rewrite F ( s ) - 5* 1 (* + 0 (» + ') Take Inverse Laplace Transform. * -l Step 2 Of 12 Substitute 1 for a .y 2 .+ iy tU=' c = iu = . e^dr .9 5 0 Find the value of J}. =.6 4 -.9 5 )e ^ ]1 / ( ') ■ -(0 .150e--'“ + yl.^ s in ( .280 for a .^ s i n ( .2 ( j * + « + 1) u .79(sin(2/))}l(/) Rearrange the equation. -0 .95«-^“ ).28«-' + /[(-4.^ » ) + ic o s ( .08PP Find the time function corresponding to each of the following Laplace transforms: (b) = (0 = (d) = (f) F (s ) = W iW (g) F(.« .2 .* Step 12 Of 12 (g) Consider the Laplace transform equation.2 j) ^ F ( s ) \^ . c -b * = .6 4 e ^ -y 2 . the time function c Step 3 of 12 (b ) Consider the Laplace transform equation.( 0 . Take partial fraction expansion and rewrite F (.28*. 6 4 e .5 -i Take partial fraction expansion and rewrite F ( 4)- r^/ \ O b C (»+l) (i+l)’ '" W 'T +sT T n T + T T T T T (2) 2 (5 ^ + 5 + 1).^ is (*+!)(*’ +4) |[l.3/cos(2/)-1.8 9 5 e .2 .1 5 0 + y i. ] for c in equation (1). ^and c. ^ / ) .6 4 -y 2 . r '( F W ) = r 'g ) . • ] for ^and .1 50-> 1.( 0 .895 for c . the time function of -------.-/4.28«-' + /[(-4 .2 jy * { s * 2 j) ‘ Step 10 Of 12 Take Inverse Laplace Transform.^ ) .2 8 « . / ( 0 = ( i. ^ -8 3 -3 9 y 20 *-4. Problem 3. 2(^-f2 )(^H -S )’ ' ( i + !) ( * = + 4 ) ’ Take partial fraction expansion and rewrite ^ ( 5)- a.6 4 « .895(-y2sin(2/)) >(') 1.8 9 5 )e -^ “ -( 0 .950 Step 9 of 12 Find the value of e.64(«-F' +*>“)+y2.64->2.))= r'(^ ] / ( ') = ^ 'W Thus.1 5 0 e ^ -yrt.^ - sin(/) •w s in (/) Thus.64(2cos(2/))-/2. -0.I 5 0 « '’' + yl. Take partial fraction expansion and rewrite F (. the time function of -------.64+>2. r'(F (. . 4 ___ 2 .6 4 -> 2 .950 for e in equation (3).^ is ( /c o s / ) l( /) • 15* +11 *----------------.895 ^^.)=tan-[i) Rewrite F ( 5) in terms of series. the time function of is Step 4 of 12 (c) Consider the Laplace transform equation.i --------.'.9/sin(2/)-1.64(2cos(2/))+. j+ iy 2^5* + lV V . F( 1-280 ^-0.150e-^“ + -« 4 . ^W=7-37^57-- Take Inverse Laplace Transform.6 4 + . 2 ( j* + j+ l) --------. .9/sin(2/)+5.95(«-^ '^“ )1 ”|-0. ^ / ) . r'(F(*))=r'[t»-g]] °'^'(7"37‘"57""] / ( 0 = i.150-.2.= +■ )' / ( 0 = ( / c o s » ) l( / ) Thus.950 for d and -4.150->1.895 ^-0.28cos(2<)]l(>)|.0 .95e'“ )]l |.^ . r'(F(*))=r W .95(2sin(2/)) ”|-0.i._. F(.950 s+\ * S -2 J ^ S + 2J { s . ^ -1 2 8 -5 7 9 j 200 = -0 . s in i( cosbt y ( » ) = ^ ^ s in /. \= A + B + C 1 = 4 --+ C 2 C=l .8 3 3 ) ^ Step 17 of 19 Apply Inverse Laplace transform. |y ( t ) = / .l ) + l y (4 )= ( 4 .5 ( 0 ) ] .2 ] .l + 4 y ( 4 ) = .2 ) ( . [4’ y ( 4 ) .j ’'( * ) = 7 d j7 4 i.5 ' 4 4+ 1 4+1 4+1 Step 12 Of 19 Apply Inverse Laplace transform. B (. f ---. (/v ri) . i+ 3 .^ s iii^ j+ ^ s iii. i^ 9 W 2^- (e) 5 ( 0 + 2 5 (0 = e*.+ 1) (4‘ + ( V 3 f ) _ i ___ 1 s ^_I_ step 14 of 19 Apply inverse Laplace transform. t 1= H 0 Therefore.3 V -0.s y ( 0 ) .5«"'-0. Use the following formula. the expression for y ( ' ) is.833e-4‘ | Step 18 of 19 A (f) Consider the equation is y ( j ) + y { t ) = t Apply Laplace transform on both sides. y4(0 + l) ( 0 ’ + l ) = 4 A=A Step 10 of 19 Determine the value of g . [4 ’ y ( 4 ) .0 .- Substitute the values of AyB ^C iD in y ( * ) expression.y ( s ) . y (0 = c » '[ c o s [ : | '. (a) 5 ( 0 + 5 ( 0 + 3 y (0 = 0.r y i ---. [4 “ r ( 4 ) .' =cosbl ‘" ( T r y ) " Therefore.le -i* -Ju. Step 13 of 19 (b) Consider the equation is 5 ( t ) + 3 y ( / ) = s in f.5 ( o ) ] + 3 y ( 4 ) = p | ^ Substitute 1 for y(0 )a n d 2 for 5 ( 0 ) 'b Ibe equation.5oosf-0.6 .^ ^ ' 4^ ( 4^ + ! ) 4 * + l 4_____ l _ ■ 4^ ( 4 ’ + 1) 4’ + l 4 '+ l __1_ 1 4 1 4’ 4’ + l * 4 ’ + l 4’ + ! = -L -2 -!-+ -J - 4’ 4’ + l 4^+1 Step 19 of 19 Apply inverse Laplace transform. Use the following formula.5 ( o ) ] + y ( 4 ) = ^ Substitute 1 for y ( 0 ) and for 5 ( 0 ) 'b Ibe equation. y(s) step 5 of 19 Apply inverse Laplace transform cL ■ . 5 (0 ) = 2 m 5 ( 0 + y ( i ) = t. the expression for y ( ' ) is.2 4 + 4 ] .j3 t Step 8 of 19 (c ) Consider the equation is 5 ( t ) + 5 ( 0 “ ®^^ Taking Laplace Transform [ s ' y ( j ) . [ s . j. y{l) = 1 .1—vsin--- 2 V ll 2 Therefore. VfT. Determine the value of . the expression for y ( 0 is.5sin/ Therefore. y (0 ) = 1.s .y ( 0 ) ] + 4 r ( s ) = 0 Substitute 1 for y^Q)and 2 for 5 ( 0 ) in the equation.y ( 0) ] = ^ Substitute 1 for y^Q)snd 2 for 5 ( 0 ) in the equation. ( 1 .S i 4+1 4’ + ! =l_ i4 . [4 > y ( 4 ) . y(»)=/-2sin»+cosf Therefore..4 p ( o ) .l + 3 . the expression for HO is. 5 ( 0 ) = .2 ] + 3 y ( 4 ) = p l^ (4 ^ + 3 )y ( 4 ) = ^ + ( 2 + 4 ) 1 2 s _ i ( 4 ^ + 3 )-(4 ^ + 1) 2 4 2 (4 ^+ 3 )(4 .5ii (<)+0..3 = 4 ( 4 .iM .. y (0 ) = I .|+ D 0 = 1.2 [ s i'( 4 ) .2 s in f + c o s f |.833«-“ Therefore.S4+0.^ s in [ 4 ^ .5c"*-0.5 ii(/)+0 . Substitute ^ = 0 ih equation (3).r + . (3) Determine the value of . |y(>) = 1.l ) ( 4 + 2)4 4 * + 3 4 -3 ( 4 -1 )( 4 + 2)4 A ^ B ^ C 4 4 -1 4+2 4 * + 3 4 .3 = C ( . y ( t) = e fc o S 'j3 l+ -^ e fs m . -3 = 4 (-1 )(2 ) .2 -1 ) -5 = 6C y (. -2 — 1 .. the expression for HO is.5 ^ . j’ + 3 j ’ + » + 4 = ^ ( » + 1 ) ( j ' + i ) + & ( 4 ’ + i ) + ( C s + D ) » ( j + 1 ) .1 S te p -b y -s te p s o lu tio n step 1 of 19 (a) Consider the equation is 5 ( 0 '* '5 ( t ) + 3 y ( r ) = O Apply Laplace Transform to the equation.4 . r' c' |=sini( U+4^ r' ( * ^|=cos4f U + i’. 4 .5 ( 0) ] + [ j r ( j ) . l+ 3 . ] .4 . s+ 3 :7 n (1) „ l ..5 ( 0 ) ] + [ j l '( s ) .i+ i (a -O ^ (^ ) —f \ *”* s -l ( ^ (2) step 7 of 19 Apply Inverse Laplace Transform.4 p ( o ) . y (0 ) = 1.4 .4 and ^ in equation (1).5 ) i+ ( 0 .i= p ^ y ( j) ( i'+ s ) = p | j^ + ( 4 + 3 ) (4+3)(s’ +i)+1 y (s ).V T T ^ i .33e' .\^ r + c o s x ^ Therefore.l) ( ( . } < » = j siiw — UsinV3< + ^ s i n ^ + c o s ^ step 15 of 19 (e) Consider the equation is.2 [ j l ' ( s ) .5sigf|.0 . y ( 0 = « 'c o s ^ + ^ « 's in ^ Therefore. Substitute 5 s -1 in equation (3). y (0 ) = 1. 4(i+l)(s’ +l) 4 *+ 3 s * + 4 + 4 4(s+l)(s^+l) A B C t+ D = — + -----. . y ( 0 ) = 1. 4*y(4) + 24y(4) = — +4 + 4 ' 4 -1 (4 + 4 ) ( 4 .5cos/-0. [ s » l '( » ) .2 5 ( / ) + 4 y ( / ) = 0 Apply Laplace Transform to the equation.y ( 0 ) ] + 3 r ( s ) = 0 Substitute 1 for y(0 )a n d 2 for 5 ( 0 ) 'h tbe equation. [ s 'y ( i ) .. y ( 0 ) = 1. [j*r(j)-j-2]+[»i'(j)-i]+3y(»)=o (s*+ff+3)K(j)*j+3 j+3 ! '( * ) = i^ + j+ 3 i+ 3 'FPf]R^ step 2 of 19 Apply partial fractions..3 = f i( l + 2) Determine the value of C ■ Substitute 4 = -2 In equation (4).1 ) ( 4 + 2 ) + B 4 (4 + 2 )+ C s( 4 .? ii 2y =e‘ ? cos--. Problem 3. Apply Laplace Transform. y Determine the value of .i 2 ^ = T Compare ^coefficients in equation (3).llif ■(w n) step 3 of 19 Determine the value of g . = s in i( ‘ " ( H i') . the expression for H O is. y (0 ) = 2 (b) 5 ( 0 .3 3 ) ^ . Use the following fomiula.V iT 'l (-y fH ) i.) = L 5 ^ 0 ^ _ 0 ^ ' ' 4 4 -1 4+ 2 = ( 1 . 5 (0 ) = 2 (e) 5 ( 0 + 2 j ( 0 = «!:.0.5 0.— ^ — r l = e~*cosbt r ' Therefore.l) ‘ + 3 ( .i ] + 4 y ( 4 ) = o y ( j) [ i'.l + 4 2 Step 11 of 19 Determine the value of Q . ■ (-y v n ) Step 4 of 19 Substitute the value of . 4 2. [ s * y ( s ) .1 ) . (4) step 16 of 19 Apply partial fractions.5 ( 0 )]+ 2 (4 r (4 )-y (0 )) = j ^ Substitute 1 for y(0 )a n d 2 for 5 ( 0 ) 'b the equation.4 + i]+ y ( 4 ) = l (4*+l)y(4) = ^ .2 5 (0 + 4 y ( 0 = 0 . ------------ S J+1 » +1 Step 9 of 19 Apply partial fraction.09PP Solve the following ODEs using Laplace transforms. Use the following formula. Substitute 4 = 1 In equation (4). 5 ( » ) + 2 5 ( /) = e ' Apply Laplace Transform. |y(0=4if(f)-2. Use the following formula.l) * .f Determine the value of g .1. ] n ( 0 ] Step 6 of 19 (b) Consider the equation is 5 ( t ) . \= A + B + D l = 4 . y(») = 4ii(r)-2. Therefore.2 ]+ 4 y ( s ) .j y ( 0) . Compare 5* coefficients in equation (3). 5 . [ 4 'y ( 4 ) .l+ 4 -2B = .4»-(0 ) . 5 (0 ) = 2 (d) 5 ( 0 + 3 y (0 = sin ». Substitute 4 s 0 In equation (4).l) * + l) = ( .. .( 0 .4 y ( 0 ). 5 ((0 = 2 (a) 5 ( 0 + j ( 0 = sin<. . step response t(sec) Figure 2: Step response of the system Hence. fl TS/ ' [0 otherwise Step 2 of 5 The shifted unit step function is. . )’. — ( r + l) * - Case (3): Find the output of the system for the / * oo.(!)= J /l(T)u(l-T)tfT =0 Since.y) » title{'step response') » xlabel('t{sec)') » ylabel('y(t)') Step 5 of 5 The step response of the system is shown in Figure 2. =0 step 4 of 5 ^ MATLAB code to plot the step response of the system: » t=0:0.47 in the textbook.1 OPP Using the convolution integral. Figure Impulse response Step-by-step solution step 1 of 5 Refer to impulse response in Figure 3. find the step response of the system whose impulse response is given below and shown in Fig.01:10. fl /S O "W=|o /<o And. » y=-(t+l). the required step response is obtained using the convolution integral. there is no overlap between the two functions u(/—r) and *(r) and the output is zero. Problem 3. Case (2): Find the output of the system for the / ^ 0. ' ^ \0 (< 0 The expression for convolution is. The function is. 0 » < 0. Step 3 of 5 ^ Case {1): The multiplication of shifted unit step function shown in Figure 1 with A(r) becomes zero for /£ 0 Recall equation (1). . j'(/)= J A(r)ii{/-r)rfr (1) -«e 0 / « -« 0 f It is known that.*exp(-t): » plot(t. /. . The output of the system is ^ W = f* A W « (£ -T )^ T s2 Step 5 of 6 The figures to illustrate convolution are u(t) b bet) i(sec) ( 2 t(stc) uCt) b ll(T) T (sec) • 2 t(s a :} h(T)ii(l-t) 1 h(T) x(sec) t 2 xfsecj h(T)u(l-T)^ 1 b (i) ^ t.5 3 lunc(soc) Step 2 of 6 ^ There are three cases to consider as shown in the following figure. o ^ t ^ 2 |0 .5 2 2. Figure Impulse response -I -OS 0 OS 1 15 2 25 3 1106 (nc) Step-by-step solution step 1 of 6 Given impulse response o f the system is ^ ri. the required step response is obtained. that is y i (i) = 0.5 0 0 5 I t. the situation is displayed in the below Bgure and shows partial overls^. r < 0 and / > 2 . Problem 3. 0. There is no overlc^ between the two functions u (£ .r ) and h (r) and the output is zero. . find the step response of the system whose impulse response is given below and shown in Fig.j) step 6 of 6 The output of the system is the con^osite o f the three segments computed above a shown in the following figure. Step 3 of 6 (b) For the case 0 £ £^ 2.11 PP Using the convolution integral. Thus. the situation is illustrated in the below Sgure. w For the case £ ^ 0. £ < 0 and t > 2 The given impulse response is *(/) -1 -0. The o u ^ u t ofthe system is s£ Step 4 of 6 ^ (c) For the case £ ^ 2. the situation is displayed in the below figure and shows total overU^. !/( /) » / . F W of a function.» ' + e .* ' + 3e-*') f S U (s) a i ^ [ l . the transform of the output is step 4 of 8 (c) The transform of the output I ' M is.5 + ) of Consider the transform.0 £ /£ l = -/. Problem 3 . Figure Plot of input signai t “ (') I I I Figure Plot of input signai «(0 1 2 3 Tim e (sec) Step-by-step solution Step 1 of 8 (a) The input signai is. .g ~ ^ ) ] S* ( j * + 2^0f. B + D ^O A + C + 2 ^ o )^ D —0 2 C (o . ' j " ( i ’ +2fflV > + “ 2) Apply inverse Laplace transform. l+ C + 2 < f l» ./.> ') + i( .2 S rS 3 Step 2 of 8 Write the formula for the Laplace transform.tt . Define the signal. / ( ' ) Ri ih^titiitp thp fi inrtinn f/t\ The transform of the output Y ( j ) is. 2 ( a . Y{s) = G{s)U(s) 3— p -----.^ f o r < ? (j)a n d * ( i_ « .c ~ * . < t^ [l-g ~ * + ( l + 3 s ) ( g ~ ^ .( . K ( j ) ■ . 0 £ / £ l =1 .1 2PP Consider the standard second-order system G(s) = - ^ + 2^Q)nS + (a) Write the Laplace transform of the signal in Fig.l) + a iD = 0 b = -K Determine C.( l + 3 5 ) e ^ + ( l + 3 j)g '* * ] 5* + 2^01^ + ) j'( i " + 2 < « v j + « ^ ) f l^ [l.3 e . l£ ( £ 2 = . (b) What is the transform of the output if this signal is appiied to Gfs^? (c) Find the output of the system for the input shown in Fig.> .( 2 f * + l) .* _ « .A + o iD ^ 0 Aoi = a i A=l Determine D.e ' * . . u{l) = t . F W of a function.F ( l + 3 5 )(g ~ ^ -e ~ ^ )] Thus. ait O A s+ C + -T + - s^[s^+2Ca>^ + a i) s + + Step 5 of 8 Equate the coefficients on both sides.-----. / ( ' ) Define the signal.2 S /S 3 Step 2 of 8 Write the formula for the Laplace transform.^ = 0 C = .---------. V j. R .5 Ine"* = ln0.W K. The output steady-state speed from the same test is found to be 2 rad/sec..K .0 . (1) + (2) Since. KJ. t f f r l = 2rad/s 100 Substitute iiiii for Vy.. j.K . v ^ (s ) 5 ( f+ a ) Where. iooa: a 'W = a . there is negligible filed Inductance.. sQ . 1= 2 ( 1 . ( s ) * R J b s e .( .0278 Therefore..39 a: B 0. From equation (2).s'*(R J> + K .s— V. (j ) g .K j:. ( 5 ) in equation (4). ^ ( j+ a ) _ioo/:|~i 1 j a L* 5 + flJ Step 5 of 6 Apply Inverse Laplace transform on both sides. 100 s{s+ a) s lOOK S0{s) •(5) s{s+ a) Step 4 of 6 Apply Final value theorem to equation (5)..V . 0.of the motor.)s K. R Jb + K ^K .13PP A rotating load is connected to a field-controlled DC motor with negligible field Inductance..0278 Vy(s) s ( s + 1. Apply Laplace transform both sides. oiep I ui o The equation of motion for a DC motor is.) l. v ^ (j) ■ R . K j . the transfer function of the motor. * + (s) = K . .K .5 -0. J A + m . . e . ( ! ) + ( * .J j^ e .. Problem 3. Step 3 of 6 The applied Input voltage Is.= K .0278 for K ^ equation (4). y . <3> Step 2 of 6 for in equation (1).k j . RJ.39) . e ..* 0 J e . ( ! ) . Step-by-step solution ste p 1 of 6 The equation of motion for a DC motor is. V . * R jb 0 .. = v . K J." . Determine the transfer function 0 W .693 fl 1 0 = 1..39) 0.0278 Step 6 of 6 Substitute 1.A test results in the output load reaching a speed of 1 rad/sec within 1/2 sec when a constant input of lOOVis applied to the motor terminals.d ..39 From equation (6). IC. 100 The output load reaching a speed of 2 lad/swithin — sec- Therefore.e . . = — ss_ and ----.39 for a and 0.J..s ' <-*A *(i)-s \0 0 K lim ( j+ « ) 100 ^ 2= (0 + a) lOOK -*2 • ( 6) From equation (5). lOOK -=2 1.* . e . — “ is !(*-H .. v . R .(/) = 100V Apply Laplace transform.( s ) = K . 1 00 ^ s Hm. 4 x ltf( n . constant force. X .( i) 2 . 9. Write the equations describing the system. i i = tape velocity m/sec {variable to be controlled).r2 K .(B s * K ) {jfS ^ + + BjS+r^Bs + r.(s ) 2 . ( j ) + B ^s^ (s ) + F ( j ) = 0 + B ^ ) $ j{ s ) + F { s ) ^ 0 F {s) = .(s ) = V . motor damping. 2 f . /C/ = 3 X 10-2 N•m/A. /C/ = 3 X 10-2 N•m/A. 4 x i y ( j + l0 0 0 ) /.{ s ) Substitute + Bjsj02{^) F {s) in equation (1). —F Apply Laplace transform on both sides.2 5 0 0 ± y 3 6 73.(s ) V -t-g ^ Bs + K ’ r.den): » sys1=feedback(sys.‘ ) i ( s + 2 0 0 ) ( j '+ 5 1 0 0 s + 2 x l 0 ’ ) 2 . ( jy * B .{s) The mechanical equation describing the system is.» ) ( 2 0 ^ + 2 x l 0 * ) /. Apply Laplace transform on both sides.» ) ( 3 x l0 .1): » step(sysl) Step 8 of 8 ^ The following is the MATLAB output for step response: Figure 1 Thus. ) f o r The transfer function of the system is. » den=conv([1 200 0]. S‘^»stitute ^ _ ^ _ j / . 2 x l 0^ f°'’ /C. Take Laplace transform on both sides. ( j ) in equation (2).. Bs [ x ^ ( ! ) . viscous damping..2QtoT B . K m B R .02(s) te » . K and B represent the spring constant and the damping of tape stretch. (a) Write the equations of motion in terms of the parameters listed below.JC. + K.’ ) ( 2 x l 0 . I JC.I.4E5*[1 1000]. /C= 2 x 1 0 4 N/m.(s ) Substitute ' “ ^ '‘"■^for A ( ^ ) .'l v + i K j( r j+ l) Here. motor-torque constant.5500 + 3.x . idler. 82 = 2 X 10-2 N m sec. F = 6 N. ( .{s )* R . BR. X . / »• . r. Find the value of current that just cancels the force.K.0 2 i+ 4 0 0 ) 2 4 0 x l0 . J1 = 5 X 10-5 kg m2. n = 2 X 10-2 m.50 from the text book.V + 1 .14PP A simplified sketch of a computer tape drive is given in Fig.K . and o)1 and oj2 are angular velocities. L . i equations of motion describing the system i Step 4 of 8 (b) Consider the equations of motion describing the system.^ ^(^)fo r J f . motor-torque constant. Apply Laplace transform on both sides.{s)-K ^ 0 . IF /_ \ IF / . (c) Find the poles and zeros of the transfer function in part (b).6739i -2.-2 0 0 . the response to unit step input is plotted. 4 x l t f ( i + 1000) / . J2 = 2 x 1 0 -5 kg m2.(i)-Ar.r^K. Step 7 of 8 (d) Enter the following code in MATLAB to find the step response of the system.[1 5100 2E7]). r2 = 2 x 1 0 -2 m. » num=2. (d) Use Matlab to find the response of x1 to a step in ia.sI. Step 2 of 8 Consider the mechanical system in Figure 3. B *\/» / * \ step 3 of 8 ation between linear and angular displacements is. respectively. The electrical equation of the system is. =A y. from your equations. ( * ) ] = f ( s) (5s+A:)[jr. ( s ) = K J . V (s) . F. ( j ) ] + i c [ j r .( J ^ '+ B ^ ) e . ( s ) _________ ( 2 x l0 .^ K . ( j ) " j ( j + 2 0 0 ) ( j ' + 5100s + 2 x l 0 ’ ) Use MATLAB to find the roots of the denominator polynomial. then eliminate the constant current and its balancing force.1 0 0 0 ) i ( i + 2 0 0 ) ( i’ + 5 1 0 0 s + 2 x l0 ’ ) Thus.Oe+03 * -2. “ 'A U ) = r. Figure Tape drive schematic Step-by-step solution step 1 of 8 (a) Consider the electrical system in Figure 3. the transfer function from the motor current to tape position is.9| and zero of the transfer function is |»1Q00|. . s ) 0 .6739i Thus. Problem 3. K.( » ) 2 x l 0-’j " + 2 x l 0.(i)fo r AT. The transfer function of the motor system is.» ( n . S = 20 N/m sec. ( j ) .50 from the text book.3.^. motor and capstan inertia. Assume positive angular velocities of the two wheels are in the directions shown by the arrows. » sys=tf(num.(j)]=F(s) (i) The mechanical equation describing the system is.'( l + 2 x l 0-’ ) i + i ( 5 x l O " ’i + l x lO '’ ) ( 2 x l 0* )( 2 x l 0-*) 1 2 x l 0' ‘ ( 2a s + 2 x l 0‘ ) i ( 5 x l 0 ’j + l x l 0 . » ro o ts ([1 5100 2E7]) ans = 1. F.’ ) ( 2 x l 0 . s r .( s ) ” * ( i + 2 0 0 ) ( l’ + 5 1 0 0 s + 2 x l 0 ’ ) Step 6 of 8 (c) Consider the transfer function.{Bs + K) s {J iS + ( 1 + + X tj) step 5 of 8 Substitute 2x10"* 2x10"* 3x10"* K.^ ) ( 2 x l0 . A positive current applied to the DC motor will provide a torque on the capstan in the clockwise direction as shown by the arrow. 5 x l0 "* f° '' •^1’ IxlO"^^®*^ 2x10'^^°'^ / j and 2x10"*^°'^ inthetransferfunction.1 0 0 0 ) ■ ( 5 x l 0 . (b) Find the transfer function from motor current to the tape position.5500 .\ T _ f r Ji . SI = 1 X 10-2 N m sec.(i)ancl r. the poles of the transfer function are |0 . I. 15PP For the system in Fig. 6 * + *(LJc‘ *2RJ>k)s+RJc‘ bKs+kK.4 V *[LJ c' *2R Jbk)s*R J^ Therefore.). s } ( j ..[ f a + * ] « .)+ B $ . 4 : .K ^ J' ' ( 4 y + 2 M j+ * ’ ) ( V + i ^ ) bK.s+kK.]j+ iM . K {‘ ) k { L ^ + R .. ) + A : . LJ>W +{2LJ>k+RJ^)s‘ + (L J :‘ t-2R. .[ t o + Jfc] j (s ) = K .( j) bK..»’ + fts + * )- [ ( B + 4 ) J ^ .^1 ( j ) ] = 0 [ + its+ (s)-[its+ k\9^ (s)= 0 [to+ k]0f (s)= [/jj*+ to+*] (s) = 3 .] * ’ + ( j^ + b s + k ) - |[ ( 5 + 4 ) « . J2S^$2 (^ ) ^^502 (<s) (-r)] * k \$ 2 ( j ) . Figure Motor with a flexible load i. the transfer function form the motor voltage to position 0 l( s ) i: « .= K . Substitute for L ^ + R^ Jt to + * J ^ { [ V + ( B + 4 ) i + * ] ( i ^ + « . ]» ‘ + ( y . ( i ) = A : . Apply Laplace transform on both sides. V / + [V i+ ( fi+ * ) i. + ( B + 6 ) s + * ] ^. K i‘ ) 1V i*’ + [V i + ( f i + 4 ) i .)(fa + * ) L^+R . Apply Laplace transform on both sides.I^ (s ) substitute W Z M M f o r / (. Step 4 of 4 On further simplification. bK. J t .+ b (e . compute the transfer function from the motor voltage to position 02. + K. J . The mechanical equation describing the system is. L . Step 3 of 4 Recall equation (2). [ V + ( B + i ) j + * ] « l ( i ) .bk)s+RJ[^ . +kL.s+kK. (2) Step 2 of 4 The mechanical equation describing the load is. ( j ) . ( » ) .s+kK. + X . + A T .-0 . / .0 .A :..K.i. ] j+ * ) ^ i. i ’ +bs+k)- 'iW { b s + k f{ L ^ + R..)+ K .-0 . V / + [ V i + ( ® + * ) i. Apply KirchhofTs voltage law.)+ k (e . Apply Laplace transform on both sides. S te p -b y -s te p s o lu tio n step 1 of 4 Step 1 of 4 Consider the electrical system in Figure 2.54 from the text book. i V + ( 2 i . + * t .']s+kR .j* + 4 s + 4 ) - [ ( 5 + 4 ) i ^ + * t ..]» ’ + (y.) ___________________________________ (V + J ? . Problem 3. 2 0 A h . Problem 3. Figure Two-tank fluid-flow system Step-by-step solution step 1 of 1 We have AA^ + 6ct^ .( S ) - .- 6a A h .16PP Compute the transfer function for the two-tank system in Fig. with holes at A and C. 5^ + s — 2* determine the following. i^ + j. the DC gain is not defined for an unstable system so the output of the system is unbounded. D C g a in -G (5 )|. C (0) = 4 Thus. the system has an unstable pole. the DC gain of the system is ED Thus. (a) The DC gain. Substitute 0 for s.1 7PP For a second-order system with transfer function G(s) = .-2 Thus. Problem 3 . Step-by-step solution Step-by-step solution step 1 of 2 (b) Determine the poles of the system. (b) The final value to a unit-step input. G (* ) = .r^ s + S -2 Consider the DC gain formula. so the final value theorem Is not applicable.j.O Therefore ^ = 1. . Step 2 of 2 (a) Consider a second order system with transfer function.2 . ------------------------------- R . ( m j* + f o + c ) A '( s ) + . and that the force exerted by the rolled material on the adjustable roller is proportional to the material’s change in thickness: Fs = c ( T . ..S Step 5 of 8 M ir j Substitute * I j for ^ in equation (1). N K .0 Apply Laplace transform. and---------— are in cascade.l. the blocks -----------.. Suppose that the motion of the adjustable roller has a damping coefficient b. I f is the field current.. Calculate the equivalent block.+ L . (m$’ + & y + c ) j ir ( s ) + ^ ( s ) + ^ ^ —^ . the inputs to the system are.x). na = c {T -x )-m g -b x -F ^ Apply Laplace transform. for in equation (4). Input voltage(v. (a) What are the inputs to this system? The output? (b)Without neglecting the effects of gravity on the adjustable roller. ( m . (^> It is known that.18PP Consider the continuous rolling mill depicted in Fig. Apply Laplace transform on both sides. From the Figure 3.I f In Figure 1. v.* L js Step 6 of 8 Draw the block diagram from the equation (7). the loop equation is.* L js .. F(s) (the force the motor exerts on the adjustable roller). K . { s ) ~ * -----.. thickness(r)andgravity(A^) and the output to the system is [ouq)utthickness(x)]■ Step 2 of 8 (b) Write the equation of motion for adjustable roller. Substitute for / (ms* +fo+c)jr(s)+— \ / X N K .(/)).. (c) Simplify your block diagram as much as possible while still identifying each output and input separately.y .If Move the summing point ahead of a block G(s) as shown In Figure 2.If (3) Step 3 of 8 From the dc motor circuit. * N K .. the inputs to the system are. Y 1 Y A T . e {s )R jr ( i) N X {s) eW Step 4 of 8 Substitute fbr e W In equation (5).. Suppose further that the DC motor has a torque constant Kt and a back emf constant Ke. Problem 3. 10 {s^. V ^ {s )^ K ^ {s ) (5) And. -<■(*) . Figure 1 Step 7 of 8 (c) 1 N K . and X(s).iVs) Draw the simplified block diagram..0 (1) Write the equation for torque in rack and pinion. Figure Continuous rolling mill Step-by-step solution step 1 of 8 (a) Refer to Figure 3. ( O = a :. Substitute for in equation (2).> | 1+ m . draw a block diagram of the system that explicitly shows the following quantities: Vs (s).//)(A:. ^ =0 sK^N .— <®> R . Is the torque constant..ifR R^ + +As+c)+(Affl:. T ^ = KF.^ ^ ^ ^ i.If m g -c T (7) R . and that the rack-and-pinion has effective radius of R. output thickness ^ x) • Therefore. I. f ] ( __ ^__ ] J v in y ^+ A y + C / r(j).51. Input voltage thickness and gravity {M g Y The output to the system is.51 in the textbook. Figure 2 Step 8 of 8 Simplify the feedback loop in Figure 2.( 2 ) Write the expression for torque of a motor. Here. the transfer function of positive feedback ioop is.19PP Consider the block diagram shown in Fig.G W W ( j) The reductions of positive feedback loop with G (s ) = i and = will be. the transfer function of positive feedback loop is. This special structure is called the “control canonical fonn” and will be discussed further in Chapter 7. the transfer function is. Problem 3.19 in the textbook for block diagram. ------. ) --------- 1 . r fj^ = — . step 2 of 7 ^ Shift the Pick-off point at j|f| to the right past the second integrator to get b^s+b 2 asshownin Figure 1. f___!___ + 0^3 + 0 2 ) 3 r(i)= 5 ( 5*+< 1.5-------------.5 + <l2) + <l} 1 5*+0.19 in the textbook. ( V + *a )s + ^ "11») S +<1. l. Figure 2 Step 5 of 7 From Figure 2.5 + <^) 1 5^5^+0. Form Figure 3.. Note that ai and bi are constants. Figure 1 Step 3 of 7 ^ From Figure 1. ' ' 1. s 1 rM — TTT I ' 1 's + a .f — 1 . I l £ i i G(s) 3 + O fS + 023 + 0^ . the transfer function. U ) 1 + 0 ^ + 02 step 4 of 7 Shift the Pick-off point at X 2 right past the third integrator to get as shown in Figure 2.5^+025 + ^ Step 6 of 7 Draw the reduced block diagram 1 y (5 )o .S*+025 + < ^ J ''^ btS^+bi3+bj 5’ +0. —5------.2~------------ 1/ ( 5) ^^5’ +0. Figure Block diagram Step-by-step solution step 1 of 7 Refer to Figure 3. Compute the transfer function for this system. r ( .G ( f) tf( j) The reductions of positive feedback loop with G(5 ) = [— ^— I—ahd will be. f— 1 - r(*) '.5 +<^5 + <l3 Figure 3 Step 7 of 7 From Figure 3.. the transfer function of positive feedback loop is.5 + 02) ff(ff*+<I.s^ + bjS+bj Therefore.5*+<^5 + <^ b.G { s ) H ( s ) The reductions of positive feedback loop with be. (l+ Q . + (^ ( l+ q C jX l+ G . G^G^Gf ( 1+ G 2)+G^G 2G^G4Gj + G 7(l+ G ^ + G 4 + ^ 0 . /Jo. R ( l+ G iG . Figure Block diagrams Step-by-step solution step 1 of 19 (fli) T h e fbUowiiig is Ihe givefi blodc diagram : Step 2 of 19 F ro m F ig m e l. * I+ G 4G 5 Step 8 of 19 ^ F igure 5 is th e reduction diagram o f F ig u re 4.G .G . th e tra n sfe r function o f th e g iv en b lo ck diagram is ~ r \ __________ 1 + G ^ G a + G 4 G ^ + G ^G aG 4Q 5___________ Step 11 of 19 (c) G iven b lo ck diagram is F ig u re ? Step 12 of 19 ^ F irst fitidtfig feedbadc padis a nd deten n in e tra n sfe r functions.) R 1+G 2+G 4+G ^G i .(l+ G ^ )+ G j 0 . R esuttant equation fo r cascade p a d il is O iOiOj 1+G . ^ . ( l + q i+ 0 4 + q .G .)+ q C ^ q G .G jG. = — ^ _ and i+qq T ransfer function fo r second feedback is H ^ = — — — . (l+ Q X l+ G . R 1+G i q + q . ' l+ G . th e transfer function becom es r qqq. q i+ q ( i) I l+ G W j q i+q step 3 of 19 F ^ u r e 2 is d ie reduction d iagram o f F ig u re l.G . G4GS R esultant equation fo r cascade p a d il is I+ G 4 ’ Step 16 of 19 ^ F ^ u r e 10 is th e reduction d ia ^ a m o f F ig u re 9.)(1+G. Step 4 of 19 ^ F rom F igure 2.) .^ .) ( l+ G . (1+G. Step 7 of 19 T ransfer fo nctioa fo r first feedback is H. d ie tra n sfe r function o f th e g iv en b lo ck diagram is R~ l+ O j Step 5 of 19 (b) T h e follow ing is d ie given blo d c diagram : F ig u re s Step 6 of 19 F irst fitMltfig feedbadc paths a nd deten n in e tra n sfe r functions.) ^ G . F ig u re 4 is d ie m odified blodc diagram o f F igure 3. Step 14 of 19 ^ F ^ u ie 9 is d ie red u cd o n diagram o f F ig u re 8.and 1+G . Feedbackl Feedbaclc2 F ig u re s Step 13 of 19 T ransfer function fo r first feedback is Hi = . T he resultant is show n in F igure 6. Problem 3. T h e transfer function is. +0.) F ig u re d Step 10 of 19 F rom th e above block diagram .G . Cascade pathl Cascade path2 F igure 9 Step 15 of 19 ^ W e sim plify d ie blo d c diagram b y red u cin g d ie parallel com bination o£ d ie contrcdler pa th (cascade path). T h e negative un ity feedback tra n sfe r fim ction is.G . G. ^ 1+G.K ^ (l+ G iG 3 + G 4G5 + 6 ^0 .G.G 5) ’ qqq.G jG .G . ) i+q+q+qq Therefore.G . T ransfer fo nctioa fo r second feedbadc is .(i+ q ) i+q ^ q + q + q q i+ q Therefore.Gi l-^G ^+ G ^+ G ^G t Step 18 of 19 ^ F ^ u r e 11 is th e reduction d ia ^ a m o f F ig u re 10.)+ G i< ^C ^Q .G .Q .( l+ G .G . G . F igure 10 Step 17 of 19 ^ F rom th e above block diagram . G i+ Q . Step 19 of 19 F rom th e above block diagram .(i+Gi)+qGiqq. 0405 ) 1+ Q G j + G 4GS+QG 3G4GS Therefore. d ie tra n sfe r funetkm o f th e g iv en b lo d c diagram is r G . th e sum m ing equation becom es .X 1+G 2+G 4+G ^G. F ig u re 8 is d ie m odified blodc diagram o f F igure 7. d ie tran sfer function becom es r G .20PP Find the transfer functions for the block diagrams in Fig. Cascade path F igure 5 Step 9 of 19 W e sinyilify d ie b lo ck diagram b y red u cin g d ie parallel cofnbination o£ d ie controller pa th (cascade padi). g|GsG.q 1+G^G2 + G 4G5 + G i 020^0^ r+q + .G3G . )(l-G . using the ideas of block-diagram simplification. Figure Block diagrams S te p -b y -s te p s o lu tio n step 1 of 14 (a) Refer to Figure 3.) G . -o r Figure 8 Step 12 of 14 Determine the transfer function of the system Y ^5^+(a.{ l-G .5 + Step 13 of 14 (d) Refer to Figure 3. ) The blocks — !— and — are in series and the resulting block forms a feedback loop with a. Figure 7 Step 11 of 14 The resultant value below the summing point is.(l.(b) is called the “observer canonical form” and will be discussed in Chapter 7. Redraw the block diagram by reducing the feedback loops. Draw the reduced block diagram.54 (c) in the text book for the block diagram. the blocks and — are in series and equivalent value is 1 .(l-G .g.G ^ . This block forms a feedback loop with and the total value of the block I below the summing point becomes s ^ * a . Figure 1 Here. ) + G . Now shift the block b^ left to the summing point.H .)+ < ^ .tf.54 (b) in the text book for the block diagram.(l-G .(l-G . The blocks b^. The blocks below the summing points.e n d are summing above the summing pointsand the resultant value is The blocks below the summing points.H . which are forming a feedback loop is --------. — . Draw the reduced block diagram.. The resultant value at the top of the summing point is * j( 4 + a .H . — . the transfer function — of the system is Step 8 of 14 (c) Refer to Figure 3. Simplily the circuit by moving the block ^ before the summing point. Figure 4 Step 6 of 14 Similarly. 62+ a 2^ +4^ Thus.\ R o. the transfer function — of the system is AB+BHD+D [+ BH + ABG+ DG +BHDG .G . + G .54 (a) in the text book for the block diagram. o. the transfer function of the system is {i-G . ^ • and i . are in series. Figure 6 Step 9 of 14 The resultant value of blocks — and a. Jt _ biS^+bjS + bj bjS + b ^ + bj Thus. -------r . By moving the block .G .H .)+ G . + G . Redraw the block diagram by reducing the feedback loops. The special structure in Fig.21 PP Find the transfer functions for the block diagrams in Fig.)(i-G . R AB’ + D r ~ u g {ab*+ d ) for g '. s+a. G . Now shift S * + <*! the block b^ to the summing point.) Thus.) + G .G . Problem 3.* G .G .s ^ + a ^ + a .( \.\+ b h ] ' “ l+ G J AB + D {\* B H ) AB+BHD+D \ + BH + ABG+DG + BHDG Thus. G .!.G .5 + 1+ s^ + a ^ s ^+ a ^+ a ^ Draw the reduced block diagram. the transfer function — of the system is J +0.)-bG . b j + ( b 2 + b ^ s )s -or s ^ + O jS ^ + O j S + a j Figure 5 Step 7 of 14 Determine the transfer function of the system.){l-G .H .) G . The blocks above the summing points are reduced by shifting the blocks left to the summing point and the blocks below the summing points are simplified by reducing the innermost loop first and proceed further. 4 (4 + a .G .{\-G . Jt g |( l.g.)( l-G . it becomes and the total block becomes The blocks below the summing point —» and a. Figure 3 Step 5 of 14 Repeat moving the blocks before the summing point for b^+bfS to get •¥b^s)s and the total value becomes above the summing point.) '(} -G ..G .54 (d) in the text book for the block diagram.H .6^-f j -F + a^b^ + b^ 4i5* + (0 |i| + ^ ) j.G fi.H .)(l-G . and — are in series and the total value becomes —. + i^ ( s + a .H .g . \+ h ( s ^ + a s + a .G . Figure 2 Step 3 of 14 Determine the transfer function of the system. ^ ( 5 + a |).* G .K ^ ) Substitute — ^ — for GC bbd — —*— for G f . Step 10 of 14 Draw the reduced block diagram. This block forms s feedback loop with and the total value becomes Draw the reduced block diagram. — .g . = Step 2 of 14 ^ Add fonvard path to and draw the reduced block diagram. H .H .) ( i. 1 b . s 1 The resultant value is • . ) ( l . 1 __ 1 ( 1 ] s{s' + a. G.)* G . UBH A \.F a .) Step 4 of 14 (b) Refer to Figure 3. Step 14 of 14 Determine the transfer function of the system.G .(l-G . form a feedback loop and the total block value becomes Draw the reduced block diagram.H . 55 in the textbook for block diagram. \*G M y Draw the simplified block diagram. !• ( .) .55 in the textbook for block diagram. The blocks and G. -Q- Draw the simplified block diagram. The blocks G^ and Gy are having the same common feedback loop through the block /fjT h e resultant blocks are given as follows. Step-by-step solution step 1 of 4 Refer to Figure 3. Figure Block diagram Step-by-step solution Step-by-step solution step 1 of 4 Refer to Figure 3. -Q- O. The resultant block for the feedback loop Gy and \*G M y Draw the simplified block diagram.22PP Use block-diagram algebra to determine the transfer function between R(s) and Y(s) in Fig. The resultant block for the feedback loop and is. The resultant block for the feedback loop G^ and Hy is. Problem 3. are having the same common feedback loop through the block /fjT h e resultant blocks are given as follows. l* G ^ H y The resultant block for the feedback loop Gy and Hy is. a = -A s 1 . Construct signal flow graph for the block diagram.-l^ AB *D J+ B H \-¥BH AB-^BHD + D ^ \. l-G . There are two fonvard paths and two loops. Here.G .54 (b) in the text book for the block diagram. 1 I .W .^ is Step 8 of 16 Determine the loop gains. /^and ly are the loop gains Determine the fonvard path gains.( I. P i+ j> . P\ ’ P i P i^^^ ^be fonvard path gains /j.-l. Problem 3. = -G|C^G.W . l. ft+ ft+ p .G . + G .< ^ f t = G. y ft+ft R ~ l-l. y _ Pl± £ i ± £ l R ~ l-l.(1 -G .)(l-G /f. Step 14 of 16 ^ Write the formula for the transfer function using Mason’s gain formula.’ / 2 = .G3+g .« . o.H .* p .ff . y ^ ft+ ft+ ft Pi ’ P i P i^^^ ^be fonvard path gains /|. ( l. 1 1 1 j +fl. 5 '* - \ + BH There are two fonvard paths and two loops. rG .) Thus. 1 .H . I-G jf f .f f . . Substitute — ---.G .g .5 +ajj+<ij byS ‘¥ b ^ -¥ b j Thus.H .54 (c) in the text book for the block diagram. ' 1+ { l-G .----vO. b.g . Construct signal flow graph for the block diagram. 1 . p.) ■(i .= C . .)^ ■ G .[^s* + a.) l-G .g. f f . R •i 11 1 1 ".) ( l. p.)+G . using Mason’s rule.( i .g. Step 5 of 16 Determine the transfer function. Y p .G . /^and ly are the loop gains Step 11 of 16 Determine the fonvard path gains. Substitute — ^ — for ( j! and — —*— for G f . . y.) + g . + G . P i' Step 4 of 16 Determine the loop gains.t . W . 1 Determine the transfer function.G . \-G . Here.. y .G . ’ ' ■ [I-G . ’ 1 . .G.G . Figure 3 There are three forward paths and three loops.(l-G .H . /. ) Step 6 of 16 (b) Refer to Figure 3. G.//.----vO.W .) + G .G ./f .G".G .-l. Construct signal flow graph for the block diagram. . = AB' Pi = D for \ + BH AB A * UBH Step 15 of 16 Determine the loop gains. Figure Block diagrams Step-by-step solution Step 1 of 16 (a) Refer to Figure 3.) ( l.= -A B ’G l^= -D G Substitute for UBH AB*G ABC l + B ff Step 16 of 16 Determine the transfer function.6^ + 4^) j + ajAj + + 4^ Thus.//0 G .tf3)(l-G . p^ and are the forward path gains /^ and / j are the loop gains Determine the fon/vard path gains. the transfer function — of the system is s +a^s +ayS+ay Step 13 of 16 ^ (d) Refer to Figure 3. Here.G.W . and py are the fonvard path gains /i and iy are the loop gains Determine the fonvard path gains.c .) l2 = -G. the transfer function — of the system is AB+BHD+D [+ BH + ABG + DG+BHDG .H .w.G . 1 Determine the transfer function. 1 I . g .)( \-G .54 (d) in the text book for the block diagram.W J _____________________ ( l.G.g . Redraw the block diagram by reducing the feedback loops.¥ B H D G Thus. Step 10 of 16 ^ Write the formula for the transfer function using Mason’s gain formula. There are three fonvard paths and three loops.23PP Find the transfer functions for the block diagrams in Fig.G . ) '( l-G .g . . = Step 2 of 16 Draw the signal flow graph.for G *. . p .^ B H ^ A B G ^ D 0 .G .4 ^ S te p 1 2 o f1 6 Determine the loop gains. the transfer function — using Mason’s gain formula is G . ft .g .G 3 g :) ( l.) ( i .//. Step 7 of 16 ^ Write the formula for the transfer function using Mason’s gain formula.I-G .-l.* p . Step 3 of 16 ^ Write the formula for the transfer function using Mason’s gain formula.G . y ft+ft R ~ l-l. .) + G.)(I-G .s+ay)'*'by(s'*-a^)+by s ^ + O fS ^ + a ^ + a ^ * { a ^ l\+ b y )s + a ^ + a ^ b 2 + l^ +OyS+ai + (a. the transfer function — of the system is Step 9 of 16 (c) Refer to Figure 3.g . ( l. g .54 (a) in the text book for the block diagram. f t = -!■ * ! ft.G . W Q l ( l .G 3(l-G . Figure 1 Here.(l-G .W J I. G . P 2 = G iG jG jG j. y(» )_____________________ g.G. (2) Thus. G .) + G . write the second loop path gain. + G .= . (5) Thus..(2). the fon/vard path gains are 2 and loop path gain is € Consider Mason’s rule for Figure 1.( G .. write the fifth loop path gain.G .G . G\ Step 8 of 10 From Figure 7.. write the first loop path gain. -H i Figure 4 Step 5 of 10 From Figure 4. = -G .. the transfer function of the system is . write the fifth loop path gain..G * /f. + G . (1) Thus..G . the third loop path gain is -GjGjGjG^Z/j Determine the fourth loop path gain from Figure 1. U =-G ...G ..G . (4) Thus..G)G5G^. write the third loop path gain. the fifth loop path gain is —G4H 2 Determine the sixth loop path gain from Figure 1. G 2 _________________G 4 G i Ge Figure 2 Step 3 of 10 ^ Refer Figure 2 and write the equation for first fonvard path gain.G.G * //. + G . -H j -H a Figure 1 Step 2 of 10 From Figure 1.. -H 2 Step 10 Of 10 ^ From Figure 9.G.(G .G . P . -H i Gi Gs Figure 6 Step 7 of 10 From Figure 6 . the sixth loop path gain is —GjJVj Consider mason’s gain formula. ( G .G .G jG */f^ (6) Thus. the second fonvard path gain is G.G.. Figure Block diagram Step-by-step solution Step-by-step solution step 1 of 10 (b ) Construct signal flow graph from Figure 3.G.24PP UseMason’s rule to determine the transfer function between R(s) and Y(s) in Fig.* P l (9) Where.) ( / /. I2 =-G .(7) and (8 ) in Equation (9). / 7 j.. G i Ge G i Gs Figures Step 4 of 10 From Figure 3. (8) Thus.(4).(7) Thus. Pi = . Step 9 of 10 From Figure 8 . .52 in textbook..P 2 is the fonvard path gains.(6).G .G. Problem 3. + G . the second loop path gain is -G jG jG ^G j/f^ Determine the third loop path gain from Figure 1. write the third loop path gain. /. (3) Thus. is the loop path gains..{5).(3).G . Determine the first loop path gain from Figure 1. I+ G 1G2G4G4/ / } ^-G iG j^tG ^/f^+G |G )G jG 4/f) + G|G}G5G4/ f 4 -\-G^H2 -¥G^H2 _______________ G |G . Determine the first Fonvard path gains from Figure 1..)_____________ 1+ / / . the first fonvard path gain is G f i 2G^G^ Determine the Second Fonvard path gains from Figure 1.+G.G . -H a Figure 5 Step 6 of 10 From Figure 5. the first loop path gain is Determine the second loop path gain from Figure 1.) Thus.G ^G .. /. G |G .. p^.. + f f .(G . the fourth loop path gain is -G fijG ^G JH ^ Determine fifth loop path gain from Figure 1.. write the second fonvard path gain..G .G . Substitute Equation {1).i¥.G jG . 1 6 3 2 ^• = 0.{ . 4 ) ( 2 ) R> 40Q Thus. step 3 of 10 (b) From the circuit in Figure 1.f “) = 0 .195 Step 10 of 10 Simplify further to obtain ^ . (d) The values of R that will result in v2 (fj having an overshoot of no more than 25%. sLC * 1 lC ~ L RyfZC 2L _ R [C ' i ' l l Thus. / ( i ) = C tF . L = 10 mH.. f-lJ! The condition to be satisfied is. find the following: (a) The time-domain equation relating i(t) and v\ {t) . Step 4 of 10 (c) Apply KirchhofTs current law to the input loop.4 2 V 10x10-“ 1 ^ ( 0 . . 4x10-* -[■ > 0 . L = 10 mH. ^ .1 9 5 f’ 1 .(i) ( 2) step 5 of 10 Substitute equation (2) in equation (1).4 4 l’ ( l .1 9 5 f’ = 0. ^ is the damping ratio which lies between 0 and 1. Rearrange the terms to obtain the transfer function. 1 9 5 ( l . > ( 0 .195 -0 . of the system is T ie Step 7 of 10 Substitute j for a>^ in 2^e>. and C = 4 fJF. . Problem 3. * J ^ > 0 . the time domain equation relating and V |(r)is . and C = 4 fJF. < -* = 0 .j — = of LC " 1 Thus.441 f = 0.* ) f ’ = 0 . 0195 ^ 1. Apply Laplace transform with zero initial conditions. Figure Block diagrams Step-by-step solution step 1 of 10 The following is the given electric circuit: L R Step 2 of 10 (a) Apply KirchhofTs current law to the input loop. M step 6 of 10 R 1 Compare the characteristic equation + s — h— with the standard second-order L LC characteristic expression.1 9 5 . y. ( of the system is f '. the voltage across the capacitor is. the values of R that will result in more than 25% overshoot are |/t > 4 0 q ] . 5* • 2^i». For maximum overshoot of 25%.3863 s C 1. assuming v1 {t) is a unit step. 0 . the transfer function V2 {s) 1^1 (s^ and the damping ratio ^ and undamped natural frequency ojn of the system. Apply Laplace transform with zero initial conditions.3863 V > -« " * . assuming v1 {t) is a unit step. v .4 j.441^1 Square the equation on both sides. Thus.S — = 0. ^ of the system is R C 2 Vl Step 8 of 10 (d) The maximum peak overshoot is.4 2>Il Substitute 4 x 1 0 “*^°*^ C 3 ^d 10x10“*^°*^ X . (b) The time-domain equation relating i(t) and v2 {t) . s*L+sR+— C 7L t s^ ~2+ s — r + —n L IC ) J £ _ ^ R * V ( 5) J£ _ Thus.4 Write the expression for damping factor. Where. the transfer function. the damping ratio.2 s = e ~ ^ —13863 = — Step 9 of 10 Determine the value of ^ . the undamped natural frequency.{s) = s U ( s ) + R I(s )+ -^ I( s ) The capacitor voltage is. the time domain equation relating and v ^ {t) is. (d) The values of R that will result in v2 {t) having an overshoot of no more than 25%.25PP For the electric circuit shown in Fig. 1. ( o = i ^ + « < ( o 4 r '( '> * Thus. (c) Assuming all initial conditions are zero. ^be equation.0 2 ) > 0 .0 . r ^ — = 0. Problem 3.537^^ = 0.59 for ^ in 2^01=2 2 (0 .T 3 2 9 4 \-C Square o n b o th sides.^ “) ^ = 0 . K r(j) s(s+ 2 ) R (s)= - i( i + 2 ) K ~ s (s + 2 )+ K K ^+2s+K T h e tnaxiim im p eak overshoot is» W h ere ^ is th e da n ^n n g ratio vrfu d i lies betw een 0 1.695' = 2. = 2 1 0.59 ^ 1 .26PP For the unity feedback system shown in Fig. 5 3 7 ( 1 . . fee gatn K o f fe e proportional controller to h a v e a n overshoot o f m o re than 10% is Io < j : < z 8 ^ .6 9 5 T h e constant K is.537 Sim plify fiirther to obtam ^ ^ 0-537 ^ 1. M. specify the gain K of the proportional controller so that the output y(t) has an overshoot of no more than 10% in response to a unit step.3 4 9 4 ^ = 0 . Figure Unity feedback system « *)o Step-by-step solution step 1 of 3 Step 1 of 3 T h e fbUowiiig is tfie b lo ck diagram a u n ity feedback system : Step 2 of 3 T h e tran sfer function o f th e system is.5 9 Step 3 of 3 ^ C otn p ate ^ + 2 s + K w ife th e standard second-order aqw essioii. F o r m axim um overshoot o f 10%. 7 3 2 9 '( l. 0 ..537 ^ = 0 .1 = 8 ^ Solve fo r ^ 2 ..87 T hus.5 3 7 -0 . ^ = 0 .^ ) ^ ’ = 0 .5 9 )0 .5 3 7 ^ ^ 1. 2^<n=2 K=at Substitute 0.7329 ^ = 0 . K =ei = 1.3 0 2 6 = Q 23026 . 873'| 0.4 J [(* + 4 6 )'+ (1 0 5 .{ y ( * ) } = 80.an d 67 for a in this transfer function.4.=46 ® '° 0 .8 73)^-80.4 * + 13 2 2 5 . b>M = 7 ’^ Square on both sides of this expression.873 0.*t)- 0.. = 115 Substitute 115 for or^ and 67 for o in the expression: =(2So+100/C) 115= = 25(67)+100Jf 100/: = 13225-1675 11550 100 K = 115.4 1 .4)' . >'(/) = 0. 105. Step 8 of 14 Consider the transfer function: iooa: + s (n + 2 5 )+ (2 5 a + 1 0 0 /:) Substitute 115. That is r ( .= - Substitute Q.-if0.316 s ( s + 4 6 ) +(105.873e-“ cos(105.4 J [(* + 4 6 )’ +(105.01:0.873 .0.l sec for t. ________ lo o y '( j+ o ) ( s + 2 5 ) + 1 0 0 X ___________ lOOK^__________ % ’ + »(o + 2 5 )+ (25a+ 100A :) Step 3 of 14 ^ Write the standard form of transfer function for the second order system. ------.4 * '+ 9 2 .25 The settling time. ^0. 11 550 = ( 4 + f l ) * ' + ( 9 2 4 + C ) * + 1 3 2 2 5 4 Compare the coefficients of 5* Compare the coefficients of s 924< + C = 0 Compare the coefficients of constant tenris.4. » impulse(sysH): The result of this MATALB code is a plot shown in Figure 3.873 for -0. 1322544 = 11550 v4 = 0. is [ i i l s l Step 7 of 14 Determine the pole location of the compensator. 100(115.4 ®.denH).4 for ^ in the expression: <i«i|.= (a+ 2 5 ) Compare the coefficient of constant terms.762. _ 4.873s 80.873 for and -$0.316 s ^ j*-h92j-Fl3225 0.6 b 46 Step 5 of 14 ^ Substitute 46 for in the expression. » plot(t.873 Step 11 of 14 Simplify further.4 * + C 8+ 1 3 2 2 5 4 Simplify further.1 sec.27PP For the unity feedback system shown in Fig. » sysH = tf(numH.4»)-0. the compensator has a pole aX s ^ .5) » W = :t i ’ + s(6 7 + 2 5)+ (25 (67)+100(115.6 f ® . M *)= i 1 .316) / 105. £ 0.25)’ = 0. Problem 3. .*exp(-46*t). in this expression.y).1 sec Write the formula for settling time. Step 14 of 14 Hence. the gain of the system.*t). the pole location of the compensator is [5 a .4»)' ste p 13 of 14 ^ Simplify further.873 0.14.6 }^ + e i Compare the denominator term: ^*+ j(a+25)+ (25fl+ 100A T ) with the denominator in the standard form: +2^Q>^+es^ Compare the coefficient of s terms. 1 rr' . Simplify further.316 Substitute 0. 4 Bs+C ^ * '° * ^ * '+ 9 2 * + 13225 Substitute 0.873.873 for ^ in the expression: *4<(>SsO 0.4) Step 12 of 14 Consider the following Laplace transform pairs. Verily your design using Matlab. and a 1% settling time of no more than 0. C o m p e n s a to r P la n t K 100 y ( j) 5+a *+ 2 S r Figure 1 Step 2 of 14 Find the transfer function for the system shown in Figure 1.873i((/)-0.4a>.873* 1 ' I * J [(*+ 4 6 )'+ (1 0 5 . » y = 0. .873+B = 0 5 = -0. 2^(8). specify the gain and pole location of the compensator so that the overall closed-loop response to a unitstep input has an overshoot of no more than 25%. y / j V _ f 0-873^ f 0-873S \ r 80.-------. y (» ) 11550 « ( s ) “ * '+ 9 2 8 + 1 3 2 2 5 Step 9 of 14 ^ The input is unit step.316 s s *+92 j + 13225 s *+92 j + 13225 0.316 for C >n this expression.873e-“ cos(105.873s 80. 1 « ‘ + ( \ a M .=46 0.*t).4») Write a MATLAB code to get the plot of H O » t = 0:0.0-873 ^ (-0 ." »c~* QO&bt (s + a ) ^ + 6 * b LT ^ « "* s in 6 / (s + a ) ^ + d * Apply inverse Laplace transform on both sides of the following expression.4 )'J i. the design is verified using MATLAB. Figure Unity feedback system C om pensator Phot Step-by-step solution Step-by-step solution ste p 1 of 14 ^ Consider the unity feedback system shown in Figure 1.4 * J [(*+ 4 6 )*+ (1 0 5 .4 )'J I 105. Clearly.*exp(-46.762e'“ sin(105. = (a + 2 5 ) 2(46) = (a+ 2 S ) 0 + 2 5 = 92 0 = 67 Write the formuia for peak overshoot. from Figure 1. X y(s) 11550 Substitute —for in the transferfunction: —r 4 = “ s------------------ * ' ’ R(s) *'+928+13225 r{ s ) 11550 1 “ * '+ 9 2 i+ 1 3 2 2 5 y ( s ) = _____ 11550______ ^ ' * (* '+92*+ 13225) Step 10 of 14 A Consider the following function: "550 ' ' *(*'+928+13225) Use partial fraction approach to detennine the response H O 11550 J Bs+C * (* '+ 9 2 * + 1 3 2 2 5 )” * * '+ 9 2 * + 13225 11550 = X (* '+ 9 2 * + 1 3 2 2 5 )+ * (« * + C ) 11550 = .316Y 105.4 Step 6 of 14 Substitute 0.^ 7 | . » numH = [11550]: » denH = [1 92 13225 Oj.4 )'J 0. . Figure 2 Verify this design using MATLAB.5 for Jf.4) (s + 4 6 ) +(105.*sin(105. 4 /. « ^ = (2 5 o + 1 0 0 X ) Step 4 of 14 Consider the following data: The peak overshoot. 4. 2^< » . I Substitute OJ25 for in this expression. the plot obtained in MATLAB is similar to the plot obtained by theoretical calculations. Re-write this expression. I (l"0-25)’ ’ ~ \ ) r ’ +(ln0. 'f .873 Substitute 0. The plot of output ^ ( / ) is shown in Figure 2. 1 + & '+ C8 11550 = 4 * '+ B * '+ 9 2 .873 for in the expression: 92*4 + C = 0 92(0. = e ^ Take natural logarithm on both sides of this expression.873tt(»)-0. w V .a Substitute for a in the expression: s ^ .5)) 11550 ” i * + 9 2 j+ 1 3 2 2 5 Write the expression for transfer function. ) = « (/) Find the Laplace transfonn of unit step input. Thus.a s = -6 7 Thus.873)+ C = 0 C = -80.*cos(105. Consider the following function: "550 ' ' * ( * '+ 9 2 8 + 1 3 2 2 5 ) Write the MATLAB code for this function.5 Thus. ^ --------=■ s* + 2 ( . . Step-by-step solution step 1 of 2 Second order sjrstem. Problem 3. G=- j.. 2 s = .^ o ^ ± c v 7 i7 _ n _ n '* o c iV r? S=-5o)„±0>4 =>— < v *p *p step 2 of 2 -IX + - . Draw the region in the s-plane that corresponds to values of the poles that meet the specification tp < tp.28PP Suppose you desire the peak time of a given second-order system to be less than tp. 7719*- Step 2 of 7 Apply squaring on both sides of the above equation.8 6 % fM 3. xl00% Substitute 17%for %PO tha equation. Step-by-step solution step 1 of 7 A certain servo mechanism system has dynamics dominated by a pair of complex poles and no finite zeros. The time-domain specifications on the rise time (tr). (a) Sketch the region in the s-plane where the poles could be placed so that the system will mee all three specifications. ts < 9.6 - 2x 3 m ] -4 rads/s Step 6 of 7 Sketch the region in the j —plane. 17%* xl00% 0M =e K .5 ^ O. Figure 2 The specific location is denoted by x figure 2 that will have the smallest rise time and also that meets the given specifications..2 s.009f’ =0. Adjecentside Hypothesis cos60® Adjecentside Hypothesis 1_ Adjecentside 2 Hypothesis Adjecent$ide(a)=l and Hypothesis=2- The dominant pole is. and the settling time (fs) are given by t.5 Step 3 of 7 Determine the damping angle. ^ 17%. percent overshoot (Mp).S 0.0 .6 .29PP A certain servomechanism system has dynamics dominated by a pair of complex poles and no finite zeros. 5 0.732 Step 4 of 7 Write the expression for rise time. . Substitute 0. £ 9.1 7 ) = .24135 f =0. 2x 0. M p< 17%.1398-13. (a) Consider the percentage peak overshoot is ^ 17%. and settling time (fs) are given by tr< 0. percent overshoot {Mp). and 0. Step 7 of 7 (b) Sketch the region in the j —plane. (b) Indicate on your sketch the specific locations (denoted by that will have the smallest rise time and also meet the settling time specification exactly.6 s. Problem 3.6sfor /j.2 sec.>/l-0. =cos"'(0. The time domain specifications on the rise time (fr). (9 .5 for ^ in the equation.^ -1. (b) Indicate on your sketch the specific locations (denoted by that will have the smallest rise time and also meet the settling time specification exactly.5* )T-tan"' (>/3) 'iT j step 5 of 7 Simplify the equation further. s = a ± Jto -l±yi. J y ll. ®. Write the expression for percentage peak overshoot.5) =60® Determine the real part of dominant pole of complex pole.6 sec.1398*^ Ra ^ V 3. /. So damping ratio ^ is 0.66 Step 2 of 5 Write the formula for rise time.8 = M 0>M :*1 step 3 of 5 Refer Figure 3. Write the formula for overshoot.9=± i °6. <p^is the undamped natural frequency.3% Thus. • Percent overshoot Wp < 16% • Settling time ts < 6. Figure 1 Thus.8 Here. 6.063 Step 5 of 5 Find the value of percentage of Percentageof x 100 Substitute 0. the value of damping ratio ^ is 0.30PP A feedback system has the following response specifications. 1.5for maximum overshoot of Refer Figure 3. locate the points where and line meet ^x)whlch Is the damping ratio.24 in the text book. Substitute 1.063x100 =6. Percentageof =0.66 for f • -irriU* =0. the percentage overshoot is |6. assuming the transfer function can be approximated as simple second order.9 • 0.20 in textbook.8 sec (a) Sketch the region of acceptable closed-loop poles in the s-plane for the system.plane as shown In Figure 1. Substitute 5. Draw the sketch of s.66. M =e' Substitute 0.5 . . the location of the pole is 30®.3%l • . Problem 3.063 for A/.9 sec • Rise time fr < 1.8 for / 1.9 for . r^is the rise time. <r is the negative part of the pole. sten 4 of 5 (b) Refer Figure 1. the simple second order acceptable closed loop poles in the s plane shown in Figure 1. (b) What Is the expected overshoot if the rise time and settling time specifications are met exactly? Step-by-step solution step 1 of 5 (a) Write the formula for settling time. is the settling time. For damping ratio ( f ) ° f 0 . Here. 948 Therefore.31 PP Suppose you are to design a unity feedback controller for a first-order plant depicted in Fig.= \i 2(2)K .i) - 4 -1 a . the value of fc can be chosen to make the coefficient of s take on any value.. ) + K K ^ . in the equation (1).3 ± J 2 Calculate the characteristic equation. s .6^0>. the transfer function is.. From the block diagram.1. the controller provides enough flexibili to place the poles anywhere in the complex (left-half) plane. Choose roots that lie in the center of the shaded region. (c) Prove that no matter what the values of Ka and a are. Step 2 of 7 (a) Refer to Figure 3. s ^ + s ( a + K K .948[- Step 6 of 7 (b) From the transfer function. + KK. ± Compare on both sides.K .)+ K K .) s(s+ a ) s’ +sa+ sKK . ( j+ ( 3 + . are 2 and — respectively.-fffl. 2 K K . the characteristic equation is.2. (As you will learn in Chapter 4. _____ 4 Step 7 of 7 (c) For the closed loop pole positions found in part (b).. 2 ) ) ( i+ ( 3 .) s ‘ + s{a + K K . the values of K m i K .= - 13 13 Therefore.57 for desired closed loop pole location.(s + K . the value of ^ correspond to the shaded region is |0. 2? (A simple estimate from the figure is sufficient.. Find values for K and Kl so that the poles of the closed-loop system lie within the shaded regions.. the value of correspond to the shaded region is 2. (1) Choose roots that lie in the center of the shaded region. s* ± -3±y2 . can be chosen so that the quantity K K .-13 K . .= 0 . 5=-3±J2 It is known that. Problem 3. K K . the configuration shown is referred to as a proportional-integral controller. 2-¥2K = 6 2K = 4 K =2 And. I J s(s+ a)*{s+ K .= 0 Substitute 2 for a and I f o r K ^ .6 Therefore.56 for unity feedback system. ^4.554^^S0. j= .) You are to design the controller so that the closed-loop poles lie within the shaded regions shown in Fig.s i n ^ sO. Step 4 of 7 Step 5 of 7 ^2 . Figure 1 Unity feedback system Figure 2 Desired closed-loop pole locations tlraCs) i r Figure 2 Desired closed-loop pole locations (a) What values of a)n and ^ correspond to the shaded regions in Fig. Step-by-step solution step 1 of 7 Refer to Figure 3. = J \3 ffl =3.K. =2 •ii'.K ^ takes on any value desired.y 2 ) ) = 0 »*+6* + 13 = 0 (2) Compare equation (1) with equation (2).6 - Step 3 of 7 Redraw the Figure 3. For this value of a value of AT. This implies that the pole can be placed anywhere in the complex plane. s ^ + s ( 2 + 2 K ) + 2 K K .){IC K .57.) (b) Let Ka= a = 2 . 0 5 )r = i^^^^ *725 ^ o .. % Finding maximum overshoot msg_overshoot = sprintf(.0. sys=tf(num.1. tp).0 5 )r = (2. now sketch these two root locations on the s-plane and derive some root locations that are likely the values for K.. Use this value iC*3. t=0:0. r is the relaxation factor. tp = t(idx): msg_peaktime = sprintf(..0 5 r Substitute the r value from equation (6).01:7. 2 = 2Ca.). both of these specifications cannot be met simultaneously by selecting the right value of K.69)(4.9 9 5 7 ) '( l. (b) Sketch the associated region in the s-plane where both specifications are met.Peak time = %3.05 =-2.21sec We have calculated two specifications Af^ and . Therefore.2f. 0 5 r From the trial and error method. Step 10 of 10 The following is the screen shot of the MATLAB: s te p R e sp o n se w ith K=3..05) = 0.. 2 -2 ^ 0 .yss)*100. A/ =5% Step 2 of 10 (a) Determine the transfer function as follows. 5 " '.f » ) 8.86 V------ 4. ^ Evaluate the natural frequency by substituting the given peak time value..5239 =4. Mp). (a) Determine whether both specifications can be met simultaneously by selecting the right value ofK.. = (0. and use Matlab to verily that the new specifications are satisfied.Time (sec).909 a 0.- s + 2 (a ^ + a . msg_title = sprintf(.. /.02.K).21)(1) =2. y=step(sys.y).34 Now substitute the values of C and O)^ which are calculated earlier in the respective equations for C and given in equation (1) to see whether these two values are equal.Step Response with K=%3. msg_overshoot). and indicate what root locations are possible for some likely values of K. = (lsec)r Here..1 sec 1- =* Substitute the value of damping coefficient in this equation to get.88 = 3. (0 .02 Substitute this value of K in the given conditions for and tp as follows.(3 ) From the part (a) we have the following conditions: 0>. ylabel{. The peak time.02 Figure 2 . num=[K]: den=[1. (0.21)(0.. Step 5 of 10 (b) Assume that the two specifications are met for the following conditions.9957 ( .69 Step 4 of 10 We know that the peak time is given e 0 > . Hence. what root locations are possible for some likely values of K. (1) Step 3 of 10 Consider the given overshoot value to solve the damping factor as follows.2 1 Step 8 of 10 Simplify equation (6) to find the expression for k • X r = - K -l a: = 4 + i Substitute the r value in this equation for the value of K: K = i+ ^ ( 2. Mp = {max(y) .4761 X ^ 0.. On the other hand.11 ».21)^ 9. 100 100 =0.0 5 )r = ( ls e c ) r .02 ^nd the following MATLAB code for plotting to verify the new specification.99 From the result it is evident that the values do not satisfy the equation. s (sec and The overshoot.34) = I 2.05) r = e ^ Further simplification yields. Equate the coefficients of powers of s and we get.3.909 -0. (0. Step 9 of 10 (c) We have the value of K from part (b).909C* = 0.(lsec)r = (2. title(msg_title). text(1.(4) (5) Step 6 of 10 Replace C nnd values in equation (3) with the values obtained from equations (4) and (5). r-2 . .y(t)..32PP The open-loop transfer function of a unity feedback system is C(j) = s(f-l-2) The desired system response to a step input is specified as peak time fp = 1 sec and overshoot Mp = 5%. -0. (2) e ^ = (0 . -0.< ) '= ( . = J k . the overshoot and peak time equations respectively are given by.t): Plot(t.9742 1 -f’ 9.den). yss = dcgain(sys). s ( l $ec)r ■ ( 6) K -\ Step 7 of 10 ^ Replace the C value in equation (2) with the value obtained from equation (5). the value of r is.Max overshoot = %3.2 . Therefore.2f..87 <•’ =0.= (0 . K=3.0 .909 -0.909(l-<-') Further simplification yields. S te p -b y -s te p s o lu tio n step 1 of 10 Consider the open-loop transfer function of a unity feedback system.909 ^ ” 2. Problem 3. xlabel{. g (^ ) 1+G(j) K __51+25_ K s‘ * 2 s * K The general expression for the transfer function of a second order system is .2.2f%%. 8 /.909C* C’ +0. 8 /.). K].nd(y==(max(y)))). Therefore. <?(*)= 5(5 + 2) The desired system response to a step input is specified by. % Finding peak time idx = max(.. (c) Relax the specifications in part (a) by the same factor and pick a suitable value for K.05)r /.. Step 10 of 13 Step 11 of 13 Substitute 20 for k and 1. ■(3) 2 b k r + —4 + — Thus.. . (4) Compare Equation (3) and (4)..(7) Where. m = mass. i= M Thus. Step 2 of 13 Figure 1 step 3 of 13 Consider the Newton’s law of equations of motion for any mechanical system.17 form in Equation (6).(5) m 2 S 6 > . y (i.5 Substitute 0.13 Thus.8 for in Equation (5).113 ^ (o o )s O .1 for ^^oo)in Equation (8)... Step 13 of 13 Refer Figure 3. 1.. Problem 3...) = 0. fflj . m is the mass of the body.. is the rise time.113-0. -G (s) f( j) Rewrite Equation Eque (2).. = .5)(1. b. Step 4 of 13 Refer figure 1 and write the equation of motion.-(/..113 for and 0. the transfer function of the system is 2 b k s^+ — s+ — m m Step 6 of 13 Substitute 0 for s in Equation (3). 1 I f i b ff s*+— s +— \ m mj 2 b k 5 *+ —j + — m m G(s)-..17 Thus. G(0) = f Step 7 of 13 Refer figure 3. b = viscous friction constant.1 >0.58(b) in the textbook..06 Thus. . Refer Figure 3.. F = n m ..1 . Step 9 of 13 Refer Figure 3.... F = n a + b i+ tc (2) Step 5 of 13 Determine the transfer function of the system.8)(6. The parameters are k = spring constant. the value of the system parameter m is |5 .1 0.58(b) in textbook...8.. t. is the undamped natural frequency. m b= = 2(0.. the value of the system parameter b is ||2. is the peak time..24 in textbook.8 for and 6. (6) m Consider general formula for rise time. The damping ratio (? ) is 0.5 for ^ . 2 G (0 ) = 0. and m7 Figure Mechanical system ■ * U u ' No b ictio (■) I r~ Step-by-step solution step 1 of 13 Refer Figure 3.58(a) in textbook to identify the direction of the spring forces on the object and draw the free body diagram as shown in Figure 1.58(b) in the textbook.33PP A simple mechanical system is shown in Fig. F is the vector sum of all forces applied to each body in a system a is the vector acceleration of each body with respect to an inertial reference frame. the maximum overshoot is 0. | 7 | .l Substitute 0.17) = 12. v l s e c Substitute 1 sector in Equation (7).. What are the values of the system parameters k. 0.1 k^20 Thus. the undamped natural frequency is 1...13.. the value of the system parameter k is I Step 8 of 13 Consider characteristics equation is . Step 12 of 13 Consider general formula for maximum overshoot. (1) Where. A step of 2 N force is applied as F = 2 * ^{t) and the resulting step response is shown in Fig.)-y H (8) Where..8^ = “ m m«6...Q6| • ... u = Ar. . Step 3 of 7 Refer Figure 1 and write the equation of motion. (1) Here. »4+2{»». (6) -(/) Step 6 of 7 Write the formula for rise time.24 in textbook.4 . l/ ( s ) = ( A & ^ + t e + * ) l'( 5 ) Substitute . 2 x 0 . the transfer function of the system is M i b k ^ ^ M Step 4 of 7 (b) Substitute 0 for s in equation (3). « ( i ) f o r 1/ ( 5 ). p is the vector sum of all forces applied to each body in a system a is the vector acceleration of each body with respect to an inertial reference frame.S fot t»|.. R{s) ^ > Since. The mass M = 20 kg and thecontrol force.. Substitute i for x io equation (4). the input is step response with amplitude x ■Therefore. A / s 0. (JI6’ + * » + * ) } '( » ) = . .5S4| respectiveiy.308 for Jt/ in equation (7).. is the rise time. and 0. M ( j ) hts^+bs*k A 4/fj’+-^5+ V Aa A M ) A T{s) = M 2 A k 4 + — 4 + ---- M hi A_ Thus. (b) Determine the values of the parameters k. [ijand |Q.. Substitute 1sec for i= M Substitute ( fo r it and l.308 4 = 0... b.and ^ is [)]. U Step 2 of 7 Write the formula for Newton’s law of equations of motion for any mechanical system.. I. So.S x l. the value of ^ is ).30$ Step 7 of 7 Refer Figure 3. u.554 Thus. m is the mass of the body. the value of 1 /(5 ) is Take Laplace transform for equation (2). (2) Write the expression for transfer function of the system.. Figure Simple mechanical system t i i z p - Step-by-step solution step 1 of 7 (a) Refer Figure 3. = l* H e r e .5fot ? . r(o )= -^ M A Consider r(o)=i (4) Consider the unit step input. The value of damping ratio (? ) is 0. is proportional to the reference input. . Step 5 of 7 Consider the general form of characteristics equation.i+a^ (5) Compare denominator of equation (3) and equation (5).g = 0.5 for maximum overshoot of 16% Substitute 0. A such that the system has a rise time of fr = 1 sec and overshoot of Mp = 16%. u = A ^ + b y + f y . the vaiue of the system parameter Jt. and zero-steady-state error to a step in r. is the undamped natural frequency.. (a) Derive the transfer function from R to Y. F = n m . Problem 3.S for in equation (6 ).59 in textbook to identify the direction of the spring forces on the object and draw the free body diagram shown in Figure 1.34PP A mechanical system is shown in Fig. (5) 5’+0.. 2 /: a: =1. 104) 0.2 f + 0 .. (1) Rewrite Equation (1).0 6 This.1 0 4 Step 3 of 20 Thus..0.45 for in Equation (6).2A: Thus. num=wn''2.( i) “ » + 0 . the value of the system parameter K is IT o il Step 17 of 20 (g) Consider the characteristic equation. /C/=0.01 the rise time is less than 4 seconds which is proved using Matlab and shown in Figure 1.104)1” Substitute 10 for • r . the final value theorem is ||9.01:150. % Finding maximum overshoot if zeta < 1 Mp = (max(y)-1)*100.4 5 Thus. . t).2 (3) r .) ' 5 ( 5 + 0 .2 ). S 0 .2 .01 for J . overshootText). (1-2) as Assume that Jm = 0. end % Finding rise time idx_01 = max(find(y<0.2f sec'.2?|- Step 8 of 20 (c) Step 9 of 20 Rewrite Equation (2).) t ) K J.2 * loge'*' . den). . = 0 .1 0 4 ) Step 12 of 20 Substitute for J' 0 .1. 0. 4=M ® .S8] Step 14 of 20 Thus.104 2ff Substitute 0.0129 /:= - 0.i): piot(t.2 j( j+ 0 .02 Vsec. the transfer function of the system is 0.(s ) R. 2. (d) Suppose feedback is added to the system in part (c) so that it becomes a position servo device such that the applied voltage is given by va = K( dr .b y .2K 5’+0. 0. 0 .0 6.104/2/wn.104 wn''2]: zeta=0.02Nm /A.1)). (a) Find the transfer function between the applied voltage va and the motor speed 9m- (a) Find the transfer function between the applied voltage va and the motor speed Om- (b) What is the steady-state speed of the motor after a voitage va = 10 V has been applied? (c) Find the transfer function between the applied voltage va and the shaft angle dm.6 0 9 ) = 5» Square on both sides.45. 0.01x10 J 0.< 2 0 % and if 1.. Substitute 4 sec for / in Equation (7).(5 ) 0.1 0 4 Step 4 of 20 (b) Step 5 of 20 Consider the final value theorem. (e) What is the maximum value of K that can be used if an overshoot M< 20% is desired? (f) What values of K will provide a rise time of less than 4 sec? (Ignore the Mp constraint. is the rise time. Are the plots consistent with your calculations in parts (e) and (f)? Figure Sketch of a DC motor S t e p . the undamped natural frequency is 0.001 fo rij.1142. 02 =e ^ Take natural log on both sides.. text( 0.1142 Thus. g. Find the transfer function between 0rand dm.5e-2].4 5 5 ) = 0.and10for R^. 104) Thus.02for JC.2 ^: . den=[1 0.1045+0..K .455 for g .(5) 0.1.2 /: 0.) Substitute0. » (1 0 )(0 .1045 + 0. overshootText = sprintf('Max overshoot = %3. 2 ^ . the undamped natural frequency is 0. idx2 = min(find(y>=0. end % Function for computing rise time fun tr = risetime(t. 0. y= step(sys.( 5 ) 5(5 + 0 . log. M . % Plotting subplot(3.1 .104 0.. Take Laplace transform. K(s) n.. (5) Compare Equation (4) and (5). tr = t(idx2)-t(idx1). b = 0. grid on.5. .0.) (g) Use Matlab to plot the step response of the position servo system for values of the gain K = 0.02 0.02 ‘ R . close all K1=[0..y) % normalize y to 1: y = y/y{length{y)).s t e p s o lu t io n step 1 of 20 (a) Step 2 of 20 Consider the equation of motion for the DC motor. Compare Equation (4) and (5). risetimeText).. t=0:0.104 • 2 (0 .01 0.2 5 + 0 .104 -1 9 .1 0 4 )'” _^ ° 0. risetimeText = sprintf('Rise time = %3.1142 for . 0.2. J.2 ^ .R. is the undamped natural frequency.01 kgm2.R . Substitute 0. and 2. ’ 5 (5 + 0 . t_r = t(idx_09)-t(idx_01).= e ^ Substitute 0.R. Problem 3.9)).2AT (4) 0 . the damping ratio is 0. ’T 7 b K . ’ 5 (5 + 0 . else tr = 0 end Step 19 of 20 The output for the MATLAB code is given in figure 1.0 . text( 0.01 Thus.1 0 4 ) ^ Step 10 of 20 (d) Step 11 of 20 Rewrite Equation (3). Mp).001 N-msec. (7) Where. I J.2 for .2 (2) K .02tbr JC. Ra = 10Q. the transfer function between the applied voltage and the shaft angle is 0. (6) Substitute 0.= 0 .9)).1)): idx_09 = min(find(y>0..U ) 0. (*)!» » Step 6 of 20 Rewrite Equation (2). 0.0 2. (8) S* + 2 ^ ^ + 6>1 Compare Equation (4) and (8).35PP The equations of motion for the DC motor shown in Fig.3.455 Step 15 of 20 Consider characteristics equation is ^ .(i) 5’ +0.2 3 Step 7 of 20 Thus. idxl = min(find(y>=0.1045+«^ Step 18 of 20 Write the MATLAB program. for i=1:1:length(K1) K = K1(i). = 0. 5.6 0 9 = * ^ ( V i V ) ( l .5.1 0 4 5 g . Ke = 0. ■ /V M ”- IM IM Figure 1 Step 20 of 20 Thus for the value of 8: < 0 . titleText = sprintf('K= % 1 . else overshootText = sprintf('No overshoot'). 5 (5 + 0.2f %'.. t_r).4 fK ).dm) where K is the feedback gain. the maximum value o f i s I Step 16 of 20 (f) Consider general formula for rise time.01x10 s) r 0 ^ 0.1 1 4 * * 0 ..45* = 0 .2*K). the transfer function of the system is 0.. wn = sqrt(0..5. Find the overshoot and rise time for each of the three step responses by examining your plots.02 x 0.5 1. 0.y). if -isempty(idx1) & ~isempty(idx2). were given in Eqs. title(titleText).2A: step 13 of 20 (e) Consider general formula for maximum overshoot. sys = tf(num..0 6 the value of M . e ) Now equation (1) becomes ^ 1 ^ [6 .. 4 ] ) .000 kg m2 B = 20. s u b p l o t ( 2 .591155)" (600000) 20000" K = (2x0....0 .. 0 . ( b) ” [ b (J s + 6 ) ] + ^ 6 (e) _ K e. ( e ) [ E ( 600000E + 20000) ] + jr 6 ( e) 1 6 7 x 1 0 -* ^ - +1.931 Step 6 of 8 (<5 t. 1 . K=400. The equations of motion are J0 + B$=Te. 1 .. K=2000.. The antenna and drive parts have a moment of inertia J and a damping B . but mostly from the back emf of the DC drive motor. 4 ] ) . n u m = [K /6 0 0 0 0 0 ]... Problem 3.93 |AT <476. a x i s (* s q u a re *) . s u b p l o t ( 2 .. 1 and Fig. s u b p l o t ( 2 .. .d e n u m ) a x i s ( [ 0 . = H < 8 0 1. s t e p ( n u m . d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] . (2..3026=. . 2 .591155) (600000) a: = 476. d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] .[s) s (J s + B ) .4 0 0 .67x10-* Step 2 of 8 (b) Given T .... (c) What is the maximum value of K that can be used if you wish to have an overshoot Mp < 10%? (d) What values of K will provide a rise time of less than 80 sec? (Ignore the Mp constraint. 2.. n u m = [K /6 0 0 0 0 0 ]. s u b p l o t ( 2 .8 ^ O L > ---- 80 IF 1. g r id .36PP You wish to control the elevation of the satellite-tracking antenna shown in Fig..4 0 0 .. 2 . where K is the feedback gain.. g r id .d e n u m ) a x i s ( [ 0 . Given values / s 600000kg. 2 . K=1000. where Tc is the torque from the drive motor.. (b) Suppose the applied torque is computed so that 9 tracks a reference command 9r according to the feedback law (b) Suppose the applied torque is computed so that 9 tracks a reference command 9r according to the feedback law Tc = K ( d r -d ). (3) Step 5 of 8 Now from equations (2) and (3) we can write 0. . 2) . these arise to some extent from bearing and aerodynamic friction. 1) .) (e) Use Matlab to plot the step response of the antenna system lor K = 200... d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] . 4) .. 3) .(s )-6 ( s )] s[Js+ B ) £l(s) [ s ( Js+ f i ) ] + ^ :6 (s ) = r e . s te p ( n u m ... 0. d e n u m = [l 1 / 3 0 K / 6 0 0 0 0 0 ] . n u m = [K /6 0 0 0 0 0 ] .. ( e) [ e (Je + 6 )]+ A T K ^ '(s ) _______ J (IF! By con^aring this with standard equation we can get B and 2 ^ ^ = ■J (2) Step 4 of 8 We know that 0.. 4 ] ) . In L ^ la c e domain this circuit is as below Js^S(s)+Bs&(s) = %{s) _____ (1) T.. a x is ( * s q iia r e * ) .591155 = 6* K =- (2 x0.. 4 0 0 .67x10-* a: step 3 of 8 (c) T akt ^ ■___ 6 .3026)* +(2. s t e p ( n u m . Do the plots to confirm your calculations in parts (c) and (d)7 Figure 1 Satellite-tracking antenna Source: Courtesy Space Systems/Loral Step-by-step solution step 1 of 8 Given motion equation J 9 + B 0 = T. 0... (s) 6 (e) _ K 6. and 2000. n u m = [K /6 0 0 0 0 0 ] ..4 ]) .1 ..591155. Assume that J = 600.. K=200. . 200 bM(MC) .400. denum ) a x is ( [0 . g r id . Step 8 of 8 LF f .591155)V __________ 20000^___________ K = (2x0.(s ) s (6 0 0 0 0 0 s + 2 0 0 0 0 ) S (s) 1.m.000 N m sec. s te p ( n u m .. 2 .. .1000..\ = e 2. = K { 0 . Find the overshoot and rise time of the four step responses by examining your plots. a x i s (* s q u a re * ) .751 step 7 of 8 w fig u r e . denum ) a x is ([0 .a x is (* s q u a re *) . 4 0 0 ... (a) Find the transfer function between the applied torque Tc and the antenna angle 6.3026)" i = 0...m^ and f s 20000N. g r id .8 ^80 =>|A:> 303. 1 . Find the transfer function between 9rand 9.sec Now equation (1) becomes f W ___________1 T . T s O. = 1. 2 . step 5 of 5 ^. s l .94 .5 0 . 5 0 .l l T ~ tr* t=0. and 1.2 =1. 1/ 1. T = 0.23 t = 0 .789 l .789 s r = 0.1.89 Step 2 of 5 Step 3 of 5 =1.2. Problem 3. 0. = 1.625 Step 4 of 5 r. t =0.5. nevertheless.0. Assume that fris the desired rise time and that 1. an ideal pitch response {qo) versus a pitch command {qc) is described by the transfer function Qa(s) _ raig(i + l/ r ) The actual aircraft response is more complicated than this ideal transfer function.1.89.38PP In aircraft control systems. = 1. the ideal model is used as a guide for autopilot design.6 •.75 Step 5 of 5 I.8 = 2.5 sec.8. .2.94 1 \ ' 1 V 1 . = 1.5 ^ = 0. Step-by-step solution step 1 of 5 For tp = 0. Show that this ideal response possesses a fast settling time and minimal overshoot by plotting the step response for tr = 0.5 \ . 024 + I) G .9 5 s+ l)(0 .54+1)(4+1) G .( * ) = 7 ■(4) ’ (0 .0 5 i+ l ) ( i ’ + i + l) Rewrite Equation (3).( 4 ) ________ .5 i + l) G .5 j’ + I .9 5 4 + 0 ( 4“ + 4 + 1) G .054+1) is the real pole in the right half plane. (0.Sm -1)(»+D ( 0 J 5 J + 1)(0 .5 2 2 5 i’ +1. This leads the system to be unstable.9 5 j + l ) ( i ’ + J + 1) Rewrite Equation (1). G .0Sj + l)(a 2 + i+ I)’ (0.554 + 0 (0 .954 + 1) is the extra pole in the equation.2 j + 1) I ( j ’ + 0 . Problem 3.55 + 1)( j + 1 ) G . (-O . So the equation can be approximated to second order system by neglecting (0. ( 4) i ( 4* + 4 + 1) Step 7 of 8 Consider the transfer function.9 5 j + 1)( j 2 + j + !)■ ( 0 .5 5 4 + 1 )(0 .054 + 1) is the extra zero in the equation.9SI+ 1)(0.5 5 4 + 1 )] In equation (4).S5f + 1)(0.5 g + l ) ( 5 + l ) Gi W = 7 (2) ’ (0.554+ 0] ( 0.5 4 + 1 )» (0 . c.2j + | ) Step 3 of 8 Consider the transfer function ( . which increases in the homogeneous response.(4)= -(5) ' (0.5s + l) ( l + l ) C4(J) = (O.05j + 1)(i 2 + » + !)■ (0. 5 5 j (1) + 1 ) ( 0 .9 5 j + 1){s^ + j + 1 ) ‘ f ifyuMu t i) C5W = ( 0 5 5 s + l) ( 0 . ' ' ( j ’ +4 + l) ( -0 .5 2 2 5 i’ + 1. This increases the rise time of the response.5* + l ) ( j ’ + 0.2 j + 1) Rewrite Equation (2). ( j ) is Step 2 of 8 Consider the transfer function.5 j ^ + 1 . (0. This increases the overshoot of the response.95 j + l ) ( j * + 0.05i + 1)(«* +* + 1) [v (» + l ) « (0.2 i+ l) Thus.954 + 1)(4’ + 4 + 1) Rewrite Equation (5).( 4 ) .5 4 + 1) Thus.( i) = (0 . ( 0 .054 + 1)(4‘ + 4 + 1) [ v (0 .954 + 1)] Step 4 of 8 Step 5 of 8 In Equation (3).95j + 1)(j2 + O J 1 + 1)’ (0. (0. (0.S» + l ) ( i .9 5 » + l)( 0 . m = (O. ( 4 ) i ( 4“ +4 + 1) Step 6 of 8 Consider the transfer function.5 l+ 1)(0. So the equation [cannot be approximated! to second order system as there is extra pole ( 0. the approximated by standard second order system <7. ^ ' (0. [ v (O . . (0. 9 5 s + l ) ( f i + S + l ) ‘ Step-by-step solution step 1 of 8 Consider the transfer function.54 + I)(0.5 » + l) ( » + l) G . 5 i+ l ) ( i+ l) <hW = (OJSj + 1)(0.5 s + | ) G i(f) = (0 .H ) G .39PP Approximate each of the transfer functions given below with a second-order transfer function.54 + 0 (0 .W = (0 .(* ) = (3) ’ ( 0 .954 + 0 ( 4’ + 4 + 1) Step 8 of 8 In equation (5).0 5 4 + 1 )(4 ® + 4 + 1 ) Rewrite Equation (4).(4)= (0.55 j + 1)(0.5 j + 1 )( j ‘ + i + 1) 1 (»’ + i + l ) Thus.954 + 1) in the system.554 + 1)(0. ( 0 .0 . the approximated standard second order system <7^ (s) is ( j ' + 0. the approximated standard second order system <?.W =' 7( 0 .0 2 4 + 1 ) G . (0.02j + 1) C5W = ( O j S j + 1)(0 .554 + 1)(0.054 + 1) <7+f4^*T (filL " ' ( 4'+ 4 + 1) (4 + 1) Thus.54+ 1)«(O . (0. the approximated standard serxrnd order system G. (0 . 40PP A system has the closed-loop transfer function m _________ 2700(1-1-25)_________ *(J) ( l+ l) ( s + 4 S ) ( l + 60)(l2 + gs+2 5)’ where /? is a step of size 7.4^^' +Cg~*“ + f t '* s i n ( 3 t + ^ ) | Step 4 of 5 (b) Settling time is set by the first pole at -1. Substitute 45 for <r in Equation (2). y(t) = l ( t ) * A e .=— .. ' 45 sO .lO sec Thus. the time function ^ ( r ) is |7 (/) + .b y . 2 .7 0 0 (2 5 ) r(o) °(1 )(4 S )(6 0 )(2 5 ) Step 3 of 5 Rewrite Equation (1).7 0 0 (^ h»25) ' ' ( i + l) ( j+ 4 5 ) ( i+ 6 0 ) ( » ’ + 8s ^ (7 ) + 25) Rewrite y ( ^ ) - „/ V 7 B C Ds + E s i+ 1 s+ 4 5 s+ 6 0 (s+ 4 )’ +9 Assign D s + E . the settling time by the first pole t. 2 . t.700(^ -t-25) r ( j) = (>»(*)) ( i+ l)( s + 4 5 ) (i+ 6 0 )( j’ + 8s+ 2 5 ) 7 Substitute — for J?(^) s 2 . (b) Give an estimate of the settling time of this step response. (a) Give an expression for the form of the output time history as a sum of terms showing the shape of each component of the response. 2. is S I Step 5 of 5 Consider the settling time is set by the second pole at -45. Substitute 1 for <r. Consider general formula for settling time. is the settling time cr is the negative real part of the pole.. S t e p . '<=T « 4 . .s t e p s o lu t io n step 1 of 5 Consider the closed loop transfer function. (2 ) a Where.F ^^d take Inverse Laplace Transform.b y .6 s e c Thus.7 0 0 (j-i-2 5 ) T (s) = (1) ( i+ l)( s + 4 5 ) ( i+ 6 0 ) ( j= + 8 s + 2 5 ) Step 2 of 5 Consider the general formula closed loop transfer function.s t e p s o lu t io n S t e p . the settling time by the second pole is | o jo | . Problem 3.* + F e *' sin ( 3 /+ ^ ) Thus. Substitute 0 for s in Equation (1)..+ B e ^ + C « . 189’ = 26. p = 240.5 sec (1 % criterion). where Figure Unity feedback system OC») -o m Find K. the value o f /< is 126.189rad/s Step 5 of 6 Determine K by substituting 5. Step 6 of 6 Verify the specifications by drawing the step response of the closed loop system using MATLAB.931 • The value of z is [ ^ . and p so that the closed-loop system has a 10% overshoot to a step input and s settling time of 1.591 Step 4 of 6 Write the formula for the settling time. B5.189 rad/s for in the relation.1. step(sys) Obtain the step response of the system from MATLAB.93 Determine the value of p by substituting 5. den=[1 p k]: sys=tf(num. z.1 =« ^ .den).= ln (0 . num= K. 0.S9\)a>.13.41 in the text book.189 rad/s for and 0.13|.3026 * * ^ ' = 2 . p=6.5 s for and 0.5 s of settling time.l) = 2. The transfer function of the closed loop system is. the specifications are met. = ( 2 )( 0 . . y (^ ) G M oM «(s) l + G(i)D(i) Substitute — ^ for G ( j ) and for G (5 )- 1 K {s+ z) K (j) j (5 + 3) s +p 1 K (s^-z) « ( j + 3 ) s +p K {s + z) 5 ( j + 3 ) { j + /» )+ A T (j+ r ) Step 2 of 6 The general fonn of the second order system with an extra pole is. The value of p is |6. Step-by-step solution Step-by-step solution step 1 of 6 Refer to the block diagram of problem 3. K = oi = 5.13 Thus. Problem 3. 2 fa . t. Write the formula for peak overshoot.I8 9 ) = 6. Substitute 1. a^p H {s )- ( j + p ) ( 5* + 2^0)^ + ) Choose 2 s 3 that results in pole-zero cancellation to reduce the transfer function into general form. Equate the overshoot to 0. y (^ ) ^ ( j+ 3 ) j ( f + 3 ) ( j + />) + A r(5 + 3 ) K s{s + p )+ K K s^ + ps + K Compare the denominator with the general second-order transfer function denominator.3 0 2 6 ^ (1 -^ *) = 0 . . Hence. K=26. = p ai =K Step 3 of 6 The specifications of the closed-loop system are 10% overshoot and 1.591 for ^ .93.591 for ^ in the relation.3 495 <• = 0.5 9 I ) (5 .41 PP Consider the system shown in Fig. 15= {O. ^3.46) ( s + 2 .46) By observing the location po les.40 ^ ( s + 4 0 ) ( s' + 4 e + 1 6 ) 3 2 0 (s+ 2 ) (s + 4 0 )[s^ + 4 s+ 1 6 ) 320 ( g + 2) Step response Y (s) = s {s + 40) + 4 s +16) 3 20(s+ 2) s (s+ 4 0 ) ( s + 2 + J3. Step-by-step solution ste p 1 of 2 Step 1 of 2 c?w = - r+ 1 1.) Then compare your answer with the step response computed using Matlab.42PP Sketch the step response of a system with the transfer function G(s) = s/2 + 1 ^ (J /4 0 + l)[(l/4 )2 + s /4 + l]‘ Justify your answer on the basis of the locations of the poles and zeros. we can say that the system is limitedly stable with S=0 . {Do not find the inverse Laplace transform. Problem 3. and the transient settling time..5x4 = 2. <2> ' ' S * + 2 ^ ^ + 6>1 Where. Step 3 of 4 Consider general formula for settling time . Problem 3. J l/..43PP A closed-loop transfer function Is given below. A f. g is the damping ratio a>^ Is the undamped natural frequency Compare Equation (1) and (2).. for this system.3 i | ...5 for ^ and 4 for ' 0 . (A) +'] H+ *] [rfr +'] [ ( i) ^ + ( 4 ) + *] [ ( A) ^ + + ' ] [® fe + *1 Estimate the percent overshoot. — 5. the percentage overshoot is | | 6 . Step-by-step solution step 1 of 4 Step 1 of 4 Consider the closed loop transfer function. = e ^ Substitute 0.5 Thus. s+2 — 2 ___ 16 4 s+2 =______ 2______ 16(s ’ + 4 j + 16) s+2 2 (1) » w = 1 6 ( i" + 4 i + 16) From Equation {1) write the second order system. 2g(&. the damping ratio g is 0. the settling time is [2. . (3) -4 Substitute 4 for in Equation (3).3sec Thus. Step 4 of 4 Consider general formula for overshoot.5.3|. = 4 . Mp.163 Thus. Step 2 of 4 Consider the characteristic equation.= e ^ -1.5 for g . y (» )= . Substitute 0.= il f® . 2 f(4 ) = 4 2 ^=i = 0.570 -g fU U = 0.. J E d i h i l i i H Simplify H ( s ) 7+1 H {s )^ 4. ts. ooo) 10. is given below: [(iro)^ (iro) + *] G(s) = [(to)^+(to) + *] [3 + *] [(iro)^+ (iro) + *] If a step input is applied to this plant. (1) Rewrite Equation (1). and overshoot to be? Give a brief statement of your reasons in each case. the percentage maximum overshoot is .000) G(s)= (2) (s^+15s"+150s+500)(s"+10s+10..den =conv([1 15 150 500]. num=500*[1 1 10000]. settling time. Step 3 of 4 Refer Equation (2) and write the MATLAB program.. 000) 500(s’ +s+10.915 Table 1 Thus.348 0. G(s).[1 10 10000]). So evaluate the problem using MATLAB code. ) sec 0 0. 000 ' ' 500(i’+i+10.den) Figure 1 shows the output for the MATLAB program. Figure 1 Step 4 of 4 Consider the values from Figure 1. .9151- . it is not easy to follow the behavior of a higher order system by the same classical method. — (j’ +j+io. However. Rise time is |Q.44PP A transfer function.step{num.0 0 0 ) 100 ' ^5' M 0.348| and Settling time is 10.)% ( t . Problem 3.000) ' (*’ +15*’ + 150s + 5 0 0 )(j* + 10s + 10. Step-by-step solution step 1 of 4 Step 1 of 4 Consider the transfer function. Maximum overshoot Rise time Settling time (M .000) Step 2 of 4 ^ In second order system it is easy to find the damping ratio using the classical method. 000 ^______ > G(i — ( i '+ 1 0 t + IO O )i(i+ 5 ) — ( j ^ + IO j + 1 0 . what do you estimate the rise-time. is E U step 8 of 11 ^ Consider general formula for overshoot.... 0.. it is not easy to follow the behavior of a third order system by the same classical method.54 Thus.541 . the settling time t. ^ Substitute 0. the damping ratio g is 0..0432 Thus.. den) damp(T) Step 10 of 11 Figure 1 shows the output for the MATLAB program. Figure 1 step 11 of 11 Consider the values from Figure 1.. 1. o>.\ ' ' 2 ( j ’ + 2 i + 2) ^ ^ S (s ) = l .6sec Thus. is the undamped natural frequency. 2 f> j2 ~ 2 1 ^ ° l2 = 0.707xV2 =4...... settling time..( i2 + 2 i + 2)’ I'M 2(» + 3) *M ' 2(lZ + 2 l+ 2 )' i ... . Consider the closed loop transfer function. Substitute >12 for in equation (5). write the second order term. is the rise time. . 2 s * » . 0.707 ^ J iW = 0... » *+ 2 t+ 2 = 0 Consider the characteristic equation.= 2 . -gxO. Substitute >12 for in equation (11). .8 — . (6) Substitute ■ n for in equation (3).45PP Three closed-loop transfer functions are given below: I'M ^ .( 11) Here. However. Consider general formula for settling time. Maximum overshoot Rise time Settling time (M .6Sl ■Rise time /^ is and Settling time /. _ l -8 '. [1 2 2]). den).2721.69 4. is m i Consider general formula for overshoot.m _ _ _________6 m (i + 3)(j 2 + 2 j + 2)' In each case. 6 ( s + 3 ) ( s * + 2 s + 2) 6 Y {s )^ R(s) ( j + 3 ) ( s^ + 2j + 2 ) (13) + 3 )^ + 2j + 2 J Refer equation (13) and write the MATLAB program..707 = 0.272sec Thus.707 for f . (3) =2 (4) Consider general formula for rise time.... . « (s ) ( s * + 2 j + 2) ^ ' (»= + 2 j + 2 ) ' ' i? W = l Consider the characteristic equation. . provide estimates of the rise time..0432 Thus..707 for g and ■Ji for in equation (12).707 Thus. Compare equation (1) and (2).. 3 /. the damping ratio g is 0.6sec Thus. Substitute 0. ' ' 2 (j’ +2s + 2) From equation (7). Consider general formula for settling time. step(num. ... ) sec ( '. Substitute 0. -gxO.707. the rise time /^ is I1.272sec Thus... ^ is the damping ratio. (5) Here.. ..)% ( / .l2 =1. i^ ta n -K w -e ta n e n liitin n Step-by-step solution Consider the closed loop transfer function. . the percentage maximum overshoot M^ \s |3.65 1. /^ is the rise time. 2 « (j) .. (10) a . Compare equation (7) and (8). (2) Here.. = 2 ..2721. ) sec 3. .8 .707 Thus. = >n Consider general formula for rise time. d /. ■(B) 5’ + 24»». Substitute 0. the rise time / is 11. the percentage overshoot is E m J o f ll ^ In second order system it is easy to find the damping ratio using the classical method. (9) e i= 2 r .f + ffl’ Here.. grid T = tf{num. _ l -8 '. the settling time t... ^ is the damping ratio.707 for g and ■Ji for in equation (6)...707xV2 =4. num=[6]: den=conv([1 3].. is the undamped natural frequency.. and percent overshoot to a unit- step input in r.707 for g . Problem 3. = . So evaluate the problem using MATLAB code. ( 12) Substitute ■ n for in equation (9).. 1. is 14.. 2 f> j2 ~ 2 1 ^ ° l2 = 0.li =1. 2««»...707.. the percentage overshoot is m u Consider the closed loop transfer function... 3Q| Step 5 of 19 eneral formula for overshoot.4m ’ + l)(P + 4 i + 40)’ 120 + 3)(s? + 4 l+ 4 0 )’ 20(1 + 2) + l)(s? + 4 l+ 4 0 )’ 3«040/40I<j 2 + 1+401) 2 + 4 l + 40)(l2 + i + 901)’ Step-by-step solution step 1 of 19 Step 2 of 19 le transfer function of the system.5 from table 9 and 10.5 sec .869 but it should be less than 5% .5 from table 1 and 2.1 sec but it should be less than 0.431 sec but it should be less than 0. 40 (1) y ^ + 4 j+ 4 0 ) Step 3 of 19 re characteristic equation _£S_ (2) + 2 fa ^ + a l amping ratio undamped natural frequency Equation (1) and (2)..i4 lercentage overshoot 3 /^ is 13541 Step 6 of 19 ition (1) and write the MATLAB program. den).32 for in Equation (3).99 but it should be less than 2. the following obsen/ations are made for the transfer ft quation (1).113 4.113 sec but it should be less than 0. Step 17 of 19 le transfer function of the system. I 40]. Hence the equati d rise time condition. -4 2 6. Hence the equr d rise time condition. Hence the equation Maximum peak overshoot condition. '([1 3].[1 4 40]).213 sec but it should be less than 0.5 sec? ransfer function(s) will meet a settling time specification of fs < 2..1 3.32 for in Equation (5).. . rved K ) % is 35. ) sec 2. Step 13 of 19 le transfer function of the system. overshoot Rise time Settling time ( '/ ) <0. 40 (7) t-l)( l* + 4 j+ 4 0 ) Step 11 of 19 rtion (7) and write the MATtAB program. sec isetime is [0 .87 s‘ + 4 s + 4 0 ) ( j ’ + i + 9 0 l ) ( i ’ + 1 + 4 0 l ) 89. rved is 4. rved K ) % is 0. grid □ obtains the step response plot den) lows the output for the MATtAB program.[1 4 40]). rved J is 1. Hence the equation sat aximum peak overshoot condition. Hence the equation required Maximum peak overshoot condition. overshoot Rise time Settling time ( '/ ) 0. 120 (8) l+ 3 )(l^ + 4 i+ 4 0 ) Ition (8) and write the MATtAB program.5 <2.87 j ’ + 4 i + 4 0 ) (j* + 2 s’ + l. '(]1 1]. Step 7 of 19 le values from Figure 1. rved K ) % is 0 but it should be less than 5% ..2 8 4 1 Step 4 of 19 eneral formula for settling time. grid den) lows the output for the MATtAB program Step 18 of 19 le values from Figure 5. ) sec ( '. 20(i + 2) (9) j + l ) ( l '+ 4 i + 4 0 ) Ition (9) and write the MATtAB program. rved is 0. overshoot Rise time Settling time ( / .f2^4v4. grid den) lows the output for the MATtAB program Step 16 of 19 le values from Figure 4.5 <2. grid den) Step 14 of 19 le values from Figure 3. 01.87 s‘ + 4 s + 4 0 ) ( i ’ + i + 9 0 l ) ( i ’ + 1 + 4 0 I ) 89.5 sec.(5) 3e time. overshoot Rise time Settling time ( / . ) sec 1. are given below: 3 . 6. 0. Hence the equation s sttling time condition. the following obsen/ations are made for the transfer ft quation (7). Hence the equr d rise time condition..5 from table 3 and 4. . den).040 4 0 l(i" + j+ 4 0 l) j ‘ + 4 i+ 4 0 )( i^ + s + 9 0 l) 89.5 sec. Hence the equation sttling time condition.99 Id values are tabulated in table 4. den). overshoot Rise time Settling time ( '/ ) <0. . L6___ ■x6.87 ( 10 ) 1 * + 4 i + 4 0 ) ( i ' + 2 i ’ + 1303s’ + 13021+361301) Ition (10) and write the MATtAB program. Hence the equation required Settling time condition.32 but it should be less than 2.32 • 0.42 sec but it should be less than 0. Step 8 of 19 irder system it is easy to find the damping ratio using the classical metho ry to follow the behavior of a third order system by the same classical me le problem using MATtAB code. (6) 6.32 ec lettlingtime is |2. the following obsen/ations are made for the transfer quation (10). ) sec ( '. overshoot Rise time Settling time ( / .8 but it should be less than 5%.53 but it should be less than 2.316 for g .5 from table 5 and 6. overshoot Rise time Settling time ( '/ ) <0. Step 19 of 19 red is 0.431 1.72 but it should be less than 2. Hence the equat e required rise time condition.5 sec. overshoot Rise time Settling time ( '/ ) <0.5 sec..316.303s^ + l . ) sec <0. '([1 1].42 3. grid den) Step 12 of 19 le values from Figure 2.5 <2. rved K ) % is 0 but it should be less than 5%..3 0 2 i + 3 6 l. the following obsen/ations are made for the transfer ft quation (8).5 sec. Hence the equation sa aximum peak overshoot condition. 0.5 <2.5 sec .5 sec. '(]1 4 40].77 but it should be less than 2.. Step 15 of 19 le transfer function of the system.5 from table 7 and 8.(3) (4) eneral formula for rise time.316 for g and 6. Hence the equatic required rise time condition.32 Id value is from the given condition is tabulated in table 6. rved J is 3. .5 sec. 3 6 . >^j« ruii* . Step 9 of 19 Step 10 of 19 le transfer function of the system.[1 2 1303 1302 361301]). overshoot Rise time Settling time ( '/ ) 0. rved is 2. ) sec ( '. Hence the equatio d Maximum peak overshoot condition.213 1. den). rved is 1. ) sec ( '. overshoot Rise time Settling time ( / .5 sec? 40 2 + 4i + 40)’ 40 + I)(j2 + 41 + 40)’ 120 -i-^tr.77 Id values are tabulated in table 2.316 lamping ratio g is 0.72 Id value is from the given condition is tabulated in table 6.. rved K ) % is 46.53 Id value is from the given condition is tabulated in table 6. Hence the equation e required Settling time condition.5 sec .[1 4 401).4 6 P P ransfer function(s) will meet an overshoot specitication of IWp < 5%? :ransfer ftjnction(s) will meet a rise time specification of tr< 0.3 0 l) 89. rved J is 3. .5 <2. Hence the equation ( required Settling time condition. rved is 1.. rved is 0.32 for in Equation (6). ) sec 0.1 but it should be less than 5%.. the following obsen/ations are made for the transfer ft quation (9). where m < n). 5 5+1 5 +2 5+3 1 9 ^ 18 10 5 5+1 5+2 5+3 It is known that. (2) 5 5+1 5 +2 5+3 Step 9 of 18 Use residue method to calculate coefficient x ■ f l) (i+ 2 ) (j+ 3 ) 2 ( » . i J 1 3 ( ^ . m zeros and n poles.' i . r.W = ( j + l ) ( i + 2) ’ (s + l ) ( i + 2 ) ( j + 3) The unit step response for Gy ( j ) i' 2 ( j. S t ^ respoase fo r G 2(t) ^1-------------------. Prove that a stable.*exp(-2. the sketch of unit response for Gy ( 5) is shown in Figure 1.+ — -+ — . Prove that a stable. 1 = ( j+ l) 5(5 + 1){5 + 2)(5 _ 3 M K fz 2 ) 5 (5 + 2 )(5 + 3) 3 (-1 -1 )(-1 -2 ) .2 ) (5 + 1){5 + 2)(5 + 3) 3 (0 -l)(0 -2 ) ( 0 + l) ( 0 + 2 ) ( 0 + 3 ) 3 (-l)(-2 ) (1 )(2 )(3 ) -I Step 10 of 18 Use residue method to calculate coefficient g . V/ \ A B C D . where m < n).^-------------------- . 2 ( » .4 e ''+ 3 e '* ' Step 6 of 18 MATLAB code to sketch the unit step response for Gy ( 5) » t=0:0.y1) » title{'Step response for G1 (t)’) » xlabel('time(s)') » ylabel('y1 (t)') Step 7 of 18 The sketch of unit response for Gy ( 5) is shown in Figure 1. ' ' 5 5+1 5+2 1 4 ^ 3 5 5+1 5+2 It is known that.2 )L .I) (.0 {-2 )(-2 + 0 ' 2 Step 5 of 18 Recall equation (1). _ i(£ z O | 5(54 2(-<-0 . (5 + I)(5 + 2)(5 + 3) r(A .+ — -+ — - 5 5+1 5+2 Substitute ( fo r X ’ .(/) = l. 3 ( -2 -l)(-2 -2 ) (_ 2 )(-2 + 1 )(-2 + 3 ) _36 ' 2 -1 8 Step 12 of 18 Use residue method to calculate coefficient / ) . i ‘■ 'f c ) " ' Apply Inverse Laplace transform. strictly proper system has an undershoot if and only if its transfer function has an odd number of real RHP zeros. » plot(t. (b) Explain the difference in the behavior of the two responses as it relates to the zero locations.*exp(-3.. \ A B C i. Step-by-step solution s te p 1 of 18 (a) The two non-minimum phase system are.' i . » yl=l-4.*exp{-2.l) I ( s + 1 )(j + 2 ) L •L 2 (0 -0 ( 0 + l)(0 + 2 ) ' 2 Step 3 of 18 Use residue method to calculate coefficient g . m zeros and n poles. ( j + I ) ( * + 2) i.4 for ^ . Let y(t) denote the step response of the system. 3'2(/) = l.( i) = . for ( 72( 5) in the above equation.y2) » title{'Step response for G2(t)') » xlabel('time(s)') » ylabel('y2(t)') Step 15 of 18 The sketch of unit response for ^ 2 ( 5 ) is shown in Figure 2.r>\ y .{ s l^ = (* + 2 )f 2 (^-0 *(*+oL 2( .*t). strictly proper system (that is.+ — _ (1) S 5+1 5+2 Use residue method to calculate coefficient x ■ _ 2 ( .l ) ( .)= _ + — .2 ) i ( i + l ) ( j + 2 ) [_ j 3 (-3 -1 )(-3 -2 ) (_ 3 )(-3 + 1 )(-3 + 2 ) "-6 — 10 Step 13 of 18 Recall equation (2).M L _ -(X I3 )[ 1 3 ( » .9 e .( .a n d 3 for ^ in the above equation.*exp(-t)+18. G .l) ( ^ .w = i ( i + l ) ( l + 2) Step 2 of 18 Apply partial fractions to simplify the function Yy(^ ) • !.l) ( » . i .2 .2 ) | 5(54 » + ') ( ^ + 3 ) L . S t ^ respoase f o r G l ( t ) tim e (s ) Figure 1 Therefore.l) . 3 ( ^ . strictly proper system has an undershoot if and only if its transfer function has an odd number of real RHP zeros. paying close attention to the transient part of the response.1s- (a) Sketch the unit-step responses for G1(sJ and G2(s^. ( e l in the above equation.l + 2 )(-l+ 3 ) ..(.' + I85‘ *'-1 0 e **' Step 14 of 18 MATLAB code to sketch the unit step response for G^ { 5) ^ » t=0:0.1:10.l) ( .*t)-10. 3 '..+ — . ( j) = . The step response is said to have an undershoot if it initially starts off in the “wrong” direction. f= ( * + 3 ) j.ii ‘ -2 —9 s te p 11 of 18 Use residue method to calculate coefficient C- . Step 8 of 18 The unit step response for <72( 5) is.l) ( ^ . 18 for -1 0 for C ill the above equation.+ ------- 5 5+1 5 +2 5+3 Substitute ( fo r x ^ . 0 (^ + 2 )JL .*t). strictly proper system (that is. The step response is said to have an undershoot if it initially starts off in the “wrong” direction.*exp(-t)+3. (c) Consider a stable.9 for q . » plot(t. (c) Consider a stable. A B C D K ( 5 ) * — + -----+ -------.( . i .l) ( ^ .( . »y2=1-9. i ‘■ 'f c ) " ' Apply Inverse Laplace transform.1:10.?ir.2 ) ' s ( j + l ) ( i+ 2 ) ( * + 3 ) Apply partial fractions to simplify the function ( 5 ).l+ 2 ) -1 Step 4 of 18 Use residue method to calculate coefficient C- C = ( s * 2 ) Y . —r — ^ r for ( 7. Let y(t) denote the step response of the system. m .!: 2 ) ' Step-by-step solution step 1 of 5 w Li this case we have ^ = 1 ■ Find H(s). Express your answers in terms of hyperbolic functions (sinh. Alternatively. F ind^(0. U [. except now C k jl. . . for a unit step ii^ut.48PP Find the relationships for the impulse response and the step response corresponding to Eq. Alternatively. we get k{t) = aj'te-"-'. HW = _________£5_________ HW = (i + { < b.:----.C is negative and the e^ onential terms become imboimded and the system is unstable..f i. Problem 3.\ Find^Cf). since we already dealt with the case of | ^ > 1 in the previous part The impulse response and the step responses are exactly the same. for a unit step input. for the cases where (a) the roots are repeated. ' 1 ____ 1 24^ r ( s ) = i + — r.where |^] > 1 For the impulse response U{s) = 1. that is ^(<) . Is negative. cosh) to best show the properties of the system response. U (s) = — By applying inverse Laplace we get Step 3 of 5 (b) Fin<)H(s).—= ^ ^ = e ~ ^ sin(<P^). 1+ Y(s) = 1 j'( s ) = i+ — !— .and using partial &action expansion.lj step 4 of 5 We can then integrate the in^ulse response to obtain the unit step response. (b) the roots are both real. .« ■ * Step 5 of 5 ^ Now we have the remaining case where C ^ negative and | ^ < 1. We can then integrate the in^ulse response to obtain the step response. (c) the value of the damping coefficient. *(0 - h{t)= . we get H O = ! .f " ' ^ ‘ 2 4 i> .) ( s + a i) Step 2 of 5 ^ For the impiilse response U(s) = 1. H [s ): 'U W H {s ) = -------------------. &nd m - _________ 1_________ __________1_________ r(r) ' I Simplify further. ^ ------------------. ) = . Thus. .)2 + <o5 ( 1 . w e get K i= l K j= A V And 2 2 ^ ( K ’ )(p=-2?avP+aw’ ) < 9 -^ y f t ) = l +Ae'*‘ +Be’* sin(cDat-0) ^ Q) (1) '' Step 3 of 5 (a) I * Term domnates Step 4 of 5 (b) “ 2^co. and assuming that ojnr = 1 and ^ = 0.7. where A= «3 -2 < a i« p + p 2 ’ P -{ P -{« * (a) Which term dominates y(t) as p gets large? (b) Give approximate values for A and B for small values of p. plot the step response of the preceding system for several values of p ranging from very small to very large. «W = . Problem 3.49PP Consider the following second-order system with an extra pole.e). (c) Which term dominates as p gets small? (Small with respect to what?) (d) Using the preceding explicit expression for y(t) or the step command in Matlab. ( * + P )(^ + 2f Show that the unit-step response is y(0 = 1 +Ae~^ +Be-^* an(coat .s+ti>«’ ) Y (s )= ^ + ^ + Step 2 of 5 Solving. At what point does the extra pole cease to have much effect on the system response? Step-by-step solution step 1 of 5 H (s) = (s+p)(s“ +2?a^s+i!^’ ) Y (s) °>»P s(s+p)(s“ +25ti>i.p P Step 5 of 5 (c) (last term becomes small as P becomes small with respect to H ^term . y(i). ^ (0 = 1 g3 ^ ~ .a^e Step 7 of 7 (c) For a given Overshoot Mp.tan" \ Hence it is proved that Step 4 of 7 (b) At peak time we know that — ^—. In general. the values o f and ^ have to be found by trial and error.2i<on5 + ai§) (a) Show that the unit-step response for the system is given by y (0 = 'c(IE((iy( + A ). I “i J Thus.2gai.(^ + A )] Where I ca^. H (A = a»g<J + 2) -I.^ = = s in ( a y . I *.z ^ + Step 6 of 7 Thus. .^ ) + .1 __ C 4 = tan" Step 5 of 7 F in d ^ + 4 .) ■ ^ + a j . we get .^ 3 tan"^ M .l .( . ^ ^ ^ 2 . .2 .) cos (a>. f l + 4 = tan" _z. = — ' JM Find = —^z^ . where f t =UD (b) Derive an expression for the step response overshoot./z’ + a>. = . -^ = = c o s (® rfi . )sin (a>^ . as ^ 2 .= 0./?) cos(iV-A-4) = o Therefore. we get ( \ 1 z . how do we solve for ^and ojn7 Step-by-step solution step 1 of 7 Consider the second order unity DC gain S3rstem with an extra zero. and find s /? + . Problem 3.^ z ^ .f .50PP Consider the second-order unity DC gain system with an extra zero. we g ett.fi. they will be different than the standard second order system values unless z is large that is the zero is far away.i - Where -1 _ L _ Simplify further.2 ■ We know that^(^) = . Mp.-z Step 3 of 7 We combine the last two terms in the argument o f the cosine term. z ( e^ + <z?'l w We write the transfer fimction in partial fic tio n form. P .^ 1 — s in (® . of this system. — 1^-1+ ijc o s (a > . Mp. dt a ^ [.a i. (c) For a given value of overshoot.^ ) + .^ ^ ( 0 = 1. z dt Step 2 of 7 Find. Mp).5 sec for a unit-step change in d r . The autopilot controller uses the pitch attitude error e to adjust the elevator according to the transfer function E(s) j+ 1 0 Using Matlab. t).3000 0 50 100 150 Time (sec) Figure 4.5 0. grid on.4f .03 j + 0 . end Step 6 of 7 The following are the plots of pitch attitude verses time for various values of X - For a : = 3.03 j + 0. risetimeText = sprintf('Rise time = %3.06) [+50A : ( j ’ + 3j ' + 2j ’ + S j + j ’ + 3j + 2j + 6 ) 50Ar(j’ +ds’ + l U + 6 ) ' r j ’ +15.5 Max overs hoot =12. comment on the difficulty associated with making rise time and overshoot uoii ly ivieiuaL.4) j +(3 0 0 A :+ 2 4 ) J Step 5 of 7 The output must be normalized to the final value of for easy computation of the overshoot and rise-time.5 0.51)j’ + (3 0 0 /C + 4 0 3 . Block diagram of autopilot.9)). risetimeText).5. else overshootText = sprintf('No overshoot') end % Finding rise time idx_01 = max{find{y<0. idxl = min(find(y>=0.3 0. g ( i) G (i)0 (i) 6 l. t_r).03 * j + 0.03].1 Mp = {max(y)-1)*100: overshootText = sprintf('Max overshoot = %3. den=[10. i+ lO J 5 0 (i+ l)(i+ 2 ) V A :(i3 -3 n ^ s ’ + 5 i+ 4 0 )(* ’ + 0 . Problem 3. The transfer function of the system is.3: K= 0.0 6 )( j + 10)+50^C( j + 1)( j + 2 )( j + 3 ) j’ + 5 j + 4 0 ( j ’ +0.51 PP The block diagram of an autopilot designed to maintain the pitch attitude d of an aircraft is shown in Fig. % Finding maximum overshoot if zeta .0 6 )( j +10) 5 0 y ( j + l ) ( j + 2 )(j+ 3 ) ” (j ' + 5j + 4 0 )( j =+0. = D{s) 4^+3) s+ W Step 3 of 7 The error signal £ ( i ) is. 5D (^+l)(^+2)_____ ^ +a0S s+O €6j The autopilot controller uses the pitch attitude error e to adjust the elevator according to the following transfer function.2 *K).y) % normalization of y to 1 y=y/y(length{y)).6 )j’ 1 [ +(550A:+17. comment on the difficulty associated with making rise time and overshoot measurements for complicated systems.0 6 )J l.5 3.104wn^2].y). text( 0.03 j + 0 . The next step is to find the value of K that will provide an overshoot of less than 10% and a rise time faster than 0. y= step(sys.. After examining the step response of the system for various values of K.03 j + 0 . else tr = 0 end Enter the following commands to plot the pitch attitude $ and also to find the peak overshoot and rise time.9)) if -isempty{idx1) & ~isempty(idx2) tr= t{idx2)-t(idx1).3.01:150.5 Max overs hoot= 17. num= wn''2. % Plotting the pitch attitude subplot(3. find a value of K that will provide an overshoot of less than 10% and a rise time faster than 0. Figure Block diagram of autopilot lAifcn ■ D ^) Step-by-step solution step 1 of 7 Consider the following block diagram of autopilot designed to maintain the pitch attitude $ of an aircraft: Figure 1.2. After examining the step response of the system for various values of K. Step 2 of 7 Consider the transfer function relating the elevator angle.1)) idx2 = min(find(y>=0.i): Plot(t. S.0 6 )( j + 1 0 )+ 5 0 K ( j + 1)( j + 2 ) ( j + 3 ) SOA: ( j ’ + 3j ’ + 2j * + 6j + j ' + 3 j + 2j + 6 ) 7 ( j ’ + 15j '+ 9 0 j + 4 0 0 )( j ’ +0.03: K= 0.12 se : 0 50 100 150 Time (sec) Figure 3.1)): idx_09 = min(find(y>0. zeta=0. in degrees and the pitch attitude. B in degrees.0 6 )Jt. 0. iiTiu a voiUc vj1 r\‘ u lai' win ^ivjviuc ai i uvci si luui ui icso u lai i i u /o ai ivj a i isc uiiic faster than 0.5: K= 3. Step response of autopilot system for K —i - For a: = 0. J + 10 ) ( 50K{s* soa : ( j + i )( j + 2 )( j + 3) j [ (( j ^+ * + 5 j + 4 0 )( j ^+0.11 sei 0 50 100 150 Time (sec) Figure 2: Step response of autopilot system for K —Z.5 sec for a unit-step change in d r .S Step 7 of 7 For K = 3 K= 3.5 sec for a unit-step change in The following is the MATLAB function for computing the rise time: %Function for computing rise time function tr = risteime(t.0 0. Rise time = 0.0300 .5000 1. title(titleText).0 3 s+ 0 . den).(s) (j " + 5j + 4 0 )( j =+0. Step response of autopilot system for K —0 3 - For a: = 0.06)( j + 10) J 5 0 g ( j + l ) ( j + 2 ) ( j+ 3 ) 1+ + 5 j + 4 0 )( j ’ +0.03 j + 0 .5.0 3 j+ 0 .0 6 )( j + 10) ] Step 4 of 7 Simplify this equation further and we get. ( 5 0 y ( j + l ) ( j + 2 ) ( j+ 3 ) J [ ( j ' + 5 j + 4 0 )(j’ +0.2f sec'. titleText = sprintf('K=%1. t=0:0.( i) ° l + G (i)D (i) S 0 ( i+ l) (i+ 2 ) J (i^ + 5 s + 4 0 )(j* + 0 . >1 Rise time = 0. t_r=t(idx_09)-t(idx_01).0000 1.03J*+(50AT+90.K). overshootText). The transfer function relating the elevator angle 5e and the pitch attitude d is e(5) 5 0 (i+ l)(5 + 2 ) W j) (sZ + 5 s + 4 0 )(l2 + 0. text( 0. wn = sqrt(0.104/2/wn. sys = tf(num.1.03 j +0. 0.06)’ where d is the pitch attitude in degrees and 6e is the elevator angle in degrees.06)( j + 10)J e. for i=1:1:length(K1) K = K1(i).2f. % Program to find peak overshoot and rise time and also to plot the pitch attitude clear all close all K1 =[3.031+0. . (a) For a first-order system. we can get 2A wfc-41 Here a -= \{\a ... show that the tim e to double is bi2 where p is the pole location in the RHP. (4) From the equation (3) |g(A))| = “ So = A ..:— ■ fai2 *2 = —. _a Let the Transfer Function as O fs) = -*----.. (3) iH if Step 4 of 4 At t s let the amplitude of the response be 2A'. Then... (2) By taking the ration of equation (2) to equation (1).........g ....:— ■ Figure Time to double Time Amplitude 2A Anqditude ______ I te w L *2 » Step-by-step solution step 1 of 4 w For the 1* order system.r Then... •(5) By taking ratio of equation (6) to equation (5). Then. we can get fe" 2 A _ te * A “ te ” _2 p Given that the increased time as Therefore ln 2 Where = ti-to Step 3 of 4 ^ (b) For the second order system.52PP A measure of the degree of instability in an unstable aircraft response is the amount of time it takes for the amplitude of the time response to double (see Fig).. theresponeis g (^ ) = gn ^ s in fa t^ +cos“* ^)= A Where cos"* ^ = cos"* |^ |+ ^ Therefore g ( < b )= -g o -4 — -s in (® A + cos"“|{|) = A .. (b) For a second-order system (with two complex poles in the RHP). tile respone is g (O = So ^ ^ ^ (<Vi Where cos'* ^ = cos”* |^|+?r Therefore g(t^) = .......... the respons e is given by taking the Inverse Ls^lace Transform o f 0{s) a g (<) = go ^ ^ sin [a ^ +cos"‘ i ) A tt let the amplitude of the response be 'A'... Problem 3.. g W = f e * = 2 A ..= -{a ^ Therefore . (1) Step 2 of 4 Now consider that at time t s ^ the an^litude is a s ' 2A’ Then.. ^ ^ ^ sin + cos"* |^|) = 2 A .. g ( g = t e < ' = A . given some nonzero initial condition.-=-------. show that lo2 *2 = —... Let the Transfer Function is 0 ( e ) = ——— s—p By taking Inverse L ^ lace Transform we can get g {t)= k e ’‘ Now consider the initial time as /g And consider at t = tgthe an^litude of the response is as ‘A’ Then. Write the Routh array for the polynomial. hence the given closed-loop system Is unstable. XG(s) S* ( j + l) First. so there are two poles not in the left hand side of the plane. it is clear that the sign changes twice in the first column. it is clear that the sign changes twice in the first column. I+ ^ G (s) = 0 5 ( i ’ + 2 j* + 3 » + 4 ) j*+ 2 j*+ 3 j’ +8s +8 = 0 To determine the Routh array. 2 x 8 . it is clear that the sign changes twice in the first column. . find the characteristic equation. beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients.cnoAd - 2 x 3 -8 x l _ 6 -8 2 =-l .53PP Suppose that unity feedback is to be applied around the listed open-loop systems.b. so there are two poles not in the left hand side of the plane.lx 0 16 ° 2 Sa-2b -1 -8 -1 6 -1 -2 4 d=b=^ According to Routh array a system is stable If and only If all the elements in the first column of the Routh array are positive. beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients.0 To determine the Routh array. 1 2 1 8 -6 8 >7 of 11 According to Routh array a system is stable If and only If all the elements in the first column of the Routh array are positive. Write the Routh array for the polynomial. ----. Step 5 of 11 (b) The transfer function of open-loop system is.. first arrange the coefficients of the characteristic polynomial in two rows. Write the Routh array for the polynomial. I-flC G (s ) = 0 4 ( j ’ + 2 j * + j +1) 1+ . From the obtained Routh array. first arrange the coefficients of the characteristic polynomial in two rows. 1 3 8 i* : 2 8 i* : a b c d Evaluate the variables a. hence the given closed-loop system Is unstable. Use Routh’s stability criterion to determine whether the resulting closed-loop systems will be stable. From the obtained Routh array. find the characteristic equation. I + ^ G ( j) = 0 5*( j + I ) + 2 ( 5 + 4 ) = 0 5’ + 5 * + 2 j + 8 » 0 To determine the Routh array.. hence the given closed-loop system Is unstable. (b) = ^ W «” <»> = Step-by-step solution Step-by-step solution (a) The transfer function of open-loop system is. Problem 3.s .----------{ = 0 s‘ (s’ +2s‘ . 30+2 15 _32 " l5 4 =4 Step 11 of 11 According to Routh array a system is stable If and only If all the elements in the first column of the Routh array are positive.l) + 4 ( s ’ +2s‘ + s + l) = 0 j’ + 2s* + 3 j ’ + 7 j ’ + 4 j + 4 . First. 1 3 4 * : 2 7 4 <4 “ i i* : *1 step 10 of 11 Solving for C| and we get. (c) The transfer function of open-loop system is. From the obtained Routh array.l ) s‘ (s’ +2s‘ .s . so there are two poles not in the left hand side of the plane. first arrange the coefficients of the characteristic polynomial in two rows. beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients. ' ' +25^+35+4) First. V . find the characteristic equation. sV 6 j“ + 2 5 -0 5 * + Os’ + + O r+ 2 5 = 0 To determine the Routh array. Write the Routh array for the polynomial.2s + 8 = 0 (d) S3 + s2 + 20s + 78 = 0 (d) S3 + s2 + 20s + 78 = 0 (e) s4 + 6s2 + 25 = 0 Step-by-step solution step 1 of 5 (a) The characteristic equation is. first arrange the coefficients of the characteristic polynomial in two rows. first arrange the coefficients of the characteristic polynomial in two rows. Step 5 of 5 ^ (e) The characteristic equation is. 1 30 344 10 80 480 s’ : 22 296 s’ : . Problem 3. *’ + i ’ + 2 0 * + 78 = 0 To determine the Routh array.5 480 s': 49 0 s’ : 4 80 Since there are two sign changes in the first column of Routh array. beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients. Write the Routh array for the polynomial.5 4 . 8 8 »>: -4 j* : 8 Since there are two sign changes in the first column of Routh array.54PP Use Routh’s stability criterion to determine how many roots with positive real parts the following equations have. it is clear that there two coefficients missing so there are roots outside of the LHP. t’ : 1 6 25 i* : 4 12 3 25 100 i* : 12. * * + &t’ + 3 2 i’ + 80s + 1 0 0 = 0 To determine the Routh array. Write the Routh array for the polynomial. (a) s4 + 8s3 + 32s2 + 80s + 100 = 0 (b) s5 + 10s4 + 30s3 + 80s2 + 344s + 480 = 0 (c) s4 + 2s3 + 7s2 . first arrange the coefficients of the characteristic polynomial in two rows. the number of roots with the positive real parts is two. the number of roots with the positive real parts is two. 1* : 1 20 1 78 -5 8 78 Since there are two si'^n Ghan*^es in the first c-oiui 1 nf Rniith arrai/ tha niimhpr nf mnt« u/ith thp positive real parts a two. s* +1 Oj ^ + 30s’ + 80s^ + 3 4 4 s + 4 8 0 = 0 To determine the Routh array.6 s": 100 Since there are no sign changes in the first column of Routh array. the number of roots with the positive real parts is two. -2 1 3 i* : 25 From the characteristic equation. beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients. Write the Routh array for the polynomial. the number of roots with the positive real parts is zero. Step 2 of 5 ^ (b) The characteristic equation is. Step 3 of 5 (c) The characteristic equation is. Since there are two sign changes in the first column of Routh array. s*: 1 32 100 s’ : 8 80 22 100 s' : 43. beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients. first arrange the coefficients of the characteristic polynomial in two rows. 1 7 8 2 -2 j’ . beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients. Step 4 of 5 (d) The characteristic equation is. beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients. . first arrange the coefficients of the characteristic polynomial in two rows. s*+ 2s’ + 7s’ .2 s + 8 = 0 To determine the Routh array. Write the Routh array for the polynomial. Problem 3.55PP Find the range of K for which all the roots of the following polynomial are in the LHP: s5 + 5s4 + 10s3 + 10s2 + 5s + /< = 0. Use Matlab to verify your answer by plotting the roots of the polynomial in the s-plane for various values of K. Step-by-step solution step 1 of 2 s’ +Ss’ +lOs’ +lOs’ SrHtN) oiep Iui ^ 8^+58^+10s^+10s^5s^^c=0 1 10 5 5 10 k 8 -5 j+ (k-25)+ 80 +8 ■1 -8 L ,.( fc - 2 5 ) ^ + 8 0 ( k - 2 5 ) l k-25+80 * k Step 2 of 2 — - i( k - 2 5 ) > 0 k+55 5^ ^ 64k+j(k-25)(lc4-55) <0 3201rtk’ +30k-(25x55) <0 =>k^+350k-1375 <0 ■ |-353.85>k>3.85l Problem 3.56PP The transfer function of a typical tape-drive system is given by __________________________ j[(j + OJ)(j+ l)(j2 + 0 ^ + 4 ) ] ’ where time is measured in miiiiseconds. Using Routh’s stability criterion, determine the range of K for which this system is stabie when the characteristic equation is 1 + KG(s) = 0. Step-by-step solution step 1 of 6 Step 1 of 6 Consider the transfer function of a typical tape drive system. , , __________g ( j + 4)__________ s + 0.5)( j + 1)( + 0.4j + 4)J Step 2 of 6 Consider characteristic equation. l + ^T G (j) = 0 .......(2) Step 3 of 6 Substitute Equation (1) in Equation (2). , K (s+ 4) 1+ f ^ ~Q 4|^(»+ 0 .5 )( 4 + l ) ( s + 0 .4 s + 4 ) ] j [ ( j + 0 . 5 ) ( j + l) ( » ’ + 0 . 4 s + 4 ) ] + ^ r ( j + 4 ) = 0 s ’ + l.9 s ‘ + 5 .ls ’ + 6 .6 j ’ + 2 s + j A :+ 4A: = 0 s ’ + 1.9s‘ + 5 .1 i’ + 6 .2 s ’ + ( 2 + A : ) i + 4 X = 0 Step 4 of 6 Thus, the characteristic equation is +1.9s*+5.L $^+ 6.2»^ + ( 2 + J l)4+4AT = 0 Apply Routh array for this polynomial Step 5 of 6 5.1 2 +K 1.9 6.2 4K F ig u re 1 Step 6 of 6 The system is stable if the equation satisfies the following conditions; • All the terms in the first column of the Routh’s array should have a positive sign. • The first column of Routh’s array should not posses any sign change. From the above statement, the stability conditions are, /:+ 3 .6 3 > 0 ^ > -3 .6 3 And - 8 .4 3 < ^ : < 0.78 Thus, the stability condition is |0<J^T < 0.781- the conditions on the system parameters (a, ^ f f ^ ) to guarantee closed-loop system stability. Figure Magnetic levitation system a Re (f+ p ) Step-by-step solution Step 1 of 6 Consider transfer function of the system. Step-by-step solution step 1 of 6 Consider transfer function of the system. 21 -y+p K, s+ p j '- o “ {s+ p )(s‘ - a ‘ )+ K K ,{s+ z) __________ K K ,{ s + z ) __________ j*+ps’ +(A3Ti, - o’ +JCKjZ- po* K K ,{s* z) Thus, the transfer function of the system is 5^ + ps^ + ( k K ^ - < ^ ] s + K K qZ - p<^ Step 2 of 6 Step-by-step solution step 1 of 6 Consider transfer function of the system. 21 5+p s ^ -a * j+ p j’ -o ’ ( » + p ) ( j’ - f l ’ ) + A ^ r ,( i+ z ) __________ K K , { s + z ) __________ j * + ps’ + (A3Tj - o’ ) z + JCKjZ- pc’ ___________ K K ^(s*z) ___________ Thus, the transfer function of the system is + ps^ + ( K K ^ - i^ ) s + K K f^ z - p ( ^ Step 2 of 6 KK(gz-pd^ K K a z -p d i Figure I Step 6 of 6 The system is stable if the equation satisfies the following condition. • All the terms in the first column of the Routh’s array should have same sign. • The first column of Routh’s array should not posses any sign change. • All the terms in the first column of the Routh’s array should be greater than zero. Therefore, for stability all the elements in the first column to be positive and we obtain the following constraints. From the above statement, write the stability conditions. p> 0 K K ^ p - K K ^ > 0 ifK > 0 ^ p > z K K ^-p a ‘ > 0 i/K > 0 = z > -^ KK^ Therefore, for stability all the elements in the first column to be positive system condition on the system parameters is \ p >zland ,V 2 - KK, Problem 3.58PP Consider the system shown in Fig. Figure Control system A J(f+ I) (a) Compute the closed-loop characteristic equation. (b) For what values of (7, is the system stable? H int: An approximate answer may be found using i+ h ' for the pure delay. As an alternative, you could use the computer Matlab (SImulink) to simulate the system or to find the roots of the system’s characteristic equation for various values of 7 and A. Step-by-step solution step 1 of 2 Y (s) s(s+ l) (a) Ae^ 1+ s(s+ l) s(s+ l)+ A e'^ CH Equation: s (s+ l)4 A (l-T s) = 0 (By taking the ^proxim ation of exponential term) s^+s-ATs+A=0 |s^+(l-AT)s+A=0| Step 2 of 2 0>) 1 A 1-AT 0 A 1-A T>0 A >0 T <— A A >0 & T <- Problem 3.59PP Modify the Routh criterion so that it applies to the case in which ali the poies are to be to the left of - a when a > 0. Apply the modified test to the polynomial S3 + (6 + K)s2 + ( 5 + 6K)s + 5K = 0. finding those values of K for which all poles have a real part less than -1. Step-by-step solution step 1 of 1 Shift the root to origin and then Routh’s Criteria. - ste p 1 o n Shift the root to origin and then Routh’s Criteria. Replacing s by s-a. For the given Equation ( s - lf +(64*;)(s-l)“ +(5+6k)(s-l)+3k=0 =>(s=-l+3s-3s’ )+ (sH l-2s)(64k)+ (5+ 6fc)(!-l)+ 5k= 0 => s’+ s' (-343c«) +s (3-12-2k+5+€k) -l+€4k-5.61d-5k=0 s*+(lri-3)!^+s(4M )=0 s’ 1 s’ Icf3 0 s’ 4k-4 s" 0 Problem 3.60PP Suppose the characteristic polynomial of a given closed-loop system is computed to be s4 + (11 +K2)s3 + (121 +K■\)s2 + {K^ + K1K2 + 110K2 + 210)s + 11/C1 +100 = 0. Find constraints on the two gains K^ and K2 that guarantee a stable closedloop system, and plot the allowable region(s) in the (/Cl, K2) plane.You may wish to use the computer to help solve this problem. Step-by-step solution step 1 of 3 Step 1 of 3 The characteristic polynomial of the closed loop system is. j^ + ( ii+ A : 2 ) j’ + ( i2 i+ ic , y + ( x , + A : ,A :2 + iio ^ : 2 + 2 io ) i+ iiK |+ 10 0=0 Write the Routh array for the polynomial. s*-. I I2 I+ £ , HAT,+100 i i + a:2 ac, + a:|A:2+iiojc 2+21 o o s^: a IIA:,+100 s : b iia :, + ioo Where. ^ ( ii+ A r 2 ) ( i2 i+ A :i) - ( A :i+ A :|A :2 + iiO A :2 + 2 io ) ( i) l I + ATj 133i + i i a :,+ 1 2 ia :2 + a :, a : 2 - a : i -AT,AC;- 1 1 0 ^ 2 - 2 1 0 11+AC2 10AC,+MAC;+ 1121 II + AC2 And, ^ a(AC| + AC|AC2 + 110AC2+2l0)-(llAC, + 100)(ll+A:2) a +AC,AC2 + iioa :2 + 210)-(1 lAC,+100)(11 +AC2 ) V. l l + AT; ) ___________________________________________ io / : i+ n A : a + ii 2 i 11+ ^2 ( ioa: , -1- 11 ^ 2 + ^ 1^ 2 -*-iioa :2-(-21o) - ( i i + a:2)(iia :,+ ioo)( i i -i- a:2) io a : , + i i / : 2 + i i 21 step 2 of 3 For a stable system, first column elements of Routh array must be greater than zero. Therefore, iu a :2 > o And, ii/:,-i-ioo>o \\K t> -m -1 0 0 Kt>- 11 Therefore, the two gain of the system are AC, and AC2 > -11 ' 11 ^ Step 3 of 3 Draw the allowable region in the ( ac„ a :2 ) plane. i 1C2 3 10 0 A llo w a b le re g io n 50 3 •lO O /ld -150 -120 -90 -AO -30 ^ 0 30 60 90 120 .................................................... 1 -50 i -100 1 1 -150 1 1 •200 1 1 -250 1 Hence, the plot is drawn. Problem 3.61 PP Overhead electric power lines sometimes experience a low-frequency, highamplitude vertical oscillation, or gallop, during winter storms when the line conductors become covered with ice. In the presence of wind, this ice can assume aerodynamic lift and drag forces that result in a gallop up to several meters In amplitude. Large-amplitude gallop can cause clashing conductors and structural damage to the line support structures caused by the large dynamic loads. These effects in turn can lead to power outages. Assume that the line conductor Is a rigid rod, constrained to vertical motion only, and suspended by springs and dampers as shown in Fig. A simple model of this conductor galloping is (»2+,2)l/2 Where m = mass of conductor, ir“fnr’c \/ortinal riicnla^om ont m = mass of conductor, y = conductor’s vertical displacement, D = aerodynamic drag force, L = aerodynamic lift force, V=wind velocity, a = aerodynamic angle of attack = — T = conductor tension, n = number of harmonic frequencies, / = length of conductor. Assume that L(0) = 0 and D(0) =D0 {a constant), and linearize the equation around the value y = ^ =0. Use Routh’s stability criterion to show that galloping can occur whenever ^+D b<0. Figure Electric power-line conductor Step-by-step solution step 1 of 2 D (a )!!^ -H a )V dx dx^ 1 \ dD d a _, . D ( a ) j: ,- L ( a ) r Step 2 of 2 Now So, -ao -1 1 da da m D { a ) x j- L { a ) V 01 PP If S is the sensitivity of the unity feedback system to changes in the plant transfer function and T is the transfer function from reference to output. We know that ^ ^ S+T= — 5 1 + 01D. T= | g + r = i| . Problem 4.i 1 + ODd s + r = 1 + GfDd 1 + GZ)„ GD. show that S + T = Step-by-step solution step 1 of 1 ^ Given that S is the sensitivity o f the unity fe edback system to changes in the plant transfer function and T is the transfer function &om reference to output. Given that S is the sensitivity o f the unity fe edback sjrstem to changes in the plant transfer function and T is the transfer function from reference to output. The transfer function is.-0 .3 5 4 l • Step 13 of 14 Refer to Figure 4. . .K ) ( . sensors should have more precision than actuators. Problem 4. * lo J d A dG Calculate ------. compute j8/ so that if /C= 10.a: ’ ( . r fA dG </ r AT’ 1 dp.1 A = - Substitute 10 for Kin the above equation.K ) ( . ( l + JC’A ) ' .6 4 2 .646 Therefore.. -(1 + ACA)’ (3AC’ ) ( i + a:A )‘ .646l • Step 8 of 14 Refer to Figure 4.642 *0. the sensitivity is [ | ] .( i + a : ’ >».' ( a : ) ] 3AT* 0 + a : a )‘ S te p 1 2 o f1 4 Determine the sensitivity.24 (a) in the text book.K ) P . Step 6 of 14 Refer to Figure 4. 1+(10)’ (0. i= /j( .3 ) ( i + a : a ). .3 6 4 2 ) 1 0 . Determine the transfer function of the system.K ) P . ia : ’ .24 (b) in the text book. 5?- 1+(10)(0.K ) ( . Determine the transfer function of the system.03 Therefore.’ = 0.24 (b) to ■ ^ J P A d G dG PPx dG d AC’ dp.q i |- Step 2 of 14 Refer to Figure 4. Compare above equation with y = -10Jt- ^10 ( i + j cPai J) ^ .24 (c) in the text book. Y (-iC )(-A :)(-iC ) R \. r = .24 (c) to . ^ d G jG d \n G k dG * “ d t/T “ d to t “ G ^ ' The purpose of this problem is to examine the effect of feedback on sensitivity. -3 Therefore.3642 for in the above equation.99|.) Which case is the least sensitive? (c) Compute the sensitivities of the systems in Fig.A : ) { . The transfer function is. R ( i + a t a A ' + ^ a J U + j^ a J AT’ (1+ ATA)’ Write the expression for the sensitivity of the system in Figure 4. JpA dG \ G ) d f i. Substitute 10 for Kand 0. Hence. (a) For each topology in Fig.099 for in the above equation.01 Therefore. ^ ) -3 K ^ -3 K ’P .K ) { . K' r= - \+K% Compare above equation with y = -10Jt- jt ’ = 10 i+ ic ’A i+ jc ’A " ‘ 10 a :’’ A . The transfer function is.{ . .099) 3 1+99 = 0. K' G= - l+K% dG dK d_ _ ^ 1 dK 'd K 1+ a a: ’/ ’a J . Step 10 of 14 (c) Determii Here.K ) P . the sensitivity 5 ^ is |Q. the value of ^ is |Q. 3 ( 1 0 ) ( 0 .K ' Determine the sensitivity.o. Figure Three-amplifier topologies Step-by-step solution step 1 of 14 (a) Refer to Figure 4.6 4 2 ^ = 0 . S? H m -3AT’ K’ 3 i + a:’a Substitute 10 for Kand 0.js r ) ( .3 a: ’ Step 7 of 14 Determine the sensitivity.a: ) ( . .fi.3 5 4 Therefore.(1 + AC>^)’ .{ . compute when G = (Use the respective j8/values found in part(a). A = - 10* _ 9 9 -l " 1000 -0 .3 6 4 2 Therefore.0 9 9 ) 99 1+99 . A AC‘ AC’ (1+AC’A ) ’ 1 + ac’ a J i + a: ’a Substitute 10 for Kand 0. (1 0 )’ (0 . The transfer function is.24 (c) in the text book.. the sensitivity 5 ^ is |~ 2 .24 (a) in the text book. S? K { -3AT’ 1 7 a:’ I ( i+ ^ A )’J i+ « A Substitute 10 for K and 0. Step 4 of 14 (b) Determine the sensitivity.(b. K ’ = 10 Substitute 10 for Kin the above equation.K ) ( . R ( i + A T A A ' + ^ A J U + J^AJ G=- ( i+ jfA ) ’ dG dK dG d AC’ dK dK .02PP We define the sensitivity of a transfer function G to one of its parameters k as the ratio of percent change in G to percent change in k.3642 for 0^ in the above equation. Using your results.\ = . the value of ^ is |0. The topology in (c) is least sensitive. ~ d p X \* K ^ P .i ) ( i + a: ’a ) " " ( a: ’ ) K‘ (i + a : ’a )' Step 14 of 14 Determine the sensitivity.24 (b) in the text book. ^ 10’ = 10.+ 3 K ‘P. The transfer function is. the value of is |o.9 2 6 4 .K ) { . The transfer function is. comment on the relative need for precision in sensors and actuators. .3 6 4 2 ) * 1 + ( 1 0 ) ( 0 . j.V io l + JTA K 1+JCA = 2.3642l- Step 3 of 14 Refer to Figure 4. . 5 ^ .a : ) ( .K ) ( . for connecting three amplifier stages with a gain of .) Which case is the least sensitive? (b) For each topology.3642) 3 1+3.1k ’ . the sensitivity 5 ^ is |-Q .0 9 9 ) * 1 + ( 1 0 ) ’ ( 0 . Y ( .K into a single amplifier with a gain of -10.3 a: ’ 7 i + a: ’ a )’ Step 9 of 14 Determine the sensitivity. Y (. 10 2. (l+ACA)’ . c) to j82 and /S3.0 dG/G dK/K Step 5 of 14 Refer to Figure 4.K ‘R Compare above equation with y = -10Jt- f i.a: ) R \.K ) ( . V = -10R. the sensitivity 5 ^ is [q^03 |.099 for 0^ in the above equation.i jc 0.K ) R l. K' G = — \+K% Write the expression for the sensitivity of the system in Figure 4.9 9 Therefore. In particular. 3AT’ 1 AT’ i + a: a )‘ J (1 + ATA)’ 3ATA l + Kp.1544 Substitute 10 for Kin the above equation. (b) For each topology. j p ^ ^ ' \G ) d p . = a : ’ [ ( .’.( . compute when G = (Use the respective j8/values found ii part(a).0 9 9 Therefore.24 (c) in the text book.Q99|.A r ) G = -p .2 . c=- Step 11 of 14 Refer to Figure 4.1544 1+ 1 0 ^ = 4 . we would like to compare the topologies shown in Fig.24 (b) in the text book. The results indicate that the closed loop system is much sensitive to errors in the feedback path than that in the fonward path. ) 3AT’ + a :’ ( o + 3a : ’ . )(K ^ ) [ J[ . p .7 r ^ . Problem 4.03PP Compare the two structures shown in Fig.. 2 S f.-F ^ . th e o u ^ nit the system iS.____ K ( SK ^ . -2 k ^h ) " W(l+7SSf.... fe e sensitivity g a in c h a n ^ fo r fee seco n d b lock is related to sensitivity o f first blo ck by..){H .f-2 {l+ K H . luc MMpuw wi » th e r e la tio n b e t w e e n . . as J i? ] is g reater fe a n zero.. F igure 1 Step 3 of 10 ^ T h e U o d c diagram o f th e second structure is show n in F ig u re 2. — { !]— Q — ^ T— ---------.1 Step-by-step solution step 1 of 10 It is given diat. 7ac’ = 7 rj(i+ J i:*7 7 j) .^ C) ' 1 + ja r . r fh iF S=- FdK W here.I (5) ' ' K Step 8 of 10 T h e m easure o f sensitivity d ue to changes in fe e am plifier g a in is.. ® Step 6 of 10 ^ C alculate fe e g am o f fe e b lo ck d io w n in F ^ u r e 2..)’ (2A:+2is:’7r... S j = — 2— . T h e eatpressio n t o m easure s e n sitiv i^ is. th e m easure o f sensitivity d ue to changes in fe e a n ^ lifie r g a in fo r fee first b lo d c is. F igure 2 Step 4 of 10 A ssu m in g fe e intetm ediate outyut from fe e felt b lo ck in F i ^ n e 1. com bine equation (1 ) a n d ^ ) to obtain fee iityut outp u t relation.)K = F . Figure Block diagrams I h p '. K .. ( 7 J .? ] a n d ? 2 - 7!?77j = (l+ JE H i)“ .. F ^ = F ^ {y + K H ^ ) 7 ? = . Use the relation as the measure. l+ K ^ H ^ Substitu tu te feevalue te fe ?2 fromequation(^)infeisequationtoobtainthe 2 _______ 2_______ i+ ( i: ^ 7 r j‘ + 2 K H . F ig u re s Step 5 of 10 ^ C alculate th e g am o f fe e b lo ck left blo d c in F ig u re 3. th e m easure o f sensitivity d ue to c f a a n ^ in fe e antylifier g a in f iv fee first b lo ck is. (i+Jsar. (fi) ' l+TSK Step 9 of 10 T h e m easure o f sensitivity d ue to changes in fe e a m p litW g a in is. step 7 of 10 n a It i» i^vcu uiot.)(77)Y 2 K (1 + K ^ H ^ ). with respect to sensitivity to changes in the overall gain due to changes in the amplifier gain. fe e m easure o f sensitivity Step 2 of 10 ^ T h e U o d c d i a g r m o f l h e first structure is fe o w n in F igure 1.. ' F [d K d ( K f— [ l+ K H j J» ] ( (l+ ^ . (F . T hus. ' 7^1 liK { (l+JC ^7r. Select H1 and H2 so that the nominal system outputs satisfy F1 = F2. and assume /CH1 > 0. C alculate th e outyut th e b lo d c diagram in F ^ u r e 3.) 2 2 s f = ■m Step 10 Of 10 ^ H e n c ^ from equations (6) a nd (7) observe fea t.)^ (7 i)"| 2 K {l+ X H . w e can o ra c h id e fea t fe e second b lo d c is less sensitive fe e first blodc.)‘ T hus.. T herefine. 7 ^ = -? ^ (1) ' \+ K H ^ C a l c u l i th e ga in o f fe e blo d c r ^ i t U o d c in F ig u re 3.( 2 K H ^ ) { k ^ ) 2K ( K ) { l+ K ^ H ^ ) T hus.) ( ^ : ) ( J i:) = F .. th e feed forw ard gain . _ s (5 + a ) ■* s(s + a) + Thus. Problem 4. tile sensitivity of tiie cbsed b op transfer fimction to chaises in tiie parameter A is obtained -I.04PP A unity feedback control system has the open-loop transfer function (a) Compute the sensitivity of the closed-loop transfer function to changes in the parameter A. (b) Compute the sensitivity of the closed-loop transfer function to changes in the parameter a.5 — --------- ^ ^ + as + A Step 2 of 5 We know that <JT(s) _ s^ + as + A . -P A rf 5(5 + a) . j ) -a l ^ (7 7 ^ 7 ^ 7 “ s ( s + a ) + j1 Thus.fiA ! + ■ (s + a) Step 5 of 5 . tiie sensitivity is obtained as s . Step-by-step solution step 1 of 5 A uni^ feedback system has the open-bop transfer functbn (a) Find the cbsed loop transfer hmction r(s) v^dch is given by r(s) = ^ ' 1 + G (s) s ( s + a) 1+ a (s + a) r ( a ) = . ^ ^ S^ + os ^°T d A A (7 7 ^ 7 7 1 7 r i. • Tda ^ a(s^ + a s + . = s ( s + a) + PA .A (s“ + as + A f dT{s) _ ^ + as ^ (s’ + OB The sensitivity of the cbsed loop transfer hmction is given by ^ _ A ^ ^ TdA' Find the sensitivity. the sensitivity of the cbsed bop transfer fimctionto chaises in the parameters is obtained sH = s ( s + a) + j4 Step 4 of 5 (0 Find T(s) r(s) = l + yS a W dT -O ^ s f [1 + ^ ' ( s ) 7 f id T j?(l + /Sffj -0 “ Tdp~ O (1 + Find the sensitivity of the cbsed b op transfer function.a) s(s + a ) + A (b) We know that Fmd — aa dT -sA (s“ + os + il) step 3 of 5 The sensitivity of tiie cbsed loop transfer fimction is given by dSf = T da Find tiie sensitivity. compute the sensitivity of the closed-loop transfer function with respect to f3. (c) If the unity gain in the feedback changes to a value of ^ 1.fiA Sf = - s (s + a ) + -p A Thus. 05PP Compute the equation for the system error for the feedback system shown in Fig..^ V l + DGH l + DGH l + DGH l +D G (H -F ) G -< -T W + .5.^ v (1) \* D G H \* D G H \ + DGH Write the output of the error detector from Figure 4. Figure Closed-loop system with sensor dynamics. E ^ R .w . (2) Step 2 of 2 Substitute equation (1) in (2).5 in the textbook for the closed loop system with sensor dynamics. . V = sensor noise Step-by-step solution Step-by-step solution step 1 of 2 Refer Figure 4. \->rDGH 1+ DGtf X ^D G H . R = reference.. Problem 4.Y . Consider the equation for mass of the car. Y = output.. U = control.-. G UDGH UDGH UDGH (X-^DGH)R-FDGR G „ DGH „ “---------------------------.^ V l +DGH " l+ D G H l+ D G H Hence..■ -tY + V l+ D G H l+ D G H l +DGH R + D G H R -F D G R G W + . the system error equation is. Kk. (c) Does the addition of tachometer feedback with positive kt increase or decrease Kv? Figure Control system Step-by-step solution step 1 of 7 (a) Refer to Figure 4. ■(3) From equation (3).25 (a) is shown in Figure 1. is inversely proportional to k. KK. K .) KK. * .(b) has the same transfer function as the system of Fig.(a). * And K% = KK. Substitute i f o r jc 'in K% = KKJc. Hence.K K . ) From the ^e denominator of the tr transfer function.= k. ' K. Therefore. K' i ( l + r ^ ) + A T '( l + * » 9 K' (2) Step 4 of 7 Compare the equations (1) and (2). i + i a : . ) — KK. s '0 + T . t . ( since fiom equation (3) w e have^ [K 'k . K K . the value of K .K . s lim ■ . ) KV. in terms of j f ' and is Step 7 of 7 (C) The velocity error constant is. it is clear that one pole exists at the origin. ■ +KKJcJa + KKJC. (b) Determine the system type with respect to tracking dr and compute the system Kv in terms o parameter fC and . Therefore. Problem 4. jr . The addition of tachometer feedback with positive k. KKJC. KKJC.25 (b) in the textbook. (a) Find values for K and so that the system of Fig. is.25 (a) in the textbook.k. Step 5 of 7 (b) The inner loop in Figure 4. Idecrwisfsl AT. s + K K J . Figure 1 Step 2 of 7 Calculate the closed loop transfer function. K K „K ^ (') KKJC. s + / a r . the type of the syst system is 1. jre.{l+K'k. s (1 + t. AA.- That is. k . the values of j f ' and are k ' K. JCy s lim4Z.(a).. ~ * [ s ( i+ r ^ ) + / a : j f c ^ ] + A M : . The simplified circuit of Figure 4.i+ r .25 (b). 1 + X K jt. = K K J .(4) KK. i KK. Calculate the closed loop transfer function from the Figure 4. (1) step 3 of 7 Refer to Figure 4. ^ i.06PP Consider the DC motor control system with rate (tachometer) feedback shown in Fig. it is clear that.K . t .25 (a) is. ■ . i) + A a r . trt _ K K . !(* ) = j( i+ r .) Therefore. s lim ^ i-** 5 ( i + r . the type of the system is jT y p e l Step 6 of 7 The value of K . what is the steady-state tracking (b) What is the system type? (c) What is the steady-state error to a ramp velocity 5. ) i + i. (» + 3) ________________ 0.)s + 1. 5 * : .5 [ « ( j) .)j+ i. R (s) O. Determine the transfer function of the inner feedback loop. Thus.5£.5 i:.5 A r. ) i + 1.5 [ i* + ( 2 0 + £ j) s * + ( 1 0 0 + 1 0 £ j+ 0 . the steady state tracking error is |Q. 0 .S£.) j + 1.J j' Determine the steady state tracking error. + 0 .) i? (* )J _______________ Q -iK .( » ^ 3 ) » (i+io) j=+(io+is:j)j ______________ O. + 0 .5 1 - j ’ + ( 2 0 + ^ r j) j^ + ( io o + iO A :i+ o .( ^ + 3) 'l ™ ® . + o .(ft-3)_________ 1 .0 .5 £ .] Thus.SK. ■ [ ' ” i ’ + ( 2 0 + £ j ) j ^ + ( 1 0 0 + l 0 £ j + 0 . s ^ + { \0 + K 2 ) s Determine the transfer function of the system. Step 4 of 5 (b) The transfer function of the system is. I M . 0 .( j + 3)___________________ “ j * [ j ’ + ( 2 0 + £ 2 ) i * + (1 0 0 + 1 0 £ j + 0 .5 Thus. — s 0 .S i ’ + ( 2 0 + + (100 + 10£. For a unit step input.5 A r .(5 + 3)_______________ i ’ + ( 2 0 + a: .(s + i) _____________ ” ( i + 1 0 )[j’ + (I0 + ^ j ) s ] + 0.5 A :. 5 i: . y (^ ) _______________ o .5 x 3 x 0 . 5 £ .( n -3 ) £ ( j ) = 0.S £ .5 A :. .5 1 - R{s) + (2 0 + K j)s^ + (1 0 0 + 1 0 ^ :.5 £ . ) i= + ( i o o + i o ir .^ [ '” .) j" + (100 + IOAr.5 £ . Determine the transfer function of the system. the system type is |type0|- Step 5 of 5 (c) The tracking error is.5| . 5 i : | Step 2 of 5 Determine the tracking error.+ o .{s + i) £M = 0.5 3 :|(t + 3) E (s) = 0. £ ( s ) = 0 .5 ir ( i) .5 £ | ) j + 1 3 A : .SK.) s ’ + (io o + iO A :. .5 A :.5 £ . 5 A : .5 £ |(s + 3) = lim— i* + ( 2 0 + £ j ) i ’ + (1 0 0 + 1 0 £ j+ 0 .5J:. ( i + 3) —> L i ’ + ( 2 0 + A :.5 £ . J » step 3 of 5 Determine the steady state tracking error.5 £ |(^ + 3) = lim^O. ) i + i .5Ar.r M l J !(.5£. « . 5 K .) j* + ( 1 0 0 + 1 0 A r j+ 0 . + 0 . = lt o i£ ( j) „ r 0 . the steady state error to a ramp velocity is 0 Observe that the steady state error to a ramp input is independent of and .J For a ramp input..5 0 .07PP A block diagram of a control system is shown in Fig.0 5— for T.) i+ i.) j + 1 .5 A : . ■ ____________________ 0 .5 £ .5a :i ( j + 3 ) £ (s) s ’ + (2 0 + A:2 ) ? + ( 1 0 0 + 1 0 A :. ) j ’ + ( i o o + i o ^ :j + o . ’ + ( 2 0 + iC . = lim j£ (j) 0 .5A:.)j + 1.) i + 1. { S * 3 ) ______________] l ' ■[ i ’ + ( 2 o + i : . (a) If r is a step function and the system is closed-loop stable.0 .(s).> r ( s ) ] M =0 5 l.5 « ( i) .5^:. i ’ + ( io + A :.5£.5A:. Problem 4. .)i 3 ) ______ 1 y (4 (^^-10) ■ i i Q 5^ . 0 .)i+ 1 . J j ' 0.5 £ . + 0 .(s * 3 ) _______________ j ’ + ( 2 0 + « .5 A :.5 £ | 1. ______________________ 0 .5AT.0 K2 = 2 and K^ is adjusted to give s system step overshoot of 17%? Figure Closed-loop system Step-by-step solution step 1 of 5 Refer to the figure of problem 4-7P in the text book. 5 K .) s + 1. = 0. r _________ O. 4“ +984“ . ^ ( « ) = lim 4£(4) from equation (12).2 s lim — 4“ +1004’ + 2004 + 200 (13) Hence. + l0 0 Refer from Figure 4.5 in fhe textbook and write the transfer function equation with respect to zero IVvaiue.. (2) and (3) in equation (6).^ (^ + 0 ... R = reference. . from 5^ ^00) Write the general formula for K. G(s) = i (1) Consider the gain of the controller. Calculate the ^ ( o o ) = H m 4 £ ( 4 ) from equation (10). . Ig -'-'-w .." 2(44-1)(4-I-100) From equation (5). the transfer function of the system from R to Y is « (4 ) 4“ + 1004“ +2004 + 200 Step 2 of 6 (b) Refer from Figure 4. U = control. . V = sensor noise Step-by-step solution ste p 1 of 6 (a) Consider the gain of the plant.2 4 ( 'I'l ^ < ” ^ ° ! a % »+1004“ + 2004 + 20ol 7 j 4’ + 984“ -2 4 ■ lim 4’ +1004’ + 2004 + 2 0 0 U 4“ +984 . Step 5 of 6 (d) Calculate the value of tracking error at unit ramp input with zero Wvalue.. s 4*+984=-24 (10) ^ W = -5 - 4’ + 1004“ + 2004 + 2001.0 8 P P A standard feedback control block diagram is shown in Fig. (b) Let R = 0 and compute the transfer function from IVto Y.2 4 fO . the transfer function of the system from Wto Yis (V(4) 4’ + 1004“ + 2004 + 200 Step 3 of 6 (C) Calculate the —“ with respect to zero Wvalue.) A(s ( M ) j* ( s + 1 0 0 )+ 2 (j+ l)(1 0 0 ) i '( 4 + 100) 2 (4 -H )( h -100) j ’ ( i+ 1 0 0 ) + 2 ( j + l)(100) . I 'M (i) W( s) 100 1+ ^ A ^ + io o J a j ’ (4 + 1 0 0 )+ 2 ( j +1)(100) 4’ ( l + 100) j ( 4 + l0 0 ) »’ ( i + 100)+2(4 + l)(l0 0 ) K (4 )_ 4 (4 ^1 00)_____ B '( i) i ’ +100j*+20as+200 I 'M 4 ( 4 + 100) Hence. the numerator part having j^term and denominator part have 5*term.. y (» ) p„(4 ) g (4) (4) « (i) l+ D „ ( i) G ( i) W ( 4 ) Substitute equations (1)..5) R{s) 4’ + 100i’ + 2004+200 y (4 ) 2(4 + 1)(4 + 100) Hence. 4’ + 1004’ +2004 + 2001 7) Calculate the e .. 5^+1005 *+ 2005+200 Calculate the value of E{s) in unit step input..5 in the textbook and write the transfer function equation — —*-4^ \iwith respect W {s ) to zero Rvalue... substitute for r ( 4 ) in equation (8).2024 . Problem 4 . substitute __ 2(5 + 1) ( j +100) for r ( 4 ) in equation (8). (c) What is the tracking emor if R is a unit-step input and W e 0? (d) What is the tracking error if R is a unit-ramp input and W e 0? (d) What is the tracking error if R is a unit-ramp input and W e 0? (e) What is the system type and the corresponding error coefficient? Figure Closed-loop system with sensor dynamics. ( 12) =(. (2) Consider the gain of sensor. From equation (5). . Substitute —for J?($) in equation (9). the value of tracking emorat unit step input with zero Wvalue is |zero|. . l i f l .24 ’ . So it is [Type-Isyslem| and velocity error coefficient is ||0Q sfc~'| . . Y = output.200 4“ + 1004“ + 2004 + 200 4’ + 984“ -2 4 V + 1004“+2004 + 200 Step 4 of 6 Calculate the value of E{s) in unit step input.. ™ * ^ + i o 0 4 ' + 2004 + 2 0 o [ 4 j ^ (*)= 0 (11) Hence. 4*+1004*+ 2004+200 2(4+ 1)(4 + I00) R (s ) 4 *+ 1004“ + 2004 + 200 2 ( 4“ + 1014+100) ° ' “ 7 + io o 7 + 2 o o 4 + 2 o o 4“ +1 004 “ +2004 + 200 . rw ( ^ ]( . the value of tracking emorat unit ramp Input with zero IVvalue Is 100 Step 6 of 6 (e) Determine the velocity emor coefficient value K. . . with the values (f4U0)' (a) Let 0 and compute the transfer function from R to Y. ■ (14) k— H Substitute equation (13) in equation (14). 1 a: »- 1 100 A:.. Conside 100 (3) . 5 5*+985*-25 (\ £ ( 4) = . = 1 0 0 se c -'. (6) (^ (i) l+ £ > „ ( s ) G ( 4 ) « ( i) Substitute equations (1). 4*+984“ .. (2) and (3) in equation (4).. (15) Determine the type of the system __ 2(4 + l)(4 + 100)__ 4*+ 1004“ + 2004 + 200 In the transfer function. Substitute -Lfor J?(5 ) in equation (9). 3 6 5 + 1 )+ 0 .( * ) G ( * ) //( j) f i+ A . .3 6 * + l)+ 0 . ( s ) H ( j) '» + 0 . 3 f a + l) j ( i + lf( 0 .7 5 * + l) J r l + 0 .w .7 5 * + l) JU J r ( i + l)‘ (0 .F ( i )0 . 7 3 G | (.7 3 (2 . requirement on the value of H ( j)s u c h that the system will remain type-l system are: ^ ^o o )= 0 w h e n is equal to 1 and /f^ ^ ^ is also equal to one.3 6 * ) . Z ) ^ ( j) = 0 . M G ( » ) g ( » ) . the value of steady-state tracking error at unit ramp input is \ + D A ^ ) G U ) flU ) .7 3 (2 .3 f a + l) «*p\ / ^ i ( j + l f (0.F M A . (11) and (12) in equation (5). iM j.7 3 ________ — I * (* + 1)‘ ( 0 .. W G W w M .7 5 ) -0 .7 3 ( 0 . 3 f o + l)+ 0 .3 6 * + l)+ 0 . Step-by-step solutiOdPP ow = 3— =« showing a lead compensation in the feedback path.7599 a: . R (s ) I g ..(5) (6) Consider the gain of the controller is.0 . 7 3 iG .5 in the textbook and write the transfer function equation with respect to zero Wvalue. What is the value of the velocity error coefficient.. 2.0 . Y = output.7 3 ( 0 . V = sensor noise QlW A *w Step-by-step solution step 1 of 7 (a) Refer from Figure 4. r i+ A .3 6 ) 'l ~[ «-33 J « .. Substitute equations (1) in equation (2).7 5 * )-0 . ( ^ ) g ( .7 3 i G.F ( .7 3 .3 6 I+ I l i t l l H = li2 1+0 73— i( ^ + l) M 0 .' Hence. ..7 3 ( 2 .3 & + 1 )+ 0 .. 1 2. ) w ( ^ ) .7 3 (2 . s f i + q .7 5 * )+ 0 1 .7 5 j+ l) a j ( i + l) '( 0 . 1 3.73 . ^ r * (* + I)^ (0 ...7 3 G | ( j ) W ( i ) .3 fa + l) + 0 .F ( s ) D A s ) G ( s ) ' Step 3 of 7 (b) Consider the gain G (^)h as single pole origin in the s plane.f ( » ) q .( s )//( s ) Hence. 1 GW = (11) 5(5 + 1)* Consider the gain of sensor.F ( i ) 0 . Substitute -L fo r J?($) in equation (3). ^ l + G „ (s )G W tfW Js l + G . Kv7 Figure Closed-loop system with sensor dynamics.36s + 1 ) + 0 . T ( i) = f ( 4 (1) l+ D „ W G W f f( * ) Calculate the —“ with respect to zero W value. ) o „ M G M ^ i ^ ’ [ l +DAs)G{s)H(s) Js Calculate the ^ ( o o ) = H m s F (i) from equation (8). ( j ) W ( j ) .3 6 ) ') ’ *(* + l)"(0 .755+1 1+ 0 . 7 3 G |( j ) W ( j ) .7 3 ( 2 ..365+1 Substitute equations (7). R {s ) ‘^ ^ ^ h *D A s )C (s )H {s ) £ (£ l 1+ D M G { s ) H { s ) . 3 f a + l) * ( s + 1)’ ( 0 .F ( i) 0 .3 6 * + 1)+ 0 . (7) Calculate the value of tracking error at unit step input with zero Wvalue. M g m Y 1 'I h £>...3 6 * + l)+ 0 . l+ Z ) „ ( * ) G W /fW JU^J = .7 3 (0 .7 5 * ) __ (11 • 5(5 + 1)* (0.755+1 H(s) (12) 0..7 5 * + 1 ) a _ | .73V .7 3 ( 2 .F ( j ) 0 .F { s ) D J s ) C ( s ) R {s )~ l + A .7 3 ( 2 .7 3 (0 . Step 5 of 7 (c) Consider the gain of the plant..7 3 (2 .3 f e + l) t ( t + i y ( 0 . R = reference.7 3 - j ( j + i y 0 . W G W i/ W J Step 4 of 7 Substitute equations (6) and (7) in equation (9).3 6 * + 1 ) + 0 .F ( ^ ) 0 . '* (* + 1 )’ (0. 7 3 G |( j ) ' foo) s lim l+ 0 .)g(..7 5 ) -0 . lim f.0 .)«m }[7) Calculate the ( « ) = l i m j £ ( i ) from equation (4). ( * ) = 3-7599 Step 7 of 7 Determine the velocity emor coefficient value J^T^from 5^ ( o o ) .7 3 (2 .7 3 (2 .7 3 i G. the value of velocity coefficient value is |j^^sQ .3 6 * ) Y i ’| * (* + l)’ (0 ( 0 .266sfC~'| ■ . ( j « ii* H = S S l+ 0 . M G M '| <3> step 2 of 7 Calculate the value of tracking error at unit ramp input with zero Wvalue.7 5 5 + 1 )U Step 6 of 7 ^ Consider the value of F ( s ) is equal to 1 in above equation.'. Substitute —for in equation (3).. from above equation. <?(5) = j G . l+ D ^ ( j) G ( j) H ( j) JW Hence.(10) 1^1 l+ 0 .7 3 G . 7 3 ( 0 .'. the DC gain of the system H { s ) is |unity|.. 1+ 0 ..2 6 6 s « .7 3 G .( 5 )//( y ) i+ 0 .7 5 i + l ) .7 3 (2 . U = control.(. ■ (15) k— H Substitute equation (14) in equation (15). Write the general formula for K . (5. Hence. 25)^ =0 i.25. .e..1 OPP Consider the system shown in Fig.25)’ k =0 Using Routh's criterion for stability. Problem 4. where Dc{s) = K (a) Prove that if the system is stable.25.5K)+(0.. s*-h ’ + s’ (K+1)+ s (1+0.875 0 s" (0 2 5 )’ K For stabili^. s* 1 K+1 (0. it is capable of tracking a sinusoidal reference input r= sin wof with zero steady-state error. it is capable of tracking a sinusoidal reference input r= sin wof with zero steady-state error.) (b) Use Routh’s criterion to find the range of K such that the closed-loop system remains stable wo = 1 and a = 0. Figure Control system (b) S.) (b) Use Routh’s criterion to find the range of K such that the closed-loop system remains stable if wo = 1 and a = 0. {Hint: Look at the transfer function from R to E and consider the gain at cuo..5 K 0 s’ 05K (0.E=limsE(s) s’ (s + l)(s ’ +<D.S. as all the first column elements should be greater than 0.25)’ K s‘ O5K+0.’ )s(s+ l)+ K (s+ a)’ (s’ +00. where (a) Prove that if the system is stable.’ ) lim = 0 ■-*" (s’ +£0. {Hint: Look at the transfer function from R to E and consider the gain at wo.25)’ K s’ 1 m ). Figure Control system cControl O f Dynamic Systems (7th Edition) Problem Consider the system shown in Fig.’ ) Step 3 of 3 (c) Characteristic equatioii is s ([s^+ lj (rH ) +K (s+0. which represents control of the angle of a pendulum that has no damping.) . find the class of disturbances w(t) that the system can reject with zero steady-state error.11 PP Consider the system shown in Fig. the closed loop transfer function is. ( j ) w (. A M _ ! _ W (s] step 4 of 6 Determine the expression for error signal. Substitute — !— for r ( s ) in the above equation. Problem 4. find the class of disturbances w(t) that the system can reject with zero steady-state error. w Step 3 of 6 From Figure 1. i ’ + A : + o . which represents control of the angle of a pendulum that has no damping. Figure Control system (1) j * + a: step 2 of 6 The reduced block diagram is shown in Figure 1. (a) What condition must Dc(s) satisfy so that the system can track a ramp reference input with constant steady-state error? (b) For a transfer function Dc(s) that stabilizes the system and satisfies the condition in part(a).. Figure Control system cControl O f Dynamic Systems (7th Edition) Problem Consider the system shown in Fig. A M =*M.. (a) What condition must Dc(s) satisfy so that the system can track a ramp reference input with constant steady-state error? (b) For a transfer function Dc(s) that stabilizes the system and satisfies the condition in part(a). ( j) ' ' i'+ A : + z ) . we can say &at &e system is of [Type l\.^ (0 . Step 2 of 2 ■S. Problem 4 .S.=lim = 2i 25 .E.E -lim sE (s) By observing the above function.S . V ^ R (s) E (0 = E (0 -Y (0 = j R (0 s +2^cd^ s+ cch^ S . Step-by-step solution step 1 of 2 Step 1 of 2 t m .1 2PP A unity feedback system has the overall transfer function R(s) ' + + Give the system type and corresponding error constant for tracking polynomial reference inputs in terms of ^ and ojn. X. S.-R (s) (s“ +2? s+ l)(s+ b)+ K (s+ a) Step 2 of 5 . K ^ O . K a<2^(K +l)| Step 5 of 5 2^(K +1) c. equation = s^+2^^+(K+l)s+Kat*0 Step 4 of 5 For Stability. a. . s and the system shoul d be of type -1. (a) Ignoring stability for the moment. S.S. Problem 4 . For the given system. what are the constraints on K.=lim sE(s)= ^ R (0 (s^+ 2^+ l) ( s ^ ) +K (s+a) Fortype -1 systemJb=0.S. sfs+b)fs^+2^s+ll . and b so that the system is Type 1? (b) What are the constraints placed on K. r^>0| a< >1 ^ |0<a<2^(K +l)| . (K+1) . a. \ ----.1 3PP Consider the second-order system ds) ? + 2 f7 + T * We would like to add a transfer function of the fonn Dc(s) = in series with G(s) in s unity feedback structure. R ( s ) = |.r *-»«(s^+24s+l)s+K(s+a) s^ Char.-----. .^-----------. 1 K+1 s’ Ka s‘ 0 s" Ka Ka>0. A sK >0. a ^ O Step 3 of 5 b. and b so that the system is both stable and Type 1? (c) What are the constraints on a and b so that the system is both Type 1 and remains stable for (c) What are the constraints on a and b so that the system is both Type 1 and remains stable for every positive value for K? Step-by-step solution step 1 of 5 s"+2?s+l D (s ) = s+b Y (s) _________ K (s+a) E (s) (s“+2?s+l)(s+b)+K (s+a) ( ! “ + 2 ? !+ l)(s+ b ) E = R -Y = . a. .E. b=0.E.=limsEfs) i-»0 ' ' s"(s"+2?s+l) 1 = lim -------.E i > 0 25 =>Ka<2^(K+l) The constraints are | b=0.Ka>0. = ^ ____ A System is of [Type -1[ Step 2 of 4 b. wi the system remain Type 2 if H f changes slightly? Figure Control system Figure Control system A »o— ^ ( i ) . (a) What is the system type? Compute the steady-state tracking error due to a ramp input r(t) = rof\{t). (b) For the modified system with a feed forward path shown in Fig. =lim sE (s) =lim .(b).(a).Y = ^ !i^ s ( ts+ 1)+ A S.0 ' * .) = R .-----x f -^1 *-»o »-»<• s ( ts+1)+A A Ka Step 4 of 4 ^ c.E.E . =>(l-H fA )=0 and[ ^ ^ S. From block dis^am . -o r V HfS Step-by-step solution step 1 of 4 T (0 = - s (ts+ l) +A R (s)= ^ E ( .S . S. we get (R E . itm usthavetw o zeros at s ^ .S. give the vaiue of H f so the system is Type 2 for reference inputs and compute the Ka in this case. (c) Is the resulting Type 2 property of this system robust with respect to changes in Hf.) Step 3 of 4 Fortype-2sy8tem.14PP Consider the system shown in Fig. Problem 4. . -Y+RHf s) A=Ys A + s(ts+ l) A + s ( ts+1) Niim erator= ts^+s -Hf As+A (1-H. = lim sE (s)=lim f *-.E. system wouldbecome |T]rpe ^ .* 0 s ( t s + l) 4 A U v S . that is. and then determine the type of system and the finally evaluate the error constants.1 5PP A controller for a satellite attitude control with transfer function G=1/s2 has been designed with c unity feedback structure and has the transfer function DcW = . By looking at the transfer function. the poles at 5 s 0 ^re usually after the Input.— 0. For a disturbance input. I . Problem 4 . s^(s+S)+W(s +2)i’J f s’ (s + S) l'] ™ [ j > (. r(s ) C(s)D(s) « (s) 1 + G (j)c (s ) 1-h 10(^ + 2) ^ * ( « + 5 ) + I 0 ( f + 2) Step 3 of 8 Calculate the steady state error.7 ^ ------ — > |_ * '( i + 5 ) + 1 0 ( i + 2 ) J 0+5 (0 ) (0 + 5 )+ 1 0 (0 + 2 ) ___ 5 _ ° 0+20 • 0. K is 0 Step 5 of 8 (b) When a disturbance torque adds to the control. 1 3 1 4 1 + a: .! ^ . the error constant. '( j * . jf . the system type is [Type 2 |. the error constant.I G ir(s ) I + GD =1— j+ 5 • 1— j ' ( i + 5 )+ IO (j+ 2 ) f^ (i+ 5 )+ 1 0 (i + 2 ) -» + 5 i ’ ( i + 5 )+ 1 0 (* + 2 ) ( » '. what is the system type and corresponding error constant with respect to disturbance rejection? Step-by-step solution Step-by-step solution step 1 of 8 (a) The transfer function of the controller for a satellite attitude control is. the system type is |t ^ Step 6 of 8 Determine the transfer function with disturbance feedback. the system has two poles at 5 = 0 - Thus. Step 2 of 8 The closed loop transfer function is. IW . Thus. . + 5 )+ 1 0 (5 + 2 )5 ’ J = lim f.l ) ( s + 5 ) + 1 0 ( j + 2 ) *lim j ’ ( i+ 5 )+ 1 0 (j+ 2 ) (0 -l)(0 + 5 )+ 1 0 (0 + 2 ) (0 )(0 + 5 )+ 1 0 (0 + 2 ) -5 + 2 0 20 _15 '2 0 3 °4 Step 8 of 8 Determine error constant K .) s+5 In order to determine a system for reference tracking.l ) ( j + 5 ) + I O ( i + 2) i ’ (j+ 5 )+ 1 0 (i+ 2 ) Step 7 of 8 Calculate the steady state error. c=4- The transfer function of unity feedback structure is. it acts as a disturbance input. (a) Find the system type for reference tracking and the corresponding error constant for this system. ' 3 4 -3 3 Therefore.25 •4 =4 Therefore. first identify the poles.25 Step 4 of 8 Determine error constant i f . Hence the system type is Type 0. (b )if a disturbance torque adds to the control so that the input to the process \s u + w. 10 ( 1 + 2 ) £ ) (. 86 Therefore.1 6PP A compensated motor position control system is shown in Fig.67 Therefore. 30 x2 K = - 2. evaluate the steady-state system error.— / IL«r«+2^l ■./ / I 4(4 + 2 ) 4(4 + 2 ) (4 + 3 0 ) + I6 0 ( 4 + 4 ) 4(4 + 2 )(4 + 30) 4+ 3 0 4(4+2)(4+30)+160(4+4) Determine the steady-state error to a disturbance input. *-*® (^4(4 + 2)(4+30)+160(4+4)J Hence. is 122. f f (4 ). give the value of the velocity constant. the velocity constant. (a) Can the system track a step reference input rwith zero steady-state emor? If yes. (a) From the block diagram. 5 ( f+ 2 )i[5 + 3 0 ( s + 2 0 ) + 2 0 (lie D ]( 5 + 4 )) ' Evaluate the velocity constant from the emor expression. Substitute i for A ^^jand —^ — 30)^°'" £W =- 1 6 0 (j+ 4 ) 1+ s(i +2)(*+30) (j+2)(i+30) i( i+ 2 ) ( i+ 3 0 ) + 1 6 0 ( i+ 4 ) Now. *W = [i-r(4 > (4 [ « l( j+ 4 ) ^ < ( 5 + 2 ) ( 5 + % ) + ie D ( 5 + 4 ) *(4 5 (j+ 2 y j+ 3 Q )+ 1 6 0 (j+ 4 )-ia i(j+ 4 ) R [s] 5(5+2)(5+30) + M 0 (5 + 4 ) 5 ( 5 + 2 ) ( 5 + 3 0 ( 5 + a » ). Step-by-step solution ste p 1 of 8 Refer to Figure 4. the sensitivity of the closed-loop transfer function is 25 ( 5 + 30 ) 5(5 + 2)(5 + 30)+160(5 + 4) Step 7 of 8 (<i) Consider the following information. H = l . 1 6 0 (j+ 4 ) .86! • . .5 ( l6 D X 5 + 4 ) ) .. Problem 4 . SA 4 [5 (5 + 4 )(5 + 3 0 )+ 1 6 0 (5 + 4 )][1 6 0 (5 + 4 )(5 )(5 + 3 0 )] " [1 6 0 ( 5 + 4 ) ][5 ( 5 + 4 )(5 + 30)+160(5 + 4 ) ] ‘ 4 5 (5 + 30) 5(5 + 4 )(5 + 3 0 )+ 1 6 0 (5 + 4) 25(5 + 3 0 ) (Since.. Step 5 of 8 (c) In order to compute the sensitivity of the closed-loop transfer function to changes in the plant pole at _2. evaluate the new expression for the error. 20 f f( 5 ) = 4+20 Determine the system type by computing ^ = Q a n d at the value. l6 0 (^ + 4 ) r(^ )= - «(« + >4)(s + 3 0 )+ I6 0 ( 2 + 4 ) Where.22. (c) Compute the sensitivity of the closed-loop transfer function to changes in the plant pole at -2 (d) In some instances there are dynamics in the sensor. Figure Control system (a) Can the system track a step reference input rwith zero steady-state error? If yes. Hence. Find the transfer function for 7 ( 4 ) • 1 4(4 + 2 ) y (^ )= - r i60(4+4)Y 1 ] 1+ I\ \.. Now by determine the transfer function. Assume that the sensor dynamics are H(s) = 1.. is |1Q. Step 3 of 8 Determine the value of velocity constant. then the system is Type 1 with respect to the reference input. 4 = 2) 5(5 + 2 )(5 + 3 0 )+ 1 6 0 (5 + 4 ) Therefore. the system cannot reject a constant disturbance.63 .(2 )(3 0 ) =10. y. Now evaluate the sensitivity of the system. . (b) Can the system reject a step disturbance wwith zero steady-state en^or? If yes..29 in the textbook. it is clear that the system is Type 1 with y (4 0 (4 R (s ) l+ H ( s ) G ( s ) y (« ) g (4 R (s ) 1 + G ( j) Write the expression for the error signal. AST T S A . the system can track a step input with zero steady-state emor.l- 4 + 20 Step 8 of 8 The velocity feedback J ^ w ill also change.(3) Step 6 of 8 Apply partial differentiation to the transfer function with respect to A. A was inserted for the pole at the nominal value of 2. Substitute — value in equation (3). e„ . the value of the velocity constant.. ' ' i +a ( s ) '• ’ (2) Step 2 of 8 . ST a( 160(5+44) 160(5+ 'I 04 “ a 4 [(^5(5 5 (5 ++y4()(5 ) (5 + 330) 0 )+ + I6 0 (5 + 4 )J 160(5 + 4 )(5 )(5 + 30) [5 (5 + 4 )(5 + 3 0 )+ 1 6 0 (5 + 4 ) ]’ o r.liin = lin ii f ----.^ '-*• ^^s(5+2)(5+30)+160(j+4)J sO Therefore. give the valu of the velocity constant.67|- Step 4 of 8 (b) The system is Type 0 with respect to the disturbance and has the steady-state emor. Repeat parts (a) to (c) for |W(*) = and compare the corresponding velocity constants. = H m j(7(5) ( ^ j( j+ 2 ) ( j+ 3 0 ) J ( 1 6 0 (j+ 4 ) ^ ™[(j -f2)(j +30)J (»60)(4 ) . The new transfer function is. give the value of the velocity constant. y /A _ y / jX Substitute for G . .( i) G .i (*+Pa)JJ Here.i‘ r c . { k . Now evaluating the steady-state error. ( i + P a ) J W . in the denominator A (z) is the characteristic polynomial.(z 4 -z . (»+zj. ( s ) 1 + G .)) (s* n .(z + p „ ))+ (A :. .. ( z + p „ ) ) j ^ n 2 .A :..( s + p .n .+■p~ii) ’ ' « ^ n rJ i(» + w )' Show that the system is of Type 0.{s) 1. ( s + P u ) } ) (» *n . ^ are the poles and zeros of the and ^ not at the origin. By analyzing this system. ) ^^ l. and Type (/1 + 12) with respect to disturbance inputs w1. we see that the system is of Type k * k .) + / :. ( g . and w3 and is asymptotically stable. n . n. n . . Step 5 of 5 Write the expression for output for disturbance ■ l + G |( j ) G j ( j r A r . ) ) ------------A W In this expression. * ' ' '' j'> Vi nn f2* . W '^ ^ ^ )+ t U 'n 3 i ( * + / > i i ) J U ^ n . ( z t r .. w . in the denominator A (z) is the characteristic polynomial.)V n .f TC.(ifL O i. Figure 1 Step 2 of 5 Consider the following data.) J = liin .n 5 i(z + z a )) ( j ' n . n .i(j+ p „ ))(» * -n . ) Y A : .( i) a n d for a . w2. Step 3 of 5 ^ Write the expression for output from the block diagram shown in Figure 1. nj. Also. . Figure Single input-single output unity feedback system with disturbance inputs Step-by-step solution step 1 of 5 Draw the diagram of single input-single output unity feedback system with disturbance inputs. we see that the system is of Type Write the expression for output for disturbance ■ i^ A s ) I 'M - l+ G .(z+z„)Y K. n. (j+p„)Ji*' nj. Type /1.(j+ z „ ))(A :.W f f ^ .(» + ? . . n : . W G . n:. in .r n . ( z + Z a ) '| i . . hence its Type 0 system. ( j + p „ ) ) ( i ' ’ n .( i) (F ..1 7PP The general unity feedback system shown in Fig.. and vv3 respectively. .[n .)]» ^ (z ) y = Iiin [ j * + i ' ’ I l . we see that there are no poles at the origin. (j+z„))(a:. ( z + P a ) A T .i(r *. Problem 4 . Now evaluate the steady-state error.(z + f» „ )„ . . has disturbance inputs w1. w. . Step 4 of 5 Now evaluate the steady-state error. Analyzing this system. ( » + r . In this expression. A By analyzing this system.g . ( » + P j.( i + z . . . w2. la.. K lim f£ ( 5 ) 5^ +kyms+mk^k2 j ' ' *-»® ^ s *kfms*mk^k2 J\sJ Iju il I + k^ms+ mk^k^ J kjkjm As the steady state emor for a unit step disturbance is non-zero this is a Type 0 system.A s the gain <7( 4 ) has a pole at origin. JS:.v * w = { . Therefore.m s \ „ .na’¥ k . ( ms Y » . £ W = n ( i) . the value of the error constant is Q]. the steady state value of the velocity v is V step 4 of 7 (c) Calculate the emor in system with respect to reference inputs. the system is a Type 1 system. ( j ) \ s * •^kyms^mkyk^) ' ^ E (s ) = K (s) s^+k^ms^ The error of the system can be compared with a system of unit gain feedback and having e nk^k^ fonward gain of <7( j ) * .B m G W mk^k2 s lim Step 5 of 7 ^ The steady state error of the system for the unit step input.j^ ^+k^ms+mk^k2 J The steady state error for a unit step disturbance input is. y h .^ -------. (c) What type is this system in relation to reference inputs? What is the value of the corresponding error constant? WVWtjat..31.----. the transfer function relating the output speed to the wind disturbance is (2) f P ( j) + ky/ns+ k^k^m Step 3 of 7 (b) The wind disturbance is a ramp function.( 0 ) + ^ ------. e„ = lim— ^-r-r y. £ ( j) *+ k^m s+ ktkim ~ m s fV {s ) 5^+ ^m s+ m ft. = limiG(j) n kikj s lim j- »-»• s +kjm s Now. .5— — . Substitute the value of K (f)fro m equation (2) in this equation. (b) What is the steady-state response of v if w is a unit-ramp function?.1 8PP One possible representation of an automobile speed-control system with integral control is shown in Fig. ( 5) lim ^ / | ’\ l + K m G (j)v )UJ 1 1 + /:. .) = W (s ) P (j) ms ( j) + m k ^ + mk^k^ Thus. From the block diagram in Figure 4. .. the steady state error of the system for the unit ramp input. k .. f ' s 0 V Substitute 0 V for V in equation (1).31 in the textbook. sO step 6 of 7 The velocity constant of the system is.V . Problem 4 .---------------W .. Therefore. the velocity emor constant is _____ Step 7 of 7 (d) Calculate the emor in system with respect to disturbance w . K ( j ) = . ' s The steady state response of the velocity v is. s ^ k^frts The positional constant of the system is.f 'W s* ■¥k^ms-¥mk^k2 -------.^ 1 = lim *-»o s lim s^ +kyms+k^k2mj k^k^ Thus. find the transfer function relating the output speed v to the wind disturbance w.— ----------I F f* ^ ^ s* ^+ k^ms+ ^ MK>^ mkykj wttk t ’ r ^ f t t ^ tr ' ‘ F (. K + .thj?iv/ie aodJ^rreRnnnriinn prrnr mnstant nf this sustpm in rplptinn tn trankinn thp (d) What is the type and corresponding error constant of this system in relation to tracking the disturbance w? Figure System using integral control Step-by-step solution step 1 of 7 Refer to Figure 4. E ( . m .) = W { s )-V (s ) 5* •i-kyms'^k^k^m» '( 4 s*■^k...+ ^ + iC \v s \m s ) V= .(1) +kyms+mkfk2 * •¥mk^k^-¥kjnis Step 2 of 7 (a) The reference input voltage. s li m — — r 1 ^ l + lim 5 G (^) (^ )v s * j 1 s + K^ ^ J_ K k^k^ As steady state error for unit step input is zero and for the ramp input is — . t \ [ The transfer function from the disturbance input 1^ ( 5 ) to the emor £ ( * ) is.. J S r. (a) With a zero reference velocity input vc = 0. -----. 5 ^ R ( s) S+2+K AsK=5.5 a ) ^ 7 OF — 5 — = S. Problem 4 .E .S . Give the corresponding velocity constant.S. Figure Control system Step-by-step solution 5tep^y-step~so1ution step 1 of 3 Y ( s) = .=lim sE (s)=lim ^ (s) c-kO *-»0 ' •^ *-»0 c-kO -jj : g+6 '• ^ system becomes type-0 1 with S. ( 7 .5 _ | _5 ‘ S+2+K s+8 S.S.=- 15 Step 3 of 3 for K=6 : S + 2 --K E (s)= R -Y = -----. find the value of a that will make the system Type 1 for K = 5.E.S.= - K .E. ..1 9PP For the feedback system shown in Fig.i E(s)=R-y= t____ 5 L ___55 S+2+K s+6 s f s + lj S. 7 =7 Step 2 of 3 for : 2 s+2— K s+F.=lim s£fs)= lim ■ H ) R (0 <-kd ' • <-k0 > System becomes type-0 witti S.S.y to a step reference for /C= 4 and K = 6. S.E.=lim sE(s)=lim s(R -y ) c^O ' ' *-»0 ' ' ix R (s ) (s+7) for type-1 system.E.E. Show that the system is not robust by using this value of a and computing the tracking error e = r .=— System is not robust.S. The acceleration error constant is. [ jr + . 4 {R -Y ) sY ( i+ a + 1 ) 4 ( /e .l) 4 /t( j+ a + l) = y [ s ( j+ f l. the system type is . the error constant is Sa . r ( j) 4 (5 -t-g + i) (3) R (s) 4 (* + f l . Thus.l)+ 4 ( 5 + g + l) 1 + G (j) (l + G (s ))4 (5 + a + l) = G ( j ) [ j( 5 + a ..l) + 4 ( s + a + l ) ] 4 ( j+ g + l)» G ( j) [j( 5 + fl-l) + 4 (5 + g + l) -4 ( s + a + l) ] 4 ( s + a + l) = G ( j)[ j( 5 + f l.y )( s + o + l) = j K ( j+ a . Recall equation (5). f4 (5 + 2 + ^ g ) = lim j (5+&j) J ^ |. To define the system type open loop transfer function is considered. the error constant is Step 7 of 8 (c) Assume that j s U d a Here Sa is the some perturbation to the plant parameter. hence the system is a type-2 system. Thus. the system type is [ |] . (1.l) + 4 ( j+ a + l) ] Y 4 ( g 4 -g 4 -l) R ^ ( ^ + f l. Substitute 1 + for a in the equation. Problem 4.4 r ( 5 + a + I) = 5 y ( « + a . For a type-1 system. (a) Find G(s) so that the system shown in Fig. hence the system is a type-1 system. ' ' s (* + l.4 Y = {s+ a )X x ^ : ^ ^ . Recall equation (5). the open-loop transfer function C {4 ) is Step 5 of 8 (b) Assume that a ^ \ - To define the system type open loop transfer function is considered..35(a) in the textbook.20PP Suppose you are given the system depicted in Fig. 1 =X J(s+a) 4 R * X . ■U ni5^G (5) * j-»0 ' ' s U m [4 ^ 'i'8 ] = [4 (0 )+ 8 ] =8 Thus. 5 ( i+ f l. (b) Assume that a = 1 is the nominal value of the plant parameter.l) + 4 ( i+ g + l) The closed-loop transfer function of the system is.4 ( 2 ^ j 4 (2 + ^ Sa 4 ( 2 + ^ f l) Thus. the error constant exists for ramp input only and the error constant is zero for step input and infinite parabola input.(b) has the same transfer function from r to y as th system in Fig. Consider the expression for left side loop. What are the system type and the error constant for the perturbed system? Figure Control system the system type and the error constant for the perturbed system? Figure Control system Step-by-step solution step 1 of 8 (a) Refer to the system in Figure 4. y (4 g (^) (4) « (s) 1+ G ( j ) Here C {4 ) is the open-loop transfer function. The acceleration error constant is.l) 4 / i( « + a + l) . Step 2 of 8 Consider the expression for right side loop.. Step 8 of 8 Find the error constant. the error constant exists for parabolic input only and the error constant is zero for step and ramp inputs.l ) + 4 ( 5 + g + l ) step 4 of 8 The general fonri of the closed-loop transfer function is. Substitute 1 for a in the equation. where 6a is some perturbation to the plant parameter. Find the open-loop transfer function <7(j)- From equations (3) and (4). For a type-2 system.. Step 6 of 8 Find the error constant. s{s+ Sa) There is only one pole at origin.l) 4 ( i + 2) i(s ) 4 ( i + 2) There are two poles at origin.ir ( s + o ) ] = iy X {s+ a + l) = sY „ sY (2) (s+ a + I) Step 3 of 8 Equate both the equations (1) and (2).! )] (5) 4 ( x + g + l) Thus.(a) where the plant parameter a is subject to variations. What are the system type and the error constant in this case? (c) Now assume that a = ^ + 5a.(a).. it. . ( a: . from equations (15). ( a: . . *’ ■— 4 s + i+ ^ r .\ UK. k . (b) Consider the value of A^.K .* :.36 (a) in the textbook for G (« ) in equation (1). Calculate the from equation (8). .) i+ A r ..(18) ‘i.36 (b) is robust. --------------i 4 £ ± i £ _ i d £ ± y j t ( .H ■.. modify equation (11)..36 (b) in the textbook for in equation (7)..21 PP Two feedback systems are shown in Fig.. •b»pv I « P 4 i ' + * + ( * . Step 4 of 4 Substitute equations (17) in equation (9). E s R —Y Substitute the gain of the system from Figure 4....] -& K .) ^ ( * ) = ------------.^ : .5 i C . i + a: .= l . 4 j+ i+ ir A ( i. = 3 ....« . and (16). ) Substitute 1 for A^ .A : .... i+ 3 ( i.) • lim — S ^ . .R(s) 4J + 1+ C A 4s + l (8) Calculate the from equation (8).. ) [ l .^ (1) ^ ^ 1 + G (s ) Substitute the gain of the system from Figure 4. + 8 a:„) e .) iir . a :„ . find the value of K .(1 2 ) Step 3 of 4 Determine the value of velocity error coefficient K .) = o . the value of is equal to |pnel • Step 2 of 4 Refer from Figure 4.H = 0 (10) Equate equations (9) and (10). (14) UK. (a) Determine values for K^. a 1... is [ ^ . . . ) = o (11) For AT^ « i .^ :. i + i :... the Figure 4.. So. = l .( \. ' ^ Substitute equations (14) in equation (3). a: . ) ( i . K2. the Figure 4. Hence.-M 4s + i + a:.. ) -----------.(\-K .p '’ Step-by-step solution step 1 of 4 (a) Refer from Figure 4...- a: . . X l i M M J (9) Consider the value of ^ (» ) • e .. ) t _ ( < » ) = ---------------------4 '^ . F orJS :. both are Type 1).(16) Hence.36 (a) in the textbook and write the emor detector output equation.. find the value of 4 . .« p H = 0 (19) Hence. and (ii) their static velocity constant /Cv = 1 when KO = 1. 4/3 for K2. . from equation (6).. (20) ■»»p V ) i + a: .. (b) Suppose KQ undergoes a small perturbation: KQ -* + 5KQ. K ^ ^ K ^ + 8 K ^ .K .a: . x .1 « 3 . + 5 a: .. .p 3 ( H . (6) Hence.... + 8 a: .. Problem 4. find the value of (® ) (oo) ■ — . (4) ’ K . ■ Write the general formula for velocity error coefficient. * For AT.. \U K . Calculate the ^ . (17) Calculate the ^ * p H = ' ™ » ^ W from equation (2).+ S A r .K . (15) Substitute equations (15) in equation (12).. (5) Substitute equations (5) in equation (3)..K . E ^ R -Y E ( s U . . * p l ) ‘™ * 4 i ’ + i+ ( A r . and K3 so that (i) both systems exhibit zero steady-state error to step inputs (that is. ( i .. 4 s + i + a :.. = .36 (a) is regardless of K^ value... ' ' Determine the value of velocity error coefficient K . s ^ ^ 1 ^ K . l ...“ 4 1 + i + a :.36 (a) in the textbook and write the emor detector output equation. What effect does this have on the system type in each case? Which system has a type which is robust? Which system do you think would be preferred? Figure Two feedback systems h p '' ^ l. i( 4 s + l ) 1 'b j-p V ) 4 s^ + s + K JC ^s' (4 s + \) ^ H = l i 2 ’ -1^ 4z sr‘ + s + K ^ i (o o )s — !— . the value of ATjis and the value of AT. (2) ^ ' A s^+s+K ^. . a: .* r j) = o K p -. j Substitute AT| value in equation (18). U K .( i ... the designers favor to [choosesystem (a ) than because of output variation is huge with small variation in input changes. ' ’ . and 3 for K3 in equation (20).■M 4 pH ' ■(13) F o r j r . ^ ( o o ) = U in s £ ( j) from equation (2). .|] ^ » pH 1+3(1+8A:. ia s 4 . e . suppose inai iiieie is some uuenuauun in uie leeuuauK paui inai is iiiuueieu oy p = u.)~| 1 (1) [ j ( i + l ) ( s + 1 0 )+ 1 0 (* ^ + * .s+*.s’J V 1) Step 7 of 8 Apply final value theorem to find the steady state error.9) s (j +1)(s + 10)+9(*. the appropriate value for error constant.)(0. e . For the system Type 0 the error for step position 1 l+ K . First choose as an integrator in the loop and evaluate the transfer function. (b) Next. + l)(“ + 1 0 ) ) ( ^ ) 1 0 (V + *. ) j^ 1 j i(i +l)(j +10)+10(*. Problem 4.i 9 From error formula we see the system is Type 0. f i = 0. (c) If = 0. j+ * . The next step is to determine the stability. Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a). it is clear that > 0 and l l ( l + * . Y I j s(s+l)(s+10)+9(t. s .. ^ ( s ( i + l) ( s + 1 0 ) + 1 0 ( * .)Js* Now apply final value theorem to find the steady state emor. f ^ (^ + i)( » * io ). ■ lim j (j + 1 )( j + 10)+9(A :. Step 8 of 8 (c) The value of feedback gain. 1 1 0 (ltl:.s+i.)Y i ^ i(s +l)(s+10)+9(*. ) 9k.)Jl. You are to design a controller for this system so that the output y(t) accurately tracks the reference input r(t). ) > 0 for the system to be stable. ( ( i + l) ( i+ 1 0 ) 10____ Y * » ^ * * ' j [(.M o ) ]] 1 ( ( i ( * + l) ( j + 1 0 )J (j(j+ l)(j+ 1 0 )+ 1 0 (* .) ) 1 [ i(s + l) (5 + 10)+ 10(t. ) ¥ 4 ^ ^ i)( » . ) / ? j ( j + l) ( 5 + 10) _r r io ( v + * .^ io (V + * .))» p(^+i)(»+io)+io(V+*. You are given the following three options for the controller Dci(s): Choose the controller (including particular values for the controller constants) that will result in e Type 1 system with a steady-state error to a unit reference ramp of less than . f + t . Write the Routh array.i+ * .)-l(l^ 11 s*: lo t. s + 9 * .)] j . f i = 0. — s Calculate the steady state error. ) ^ J '■ i ( s + l ) ( s + l 0 ) + 1 0 ( * ..5 + * .s+*. first evaluate the transfer function and then determine the steady-state error.) _ 5(s + l) ( ^ + IO) 1 0 (A ^ + Jfc. (a) In order to choose a controller that will result in Type 1 system.) JTfil (s{s ^ f +l)(s+10)+9(*. (a) Let f3 =1. beginning with first and second coefficients and followed by the even numbered and odd-numbered coefficients. .s+9*..9 The input is.^ -9 a: = -1 0 Therefore. and see if all the poles are on the open left half plane.)-1 0 (1 .) 1+ A j ( 5 + !)(« +10) Step 2 of 8 Now evaluate the system error. s + 1 . is |» | q | .j s (s + l ) ( j + 1 0 ) + 9 t .35 in the textbook.)J i f ( i + l ) ( i + 10) ) s lim [ i ( i + l)(i+ 1 0 )+ I0 (* . 9 l + K^ \ + K .)/>-io(V+*. ) ( l ) Js' s (5 + l ) ( j + 10) ^ I ■t [^j(5+l)(i +10)+10(*^+*. t i l s ’ 4 . where the feedback gain jS is subject to variations.)J L 5 *y Therefore.9.. (d .9 in equation (1). From the Routh array. (2) Step 4 of 8 From equation (2).1 0 ( t .... [ i ( i + l) ( s + 1 0 ) + 1 0 ( * ^ + t . use the Routh criterion The characteristic equation of the closed-loop poles is. the system is no longer Type 1. = ---------] j .) JU’J l)(j +10)+9t. Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a). iNexi. suppose that there is some attenuation in the feedback path that is modeled by p = 0. io(*^+*.)^ step 3 of 8 Substitute ^ = t in equation (1). we first arrange the coefficients of the characteristic polynomial in two rows.1 0 * . s’ : 1 f’ : 11 KM. .) 5 ( j + l ) ( j + I0 ) 1— 10(*.-m) U io)A ' * ^ ( ( .22PP You are given the system shown in Fig.) j\s) j { 5 + l ) ( j + 1 0 )-ik ^ j-ifc .9 The input is.-10*. Step 6 of 8 (b) The value of feedback gain. it is clear that 2 1 0 which meets the steady state specifications. what is the system type for part (b)? What are the values of the appropriate error constant? Figure Control system X S te p -b y -s te p s o lu tio n step 1 of 8 Refer to Figure 4. ^^j(i + l)(s + 1 0 )+ 1 0 (t^ + * .. ) j( 5 + l) ( « + 10)^ j ( j + l ) ( « + 10) 1- j ( j + l) ( i+ 1 0 ) + 1 0 ( * . = lim s [£ ( j) ] |^5 (5 + l)(j+ 1 0 )+ 9 (A :^5 + ifc .1 0 * . s + t .9. j + * . s + * . ) ( 0 .s. =.)/jJ J « ’ fi 1 0 (V + *. Substitute f i = 0. the steady-state emor due to ramp input is 0 Since the steady-state error due to ramp input is infinite. s + t . 9 ) .to s [£ ( j) ] ^ 4 (4 + l)(j + 10) + 9(jfc^J+*.s+t.^+t.l o ( i ^ + 1^) = 0 Step 5 of 8 To determine the Routh array.) J ^10 ‘ lOJfe.s+ * .)^ -io (V + * . at least (n + 1) zeros out o f *q’ should present at the origia .E (s )_ n ( ^ ‘) step 2 of 2 S. (b) For this system to respond to inputs of the form r(t) = fn1 {t) (Where n < q ) with zero steady- state error. • • •.»iXi+ f t ) " ' C*+/>«) Step-by-step solution step 1 of 2 Y (0 _ G R (s ) 1+G f E (s )= R (s )-Y (s ) . what constraint is piaced on the open-loop poles p1.S. pg? Figure Control system («+. p2.E.=limsEfs) t-»o ' '' n ( '+ P i ) „i sn(s+pi) A nd »i_______ is required to be equal to zero. Problem 4.23PP Consider the system shown in Fig. Hm-------------------- For this condition to be True. (a) Find the transfer function from the reference input to the tracking error. R{s) + k ^ + k . = 4 3 % . e(t).4 1 4 .. from part (d) is small?.) ’ » 0 4 tt. » ( * * .jk+ k....----. Step 4 of 10 (b) Write the denominator part of the equation (2).. kl) to ensure that the amplitude of the transient tracking error.must be satisfy the 4 j M : .) in equation (1)..(* * .(10) Write the transfer function of the entire system. k Hence..| = « '* 's i ii i< (18) Apply inverse Laplace transform in equation (17). the value of peak overshoot is \M ^ =4.. the roots of the closed loop poles from the transfer function from J?(5 )to is ..k ^ ± y l{ k ^ y .707 for ^ and 1 for <o^ in equation (14). f+ l Determine the peak overshoot value from damping ratio value and natural frequency value («>.. .. Step-by-step solution step 1 of 10 (a) Refer from Figure 4..Jt = 1.)’ 8in J s O 2 V. the overshoot time is finite and rather small for practical purposes... Problem 4.39 in the textbook for in equation (11).+ 4 )-S L i. 5’ -f 2Co ^ . the transfer function of the system from J?(. K (. W . kl...« H = o (3) Hence......707 for ^ and 1 for <o^ in equation (7).-(tt.) .3% |. 7 0 7 ) ( l ) f + ( l ) ’ * = »’ + 2 ( 0 . k » 0 . (21) 2 Step 9 of 10 ^ Hence.-(**. e(t). the relation must satisfy the condition |4jbfc^ . ■■ 2 Hence. E = R -Y I E{s) = .. ( 1 6 ) s ^ + k fk s + k . * .. (5) Find the roots of the above equation.k Substitute -L fo r in equation (16). and a unit-ramp reference input signal. r ' f . Write £ ( j) fr o m equation (2). In PI controller the relation between parameters ifc^and ifej. if the reference input to the system..<?a fiinrtinn nf timp if thp rpfprpnnp inniit tn thp svRfpm (d) Find the tracking error signal as a function of time. is very large? Does the unit-ramp response have an overshoot in that case? Figure Control system diagram GmtfDller Haat (*.) ’ . (15) Hence. the value of tracking error signal as a function of time e (/)a t unit ramp signal is ^ 4 tt.. e(.7 0 7 ) *.. ■i-kpks+kk. |effj|. j ^ + * ^ + * ..M l « (•) s^ + k.k Y (s)_ k ( k ^ * k .)_ 0 {s )D A s ) R(s) 1+G (*)D c ( 4 Step 6 of 10 ^ Substitute the plant gain of the system and controller gain from Figure 4.)to £ ( j ) is £ W = .k Step 2 of 10 ^ Calculate the = from equation (2)... > « (*) (1) 1+ G(5)D c (») Substitute the plant gain of the system and controller gain from Figure 4.1 — .. 1 E {s )= - 1+ (‘... (9) * .414. » ( J M : ^ f a n d ^ > 0 condition. (4) Hence. (a) Compute the transfer function from R(s) to E(s) and determine the steady-state error {ess) fo a unit-step reference input signal... fj4tt.39 in the textbook and write the error detector output equation. * = j '+ 2 ( 0 .innpl a.k ) Step 5 of 10 (c) Write the characteristic equation of the second order system. s * + k fk s + k fk ^ s ^ + 2 f y a ^ + t o * . < > h- .... -^kfks-ykfk + k ^ + k .. ) ’ (20) ^ > 0 .. M . (f) What is the transient behavior of the tracking error.39 in the textbook for G {..... s ^ + k ^ + k . Substitute equations (9) and (10) in equation (12).707 and ojn= What percent overshoot would you expect in y(t) for unit-step reference input? fdXJFinrlthft.t3 ) i* + l. (b) Determine the locations of the closed-loop poles of the system.y Hence....( « . r(t).4 ( k ..Jt = 2 (0 .trackinn prmr <.. Hence.(* * . = lim - *-*® s^+ k p k s + k fk = 0 . % A /.. 7 0 7 ) i + I ... Step 7 of 10 (d) Determine the tracking emor signal as a function of time e ( /) at unit ramp signal. (6) Equate the equations (5) and (6)... the steady-state error for the unit step input is signal is |e_ ^ (oo) = 0 |. = e”* ''^ x l 0 0 (14) Substitute 0. Step 3 of 10 ^ Calculate the from equation (2).) Write the general formula for percent of peak overshoot..4 ) 1+ R(s) s ^ + k jM + k . (c) Select the system parameters {k. kP.k ■(2) J ? (i) s* + k ^ + k . . £ (4 ) = ^ R {s ) .. for a unit-ramp reference input if the magnitude of the integral gain.K i> / » 0 . the steady-state error for the unit ramp input is signal is |e ^ (oo) = 0 |.(17) Consider the general form of inverse Laplace transform.. is a unit-ramp. Step 10 of 10 (f) Determine the transient behavior of the tracking emor for unit ramp input from equation (19). . (7) Substitute 0.(8) Write the coefficients of the equation. * = ! ..24PP Consider the system shown in Fig.. iif i..k E(s)~ . kl) such that the closed-loop system has damping coefficient ^ = 0. ' _ r —n From equation (22).= e -^ 4 ” ’l ' ' ' i ^ x l 0 0 M . -2 1 »-»» s ^ + k fk s + k jk s s .)= ^ Sin (19) V 4 tt.. ) ’ 2 ' ^ / Step 8 of 10 (e) Write the small value of |e(/)| from the equation (19). (e) How can we select the PI controller parameters {kP.. 0/n. Ki where dm is measured In radians.With a tachometer we can measure the motor speed Consider using feedback of the error e and the motor speed ^ in the form where K and TD are controller gains to be determined. on comparing the coefficients.\ s^ + s(a^ + 1 For^.« we get 5 s „ = lim s 5 (s) K=o| Step 5 of 5 (e) We know that 0 Qi s* + TjfKbft) + Kbfi Find the response to torque. Problem 4. If there is no load torque (w = 0).With a tachometer we can measure the motor speed 6m- With rotating potentiometers.. using the results of Chapter 3.0. (a) Draw a block diagram of the resulting feedback system showing both dm and ^ as variables in the diagram representing the motor. w e obtain Sm+ 0\4m — ^Mfl + where „ KtKe . ^ I With rotating potentiometers. what speed {in rpm) results from va = 100 V? (c) Using the parameter values given in part (b).5 . and cO = 10.05 s Find . we get ^ = 2938 rpm . 0 „ = lim s0{s) " j-»o '■ ' 0 „ = lim s0{s) " j-*o ' . = 0. Step 2 of 5 (b) If V a = constant the system is in steady state &= ^K - Find 6 which is given by ^ 2 0 0 x 1 0 0 60 rad s-’ p= - 65 27T rpm 0 = 2938 rpm Thus. = 3 . (b) Suppose the numbers work out so that a1 = 65. bO = 200. £ (s) = 0 ^ -0 s ( s + a . it is restated here.05 sec.T ^ IC b . a step change in dref with zero load torque results in a transient that has an approximately 17% overshoot and that settles to within 5% of steady state in less than 0. Thus.0/n. it is possible to measure the positioning error between 6 and the reference angle 0ror e = d re f. Ki = Mb + ^W.25PP A linear ODE model of the DC motor with negligible armature inductance {La = 0) and with a disturbance torque wwas given earlier in the chapter. let the control be D = kP + kDs and find kP and kD so that. = 0. Step 3 of 5 (c) Find — 0 0y Conq>aring with standard second order equation 0 aj 0^ When = \1% we get ^ '= 0 . it is possible to measure the positioning error between d and the reference angle drox e = dref . (e) Derive an expression for the steady-state error to a constant disturbance torque when dref= and compute Its value for your design in part (c) assuming w = 1. (d) Derive an expression for the steady-state error to a reference angle input and compute its value for your design In part (c) assuming dref= 1 rad. K. Step-by-step solution step 1 of 5 Sketch the block diagram of the resulting feedback sjrstem. 8 xlQT^I step 4 of 5 (d) Find the steady state error. in slightly different form.05 k = 120 Thus. Dividing through by the coefficient o f’^ . the required block diagram is sketched. we get i : = 72| t. 0 S C ( s ) .0 1 ) y (s )_ 0.0 2 ) F .01 *. J+ 0 . ■(13) Substitute equation (13) in equation (10). . ( 10) ^ ’ J + 0 .000kg.K j u .....(12) Hence. (15) Compare the equation (15) with denominator of equation (14)..j+<a. A and B... U is in degrees.D v .0 .. X is velocity. 10 U (s) (1..OOOI].0 2 ' ’ j+ 0 .0 2 s + t. Problem 4...0 5 2 | « ..(14) j ’ + 0 .. > 0..( i) .. Design a proportional control law U = -kP V that will maintain a velocity error of less than 1 m/sec in the presence of a constant 2% grade. . acceleration and force.0 2 + i t. s+ 0. 0 2 + t..0 2 + * . ■ (s + 0 . the transfer function of the automobile system is t/( s ) ( s + 0 . 0. .. 0.0 .0 2 ) l( .0 2 s+0. 0. ( s ) ..= 0 .O.. Consider the velocity error of less than 1m/sec in the presence of a constant 2% grade. Step 6 of 6 (d) Calculate integral constant value. £ (s ) = » . Vis velocity. U is in degrees.. A and 6 in equation (8).. ^ 7 1.0S £ (s ) = i. the value of Integral constant value is [*. x is acceleration.0 S jG ( t) .0 0 0 (s +0.0 2 Step 4 of 6 Compare equation (8) from equation (9) and find the parameters a.. =0. (5) Where..U {s)-D V {s) m sy (s )+ D y (s ) = /C.1 <1 0 .1 ( 11) 0 .’ .02 <0. . Substitute a.000(0.02 s+ 0 .. no proportional tenn) is advantageous.( s ) .02 i + * . (4) Where.0^ Step 5 of 6 (c) The clear advantage of integral control is [zerosteadystateeiTor| for step input. ( s / ^ ) | = Ums J + 0 .01) y (s) 0. (16) Write the damping ratio at critical damped condition. select the feedback gain so that the roots have critical damping (C= 1)• Step-by-step solution step 1 of 6 (a) Consider the mass of the car. j * + 2 « ) ^ + c i ) / s j * + 0 . * ’ +2t«>..) F .05 = > . Modify equation (4).. where V is given in meters per second. (8) 1+—^ 1+—^ V.D x . 0 8 .0 1 ) Step 2 of 6 (b) Draw the general block diagram of automobile speed control system.=0... vis acceleration. (17) Determine *^ from equations (16) and (17).( 5 ) ..... The value of ( jf ^ ) | lesser than 1.. Assume F ^ (j) is equal to zero.(2) Consider the air drag provides a friction force proportional to velocity.0 5 G ( s ) S + 0 . and W is the percent grade of the road..(*)- t + 0 . ( s ) ... u is displacement. * ... the proportional constant value is * . > s + 0 ..’ . 0 2 + ^ ..0 2 ) i.0 2 + * .) *- 1+ 1+- ».0.F . ffl=l.0 2 + * .0 0 0 1 Hence.. Substitute equations (1).. (2) the accelerator is the control U and supplies a force on the automobile of 10 N per degree of accelerator motion. =10Nperdegreeofacelerator. * ’ + 25 o>.(s )-F (s ) f M J £ ( s ) = i. » ’ + 0 . .. and W is the percent grade of the road. s+ a s+ a Consider the velocity changes of the system. (2) and (3) in equation (6).0 2 .4+<i>.02 0.0 2 + * . j .. (3) Consider the relationship between the mass. (d) Assuming that pure integral control (that is. m r = X .U(s) (ms + D )y (s ) = /C. C = 1 . (c) Discuss what advantage (if any) integral control would have for this problem.U(s) (6) 1/ (s) (»« + D ) Calculate the transfer function of the automobile system. ■ j(j+ 0 .0 2 + t. Assume that (1) the car has a mass /n of 1000 kg. (1) Consider the accelerator is control U and supplies a force on the automobile..02 «+ao2 where V is given in meters per second.. S+0-02 s+0-02 G (.000s-M O ) l'( s ) 1.sec/m. * .0 5 g ( ^ ) i+ 0 .( s ) . and (3) air drag provides a friction force proportional to velocity of 10 N • sec/m . Write the general characteristics equation of the second order system.01 (7) u ( s ) ” (i+ 0 . 2<o. j. m sy (s) = K . — !— LF(s)+ HW . m v . ( s + 0 .0 .^ ’ n a = K ^ . (b) Assume the velocity changes are given by s+0. Apply Laplace transform in equation (5).0 2 j + * .(9) ' ’ j+ 0 .< : . (a) Obtain the transfer function from control input U to the velocity of the automobile.0 . fy Figure 1 Step 3 of 6 Write the output equation of error detector from Figure 1.01 Hence.0 2 + * . D=10N...0 .(s ) -0 . Calculate the = l i m j £ ( j ) — from equation (10). F ( i ) . .02+A . Design a proportional control law U = -kP V that will maintain a velocity error of less than 1 m/sec in the presence of a constant 2% grade..1 s lim J+0.26PP We wish to design an automatic speed control for an automobile. ( » + 0 .S .. Consider the value of proportional constant.01) I / ( s ) ° l. > 0 .. " Substitute A+ S A ^^^ ^ in equation (28). = 0 ) Calculate the value of lim « y ( j)fro m equation (1). A H .A H . r B Ak.. 1 (29) Substitute equation (2) in equation (29). ' ' s + a + k ..^ ( j ) + — (5) r-« w ^s+a+k^A N ^ ' ^ s+ a+ k^A H ^ ' ^ Substitute 0 for /f^and 0 for i r ( s ) in equation (5).... .. Step 5 of 8 Calculate the value of H*ity_^(f)from equation (1).. r ( . Step 6 of 8 (d ) In feedfonward system. (28) ’ \a * k .' ^ (o o )= [-----. ... ! ™ y ^ ^ W = [ f « 4 + f 'o ] B U m > -. Calculate the value of Hm>’(f)from equation (1). Substitute 0 for Hj.H . H .. . effect of disturbance can managed by chooses [largevalii^ of k^. Consider the value of Km AkM . . r B ( A + & A )k .. (11) ^ \_s+ a+ k^A H y ' Substitute equation (4) in equation (1 lim y (/) = Ii m[— — .— H. ( 12 ) f k ^ [s + a + k .‘. (b) Assume that the desired speed is a constant reference r.A H . =r. H ..A ..— H. .^ ^ ^ r . (c) Repeat part (b) assuming that a constant grade disturbance W (s) = ^ is present in (c) Repeat part (b) assuming that a constant grade disturbance ^ is present in addition to the reference input....A H . ^ '( ” ) a + k.AH.(15) s Calculate the value of K m ^ ^(f)fro rn equation (1) ! ™ y r ( ') = 'i2 ^ JU ^ ' —— *(*)] s+a+k.(18) Substitute equations (4) and (15) in equation (18).. the value of H . from the above equation concludes that |10% o f errorl is raised in the A value.(1) ^ ' s+ a+ k^AH .. the value of //^ in feed forward system is k = — Ak^ ' AH.. -----------H5..^ R ( s ) . B lim y ^ f f ) = lim 5 . calculate the y y (oo) from equation (1). AkH . Compute values of the gains kP.. . So the feedback control is not necessary in this system. In particular. (14) Hence.2-----L . — H. ■■■ W„ + ---------— } B a+k. (30) Ra H ' .. = 0 .. zero value of f f = 0 step 3 of 8 Calculate the value of at feedback path.\n feedback system is: and the value of Ak^ at H 960 ' H ^ A -A H . —— ^ w l S + a + ki. •-* s + a ‘ lim— -— r.. so that j{(j) = & Assume that the road is level..(oo) = lim j[------. R fH ' .. S The feedback gain value is considered as zero means the system belong to open loop system. Jk H I i m 5 y ( 5 ) « l i m . A kM .j t ( $ ) ] [a + * .. A k ^ U ^ . (6) Substitute equation (4) in equation (5).A H .A H .H . ' ' Step 2 of 8 (b) Consider the value of road level. Km ■(13) s+a+ k^A H ^ * j Write the value of H^trom equation (13).(24) ' \a * k . ^^ r lim y ( / ) s lim s ------------------.40 in the textbook and write the transfer function from IP^^Jand from J?(5 ) to y ( i ) .. ' Ak.27PP Consider the automobile speed control system depicted in Fig. \+ k . How should the gains be chosen in this case to reduce the effects of 6A7 Figure Automobile speed-control system ^— 0 - S te p -b y -s te p s o lu tio n step 1 of 8 (a) Refer from Figure 4. B .from equation (8). and Hy to guarantee that Include both the open-loop (assuming Hy = 0) and feedback cases {Hy ^ 0) in your discussion.. liin fK (f)s lim j[------. (7) Ak H Consider the value of |im ^ ' j. .A H .--— r ( » ) + . ' ^ Substitute 0 for in equation (17).. a + (A + S A )k ^ ffy f (^+ & 4 ) a + k ... k .. the same of % error is tracking..(16) a * k ... Consider the feed fonvard ( H . y. j i n i y ..— -----.. V A k . ...— « ( . ..A H . • a+ k.. « y > (“ ') = :a * k .( /) = » --------------... Hr..A H .— -----. (d) Assume that w(t) = 0 and that the gain A undergoes the perturbation A + 5A. . (2) Consider the gain of the feedback path.(23) Hence. Problem 4.^ ( 4 ) + --------------. ' ’ Substitute equation (9) in equation (16).... • a+ k. ^ ^ a+ k. ^ ( „ ) .H . (26) Step 7 of 8 Substitute equation (9) in equation (26). id ± 5 :4 * c 5 L r . Determine the error in speed due to the gain change for both the feed fon/vard and feedback cases. + ------.A H A &A BA Hence..R(.. ' s*a ‘ j+ a * y ( j ) = ------------------. B . Km y ( / ) s lim s ---------£ ---------iy ( s ) + ---------— — R (s ] ( 10 ) MA ' ' m O s+a+ k^A H y ' ' s*a ^k^A M ^ ' ^ Substitute equation (2) in equation (10). B .. (a) Find the transfer functions from W(s) and from R(s) to Y(s). (21) Substitute equation (14) in equation (21). J r. ' ' a+ k.+ Ak„H.... (27) Hence..fy(s)+ -----. . ' ' J /V ^B w. (3) Consider the value of reference input... find the variation in speed due to the grade change for both the feed forward and feedback cases.) . . Step 8 of 8 In feedback system. w { s ) = 0 .A H .-----.(17) a+ k..d M : . Use your results to explain (1) why feedback control is necessary and (2) how the gain kP should be chosen to reduce steady-state error. a+ (A+ SA )k.. Step 4 of 8 (c) Consider the value of disturbance. ■(9) A k.A H .. no possibilities available to [reduce the effect o f ftishirbanc^ in feed fonward system. so w(t) = 0. ..A H .and A+SA^^^ in equation (25). " Substitute equation (14) in equation (29). calculate the y ^ (oo) from equation (1). ’ Hence.A H . . ( / ) = s ------------.. the transfers function from JP(j)andfrom J?(f)to y ( * ) is i'W = s + a + k . -----. ..A H . ..) = ■R(s) U k ..(0 = -« 4 + < i (19) B «)>(«>) = — Wo (20) Hence. Hence. (y4 + &4)Jt. H. ) . ' ’ s+a+k„AH .y W . ( A * iA ) k . " Substitute equation (2) in equation (24). — . ? .— i y ( i ) + ------------ a + k AH ' ' a + k AH ' ’ lim . ' ’ a * k .ff..^ W + — — /f(s)l f ys+ a+ k^A H ^ ^ ' s+a+ k^A M ^ ^ 'J A s [ B /V Ak„H. Ak. ( 8) *-“> s+ a Write the value of jyj. the equation concludes that the tracking parameter variation is reduced by [largev a lii^ of . Kmj y (g) = Iim 5 ^ -^ ^ ^ j . Problem 4.28PP Consider the multivariable system shown in Fig. Assume that the system is stable. Find the transfer functions from each disturbance input to each output and determine the steady-state values of y1 and y2 for constant disturbances. We define a multivariable system to be type k with respect to polynomial inputs at wi if the steady-state value of every output is zero for any combination of inputs of degree less than k and at least one input is a nonzero constant for an input of degree k. What is the system type with respect to disturbance rejection at vv1 and at w2? Figure Multivariable system Step-by-step solution step 1 of 4 Sketch the given figure o f the multi variable system. -o n step 2 of 4 We know that „ 1 _ s — s(s + 1) « ^i = T- ~T~.-----T7 ~r~,----- T7 ^ 5 + ff + l e + ff + 1 s +s+\ Consider that there is constant disturbance condition. Obtain the es^ression for the diff^erent variables in the equation t V „ + ( s + \)W :„ + S+1 Step 3 of 4 Let U2 be the signal coi^ling both the sjrstems. Thus we get (s + l ) ( j ^ - y a ) + s ( s + l ) » ; ff + S + 1 Obtain the eg ressio n for Y2 . y S. I (^ + 1 )^ . ’ TTlT+l +3s+ 2 ( s + l ) ’ ( -ira ) + s ( s + l ) V a (s’ + 3s + 2 )(s’ + s + 1) Step 4 of 4 ^ Obtain the different system types with respect to disturbances: with respect to Type 1 with respect to Type 1 y 2 with respect to Wj Type 1 y^ with respect to Type 0 Problem 4.29PP The transfer functions of speed control for a magnetic tape-drive system are shown in Fig. The speed sensor is fast enough that its dynamics can be neglected and the diagram shows the equivalent unity feedback system. (a) Assuming the reference is zero, what is the steady-state error due to a step disturbance torque of 1 N m? What must the amplifier gain K be in order to make the steady-state error ess < 0.01 rad/sec? (b) Piot the roots of the closed-loop system in the complex plane and accurately sketch the time response of the output for a step reference input using the gain K computed in part (a). (c) Plot the region in the complex plane of acceptable closed-loop poles corresponding to the (c) Plot the region in the complex plane of acceptable closed-loop poles corresponding to the specifications of a 1% settling time of ts < 0.1 sec and an overshoot Mp < 5%. (d) Give values for kP and kD for a PD controller, which will meet the specifications. (e) How would the disturbance-induced steady-state error change with the new control scheme i part (d)? How could the steady-state error to a disturbance torque be eliminated entirely? Figure Speed-control system for a magnetic tape drive /-aiokra* S t e p - b y - s t e p s o lu t io n Sketch the given figure. Figure 1 (a) Consider the reference is zero and the torque is 1 N.m. The given steady-state error is ^ 0.01nd/s• Determine the transfer function — for disturbance. W 1 Js + b {J s + b 0.5j + lj Determine the steady-state emor, • (stepin ------- Thus, the steady state error with a step due to step disturbance is I + 1 0 *. £ 0.01 we get AT, ^ 9.9 Thus, the amplifier gain lAT^ lOl. to make the steady state error ^ 0.01 nd/s• (b) r(s ) Find the expression for 10 *. 1 y ( j) +b'0.5s + 1) Substitute 10 for 1 for ^ and 0.1 for J in the equation. 10(10) . y ( j) 0 .5 J -H o . i j + i l^ O .b + 1 0.5j + i J 100 (O.U+1)(0.5j + 1)+100 ______ lOO_______ "(O.I)(0.5)[(j +10)(j +2)+2000] 2000 5^ + 12.^+20 + 2000 ‘)(W I 5^ + 125+ 2020 Compare the expression with the standard system response and find and^ Thus, weget s V2020 2V 2^ Approximate the values to the nearest whole number to get a>^ s 44.94, ^ s0.13- Hence, the roots are undesirable because, the damping is too low and there is high overshoot. The roots of the closed loop are plotted and sketched. 4> •3 I Figure 2 The time response of the system for a step input is sketched. 1 I Figure 3 (c) Given that there is 1 % of settling time of ^ 0. 1secand an over shoot of We know that for i 0.1, we have ^ ^ 46- Similarly for S 0.05, ^ ^ 0.7 Step 8 of 11 ^ The region of acceptable closed loop poles is sketched in the complex plane. Lines ofdam ping and (d ) For a PD controller larger values of and^ are required. This can be achieved by increasing and adding derivative fed back. y ( i) finri ' ^ lOAT,_L y (» ) 0 . 5 i- f l J s * b w k , ( k ^ + \) (o .5 j + i) ( y * + ft) y ( j) 200X. n ,(s ) + ( 1 2 + 2 0 O * :,* r„ ) i + 20(1 + 1 0 *:,) r(s ) 200 * : , Hence, the transfer function is a ,(s ) s* + 12 + 2 0 0 * :,* :.) s + 20(1 + 10* : , ) By choosing the suitable values for Kp and KD any values for ojj,, and^ can be achieved. Step 10 of 11 (e) Find the transfer function — for the disturbance. 1 Js + b IT 1 io * : , ( * : „ f + i) J s + ft (0.5» + 1 ) 20(0.5t +1) ly ” j ’ + ( i 2 + 200 * : , * : . ) j + 20(1 + 10* : , ) The error when a step input is given is e,(step m W) = — 10*:, Step 11 of 11 The derivative feedback affects the transient response only. To eliminate the steady state error an integrator is added to the loop.. This can be represented by replacing AT^with AT^ + in the fon/vard loop and still keeping PD control in the feedback loop. Thus, we get y 20(0 .S s + l ) i ly " s’ + (1 2 + 200 * : , * : . )s ’ + (2 0 + 200 * : , + 200 * : , * : . )s + 200 * :, Hence, the steady state error to the disturbance will be [c , (step in W) = O] - Thus, the steady state error to the disturbance torque is eliminated entirely. Problem 4.30PP Consider the system shown in Fig with PI control, (a) Determine the transfer function from R to Y. (b) Determine the transfer function from W to Y. (c) What are the system type and error constant with respect to reference tracking? (d) What are the system type and error constant with respect to disturbance rejection? Figure Control system Figure Control system 777T Step-by-step solution step 1 of 8 (a) Refer to block diagram in Figure 4.43 in the textbook. To calculate the transfer function from r to y , equate disturbance rejection to zero. The modified block diagram is shown in Figure 1. Step 2 of 8 Calculate the closed loop transfer function from R to Y. 10 I t S) [ s JU - + S + 20J R(s) 10 y + f+ 2 o j (V-t-t,)(io) 5 ( i * + j + 2 0 )+ ( + it;) (10) io ( V + * ,) »’ + j ’ +20s + 10*,i + 10*, I0 ( * ,s + t ,) »’ + f ’ + 1 0 (*,+ 2 )f+ IO *, Therefore, the closed loop transfer function from R to Y is s’ + i ’ + 1 0 (* ,+ 2 )i + 10t, Step 3 of 8 (b) To calculate the transfer function from fp to y , equate reference tracking to zero. The modified block diagram is shown in Figure 2. 10 J ‘ i'+ i+ 2 0 V +*/ s Figure 1 Step 4 of 8 Calculate the closed loop transfer function from jp to Y. R(s) 10 1+ (“ )(7+ S + 2 0 J 10 , _______ s i + j + 2 0 _ _ ^ j ( j * + s + 2 0 ) + 1 0 ( i , j + i, ) i ( i ’ +s+20) IQs s’ + j ’ + 20s+10*^ + l0*, lOs ” s’ + 4 ’ + 1 0 ( t ,+ 2 ) s + 10t, \0s Therefore, the closed loop transfer function from W to Y is s’ + s ‘ + 1 0 ( t, + 2 ) j + 10t, Step 5 of 8 (C) The GharacteristiG equation is, j ’ + s ’ + 1 0 ( * , + 2 ) i + 1 0 * ,= 0 . Apply Routh’s array criteria. 1 I O ( * ,+ 2 ) I lOit, I0 (* ,+ 2 -* ,) m , For stability. 10jlr,> 0 k ,> 0 And, 10( * , + 2 - * ,) > 0 * , + 2 -* ,> 0 * ,> * ,-2 The open loop transfer with respect to reference tracking is. ” ''>■18 10 + S + 2 0 ) ( V + t,) ( 1 0 ) Therefore, the system is Type 1 with respect to reference tracking. Step 6 of 8 Calculate the emor constant. ATy s lim 4G (4) r (v + * ,) ( io ) ' s lim j ^ » ( s * + i + 20) •-••I i^ +s i^ + s ++ :2 0 J ( t,(0 )f* ,) (1 0 ) (0 )^ + 0 + 2 0 lOA; 20 k, 2 Therefore, the velocity constant, with respect to reference tracking is I Step 7 of 8 (d ) The open loop transfer with respect to disturbance rejection is. s Therefore, the system is Type 1 with respect to disturbance rejection. Step 8 of 8 Calculate the en^or constant. = Um sG (5) =lim(V+*;) =* , ( 0 ) + t , -k , Therefore, the velocity constant, AT, with respect to disturbance rejection is 0 Problem 4.31 PP Consider the second-order plant with transfer function (f+IX5f+I)’ and in a unity feedback structure. (a) Determine the system type and error constant with respect to tracking polynomiai reference inputs of the system for P [Dc = kP\, PD [Dc (s) = kP + kD s], and [Z>c(s) = + ^ + controllers. Letk P = ^ Q ,k l= 0.5, and (b) Determine the system type and error constant of the system with respect to disturbance inputs for each of the three regulators in part(a) with respect to rejecting polynomial disturbanceswffj at the input to the plant. ueiermine me system type ana error constani or me system witn respect to aisturoance inputs for each of the three regulators in part(a) with respect to rejecting polynomial disturbanceswffj at the input to the plant. (c) Is this system better at tracking references or rejecting disturbances? Explain your response briefly. (d) Verify your results for parts(a) and(b) using Matlab by plotting unit-step and -ramp responses for both tracking and disturbance rejection. Step-by-step solution There is no solution to this problem yet. G et help from a Chegg subject expert. ASK AN EXPERT Problem 4.32PP The DC motor speed control shown in Fig. is described by the difTerential equation y+ 60y= 600va-1500w, where y is the motor speed, i^a is the armature voltage, andw is the load torque. Assume the armature voltage is computed using the PI control law Va = - ^*j>e + t / j T e d t^ , where e = r - y . (a) Compute the transfer function from W toYas a function of kP and kl. (b) Compute values for kP and k l so that the characteristic equation of the closed-loop system (b) Compute values for kP and k l so that the characteristic equation of the closed-loop system will have roots at -60 ± 60/. Figure DC Motor speed-control block diagram Step-by-step solution step 1 of 4 (a) Consider the differential equation of the DC motor. +60;>=600v, - l.SOOw Apply Laplace transform on both sides. jK(j)+60y(s) =600K.(5)- 1.500FK(5) Consider the armature voltage value in PI control. Apply Laplace transform on both sides. Step 2 of 4 Substitute for in the Laplace transform of the differential equation. (i +60)r(i)-600^-*,,£(j)-^£(j)j-l,500»'(s) {s + 60)Y(s) = - 6 0 0 ^ ^ * , j £ ( j ) j - l ,5 0 0 » '( i ) ( j + 6 0 ) r ( j ) + 6 0 0 ^ ^ * , + ^ j £ ( j ) j = - l , 5 0 0 » '( j ) ^ s + 60 + 6 0 0 t , + 6 0 0 ^ j £ ( s ) = - l, 5 0 0 » '( » ) E (s) -1 ,5 0 0 + 6 0 + 600*, + 6 0 0 ^ j - l,5 0 0 i ” ( i “ + 6 0 j + 6 0 0 * ,* + 60 0 * ,) Consider the input R as zero. The error function is, £(s)=-r(s) y (j) ________ l,5 0 to __________ W{s) « '+ 6 0 ( l + I O * ,)s + 6 0 0 * , 1,5005 Thus, the transfer function from 1/Yto Y is *’ +60(1+10*,)i +600*, Step 3 of 4 ^ The roots of characteristics equation are - 6 0 + 6 0 y ,-6 0 -6 0 y Write the characteristic equation from the roots. ( s + 6 0 + 6 0 y ){ i+ 6 0 - 6 0 y ) = s’ + 120*+7200 The characteristic equation of the system is, , * + 6 0 ( l + 1 0 * , ) i + 6 0 0 * ,= 0 Compare the two characteristic equations. 600Jt,»7200 60(U10Jfcp)»120 Calculate integral constant kf ■ _ 7200 600 -12 Thus, the value of kf is m Step 4 of 4 Calculate proportional constant kp ■ 60(1+ 1 0 *,) = 120 1+ 10*, = 2 * ,= 0 .1 Thus, the value of kf, is I P Problem 4.33PP For the system in Fig., compute the following steady-state errors: (a) to a unit-step reference input; (b) to a unit-ramp reference input; (c) to a unit-step disturbance input; (d) for a unit-ramp disturbance input. (e) Verify your answers to (a) and (d) using Matlab. Note that a ramp response can be generated (e) Verify your answers to (a) and (d) using Matlab. Note that a ramp response can be generated as a step response of a system modified by an added integrator at the reference input. Figure DC Motor speed-control block diagram Step-by-step solution Step 1 of 16 Refer to Figure 4.44 in the text book for the block diagram of a DC motor speed-control. The controller transfer function is, D ,(5 ) = * , + i Consider the differential equation of the DC motor. + 60;> = 600v, - l.SOOw Apply Laplace transform on both sides. » r(i)+ 6 0 1 '(s ) = 6 0 0 F .(i)- l,500»'(j) Consider the armature voltage value in PI control. Apply Laplace transform on both sides. Step 2 of 16 Calculate the transfer function of the PI controller. ( i + 6 0 ) r ( i ) - 6 0 0 ^ - * , , £ ( j ) - ^ £ ( j ) j - l , 5 0 0 » '( s ) {s + 6 0 )Y (s ) = 6 0 0 ^ ^ * , j £ ( j ) j-1 ,5 0 0 » '(* ) ( j + 6 0 ) r ( j ) + 6 0 0 ^ ^ * , + ^ j £ ( j ) j = - l,5 0 0 » '( j) ^s + 60 + 6 0 0 t , + 6 0 0 ^ j £ ( s ) = -l,5 0 0 » '(» ) -1,500 ^ i+ 6 0 + 600*, + 6 0 0 ^ j -1.500s ( j “ +60s + 600*,j + 600*,) Consider the value of R as zero and write the value of error detector. £ ( » ) — l'( s ) 1,500s »'(s) s '+ 6 0 (l + 101:,)s+6001:, Step 3 of 16 The roots of characteristics equation is - ^ + 6 0 J y - 6 0 - 6 0 J ■ Write the characteristic equation from the roots. (s+60+60y){s+60 -6 0 y ) = s’ + 120s+7200 Write the general characteristic equation. 5*+2C o ^ + o),* « 0 Calculate natural frequency from equations (7) and (8). « ),= V 7 i2 0 0 < t t= i0 j2 Step 4 of 16 Calculate damping ratio, ^ . 120 2 ( 6 0 j2 ) 5 = 0.707 Calculate integral constant k, ■ 600k, -(6 0 > ^ )“ k,^12 Step 5 of 16 Calculate proportional constant k ^ . 60(1+1 Ot, ) = 2(0.707)(60V2) 60 2 (0 .7 07 )(6 0^ ) ^ lO/t, 60 * , = 0.1 Step 6 of 16 (a) To determine the transfer function with reference input, take disturbance input, » '( s ) as zero. Determine the transfer function from the output to the reference input. 1+600 6 0 0 ( V + it ; ) s(s + 60) ♦ 600(^^ + k,) Step 7 of 16 Determine the steady state error. £ (s ) = « ( s ) - l '( s ) I y (» ) £ (s ) £ (s) I 600(*,s + * ,) s(s+60) + 6 0 0 (*,s + *,) s(s + 60) s (s + 60)+600( t ,s + * ,) j ( j -f 60) £ (s ) = g(s) s(s + 60) + 600(k^ + k,) Step 8 of 16 The unit-step reference input is, I £ (s ) = l Determine the steady-state emor to unit-step reference input. = lin i5 £ (^ ) + 60) 1 s lim j 5 (j + 6 0 ) + 6 0 0 ( it^ + /t,) 5 Thus, the steady state error to unit step reference input is . Step 9 of 16 (b) The unit-ramp reference input is. Determine the steady-state emor to unit-step reference input. g |im j£ ( f ) g (j+ 6 0 ) 1 s lim ; g ( g + 6 0 ) + 6 0 0 ( it^ + * ,) 4 ^ (j+ 6 0 ) = lim - • 5 (g + 60) + 6 0 0 ( + Jfc,) 60 600A;, I " lo t , The value of k, is 12. 1 10(12) 1 *120 Thus, the steady state error to unit ramp reference input is Step 10 of 16 (c) The transfer function from the output to the disturbance input is, £ ( i) -1,500» B '( i) « '+ 6 0 (l + 10*,)s+600*, Substitute 12 for k, and 0.1 for k^- £ (£ )= . -l,5 0 te ^ (s ) j ’ +6 0 (l + 10x0.1)f + 600(12) -l,5 0 0 i j’ + 120j + 7200 The error function is. E (s ) = ~i-------------------fY(s) g’ +120g + 7200 The unit-step disturbance input is. Step 11 of 16 Determine the steady-state emor to unit-step reference input. e „^ \m sW {s) -hSOOs 1 slim g- g* + 120g + 7200g =0 Thus, the steady state error to unit step disturbance input is . Step 12 of 16 (d) The unit-ramp disturbance input is. Determine the steady-state emor to unit-step reference input. e„ = \m sE {s) -USOQg 1 s lim g - r <->a g* + 120g + 7 2 0 0 g * .. -1,5 00 ^ ................. « - » i' + 120f + 7200 1500 = -0.208 Thus, the steady state error to unit ramp disturbance input is |-0.2081- Step 13 of 16 (e) The error function is. g(s + 60) E (s) = s (s + 6 O )+ 6 O 0 {k p S + k , j( g + 60) ■ , ’ + (6 0 + 6 0 0 t , ) j + 600*, Write the MATLAB code to verify the steady state emor for unit step reference input. ki=12; kp=0.1; n=[1 60 0]; d=[1 60+600*kp 600*ki]; sys=tf(n,d); step(sys) Step 14 of 16 Get the MATLAB output for the step response. Step 15 of 16 The transfer function to disturbance input is. y ( j) -l,50to » '( i) +120s+ 7200 Note that the ramp response is obtained by added an integrator tenn to the transfer function. y (s ) -1.500s B '( s ) " s ( s ’ + 120$ + 7200) Write the MATLAB code to verify the steady state emor for unit ramp disturbance input. n=[-1500 0]; d=[1 120 7200 0]; sys=tf(n,d); step(sys) Step 16 of 16 Get the MATLAB output for the ramp response. Hence, the result is verified. Problem 4.34PP Consider the satellite-attitude control problem shown in Fig. where the normalized parameters are J = 10 spacecraft inertia, N m sec2/rad dr= reference satellite attitude, rad. 6 = actual satellite attitude, rad. Hy= ^ sensor scale, factor V/rad. Hr = 1 reference sensor scale factor, V/rad. w = disturbance torque. N m. (a) Use proportional control, P. with Dc (s) = kP. and give the range of values for kP for which the system will be stable. use piupuiuuiieii ia ^ (o/ —A/~, eiiiu ^ive me teiii^e l veiues i« system will be stable. (b) Use PD control, let Dc(s) = {kP + kDs), and determine the system type and emor constant with respect to reference inputs. (c) Use PD control, let Dc(s) ={kP + kDs), and determine the system type and error constant with respect to disturbance inputs. (d) Use IP control, let D c(s) = {kp + ^ ) , determine the system type and error constant with respect to reference inputs. (e) Use IP control, let Dc(s) = {kp + y ) , determine the system type and emor constant with respect to disturbance inputs. (f) Use IPD control, let Dc{s) = {kp-\- ^ ^nd determine the system type and error constant with respect to reference inputs. (g) Use IPD control, let Dc{s) = {kp + ^ + kps) >^^d determine the system type and e constant with respect to disturbance inputs. Figure Satellite attitude control -O 0 S te p -b y -s te p s o lu tio n step 1 of 9 J=10, H y=l, H j= l, ‘W=Disturbance torque Step 2 of 9 (a) From the block dis^gram of the sjrstem, Js’0(s)= ((6t.e)D + W ) >e(s)= '■ ' D+Js" ^ D+Js" E(s)=%(s)e(s) D+Js^ D+Js^ Step 3 of 9 Characteristic equation is D+Js^ = 0 i.e., 10s^-HCp=0 (Proportional control) By routh's criterion for stability, [kp > 0| and system is marginally stable with roots on jo) axis. Step 4 of 9 (b) D (s)^p-H cpS (PD control) c .= lim sE (s) lOs^ 1 =lim s - k ,4t„s+ 10s"^ 7s- 12. 1 But,as e,f= — (k^=Accelaration const.) ^"T o step 5 of 9 (c) D(s)=fcp4fcps E (0 _ -Y (0 W (s) W (s) = i(!l w (0 kfl+k^s+lOs >S3Tstemis of |Type-0| [Bror const: 4Cp j Step 6 of 9 D (s) ^ p + — (PI control) s lOs' E (0 = ^ (0 kp+ — +10s' s 10s’ T ><r(0 kj-HtpS+lOs' e_=lim sE(s) * S-kO ^ ' 10s- 1 10 s lim s k|-HcpS+10s^ System is of [Type -3] and error const | ^ . Step 7 of 9 W D ( s ) = k ,+ ^ E ( 0 _ _____ L k ,+ ^ + 1 0 s ’ k|-HcpS+10s [Type -Ij "With error const = | Step 8 of 9 (£) D ( s ) ^ p + — "Hc^s (PID control) s lOs^ — k ---------------7 rW kp+— lOs' sE (s)----------------- >-------w ^ ( s ) ^ ^kpS+4ki-HcpS^+10s^ ^ ‘ [Type -31 witti error const = — Step 9 of 9 (g ) D (s )^ p + -^ -K tp S ECO ________ ^C O k ,+ ii-H c„s+ 10s’ s [Type ^ with error const. = the differential equation of the system without feedback is ^ + 0 J 4 I ^ . denDc=(s+2. (10) Consider the value of b.0 0 2 5 * ’ *‘ +0. step(sysCL). -0 ..0 . 1)..3 2 ) H e a d in g ^ *’ (* + 0 .46 in the textbook and write the differential equation without consider the feedback path..0 ..0 .0 1 9 6 8 * + 0 .0 *‘ + 0.009)..0 6 4 tfl <** dr’ dr’ dr’ dr ) (3) Hence.. Are these reasonable values for a large tanker? Figure Ship-steering control system Rudder —N.0. (11) Consider the value of K.. Is the closed- loop system stable as shown?( Hint: use Routh’s criterion.(4) Substitute equation (2) in equation (4). 1 2 — -0 . -0 . with the plant transfer function relating heading changes to rudder deflection in radians. *=2 .0 0 2 5 * ’ .-----.) * ’ ') ^ sO (8) [+ 0 ..0 1 9 6 8 A :^ + 0 ...0 6 4 ) G (* ) = (2) i V 0 .2K j . (11) and (12).I 6 4 ( j ’ ..0 1 0 4 % „ 1+ JC. sys=tf(numP*numDc/(denP*denDc)).0 6 4 s ] dt* d t^ d t^ dt ' Step 2 of 8 (b) Write the general characteristic equation of the system.2 5 ) ( j.1 6 4 * ’ + 0. Step 6 of 8 Therefore only the proportional constant is |not sufRcientI bring the system from unstable to stable. 8 .0 ..1 2 ^ . .. .0 1 9 6 8 * + 0 . .I6 4 ( s ’ .1 6 4 ( t + 0 .. the closed loop is step response and its rise time is very short..0 . {10).164*(s+0. (12) Write the MATLAB program for design the dynamic controller of the form from equations (2). Figure 2 Step 8 of 8 From figure (2).25)*(s-0..0 .1 6 4 * ’ + 0. .0 .1 6 4 f^ .241*’ -0 . numP=-0. .0 0 9 ) Multiply out the denominator and numerator terms in equation (1).. So the system is lunstablel step 5 of 8 (c) Write the closed loop characteristics polynomial equation. sysCL=feedback(sys.0 ..8)''2.0 1 0 4 9 6 8 :.0 .0 ..0 1 9 6 g * + 0 .* + 0 .0 0 2 5 * ’ Substitute 1 for f f ( s ) in equation (5.0 .0 0 2 5 .002)''2. .0 1 0 4 9 6 i:. 0 0 2 5 ^ = -0 ... all values in first column of Routh’s array are not positive.. 2 4 1 s ’ -0. ^ + 0 .0 0 2 5 * ’ *‘ +0.. ' *‘ +0. Step 7 of 8 (d) Consider the value of 0 ^ ( 4 ) • Consider the value of a.0 1 0 4 9 6 ( 1 ).1 2 * .. 0 0 2 5 ^ ~ ... Apply Routh-Hurwitz criteria in equation (6).1 2 j . numDc=25*(s+0. 2 4 1 ^ ..0 .0 1 0 4 9 6 „ .. g ( * ) = o ...2 4 U ’ -0 .. • The first column of Routh’s array should not possess any sign change...O ..241** -0 ... 3 2 ) (1) » * ( * + 0 .1 6 4 * ’ + 0. all coefficients of characteristics equation are not same sign.0 .1 6 4 8 :...009) r * L _ Step-by-step solution step 1 of 8 (a) Refer figure 4. .ff(t) = 0 -(5) *‘ + 0.1 6 4 * ’ + 0 ... (a) Write the differential equation that reiates the heading angle to rudder angle for the ship without feedback. K ^ 2 S . angle . From equation (8). • All the terms in the first column of the Routh’s array should be positive sign.0 .(7) -0 .0 . Figure 1 Step 4 of 8 From the Figure (1).. Hence these values are [nnfealistic for the large tanker]. = 0 ['*‘ + 0. . Problem 4.0 1 9 6 8 8 :.* ’ + 0 . The output of MATLAB program is given below.) (c) Is it possible to stabilize this system by changing the proportional gain from unity to a lower value? (c) Is it possible to stabilize this system by changing the proportional gain from unity to a lower value? (d) Use Matlab to design a dynamic controller of the form Deis) = K ^ closed-loop system is stable and in response to a step heading command it has zero steady- state error and less than 10% overshoot..1 6 4 A :. ... I 6 4 ( ^ . ■ .. denP=s''2*(s+0. i + a: ..2 5 X * . Such a control system for a large tanker is shown in Fig..0 1 0 4 9 6 = 0 -(B) Step 3 of 8 The system is stable if the equation satisfies the following condition.2)*(s-0..0 0 2 5 * ’ ^ (4 -® . a = 0.0025»* Write the differential equation that heading angle to ruder angle for the ship without consider the feedback path by apply Laplace transform.I 6 4 ( j -f 0.0 6 4 ) *’ + 0 ... 2 ) ( j . 1+.002 ..0 .. (b) This control system uses simple proportional feedback with the gain of unity.0 .32).0 . l + G ( s ) / f ( s ) = 0 . .241*’ -0 .01968* + 0 .241*’ -0 ...241*’ + ( .0 .241** -0 .0 0 2 5 * ’ -0 .35PP Automatic ship steering is particularly useful in heavy seas when it is important to maintain the ship along an accurate path. ^ = 1 .2 Step 8 of 8 For PID control.(b).5 .8 7 5 1.556. The time delay and slope of the transient response may be determined from the figure. K^=3.(a) where the input into the system is stock flow onto the wire and the output is basis weight (thickness).2 For PID control. k p = ^ ^ = r6 ] Step 3 of 8 For PI control. and PID-controller parameters according to the Ziegler-Nichols ultimate sensitivity method. We get. PI-. Figure Paper-machine response data II ic —i^ioi luio ului IIdle oci loiuviiy 111011100.28125 “ 8 The control is given by D (s)=Rp 1+— +Tps V.K. (b) Using proportional feedback control.25 (From unit impulse response g r ^ h ) Step 6 of 8 For proportional control.36PP The unit-step response of a paper machine is shown in Fig.=5.3 Step 4 of 8 1. Figure Paper-machine response data m e (mc ) H m efee (I) 0>) Step-by-step solution step 1 of 8 (a) From the step response vs.2 ' RL 1 ^ 5L=0.278 Step 7 of 8 For PI control. Determine the proportional-.6.45. Rp=0.=8. k„= — = ls!^ T i= — ^ ‘ 0. control designers have obtained a closed-loop system with the unit impulse response shown in Fig. =0. PI-. ) . k p * —^ « 7 .= ^ = U 2 5 T „ = is. time graph. K^=4. Problem 4. When the gain Ku = 8.= .8502 T. and PID-controller parameters using the Ziegler-Nichols transient- response method.556 Pq = 2. 6 L sl Step 2 of 8 For proportional control. RpsO.1336 T . Rp=0.5| Step 5 of 8 (b) K. (a) Find the proportional-. the system is on the verge of instability. t= 3 .044 as shown by the unit impulse response in Fig. Find the optimal PID-controller parameters according to the Ziegler-Nichols tuning rules.0201--------. R b=0.5I^|i] Step 3 of 3 (b) K„=3. T .1------- 1— r n Figure Unit impulse response Step-by-step solution Step 1 of 3 A = l.=S.37PP A paper machine has the transfer function <Hs) = 3 s+ l' where the input is stock flow onto the wire and the output is basis weight or thickness.6K„H1-82641 T. Problem 4.044 For PID control.= 2 L = 0 T„=0. Figure Unit impulse response 0. (a) Find the PID-controller parameters using the Ziegler-Nichols tuning rules. z < 3s=2 r= A=1 z 3 L=z<i^2 Step 2 of 3 (a) For PHD control.1--------.=lT75l 4 Tb = | = I . (b) The system becomes marginally stable for a proportional gain of Ku = 3. £ (»)+G -'(0 )J ? (» )] (6) Substitute equation (5) in equation (6).01:5.2551 .. Figure 2 Step 7 of 7 The figure 2 clarifies the influence of feedfonward control in eliminating the steady-state tracking error for a step output disturbance.t .. ( ..G ( i ) t . (5) step 2 of 7 Refer from Figure 4.. f(4 ) = J?(4)-y(5) .123 Consider the gain of the controller.123 G " ( 0 ) = i59. 15. the dc gain of the closed loop system is |zero|.(2) Consider the gain of sensor.. w..'( o ) » '( s ) ] y ( s ) .. t=0:. ylabel{'$y(t)$'. G W ] y ( s ) = [ i.49 in the textbook and write the transfer function equation with respect to zero Wvalue..'( o ) G M ] (12) W (s)~ [l+ t. t .G (j)] Step 6 of 7 Write the MATLAB program for disturbance rejection response with feedfonvard systems from equations (1). r ( j) = iy ( i) + G ( s ) [ * ... (b) Also add feedforward control to eliminate the effect of a constant output disturbance signal.01:5. yw1 =step(Tw1 . l ' ( j ) + G ( j ) G . xlabel{'Time (sec)'). ) = « W [ G W * . ) = » '( * ) + G W [ * .G . clc.( o ) « ( * ) ] r ( j ) = G (f ) * .( o ) G ( * ) ] iy W r(4 [ i... G=59.-fG -'(0 )lG (» ) (7) « ( i) l+ * . ylabel('$y(t)$'. xlabel('Time (sec)'). G ( j ) r ( s ) . ) y ( ..123). Figure 1 Step 4 of 7 The figure 1 clarifies the influence of feedfonward control in eliminating the steady-state tracking error...'interpreter'..) = (1) i"+6.y1).t): plot(t.. and (7). £ ( 4 ) = . nicegrid The output for MATLAB program is given below. Hence.r W ) + G .'( o ) G ( s ) ) r ( j ) [ N ...yw1). Tw1=(1-1/dcgain1*G)/(1+kp*G).G . kp=3. (a) Add feedfonward control to eliminate the steady-state tracking error for a step reference input.'latex'): The output for MATLAB program is given below.292/(s''2+6.. the addition of feedfonward control results in zero steady-state tracking error for a step reference input and the dc gain of the closed loop system is Iunity!. with proportional control.'( o ) J iW ] i'( 4 ) = G ( 4 ) [ M ( 4 ) . (10) Substitute equation (10) in equation (9).292 C -'(0 ) = 0. s=tf('s')..978s + 15.y M ) .. y1=step(T1. and (12).292/(s''2+6. 59.'latex')..G .t .r ( 4 ) . H{$) = \ . Figure Block diagram Step-by-step solution step 1 of 7 (a) Consider the gain of the plant. y ( . dcgain1=dcgain(G). ( « M . « ( j ) .'( o ) » '( j) ] (8) Write the output equation for error detector.t). kp=3. figureO plot(t. i'( 4 ) ) + c r .... £ ( j ) = j t ( j ) .49 in the textbook and write the transfer function equation with Wvalue.. £ ( j) + G ..38PP Consider the DC motor speed-control system shown in Fig. (4) Write the output equation for error detector.. dcgain1=dcgain(G). Hence. (9) Consider the value of J?(4 )- = 0 .G . = 3 . s=tf('s'). T1 =G*{1/dcgain1 +kp)/(1 +kp*G). G=59..123).'( 0 ) ] y(») f t ..G ( i) Step 3 of 7 Write the MATLAB program for tracking response with feedfonward from equations (1). .292 C ( . + G W G . (11) Substitute equation (11) in equation (8).' {0)R(s) ( l + G ( * ) * .'interpreter'. on the output of the system.978*s+15. Step 5 of 7 (b) Refer from Figure 4. y ( » ) = G W [ * . t=0:. y (« ) = G (» )[*.978*s+15.y ( 5 ) . Problem 4.. ( y ( i ) . (3) Determine the DC gain of the plant. 4 0 + 0 .7 8 (11) £ (z) z ’ .+ 0 .3 8 ) ( 4 0 .2 (z ’ .6 ) I /( z ) 4.l .3 0 ) ( 4 0 .^ .9 ) £ (z)” 4 4 ( z .1 9 9 .516.3 9 .5 9 9 7 ) 4 .801 z . 2 z -\ * “ 0 .909Z-1.1 .5 ( .0 .0 0 0 . Step-by-step solution step 1 of 4 (a) Consider the sample period.1 .7 ^ l.2 z ^ -3 .8 9 8 z + 0 .2 z * .0 5 s e c .01) l/(z ) Calculate — “ from equation (4) and (10)..6 Step 4 of 4 (d) m .2z* -7 . (1) Write the value of Z ).(10) ’ E {s ) (z + 10)(j+0.4 0 + 1 0 z + 10j ^ 4 0 z .72 ■(7) £ (z) z . l + GDci = 0.4 0 + 2 z + 2 j ^ 4 0 z .1 9 9 .5 9 9 7 I/(z ) 4 .8 Hence.1) [O) — J(..2z z+ +2 2" \j ^ 4 0 z-4 0 H =2 4 0 z -4 0 + 4 z + 4 j z+l 42Z-38 =2 4 4 Z -3 6 I /(z ) 4 2 ( z ..78 z ^ .6 z + 0 .05 sec in each case.0 .+|0)(s+ 0..^j)o r .1 9 9 .9 ) 2 .909Z-1. the value of discrete equivalent for given transfer function is 4.1 .8 z + l. £ (z) £ (£ l = 5 ± tij £ (z) z+10L.6 ) 4..0 5 z + l z = 4 0 i 4 ._ffl«-i +2 +4 + 2...6 z + 0 .5 9 9 7 .3 9 .5 z’ -3 . z +1 Substitute equation (1) in equation (3).5 9 9 7 Hence. Let Ts = 0.2 ( z . z tl 2 .. Problem 4. the value of discrete equivalent for given transfer function is 4.1 .1 ) I £ (z) ( i+ 1 0 ) ( i+ 0 .5.1 z .0 |) - Eq..5 (z ‘ .0 .2Z-3.9 7 z + 3 . the value of discrete equivalent for given transfer function is 1.6 Hence. (=) D c s W — fd) D ^ ( s ^ = 5 (j +2)(5-44). (3) t.9 ) ° ’ ( 5 0 z .. ( 4 2 z . the value of discrete equivalent for given transfer function is z+l Step 2 of 4 (b) £ (z ) ■(6) ’ £ (z) z+4 l/(z ) Calculate — “ from equation (4) and (6).0 1 j z+l . .2 £ (z) ’ 2.6 z + 0 ..„>+ +2 'F i l +10 l '4 0 z .6 z + l.M l# l/(z ) Calculate — “ from equation (4) and (8)..n rv e ( 5 + 2 ) ( 5 + 0 ... £ (z) £ l £ i = 2 £ ± l| £ (z ) z+4|.« 0 .0 .6 z + 0 . (b) D d (s ) = 2 ^ . (a) £>c1 {s) =(s + 2)/2.0 .1 0 ) ( .9 ) z ’ . (5) £ (z) z+ 1 2 IZ -1 9 Hence.8 ) I /( z ) 1..8 Step 3 of 4 (c) Write the value ■(8) a .0 1 z + 0 ..('*> Z+ 1 U (z ) Calculate — “ from equation (2) and (4).1 z + 0 .0. Write the formula for discrete operator..7 ..0 .0 .39PP Compute the discrete equivalents for the following possible controllers using the trapezoid mle of Eq.6 8 4 .0 .72 z .905) ( z .1 ) (d) O c4 W .0 0 0 .0 1 z .9 7z+ 3.6 8 4 ..1 .2 (z ’ .801 (9) £ (z)° z .. ? .9 9 ) t/( z ) l..4 0 + 2 z + 2 'l _____ £±1_____ L r 4 0 z -4400++l< 10z + 10j I z+l 4 2 Z -3 8 =5 SO z-30 I /( z ) 4 2 (z-0 .1 j ° ^ ^ 4 0 z .2Z-3.0 1 ) L _ ^ ^ 4 0 z .4 0 + 0 .9 0 S ) £ ( z ) " ’ 5 0 ( z . £ (z ) . £ (z) t/( z ) z -f2 | 2 4 0 z-4 0 + 2 z+ 2 =____ thJ____ 2 4 2 Z -3 8 ..8 9 8 z + 0 .01)* /. in Appendix W4. £ (z) U (z ) ^ (z + 2 )(z + 0 . 2 e ( i t ) .1 ) a ( i t ) = 0 . the difference equations to the discrete controller are |a(ifc) = 0 . respectively. H £ i= ll£ z li (1) E (z ) z+ 1 ' ' Modify equation (1).2Z-3.2 -7 .9 0 9 e (* )-1 .'t/ (z ) = 1.8 0 1 z . l/( z ) z ( 4 . Let Ts = 0.801 (7) £ (z )~ z .0 .0 ..78 ■(9) £ (z) z '-1 .1) = 4 . 8 0 1 z .') E {z ) ~ z ( l + r .. » ( * ) + u (t-l)= 2 1 « (* )-1 9 e (* -l) ii(* ) = K ( * .909£(z) .') U ( z ) ( l. 6 z ”') t/( z ) ( 4 .8 ' ' Modify equation (5).7 2 z . U {z ) 4. 2 £ ( z ) .6 Modify equation (7).8 z .6 a ( it.2 z (it)-7 .8« (* .2 .6 z -'£ /(z )+ 0 .3 . in Appendix W4. (a) £>c1 {s) =(s + 2)/2.5997z-=)£/(z) = (4 .0 .0 .8 0 1 z . 6 a ( i t ..3 . U {z ) 4.1 .0 .9 7 z+ 3.l) -0 .39P and write the discrete equivalent equation for the controller.9 7 z -'+ 3 .') £ ( z ) .7 8 e (it-2 ) Hence.9 7 z-'+ 3 .2 -7 .0 .1 .2z* -7 .2 e (it)-7 . Problem 4.l) + 0 .') = ( l.5 9 9 7 a (il-2 ) = 4 .l ) Hence.7 2 z (* -'i)|.0 .78z-^) £ ( r ) “ z* (l-1 .') U ( z ) ( l .5997z-=) (l.3 .0 .7 2 z-‘) E (z ) z ( l .8 0 1 Z -') £ (z)~ ( l . the difference equations to the discrete controller is |ir(it) = -ii(lfc-l)+ 2 1 e(lfc)-1 9 e(it-l)| Step 2 of 4 (b) Refer part (b) in the solution of 4.7 8 z (t-2 )] .2 ..5 9 9 7 z -'U (z ) = 4 . 1/ (z ) .l ) Hence. I / ( r ) _ 1. (b) Part 2.2 £ (z )-7 .6 B (* -l)-0 . Step-by-step solution step 1 of 4 (a) Refer part (a) in the solution of 4. (6) Convert z to /f form from equation (6).5 9 9 7 Modify equation (9).8 0 I z . write the discrete equivalent equation for the controller...8 a ( it-l)+ l. l/ ( z ) = « l ( t ) (4) Substitute equations (3) and (4) in equation (2). U {z ) z(l.... (8) Convert the equation from z to /f form.7 2 z-'£ (z) » ( * ) .9 0 9 e(t) .0 .2 z (£ ) .6 z-'+ 0. the difference equations to the discrete controller is |a(ifc) = l. (c) Part 3.9 7 e (it-l)+ 3 .5 9 9 7 a (it-2 )+ 4 .1 .2 .. (a) Part 1. (8) Convert z io k form from equation (8).9 0 9 .1 9 z ... 6 z .6 z -' + 0. I /( z ) z ( 2 l.72 £ (z) z . Eq l+ G D c / = 0.l ) + 2 1 « ( * ) .1 9 e ( * .1 . 6 z .1 ) + 1 . (d) Part 4 (d) Part 4 Compute the discrete equivalents for the following possible controllers using the trapezoid mle of Eq.7 8 z -’ ) £ (z) (l.l ) = l.05 sec in each case.1 9 z . l / ( z ) .6 z .8 z .9 7 e (il-l)+ 3 . 8 a ( * ..7 8 z -'£ (z ) U ( z ) ( l. the difference equations to the discrete controller is |a (it) = 0 .6 z + 0 .9 7 e (il-l)+ 3 .l ) (3) Consider the general form of U {z ).9 7z-'+ 3.'£ / ( z ) = 4 .5.2 z (it)-3 . (b) D t i( s ) = 2 ^ . a ( * ) .909z-1.1 .9 7 z -'£ (z )+ 3 . 6 z .3 .6 z -' + 0...5 9 9 7 a (it-2 )-t-4 .') = ( 4.40PP Give the difference equations corresponding to the discrete controliers found in Problem.7 2 « (t-l) a (* ) = 0.801e(lt .^ j .3 .2 -3 .9 0 9 -1 .39P.l ) 4 4 .0 .3.6 B ( il.') = ( 4 . (=) D c3( s) — 5 .7 8 e (it-2 ) a ( * ) = 1 .8 z .39P and write the discrete equivalent equation for the controller.') ( £ ( z ) ) (2) Convert z to /f form: Consider the general form of • z -'U (z ) = u { k .') l /( z ) = ( 2 1 .') 2 1 -I9 Z -' 1+Z-' ( l+ r .0 .1. Step 3 of 4 (c) Refer part (c) in the solution of 4.') £ ( z ) .39P and write the discrete equivalent equation for the controller.') £ ( z ) .2 e (it)-7 .6 a (it-l)+ 4 .2 -7 .7 8 z ^ )£ (z ) t/(z )-1 .9 0 9 z (* )-1 . U {z ) z’ (4 .5997z-^ ) U {z ) (4 ..8 0 1 e (it-l)| Step 4 of 4 (d) Refer part (d) in the solution of 4.0 .') £ (z) z ( l . 8 0 1 z ( i t .'£ ( z ) a ( it) .72«(* -1 ) Hence.6a ( £ . a * 4 V (4 ) + d<it^4*c(4)+ ^iky4c(4) * =- r b = s d {s)+ ^ ik ^ c (4 )+ A k. |N) versus j)arameter T. (5) Compare Equation (1) and (5)... the required parameters are....^ s + ^ j = 0 . ) + ^ | G] (gs(s)) = 0. a(s).. /.. = ( * + 7 ] '( ® ) Therefore. ^5 + —^ 0 ( 4) + ^ 4 * ^ c ( 4 ) j k ^=kk^A ^A b = * ( 4 + 7 : ] cW Therefore. ^+1 (2) Compare Equation (1) and (2). ^ 4 + i j c ( 4 ) sO ■ (9) ^4 + .3+iA (T s+1) = 0 (i) versus parameter/A. ( j + c )’ + y<(7i + l ) = 0 ( j + c ) ’ + . Give L(s).c(s)+ v4ifcfl4*c(4) .sc[s) b ^ s d ( s ) + A k ^ ( s ) + A k . [4</ ( 4 ) + ^ jfc.. Say why you can or cannot. the required parameters are. Problem 5. (4) Compare Equation (1) and (4). ) =0 'd ( s ) s d (s ) +— Akf + —j c ( j ) + — s^Ac{s [ 4 + . (i) versus parameter/A.. if possible. the required parameters are. Step 8 of 9 (iii) Rearrange Equation (7).. 4^4 + i j < / ( 4 ) + ^ * . (i) versus kp (ii) versus kl (iii) versus kD (iv) versus t Step-by-step solution step 1 of 9 (a) Write the general formula for the characteristics equation. 4 ( 7 i+ l) = 0 ( 5 + d ) + A T s + A —0 [ ( j + c ) ’ + 4 l]+ > l7 i = 0 .^ ^ 4*0 ( 4 ) Compare Equation (1) and (9). (1) The characteristic equation is. Therefore.. + —^ 0( 5) + ^ jH =0 (7) Compare Equation (1) and (7)... (iii) versus the parameter c. ] < /( * ^5 )+ Ak. <i = ( j + c ) k^A T 1 _____ T Step 4 of 9 (ii) ai doici isuo CLiuduui i ( s + c ) ’ + . Be sure to select K so that a(s) and b(s) are monic in each case and the degree of b(s) is not greater than that of a(s). Step 6 of 9 (d) (i) The characteristic equation is.. the required parameters are. = 0 . (a) s + {^/T) = 0 versus parameter r (b) s2 + cs + c + 1 = 0 versus parameter c (c) (s + c. (3) Compare Equation (1) and (3). a = (5 + c )* + i< k = AT bss Step 5 of 9 (iii) The parameter c enters in a nonlinear equation. A ^ ^ . d (s ) 'A ( s ) s d (s ) ( ts + 1)A( s ) U A * ^ ^ .jc ( 4 ) + . (8) *Ak...c { s ) + A k p S ^ c { s ) Therefore.4c( 4 ) + A k . k -e Therefore. the required parameters are. a^s * = i Step 2 of 9 (b) The characteristic equation is. Step 9 of 9 (iv) Rearrange Equation (7). (ii) versus parameter T. a^s r Therefore.. Ak„ k- T b ^ s ^ c [s ) Therefore.. = 4^4 + —j r f { 4) + ^ 4 * ^ c ( 4 ) + Ak^ ^4 + —^ c (4 ) k^A k. = ^4^5 + —j< /(4 ) + Ak. Assume that G {s) = w^i®re c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s).4c(4)] Compare Equation (1) and (10). a + * * = 0 .. so the standard root locus does not apply. a *(^ + c )’ k = AT b = s+ — T Therefore. and b(s) and the parameter K in terms of the original parameters in each case. (6) Substitute in Equation (6)..01 PP Set up the listed characteristic equations in the form suited to Evans’s rootlocus method. the required parameters are. Compare Equation (1) and (8). a = s^d{s)+ A k ^ ^ c [s )+ Ak.. c { s ) + A k^s^c ( 4 ) ] — *0 (10) + [ 4^ ( 4 ) + i< *^ * c (4 ) + i< *.4 7 . using a polynomial solver. Step 7 of 9 (ii) Rearrange Equation (7). a = s’ + l k -c b -s + l Step 3 of 9 (C) (i) The characteristic equation is.. a = ( 4 + c ) ’ + 4( k^A T b=s Therefore. the required parameters are. 5 * + c f+ c + l* 0 j ' + l + c ( i + l ) = 0 ./W ( . Can a plot of the roots be drawn versus c for given constant values of A and T by any means at all? (d) 1l + [| ^^ + **.. the roots are plotted versus c... the required parameters are. 667 Angles o f asymptotes are ±60 and -1 8 0 Angle o f departure is -90 Break-in(s') is -2. Step 3 of 6 w a (s) = s^ b (ff) = (ff+ 1) b(s) = ( s + l) Break-in(s’) is . Step 6 of 6 (f) a (s) = + 3 e ^ .5 i>(s) = s+1 Center o f asymptotes is -0. a rough evaluation of arrival and departure angles for complex poles and zeros. Step 2 of 6 (b) fl(s)=s'+0. Note that in Fig.{c) there are two poles at the origin.586 Sketch the root locus for the given pole zero plot Real Axis Thus.06 Breakaways) is 0:503 Sketch the root locus for the given pole zero plot Thus.97 Sketch the root locus for the given pole zero plot Thus. without the aid of a computer. so the results partly depend on what was dre amed up.3 Sketch the root locus for the given pole zero plot Thus. a ( j) where the roots of the numerator b(s) are shown as small circles o and the roots of the denominator a(s) are shown as ’<’s on the s-plane. and the loci for positive values of the parameter K. the root locus for the given pole zero plot is sketched.2 Sketch the root locus for the given pole zero plot Real Aids Thus. Problem 5. . Figure Pole-zero maps Step-by-step solution step 1 of 6 We had to make some numbers to do it on MATLAB.2f+l i(s ) = s+ l Angle o f departure is 135. the root locus for the given pole zero plot is sketched.634 Sketch the root locus for the given pole zero plot Thus.37 Breakawa3r(s) is-0.7 Break-in(s) is -4.43 Breakawayfis) is -0. the root locus for the given pole zero plot is sketched. the root locus for the given pole zero plot is sketched. Show your estimates of the center and angles of the asymptotes. Each pole-zero map is from a characteristic equation of the form I + i C ^ = 0. Step 4 of 6 (d) a (s) = +5e+ 6 i( s ) = Break-in(ff) is -2. w a(ff) = + s And b (s)= s+ l Break-in(ff) is -3. Step 5 of 6 (e) a ( s ) = s ^ + 3 s ^ + 4 s -8 Center o f asymptotes is -1 Angles o f asjm^totes are ±60 and 180 Angle o f departure is -56.02PP Roughly sketch the root loci for the pole-zero maps as shown in Fig. the root locus for the given pole zero plot is sketched. but the idea here is just get the basic rules right. the root locus for the given pole zero plot is sketched. 0. ■(1) The roots of the general form of an equation by the root locus method is. To find poles put denominator D (» ) = 0 j “ (** + l) ( i+ 5 ) = 0 .64325 Substitute the value -3.65)’ -5(-0.. is |±4S^| and is |±13S^|.1I25 = -0. 4 for n and 0 for m in the equation...65)" -6(-0. Substitute for r.85 for s in Equation (5). Step 4: Calculate the value of centroid (<r^)- _ (sumoffinitepoles)-(sumoffimtezeros) * (numberof fioitepoles)-(nuiiU>eroffiiutezeros) {“ ) .65 for s in Equation (5). the real axis is . So. • Locate the centroid on the real axis.. To find zero put numerator jv W = o . a: = -(-0. Step-by-step solution (a) Consider the characteristics equation.65)’ --0...85. Recall Equation (1). -1 .03PP For the characteristic equation 1 + ^ . ^ = 0 ds — ( j ^6 j *5 s* ) = 0 4 j ’ + 185*+ IO j = 0 s (4 s ’ + I 8 j + I0)=*0. -0. (3) The roots of the equation are 0. (6) The roots of the equation are 0. * 4 -0 = ±135« Thus.85l- Step 8 of 11 ^ Step 6: Procedure to draw root locus plot for the uncompensated system: • Take real and imaginary lines on X axis and Y axis respectively.. » sys=tf(num.64775 . Numberof path « Numberof poles »4 Step 3: Write the expression for the angle of asymptotes. • Locate the breakaway point on the real axis. and draw the asymptotes from centroid at an angle of 45 * and 135*.. (2) The roots of = 0 are called the zeros The roots of D(s) = 0 are the poles Step 1: Compare Equation (1) and Equation (2). 4 for n and 0 for m in the equation. the root locus diagram for the given system is verified Step 10 of 11 (d) MATLAB program for the root locus with the characteristic equation: » num=[1]. (5) Differentiate the equation with respect to s and equate to zero. {±1)1802 ^ 4 -0 = ±45* Substitute ±3 for r. K 1-f =0 j'+ 6» ’ +5i' K =-l i* + 6 s ’ + 5 j* X = .. 6. Thus....± 5 ..65)* -6(-0. Thus. the breakaway points are 0 and -3. the angle of asymptote.65)’ -5(-0.( 0 ) “ 4 • -I.2.. (d) Verify your sketch with a Matlab plot. F^nre2 Thus.. Thus. Thus.1785+1. the breakaway points are and |»3.5 Thus. . (b) Sketch the asymptotes of the locus for K - (c) Sketch the locus (d) Verify your sketch with a Matlab plot. there is no zero in the transfer function.65 and -3. • Mark the poles on the real axis.5807 The value of K is positive.. the root locus diagram for the given system is verified by MATLAB.( j * + 6 j ’ + 5 j = ) .den): » rlocus(sys) Step 11 of 11 The root locus plot is shown in Figure 2. 0^ _rl8(P a . and -5. K = -(-0. 1+.. (4) n —m Here.^ _____ = 0 . (b) Step 2: Consider the formula for the number of paths. l2 ( i+ l) ( j+ 5 ) (a) Draw the real-axis segments of the corresponding root locus. the value of centroid is b U (C) step 5: Consider the breakaway points.. Problem 5.65)“ -48. Number of poles is n Number of zero is m r = ± I. • Draw the root locus. » den=[1 6 5 0 0].( j ‘ +6»’ +5s=) X = .85 Substitute the value -0.± 3 . there are no arrival and departure angles in the transfer function. .10. .3. Hence. The roots of £>(a) = 0 are the poles.. Hence. ^ l80°-t-360»(/-l) n -m Where. Compare the Equation (2) and the Equation (3).. Step 24 of 26 The root locus plot is shown in Figure 7. • Mark the poles on the real axis. f8 0 ° -t-3 6 0 ° ( 2 .r = 0 (6) z(z+l)(j+5)(z+I0) Consider the number of poles and zeros from the characteristics equation. Sketch the root locus with respect to K for the equation 1+ KL(s) = 0 and the listed choices for L(s). Step 9 of 26 Consider the formula for the asymptotes. . and draw the asymptotes from centroid at an angle of 6cr • Draw the root locus. There is no complex zero or pole in the given function.1. After completing each hand sketch.9(y>|. . (7) 4(z+1)(j+5)(j+10) Consider the number of poles and zeros from the characteristics equation. The roots of the equation (4) are 0.l) 4 -0 = 135» Substitute 3 for /. Step 7 of 26 Thus. . Be sure to give the asymptotes. . the four poles are 0. . 4 for n and 2 for m in equation (5). Compare the Equation (6 ) and the Equation (3). Thus. . • Locate the asymptotes on the real axis. 4 for n and 0 for m in equation (5). . The roots of the equation (4) are 0. Step 18 of 26 Consider the following given function. Step 23 of 26 ^ Procedure to draw root locus plot. . • Take real and imaginary lines on X axis and Y axis respectively. the root locus is verified from MATLAB output.. I8 0 °+ 3 6 0 °(I-I) 4 -0 = 45* Substitute 2 for /. . l* K ^ = 0. . Step 17 of 26 ^ Procedure to draw root locus plot. there are no arrival and departure angles in the transfer function. Step 25 of 26 Consider the following given function. the four poles are 0. 4 for n and 0 for m in equation (5).33 Thus. To find poles put denominator D(s) = 0 .zeros a .5 and .10. f80°+360o(l-l) 4-1 =60» Step 10 of 26 Substitute 2 for /.1.1. 4 for n and 0 for m in equation (5).10. Number of poles is n Number of zeros is m / =i. Turn in your hand sketches and the Matlab results on the same scales. Root Locos Figures Hence.. rlocus(sysL) The root locus plot is shown in Figure 8 .4. • Locate the asymptotes on the real axis. Hence. the asymptotes is E H step 16 of 26 A Substitute 1 for /. sysL=(s+2)*(s+4)/{s*(s+1)*(s+5)*(s+10)). Thus. rlocus(sysL) The root locus plot is shown in Figure 4. 180°+360°(3-l) 4-1 =300“ =-60» Thus.10. | | 8Qe|and | .10. s=tf('s'). To find zeros put numerator JV(j) = 0 Step 15 of 26 A Thus.l) 4 -0 -3 1 5 « = -l3 5 » Thus. Thus. Step 4 of 26 s\- Procedure to draw root locus plot.60^1 ■ There is no complex zero or pole in the given function.5 and . i ( z + l) ( s + 5 ) ( z + 10) ua : 0. Step 21 of 26 (b) Substitute . there are no arrival and departure angles in the transfer function. for A (j)in Equation (1).10. the root locus is plotted for the given transfer function and it is shown in Figure7.+*f)'J2SjS.1. (7) z (z + l ) ( j+ 5 ) ( j+ 1 0 ) Consider the number of poles and zeros from the characteristics equation. [ 135**!. rlocus(sysL) The root locus plot is shown in Figure 2.5 and . The root locus plot is shown in Figure 3. . the root locus is plotted for the given transfer function and it is shown in Figure3.: -------------------------. the root locus is plotted for the given transfer function and it is shown in Figures..5 and . The roots of the equation (4) are 0. 4 for n and 1 for m in equation (5).1. and draw the asymptotes from centroid at an angle of 9 tt • Draw the root locus. j (« + 1)(4+5)( j + I0) = 0 . (4) The roots of the equation (4) are 0.(i) = 0 . the two zeros are . there is no zero in the transfer function.. The root locus plot is shown in Figure 5. the asymptotes is E 3 Consider the formula for the angle of asymptotes. • Mark the poles on the real axis. the angle of asymptotes are |9(y>|and | . . • Take real and imaginary lines on X axis and Y axis respectively. Consider the formula for the asymptotes. V " * ’’ a — 4J3 VV Vv ■Ov ■*45 •30 -2S -20 -IS -to -5 S to IS XnlAiit(ftc<aO>4) Figures Hence. . |-4 5 ° |and | . i ( z + l)(x + 5 )(z + I0) 1+ J f .04PP Real poles and zeros. and draw the asymptotes from centroid at an angle of 4r • Draw the root locus.10. the asymptotes is E 3 step 22 of 26 ^ Substitute 1 for /.2 and . the root locus is verified from MATLAB output. . rlocus(sysL) Step 19 of 26 The root locus plot is shown in Figure 6 . Step 14 of 26 (c) (s+2)(j+4) u a : =0. 1+ A 2. Problem 5. the one zero is . Hence. n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoles)-(numberoffinitezeros) (0-l-5-10)-(-2-6) (4)-(2) = -4 Thus. Step 8 of 26 (b) (4 ^ 3 ) for L ( j) ir i Equation (1). 4 for n and 2 for m in equation (5). • Take real and imaginary lines on X axis and Y axis respectively.135**|.Ceam of the «yni|]Mes yy y/ . the root locus is verified from MATLAB output. • Locate the asymptotes on the real axis.10.. the root locus is verified from MATLAB output. 180°+360«(1-1) 4-2 = W> Substitute 2 for /. 4 for n and 1 for m in equation (5). .4 ) ------- ' ' i( i + l ) ( s + 5 ) ( s + 1 0 ) Write the MATLAB program to obtain root locus.9(y*|. Compare the Equation (6 ) and the Equation (3). (a) Us) - (b) Us) = (<-■) Us) = .5 and . Figure 1 Hence.5 and . .. Figure? Hence. £ (*) = - z(z+l)(s+5)(s+10) Wnte the MATLAB program to obtain root locus. Thus. venty your results using Matlab. To find zeros put numerator JV(j) = 0 Thus. Figure 4 Step 13 of 26 Thus. Step 26 of 26 Thus. = 0 . . there are no arrival and departure angles in the transfer function.1. and draw the asymptotes from centroid at an angle of • Draw the root locus. l8 0 * + 3 6 0 » ( 4 . n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoles)-(numberoffinitezeros) (0-l-S-10)-(-2-4) (4)-{2) =-5 Thus. . 180“+360“(2-l) 4-2 =270“ —90“ Thus.5 and . (3) D(s) Where.. i8 0 » + 3 6 0 » ( 3 . the angle of asymptotes are |6()v|. i ( j ) = ------------------------------- ' ' i(i+l)(s+5)(s+10) Write the MATLAB program to obtain root locus. Step 12 of 26 Consider the following given function. n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoies)-(nuinberoffinitezeros) (0-l-5-l0)-(0) (4)-(0) =-4 Thus. Consider the formula for the asymptotes.<. • Take real and imaginary lines on X axis and Y axis respectively.1. IUMLocm 30 K-poles o . Compare the Equation (6 ) and the Equation (3).2. s=tf('s'). 4 for n and 2 for m in equation (5). • Locate the asymptotes on the real axis. To find zeros put numerator N(s) = 0 Thus. • Mark the poles on the real axis. the angle of asymptotes are |90<>|and | . Step 6 of 26 Consider the following given function. 180“+360“(1-1) 4-2 =90“ Substitute 2 for /. Step 11 of 26 -rv Procedure to draw root locus plot. sysL=(s+3)/(s*(s+1 )*(s+5)*(s+10)). A V . V (s+2)(i+6) r . s=tf('s').. (s*2){s+ 6) ' ' i(i+l)(s+5)(s+10) Write the MATLAB program to obtain root locus.. s=tf('s').1.. To find zeros put numerator N(s) = 0 Thus. 180»+360°(2-l) 4-2 -270“ — 90“ Thus. 180°+360°(2-l) 4-1 =180» Substitute 3 for /. • Mark the poles on the real axis. (2) a(5+I)(j+5)(a+IO) Consider the roots of the general form of an equation by the root locus method. the four poles are 0. sysL=(s+2)*(s+6)/{s*(s+1)*(s+5)*(s+10)). . sysL=2/{s*(s+1 )*(s+5)*(s+10)). Step 2 of 26 ''V Consider the formula for the asymptotes. n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoies)-(numberoffinitezeros) (0-l-5-l0)-(-3) (4)-(l) =-4. the root locus is plotted for the given transfer function and it is shown in Figuret. 4 for n and 0 for m in equation (5). the angle of asymptotes are |4 S«»|.. . a(a+l)(s+5)(a+IO) 2 l+ J t. 4 for n and 2 for m in equation (5). Step 20 of 26 Thus. the two zeros are .2 and . i ( j ) = ------(4 + 2 ) ( 4 -i.6 . and the arrival and departure angles at any complex zero or pole.ji-m Step 3 of 26 Substitute 1 for /. The roots of JV(j) = 0are called the zeros of the problem.ii)) Step-by-step solution step 1 of 26 (a) Consider the general form of characteristics equation.5 and . There is no complex zero or pole in the given function. 4 for n and 1 for m in equation (5).l) 4 -0 «225« = -45» Substitute 4 for /. the four poles are 0. Step 5 of 26 The root locus plot is shown in Figure 1.(1) Substitute for L ( i) in Equation (1). Consider the number of poles and zeros from the characteristics equation. There is no complex zero or pole in the given function.. the asymptotes is E n a Substitute 1 for /. s^ 1 0 s’ 8 3K s’ 8 E ’. > 0 8 Therefore. . . W e have to substitute K = 0 into the equation.2 4 Characteristic equation: s*+ S ^+ K = 0 As all the coefficients o f the characteristic equation are not present. a Asymptotes: 45®. Step 15 of 23 L (S ): s ^ (s + 4 ) Step 16 of 23 K = 0 points: 0. 0. t ( j) = '5+S) r (1+1)^ . 0.i. . a. 0. a. in order to get point at w hich the root loci intersect the imaginary axis. 180®. 108®. 0.6 .3 ^ 0 8 s“ 3K Step 12 of 23 For system stability. a. ot. 300® Centroid = = z l = -0 . 0.e . a Asymptotes: 60®. 225®. let us substitute K = 4 into the equatioa — s’ + i r = 0 2s’+ 4 = 0 s’ = . P r o b le m iltiple poles at the origin. .3 Asymptotes: 90®. a Asymptotes: 90®. Turn in your hand sketches and the Matlab results on the same scales. 5 Step 9 of 23 s+2 U s ): ^ (s + 8 ) Step 10 of 23 K = 0 points: 0. Sketch the root locus with respect to K for the equation 1 + KL(s) d the listed choices for L(s). 2 Characteristic equation: s ^ ( s + 4 ) + A r ( s + l) ^ = 0 f f * + 4 s ^ + J ^ ( s ^ + 2 s + l) = 0 Step 17 of 23 ^ s* 1 K K s’ 4 2K s’ AK-2K K 4 s’ E ’-4 E 0 XI2 s“ K Step 18 of 23 > 0 > AK K >4 ( 2) The condition which satisfies both the equations (1) and (2) is K >A Now. 8 s“+3JT = 0 8s“ = 0 s = 0 Therefore. Step 13 of 23 A s+3 L (S ): ^ {s + A ) Step 14 of 23 K = 0 points: 0.2 i! : s + ji: = o s’ + 2 0 s * .j!:) > A : ( 2 0 o o . 180®. 0. 0. . 0. 135®. -1 0 K = apoints: -1 . we can say that. a.2 i : s + j i : = o step 21 of 23 s’ 1 100 2K s* 20 K K 2000K-K AOK-K 20 20 s’ 200K-K^-ia0K K 2000-K s’ XY-WZ 0 r s" K step 22 of 23 2000-K ■Where fT = 20 X : 40K-K 20 r : 2000K-X^-7S0K 2000-K Z Step 23 of 23 From the above array. Step 19 of 23 s’ (s-l-lO) Step 20 of 23 K = 0 points: 0. a.3 3 3 3 Characteristic equation: s*+ A s^+ K {s+ 3 ) = 0 s*+ A ^+ K s+ 3K = 0 As the coefficient o f s^ term is absent. 180®.8 K = apoints: a.i: ) ^ 4 7 5 8 0 ^ -3 9 1 !:’ > 4xl0 * + J!:’ -4 0 0 0 4 0 i : ’ -5 1 5 8 0 ii:+ 4 x l0 ‘ > 0 . 0.i . 180®. Centroid = > z2 = _ . -1 0 . a Asymptotes: 60®.3 Asymptotes: 60®. ot.io o s ’ -i-Ji:s’ . .0 . . . we can say that the system is not stable. . the ro o t loci start &om the origin and they do not intersect the imaginary axis anywhere. verify your results u itlab. Be sure to give the asymptotes and the arrival and departure gles at any complex zero or pole.2 . a Asymptotes: 36®. 300® (-lO -lO )-(-l-l) Centroid = ' = -6 3 Characteristic equation: s’ ( s + i o ) V j i : ( s + i ) “ = 0 s’ (s’ -M 0 0 + 2 0 s )+ i:(s ’ +l-l-2s) = 0 s’ + io o s ’ + 2 0 s *-i-J i:s ’ . 324® Centroid = — = . 270® ^ -8 -(-3 ) Centroid = -----. 0.5 2 Characteristic equation: s ^ ( s + 8 ) + i: ( s + 3 ) = 0 s^+S s^+K s+3K = 0 Step 11 of 23 A.8 K = apoints: a.2 s= ±J^ . in order to determine. we can say that the system is unstable.i— i = -2. . After completing each hand sketch. K >0 39K^{\220-K)-K{2000-K) > 0 3 9 a : ’ ( 1 2 2 0 . K >0 — > 0 i.4 K = apoints: a. 270® -(-1 -1 ). As the roots are not complex conjugate. a. 0.1 .1 . angles o f departure o r arrival need not be calculated. 300® Centroid = — = . 315® Centroid = — = . the point at which. a.i.6 6 3 Intersection o f root loci on the Imaginary axis: Characteristic equation: s^+S s^+K = 0 Step 3 of 23 1 0 s’ 8 K s’ -K 0 8 s“ E Step 4 of 23 As the necessary condition o f Routh’s criterion itself is not satisfied. Step 5 of 23 I £ (S ) : ff^ (s + 8 ) Step 6 of 23 K = 0 points: 0. 0 .8 K = apoints: ot» ot* .4 K = apoints: -1 .? c rn « Step-by-step solution Step 1 of 23 U s ): ^ (s + 8 ) Step 2 of 23 K = 0 points: 0. Step 7 of 23 s*(s+ 8 ) Step 8 of 23 K = 0 points: 0. we can say that the system is unstable..1 . a. the root loci cross the imaginary axis. 252®.8 K = apoints: a. 56“ —24. Step37of37 Thus.ji.'^ ij -135“ 61-90“ 6. Besuretogivetheasymptotesandthearrivalahddeparture anglesatanycomplexzeroorpole.theangleofasymptotesare .. .l-tan-[i] -82. sumofangleofvectorto angleofarrival [ . Tofindzeros.(1) . •Locatetheasymptotesontherealaxis. 6.(6) step9of37 XV Comparetheequation(6)andequation(3). z*(z+2)(x+3) Comparetheequation(12)andequation(3). +360“(3-l) -225“ —45“ Substitute4fori. ^ ^ (x +1+7K x + 1 . sumofangleofvectorto angleofdepature[ 180“. -2-9J and-2+9J- Thus. riocus(sysL) TherootlocusplotisshowninFigure2. Root Locus Real Axis (seconds'*) Step 35 of37 XV Hence.74“ Considerthedepartureanglesinthetransferfunction.Aftercompletingeachhandsketch. TofindpolesputdenominatorDis)=0 s’(s+I0)(s+3+4/)(s+3-4y)=0.+«.0. TofindpolesputdenominatorD(s)=0 s’(s+IO)(j+2+9/)(i+2-9y)=0 .. —3—4yand— 3+4y Thus.^44l+6a) Step-by-step selution (a) Considerthegeneralformofcharacteristicsequation.93“ 6(=180“-lan-'^|j -126. . x*(x+2)(x+3) [(4 +l)‘ +l] ‘ ■^^x’V + 2 )(x + 3 ) “ ..5fornand2forrrrinequation(5). _________ s j j __________ s*(s+10)(s+3+4y)(s+3-4y) ■T=0. riocus(sysL) TherootlocusplotisshowninFigure8.5forrrand1forminequation(5).(9) Therootsoftheequation(9)are0.74“ 61-90“ 61-90“ Considerthedepartureanglesinthetransferfunction.. n-m (sumoffinitepoles)-(stimoffinitezeros) (mimberoffinitepoies)-(mimberoffinilezeros) -2-3-1-1+7-1-Jl 2 -5+2 2 Thus. TherootlocusplotisshowninFigure9.anddrawtheasymptotesfromcentroidatanangleof 60- •Drawtherootlocus.. l. ConsiderthefollowinggivenfunrXion.56] -431..t t n . 61=180“-lan-'^ij -126.therootlocusisplottedforthegiventransferfunctionanditisshowninFigures.(2) 5(s-l.ofor7(«1inE^quation(1). n-m (sumoffinitepoles)-(stimoffinitezeros) (ntimberoffinitepoies)-(mimberoffinitezeros) -IO-2+9J-2-9y-[-3-2+8y-2+8y1 3 “ 2 Thus.. putnumeratorJV(a)=0 Thus. sysL=(s+3)“2/(s“2‘(s+10}*(s“2+(6‘s}+25)). 61=180“-[90“]+[l35+135+45+26. — 3— 4yand— 3+4y.thearrivalangleinthetransferfunctionis ats=— 2+87 step27of37 xv Proceduretodrawrootlocusplot: •TakerealandimaginarylinesonXaxisandYaxisrespectively.34*for 6( and 26. 61=180“-[102. Real Axis (seconds'*) Figure3 Hence.2 .. Thus. ( .therootlocusisverifiedfromM ATLABoutput..4forrrandIforrrrinequation(5). 61 =180“. .0.). Therootsof Dis)=0arethepoles.theasymptotesis EEI! Substitute1for). Considerthefollowingequation.+2f+2) (c) u n = .5forrrand1forminequation(5). Determinethenumberofpolesandzerosfromthecharacteristicsequation.2“ Thus.fio thecomplexpoleA fiomacomplexAJ Jsiimofanm otherpoles gleofvectoistothe] [complexpoleAfiomzeros j 61 = 180“-[61+61+61+61]+61+61 Substituteall ffvaluesinequation.4fornand2forrrrinequation(5). •M arkthepolesontherealaxis. l80°-i-360°(f-l) n-m W here.86“]+[l04.y ) Considertherootsofthegeneralformofanequationbytherootlocusmethod.m Substitute1fori. . — 1—j and— l-i-j.+6!.56“ « .theasymptotesis E3 Considertheformulafortheangleofasymptotes.'^ i j = 126. 61=180“-[90“+82. Step5of37 -o Proceduretodrawrootlocusplot: •TakerealandimaginarylinesonXaxisandYaxisrespectively. ConsiderthefollowinggivenfunrXion. 180+360(2-1) 3 3 -180“ Substitute3fori. TofindzerosputnumeratorJV(x)=0 Thus..5fornand3forrrrinequation(5).. D(s) . (*+3M.. 180“+360“(1-1) 4-2 =90“ Substitute2for).[i] Considerthedepartureanglesinthetransferfunction.93“ 61-90“ 9 . TofindpolesputdenominatorDis)=0 Thus. 180“+360”(1-1) 5-1 =45“ Substitute2fori. Considerthefollowingequation.-10. ConsiderthefollowinggivenfuntXion.]+61 Substituteall ffvaluesinequation.theangleofasymptotesare1^^.h m .93“ 6(=180“. s‘(s+10)(s’+6t+25) s*(s+10)(s“+6s+25) .03“ 61-90“ 61-90“ 61-90“ f.. 180“+360“(2-l) * “ «_i Substitute3fori.5fornand2forrrrinequation(5). r(44*)’ 4 i] s‘(s+2)(z+3) M ATLABprogramtoobtainrootlocus: s=tf('s').-ten-[i] -26. (j +3)( x4+4x +68) i( x ) .(4) Therootsoftheequation(4)are0. sysL=(s+3)/(s*(s+10)‘(s''2+(2's)+2)).5fornand2forrrrinequation(5).thedepartureangleinthetransferfunctionis I0L5^ats=— 3+4y step12of37 xv Proceduretodrawrootlocusplot: •TakerealandimaginarylinesonXaxisandYaxisrespectively.5 .-10.60“ Substitute2fori.thecomplexpoleA fiomacomplexAJ Jsiimoffio motherpoles angleofverXoistotheJ [complexpoleAfiomzeros j 61=180“-[«.therootlocusisverifiedfromM ATLABoutput. 180“+360“(2-l) 4-2 540“ 2 =270“ =-90“ Thus. r’(j+10)(j’+4j+85) M ATLABprogramtoobtainrootlocus: s=tf('s'). = B n ... Considertheformulafortheasymptotes. 61=180“-lan-'^|j -102. Therootsof jy($)=Oarecalledthezerosoftheproblem.SketchtherootlocuswithrespecttoKfortheequation1+KL{s) =0andthelistedchoicesforL(s).anddrawtheasymptotesfromcentroidatanangleof 9V •Drawtherootlocus.therootlocusisplottedforthegiventransferfunctionanditisshowninFigures. 1— +4s+68)-forDs)inEquation(1).0 7 P P 5 . ComparetheEquation(2)andtheEquation(3).4fornand2forrrrinequation(5).-10.fio thecomplexpoleA fiomarxnnplexAJ Jsiimofanm otherpoles gleofvernoiatothel [complexpoleAfiomzeros J 6[.[l] -29.+61+61] Substituteall ffvaluesinequation. Step22of37 . 61 =180“.theasymptotesis j^^33j.-2and— 3. i(.'^ l j -135“ 61=180“.03“ 61=180“. ___ s+3___ '*'“s(j+10)(s’+2s+2) M ATLABprogramtoobtainrootlocus: s=tf('s').0.thedepartureangleinthetransferfunctionis|^^I002^ats=— 2+9y... 61=180“-tnn-'^ij . . sysL=(s+3)/(s''2‘(s+10)'(s''2+(6*s)+25)).-10. 61„=180“-[126. TofindpolesputdenominatorDis)=0 s(a+10)(s-i-l+y)(s+I-j)=0. riocus(sysL) TherootlocusplotisshowninFigure6. 180“+360“(4-l) 5-1 =315“ =-135“ Thus.03“+90“+45“+(-90“)] =260.anddrawtheasymptotesfromcentroidatanangleof 60- •Drawtherootlocus.therootlocusisverifiedfromM ATLABoutput. TherootlocusplotisshowninFigure3.-10.. ConsiderthefollowinggivenfunrXion.. Root Locus Real Axis (seconds'*) Figure1 Thus.]+[61+61+61+61+«. Considerthefollowingequation.56“ Thus.. .(j+3) 77—.]+6) Substitute 135“for61. Considertheformulafortheasymptotes.4forrrandIforrrrinequation(5). 180“+360”(1-1) 5-3 =90“ Substitute2for).(8) 5*(s+10)(s+3+4y)(s+3-4y) Comparetheequation(8)andequation(3).. •Locatetheasymptotesonthelealaxis. .52“+90“+48.Jo+iS!S+iSi+ai .0.thecomplexzeroA fiomacomplexAJ Jsumoffio motherpoleszeros angleofvectorstothe[ [complexzeroAfiompoles J 61 =180“ -[61]+[61+61+6.74“]+90“+90“ Thus.4forrrandIforrrrinequation(5). sumofangleofvectorto angleofdepature1 180“. j ’ (i +I0)( s’ +6!+25) (443)’ l+ K - s'(s+10)(s“+6s+25) ...la n .52“ l%=180“-tan-'[|j -102. (0) Substitute (443)’ forL(s)inEquation(1).thepolesare0. •Locatetheasymptotesontherealaxis.34“]+26.52“+102. .anddrawtheasymptotesfromcentroidatanangleof 9<r •Drawtherootlocus.therootlocusisverifiedfromM ATLABoutput. (x+3)(x* Substitute ---. TherootlocusplotisshowninFigure5.therootlocusisplottedforthegiventransferfunctionanditisshowninFigurel.-10. |— 45“|andj_^y3^.anddrawtheasymptotesfromcentroidatanangleof 45- •Drawtherootlocus.87“ -45“ Considerthearrivalangleinthetransferfunction. d+3 j(s+I0)(s+l-ky)(5-f1-y)for inEquation(1)./ ) .93“+90“+29..(7) Therootsoftheequation(7)are0.25i Substitute1fori.thecomplexzeroA fiomacomplexAJ Jsumoffio motherpoleszeros angleofvetXoistothe[ [complexzeroAfiompoles J 61=180-J61+«.thefourpolesare0.-n n .la n .-10. Substitute —zc-. — \—j and—1-t-y.(3) W here. TherootlocusplotisshowninFigure1.'^ i j -104.[ i] -26. Thus..5forrrand1forminequation(5). . ConsiderthefollowinggivenfuntXion.10)(j +1 -hy)(5 +1 ..03“+104. •Locatetheasymptotesonthelealaxis.tt n .0. 180+360(3-1) 3 _300 3 --60“ Thus.65“ -48.theangleofasymptotesare |9Q^and .. Substitute1fori.%= B n . Step34of37 XV Proceduretodrawrootlocusplot: •TakerealandimaginarylinesonXaxisandYaxisrespectively..therootlocusisverifiedfromM ATLABoutput. TofindzerosputnumeratorJV(s)=0 Thus. — 2+8yand— 2—8y.. sumofangleofvectorto angleofdepature[ . (a) — .thezeroare-3. •Locatetheasymptotesontherealaxis. 61 = 180“-[126.56“for 6( intheequation. Z ic lfL n-m (sumoffinitepoles)-(sumoffinitezeros) (ntimberoffinitepoies)-(numberoffinilezeros) -10-3+4y-3-4y-(-3) 4 -13 4 Thus. sysL=((s+1}"2+1}/(s“2'(s+2}'(s+3».-ten-[l] -4 5 “ 6. . n-m (sumoffinitepoles)-(sumoffinitezeros) (ntimberoffinitepoies)-(mimberoffinilezeros) -10-3+4y-3-4y-[-3-3] 3 -10 3 Thus. jlgg^and Considerthefollowingequation.thezeroare-3and. r.j £ ± 3 i i l 3 L .'^ l j -126.93“+126.-180“-[6!+6)+6!. _ j(n. 180“+360“(3-l) rr_o eangleofasymptotesarejgQ^.thepolesare0. a .therootlocusisplottedforthegiventransferfunctionanditisshowninFigures. Considerthefollowingequation.0. jlgg^and Considerthefollowingequation..theasymptotesis l-3. -----L j’(s+10)(s’+4i+85) = (4+3)(x*+4x+68) s'(i+10)(s“+4!+85) (x+3)(l+2+8y)(x+2-8y) j’(x+10)(s+2+9y)(s+2-9y) .] Substituteall ffvaluesinequation. sysL=((s+3}*(s“2+(4‘s)+68})/(s''2‘(s+10)‘(s''2+(4*s)+85))..thezeroare — 1+yand— 1— j. sumofangleofvectorto angleofarrival [ . Considertheformulafortheasymptotes.56“ Considerthearrivalangleinthetransferfunction.5fornand3forrrrinequation(5).. RootLocus RealAxis(seconds *) step20of37 xv Hence. .----.io(W) K**. •M arkthepolesontherealaxis.] Substituteall ffvaluesinequation. Thus.0 7 P P Mixedrealandcomplexpoles.'^ |j = 104. —3—4yand— 3+4y Thus.. Thus.-10.thezerois-3.. ..thepolesare0.fio thecomplexpoleA fiomacomplexAJ Jsiimofanm otherpoles gleofverXoistothe) [complexpoleAfiomzeros j 61„=180“-[61+6l. •M arkthepolesontherealaxis.74“]+90“ Thus. TherootlocusplotisshowninFigure7. riocus(sysL) TherootlocusplotisshowninFigure10. 61.52“ 61-90“ 61-90“ 61-90“ -83.78“ Thus. 90“for6(.thezerois-3. 180”+360“(2-l) ft “ 5-3 X X 540“ 2 =270“ —90“ Thus.thepolesare0. -2-9J and— 2+9J. Stunofangleofvectorto angleofdepature1 180“.k if- s-l-3 = » . 180“+360”(1-1) ^' k_o =60“ Substitute2fori. Considertheformulafortheasymptotes.theasymptotesis m Substitute1for)..93“+126.135“ ^-90“ .thearrivalangleinthetransferfunctionis|^^7L5^atJ=— 1+y.. 180“+360“(2-l) 5-2 540“ 3 -180“ Substitute3fori.. .'i -29. 6l=180“.(11) Therootsoftheequation(11)are0..+jC_--. n-m (stunoffinitepoles)-(sumoffinitezeros) (ntimberoffinitepoies)-(nuinberoffinilezeros) -10-l-y-Hj-(-3) 4-1 *3 Thus.thedepartureangleinthetransferfunctionis 13^^atx=— 3+4y step19of37 XV Proceduretodrawrootlocusplot: •TakerealandimaginarylinesonXaxisandYaxisrespectively. .5forrrand1forminequation(5). Numberofpolesisrr Numberofzerosisrrr / • l.65“] Thus.theangleofasymptotesarejg^andj^go^. 180+360(1-1) 3 3 .36“]+[90“+90“+83.93“ 61-90“ 61-90“ 9.t a n .+6l]+[«. riocus(sysL) TherootlocusplotisshowninFigure4. TofindzerosputnumeratorJV(x)=0 Thus. Considertheformulafortheasymptotes. TofindzerosputnumeratorJV(s)—0 Thus.ta n . RootLocus RealAxis(seconds *) Figure7 Hence. •M arkthepolesontherealaxis.0.=180“-[l35“+90”+6.3. (44-^) ' ’ s’(s+10)(s’+6s+25) M ATLABprogramtoobtainrootlocus: s=tf('s').thedepartureangleinthetransferfunctionis 24TO^ats=1+y.93“+90“+29.TurninyourhandsketchesandtheMatlabresultsonthesamescales.36“ Considerthedepartureanglesinthetransferfunction. — 3—4yand— 3+4y.' ' Comparetheequation(10)andequation(3). (443)‘ ' ' r ’ (j+ 1 0 )(j’ +61+25) M ATLABprogramtoobtainrootlocus: s=tf('s'). l-hiX(s).verifyyourresultsusing < Matlab.+61+«. jl3^. .0.therootlocusisplottedforthegiventransferfunctionanditisshowninFigure7. •M arkthepolesontherealaxis. Step 30of37 Thus. Substitute [(j+ L' Ox’ + ilJ forlis)inEquation(1). TofindpolesputdenominatorD(s)=0 s’(s+I0)(s+3+4/)(s+3-4y)=0. (j)• num=[1 -1]: den=[1]: sys=tf(num. = ..3.l) 3 = 120” Therefore.-™ n —m . verify your results using Matlab. = 1. = . 360”(1-1) "■ 3 = 0” .. P.l) [ ( » + 2 ) . = 1 . Step 7 of 17 (c) Write the expression for £ (s)- There are 2 poles and one zero.3 .6 8 * ) = -92.„ n -m ...l ) * ” 3 Therefore. . 360”( 2 . = 1. Step 9 of 17 ^ Write the MATLAB code to obtain root locus for Z.s te p s o lu tio n step 1 of 17 (a) Write the expression for £ ( i) - r/ \ ^+ 2 1 There are 3 poles and one zero.2 .-™ n -m .1 0 -(-2 ) 2 Step 3 of 17 Write the MATLAB code to obtain root locus for i ( j ) • num=[1 2]. Calculate the number of asymptotes.. » '. sys=tf(num.2.1 0 + l-l-(-2 ) 2 Write the MATLAB code to obtain root locus for L { s y Step 6 of 17 num=[1 2].. + ! .3 . P.1 5 * 0. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole.0.73y.den): rlocus(sys)... n represents number of open loop poles. Calculate the number of asymptotes. Write the MATLAB code for £ ( j ) • Step 14 of 17 ^ num=[1 2]. den=[1 10-1 -10 0]. Root Locus 0. What is the largest value that can be obtained for the damping ratio of the stable complex roots on this locus? ..2 0 .den): rlocus(sys). 360”( 2 . = l80*-140. Turn in your hand sketches and the Matlab results on the same scales.2 . den=[1 10-1 -10].. Calculate the number of asymptotes. * = 5 5 M . Calculate the number of asymptotes. the asymptotes are 120* apart from each other. Determine the intersection point of asymptotes on the real axis. 360”(1-1) 3 = 0” . P = 3 -0 -3 Determine the angle of asymptotes. a . P.1 • h .-™ n -m 360°(1-1) 2 = 0° 36 0 °(2 -l) 2 = 180” Therefore. sys=tf(num. the asymptotes are |g()® apart from each other.. (i-l)((i+ 2 F + 3 ] S te p -b y . P = 4 -l -3 Determine the angle of asymptotes. Root Locus Step 13 of 17 ^ (e) Write the expression for £ (s)- ' ' j(i-l)(i+ 6 )’ There are 4 poles and one zero.den): rlocus(sys)..^ t a n ..3 . .0 5 • .3 . After completing each hand sketch. P..RHP and zeros.68* The angle of departure from the conjugate pole l —J. 360”(1-1) 3 = 0” . P =3 -l s2 Step 2 of 17 ^ Determine the angle of asymptotes.den): rlocus(sys).. P=« -« Here..'^ ^ j + 9 0 * j = 180 *-I20* = 60* The angle of departure from the conjugate pole —2 —1..73J ■ = 1 8 0 * . P. den=[1 38 322 -720 800 0].. den=[1 11 24 -36 0].15 • -0.05 ‘ I ^ . the asymptotes are 120* apart from each other.+ 3 ] There are 3 poles and no zeros. Determine the intersection point of asymptotes on the real a)s n -m 3 . Step 15 of 17 Write the expression for £ (s)- I L {s ) = ( i.den): rlocus(sys). (c) L(f) = i (d) L {s) = —j— .2. 360”( 2 . = -60* Step 17 of 17 Write the MATLAB code for £ ( j ) • num=[1). P= 2 -I si Step 8 of 17 ^ Determine the angle of asymptotes..1 6 -3 .2 .1 • -0.45*-h 53. 360®(I-I) 3 = 0® Determine the intersection point of asymptotes on the real axis.13* =92. sys=tf(num.2. (a) L(s) = model for a case of magnetic levitation with lead compensation. P. Step 16 of 17 ^ Determine the angle of departure from the pole —2+1.-™ n -m . ^^ = -(9 2 .l6 . Calculate the number of asymptotes.. the asymptotes are 120* apart from each other.. P = 5 -2 s3 Step 11 of 17 ^ Determine the angle of asymptotes. . the asymptotes are 120* apart from each other..2 0 + \+ l. . = . .l ) "■ 3 = 120” Therefore. sys=tf(num. sys=tf(num.2. (b) L(s) = magnetic levitation system with integral control and lead compensation. = 1 . Calculate the number of asymptotes. Root Locus Step 4 of 17 (b) Write the expression for £ (s)- K \-_ £ ± 2 _ I There are 4 poles and one zero.IS lId L n -m 0 -2 0 ..{ -l-\) 3 = -I2 Determine the angle of departure from the pole i+ y .3. * = 5 5 M . ^ . m represents number of open loop zeros. = 120* Determine the intersection point of asymptotes on the real axis..2 s + 2 ) There are 5 poles and two zeros.^. n -m 0 .-™ n -m . den=[1 3 3 -7).68* Step 12 of 17 Write the MATLAB code to obtain root locus for i ( j ) • num=[1 2 1].. P = 4 -l »3 Step 5 of 17 ^ Determine the angle of asymptotes.den): rlocus(sys). 360”(1-1) 3 = 0” 360”( 2 . A = = i8 0 ” .l ) 3 = 120” Therefore.5 0 Real Axis (seconds**) (d) Step 10 of 17 ^ Write the expression for £ (s)- L( 1 +2s+l '* ^ ” i ( i + 2 0 ) ' ( j ’ .. *exp(-0.039 s 5+0.5 Step 16 of 23 MATLAB code to plot step response when a —0.t-(f) = 1 . estimate the closed-loop pole locations.01:5. j* + (2 -» -5 a )s + 5 s ’ + [ 2 + 5 ( 0 .den): step(sys).. in root-locus fonn with respect to the parameter a.5567. Figure Control system -O K s(s+2) S te p -b y . plotfty) title('step response when a=0'). for R{s) in the above equation.. 5 i+ 5 « 0 j .h 5 o ) s -i-5 s* + [2 + 5 (0 )]s+ 5 s* + 2 s -1-5-0 S i^ = . Step 18 of 23 ^ Calculate the closed loop pole locations when a —2 - + (2 + 5 a )5 + 5 ff^ + [2 + 5 (2 )]s + 5 j *+12 j +5 -0 s . y=1.4 3 2 )( s +11. 1+ J C ^ = 0 . 1 + G ( s ) f f { i) = 0 f ( « + 2 )+ 5 (1-I-as) s 0 s*+ 2 s+ 5 + 5 as = 0 (1) s* + 2 s+ 5 ' The above equation is in the form.4 3 2 . s* + 2s + 5 Therefore.5 . ^ (f)= « (/)-e"'co s(2 /)tt(f)-0 .*t). Step 11 of 23 The step response when ^ s 0 shown in Figure 3.den): rlocus(sys) Step 4 of 23 The root locus plot is shown in Figure 1.0404.'‘^ ' ] h (») step 20 of 23 ^ MATLAB code to plot the step response: t=0:0.j= .*exp(-2.SPtP |.2 . den=[1 2+5*a 5].-II. a ( s ) » s * + 2s + 5 A (s )> Ss Step 3 of 23 MATLAB code to plot the root locus: num=[5 0].5').0015-1. The step response is shown in Figure 2. Step 9 of 23 -A. the closed loop pole locations when a —0 ET±7g Step 7 of 23 The closed loop transfer function is.5 ) ] j + 5 l ’ + 4 .01:5.432.5.*sin{2. I 'M - Calculai the step response when a —0 - Calculate 5 I 'M - j ( 5 + 2 ) ( j + 2. s te p resp o B se w faea a= 0 Step 10 of 23 MATLAB code to plot step response when a —0 - a=0 . Step 15 of 23 The step response is shown in Figure 4. and 2. s (s + 2 ) rW - 5 s(s+ 2 )+ 5 + 5 a s 5 s*+ 2 s+ 5 + 5 as 5 s * + (2 + 5 a )s-f5 Step 6 of 23 ^ Calculate the closed loop pole locations when a —0- s* + ( 2 . /f(i) = l + Oi The characteristic equation is. Step 12 of 23 ^ Calculate the closed loop pole locations when g s 0 . and sketch the corresponding step responses when a = 0. s 5 r(3 )= j( i^ + 1 2 s + s ) Calculate the step response when a —0 ■ K (5 )- 5 ( j + 0 . num=[5].Se~^u ( f ) + A e '^ u (/) Step 14 of 23 MATLAB code to plot the step response: t=0:0. a(s).2 . Step 23 of 23 The step response when a —2 shown in Figure 7. title('step response when a=2').5. the values of £ ( s ) . I 'M 5 « ( i) 3’ +12 s+5 Substitute i. and identify the con'esponding L(s). y=1-exp{-t). den=[1 2+5*a 5].5. 5 Figure 5 The step responses in Figure 4 and Figure 5 are same.*t)+0.0 4 0 4 « . a ( s ) and ^^s) are. The step responses in Figure 2 and Figure 3 are same. den=[1 2+5*a 5].den): step(sys).039« .5|- Step 13 ot 23 The closed loop transfer function is.5'). stq > r e s p o n s e w k e n a= 0 . and b(s). sys=tf(num.5S67|.y) title('step response when a=0.*t)+4.*t): plot(t. 0.j= -0 .l± J 2 Therefore.. Step 21 of 23 ^ The step response is shown in Figure 6. 0 0 1 5 .0015 1. Sketch the root locus with respect to the parameter a.den): step(sys). sys=tf(num.*t): Plotfty) title('step response when a=2').01:5.52 in the textbook. y=1-5. (2) a{s) Step 2 Of 23 Equate equation (1) and (2).5 Apply inverse Laplace transform.5 s + 5 ! — for X (s) in the above equation.5) _ I ^ ___5 ^ 4 s s+2 5+2. sys=tf(num.l .5 e"'sin (2 r)tf(/) step 8 of 23 MATLAB code to plot the step response: t=0:0.039.432 5+11. the closed loop pole locations when a —2 |-Q. n s )_ s R{s) s*+ 2s+ 5 Substitute — for X (s) in the above equation. s te p re s p o n s e w h e n a= 2 Figure 7 The step responses in Figure 6 and Figure 7 are not same. a ( s ) = s* + 2s + 5 A(s) = 5s The open loop transfer function is. R e a l A xis (secoods*^) Figure 1: Root locus Step 5 of 23 ^ Calculate the closed loop transfer function.*cos(2.5 Therefore.*exp(-11.039e‘ " '“ ’'i((() = [ l .. —2.. title('step response when a=0').5 - a=0.*t)-0. Use Matlab to check the accuracy of your approximate step responses.432>-11. title('step response when a=0.*exp(-t). den=[1 2 5].Put the characteristic equation of the system shown in Fig.5567) . s te p re s p o n s e w h e n ^ 0 .5 5 6 7 Therefore.5567 Apply inverse Laplace transform.*exp(-2. s Calculate the step response when a —0 - s ^ s*+ 2 s+ s) 1 s+ 2 ( s + I ) " + 2* 1 s+ 1 1 s (s + l) * + 2^ ( s + l)* + 2 * Apply inverse Laplace transform. num=[5].0404 0.2 . ) = « ( /) .s te p s o lu tio n step 1 of 23 Refer to control system in Figure 5.5 is shown in Figure 5.5.0404«-*""ii(t)+ 0. s t ^ r e s p o n s e w h e n a= 2 Step 22 of 23 MATLAB code to plot step response when a —2 - a=2 . Step 19 of 23 The closed loop transfer function is.0 0 15ii(/)-1. 5 J l( j) j* + 4 . sys=tf(num. num=[5].^ +0. .. . Step 17 of 23 ^ The step response when g s 0. the closed loop pole locations when f f = 0. .5. rltool(sys) Step 2 of 4 ^ Execute the code at the MATLAB command window and study the behavior of the root locus. w/ \ ________ Js+a T « ______ ' * ^ " i ( j + l ) ( i ’ + 8 s+ 5 2 ) Determine the characteristic equation of the system.5:10. r r n n OotidleeedutoV c r n r r □ n OotidUid^iev n r n r r n n OotidImp ntoir n r: n r' r' n n Oodiloopl n r r r r n n ComptAtMotC nn n r r n n n r n r r n n RIMO r n r r r n n SmotH Step 4 of 4 Get the MATLAB output for the pole-zero plot. Select Pole/Zero and Response as Closed Loop r to y. paying particular attention to the region between 2. Problem 5.[1 8 52]). a=0:0. CenlanlsafPWs---------------------------------------------------------------- nod Response 1 2 3 4 s 6 AM B r: r r n r n OoMdloMrtev n r' n r r n n OotidLoMrtOM Q r r. Verily that a multiple root occurs at a complex value of s for some value of a in this range. den=conv([1 1 0]. sys=tf(num. Observe from the pole/zero plot that there is a multiple complex root in the region of a between 2. num=[1 a]. Click on Show Analysis Plot. Step-by-step solution step 1 of 4 Step 1 of 4 The loop transfer function is.(i) = 0 I+ if - j ( i + l ) ( j * + 8 j + 52) Write the MATLAB code to draw the root locus when the parameter a is varied from 0 to 10 using ritool function.1 OPP Use the Matlab function ritool to study the behavior of the root locus of 1 + KL(s) for (i + o) U s) = j (j + 1)(j 2 + 8j + 52) as the parameter a is varied from 0 to 10. I + X Z .5.den).5 and 3.5 and 3. Step 3 of 4 Go to Analysis Plots. num=[1 2]. num=[1 1 2]: den=conv([1 11 30 0]. the system is unstable for all values of . Step 8 of 9 The characteristic equation is. I)[(A: + 2.46)’ + 39.. den=conv([1 -2 0]. (1).. sys=tf(num.16l- Step 4 of 9 The characteristic equation is. Problem 5. The open loop transfer function is.[1 2 10]).. The characteristic equation of the system is. l + G (s )H (j) = 0 1+^-..[1 2 1]). sys=tf(num. Step 6 of 9 (b) Refer to Figure 5.den).1 tfU-X A u u } ~iiP*9t72P~>«7ttOr*lS13S120 161t step 3 of 9 For the system to be stable. the range of AT for stability is |Q< JC<484. The characteristic equation of the system is..6’ < 0 K < -m ..( 1 ) -13A:’ + %72Ar*-67860Ar + 1513SI20 (6 1 8 -A :)’ (K + 739.I5 < 0 ( a: + 2. Thus..54 (a) in the text book for the block diagram of a system. the system is unstable for all values of IC- Thus.2 ) ( i ’ + 2 i + 10)+AT(s + 2) ■ 0 * ' + 6 s’ + ( A : ... ..11 PP Use Routh’s criterion to find the range of the gain K for which the systems in Fig. Figure Feedback systems Step-by-step solution step 1 of 9 Step 1 of 9 (a) Refer to Figure 5. 0 < ^ < 4 8 4 .»r) < ii-r iX 30*x)-2r 390*nr 13 13 13 ’ 13 — ^. and use the root locus to confirm your calculations.2 ) ( j * + 2 j + 10) Draw the root locus using MATLAB to find the range of the gain for stability. = 0 5 ( j .46)’ +39.16 (3) a: < -8 0 . **: 1 30-fr » 2t »xS3-(7i. the root locus result agrees with the Routh’s result. 2K>0 K>0 .g iT T --------------------— 13 ( • r* *«M:*3<K>»Y3w*iig'| . The open loop transfer function is. 1 + G (j)ff(* ) = 0 j’ + j+2 1 + iT .\5 (2) -AT’ +404A: + 38808 6 18-A : ^ (A :-484. / 1 6 2g s ’ 0 AT-20 s’ — 0 As the first column element is infinity. are stable. . (3) and (4) is. (2). (4) The range of satisfying all the four conditions.16)> 0 A :<484.6’ ] > 0 The conditions are. all elements in the first column of the Routh table should be greater than zero.1 6 J i:< 6 i8 .den)..— — t :t = 0 s(s-2)(s’+2s+t0)' i ( j . i+ ic —— ----------. Observe from the root locus that the gain should be less than 482 for the system to be stable. rlocus(sys) Step 9 of 9 Get the MATLAB output for the root locus..54 (b) in the text book for the block diagram of a system.2 0 ) i + 2A: = 0 Step 7 of 9 Construct the Routh table. rlocus(sys) Step 5 of 9 Get the MATLAB output for the root locus. the root locus result agrees with the Routh’s result.! 6 Thus. Hence.. . r=0 j (j + 5 ) (5 + 6 ) (5 * + 2 j + 1) j ( j + 5 ) ( j + 6 ) ( j = + 2 i + 1 )+ JC (s’ + j + 2 ) = 0 I * + 1 3 j ‘ + 5J s ’ + ( 7 1 + ^ :) *’ + ( 3 0 + a: ) j + 2 ^ : = 0 Step 2 of 9 Construct the Routh table.16)(A :+ 80. Observe from the root locus that the locus does not touch the zero-axis. „ s’ + f + 2 r=0 s (s + 5 ) ( i + 6 ) ( i ’ + 2 j + 1 ) Draw the root locus using MATLAB to find the range of the gain JC for stability. A: + 7 3 9 . the asymptotes is E O - Consider the formula for the angle of asymptotes. To find zeros put numerator J V (j)= 0 Thus.5. Step 3 of 5 ^ Procedure to draw root locus plot: • Take real and imaginary lines on X axis and Y axis respectively.2. Number of poles is n Number of zeros is m / = l.275l when the value of gain K is [|Q.1 2PP Sketch the root locus for the characteristic equation of the system for which and determine the value of the root-locus gain for which the complex conjugate poles have the maximum damping ratio. Compare the Equation (2) and the Equation (3). the three poles are 0. (4) The roots of the equation (4) are 0. .Ji—m Substitute 1 for /. The roots of D ( j) = 0 are the poles.0.. s=tf('s'). Thus. the maximum damping ^ is 0. To find poles put denominator D (s) = 0 • »“ ( i + 5 ) = 0 . Figure 1 Step 5 of 5 From Figure 1....0.275 when the value of gain K is approximately 10.. the ahgle of asymptotes are |9(y>|ahd |27Q**|.l) 3 -1 = 270» Thus. the maximum damping ^ is |Q. • Locate the asymptotes on the real axis. • Mark the poles on the real axis. and . and draw the asymptotes from centroid at an angle of gffana 270" • Draw the root locus. 3 for n and 1 for m in equation (5). Consider the number of poles and zeros from the characteristics equation. Step 2 of 5 Consider the formula for the asymptotes..2 .5 Thus.663 + 2... ^ l80°+ 3 6 (y * (/-l) n —m Where. I + AX(i) = 0 (1) ( s + 2) Substitute for I / j ) in Equation (1). 180” + 3 6 0 ° ( 2 . Problem 5 .. sysL=(s+2V(s''2*(s+5)): rlocus(sysL) [K]=rlocfind{sysL) Step 4 of 5 The MATLAB output for the value K. The roots of A f(» )= 0 are called the zeros of the problem. (3) D is ) Where. Step 1 of 5 - Consider the general form of characteristics equation.7 The root locus plot is shown in Figure 1. Select a point in the graphics window selecteu_point — -0. Thus.. a’ ( ^ + 5 ) 1 + a: (2) a* ( a + 5 ) Consider the roots of the general form of an equation by the root locus method. 3 for n and 1 for m in equation (5).7 | and the root locus is plotted for the given transfer function and it is shown in Figurel. 180°+360»(1-1) 3 -1 = 90“ Substitute 2 for /. and . the one zero is .. n -m (sum o f finite p oles) . Consider the following given function. ..5.32i K= 10. ' ’ * ’ (s + 5) Write the MATLAB program to obtain root locus.(sum o f finite zeros) (n u m b e ro f finite poles )~ (n u n d > e ro f finite zeros) (°-S )-(-2 ) ■ (3)-(l) = -1.7. What is the approximate value of the damping? Step-by-step solution step 1 of 5 Consider the General form of characteristics eouation. Step 6 of 10 (b) Draw the root locus for i ( * ) . RULE 3 : calculate the centre of asymptotes. root locus has been implemented.9 0 '.68Sl- Step 10 of 10 (d) Enter the following code in MATLAB to plot the response of the closed loop system to a reference input.'^ Y j + W +90‘ -|^ 9 0 '+ 9 (r + 90‘ + tan-' = 8 4 . Step 2 of 10 Procedure to draw root locus: RULE 1 : There are five branches to the locus.[1 0 100]). ( * + ! ) ( * ' + 8 l) i( * ) = ’ * '( * + 1 3 ) ( * '+ 1 0 0 ) Enter the following code in MATLAB to draw the root locus and asymptotes at ^ s 0. it is observed that the asymptote at ^ = 0. » num=conv([1 1].8 7 ] ” = 103.[1 0 81]).5 cuts the root locus at a point of the form. » sys1=feedback(sys. 5 -3 _ -12 2 s -6 The angles of asymptotes are at .1). Figure 5. £( 5) has both poles and zeros near the imaginary axis and should expect to find the departure angles of particular importance. G(») r ( j) = l + C ? (i)//(j) j.685 Thevalueof jfth a t yields closed loop poles with the damping ratio ^ = 0. Therefore.5 .707 is |lQ3. 57 ) ' + 8 I) [V(-5. + '5 '. » step(sysl) The MATLAB output for the step response is shown in Figure 5.5y+13) K = ( .1 8 ( r = -22 3 . (b) Is there a value of K that will cause all roots to have a damping ratio greater than 0.6417302144823378589757341357966-2. Figure 2: Measuring the angle o f departure Step 3 of 10 The root locus condition is. ( . Substitute value of 5 In equation (3). Consider the magnitude condition 1 ii( * ) i ■ * ' ( * ’ + 100 )(*+13) K = (3) ( * + ! ) ( * ' + 8 l) To find value of x that yields closed loop poles with 0. cos$=^ 9 ■ cos"' ^ = cos"'(0. Remaining four roots are complex roots which are not valid for breakaway points.5y+ 1)((-5. three of which approach to finite zeros and two of which approach to Infinity. There exists a value o f X ^bat causes roots to have damping ratio greater than 0.6 + 6.2 T Step 4 of 10 RULE 5: Find the breakaway points. calculate the angle 0 .5 .707. Thus. the response of the closed loop system using MATLAB is plotted.1 D' +(72!iF][>/(7.5)' ] ’ [V(89. ■(1) Equation (1) is of the fonn 1+ K L (s) = 0 .707. Problem 5. » den=conv([1 13 0 0]. »solve('x'‘7■^16*x^6+350*x''5■^2364*xM+20412*x''3■^129600*x''2■^210600*x').fl V J*+81 U + 1 3 j [ i ' ( f ‘ + 100) \s + \3 ) The characteristic equation of the system is. j* + 8 l \-¥K\. » sys=tf(num.3 7 . Draw pole-zero plot for the transfer function. Closed loop transfer function of the system is. d value is to be found. * = v . = l a n .i 8 c r .13PP For the system in Fig.den).6 + 6. or _ _ Figure Feedback system j*+«l 100> ____________ S te p -b y -s te p s o lu tio n step 1 of 10 (a) Obtain the characteristic equation for feedback system shown in Figure 5.[1 0 100]).1 D ' + ( 7 2 . As Rvalue is known.. + H « . Figure Feedback system Ul I .3112175660279159755982613661971 All the three roots are valid break-away points since the real-axis segment defined by . (i’ + 8 U + 8 1 )(5 j * + 52i’ +300 j ’ +2600 i )- ( i ’ +13 « V 100^*+130fts’)(3*’ + 2 i +81) ds + s * + 8 U + 8 IJ dK Equate -----to zero. where ^ value is in the range 0 < ^ < 1 • From the value of ^ it is clear that system is under damped. In under damped system.5 • » num=conv([1 1].( ^ + A + ^ i ) . sO U + 1 3 j [ j ’ ( i ' + 100)J j * ( s + 1 3 ) ( s * + 1 0 0 )+ A :( j * + 8 1 ) ( j + 1) = 0 Solve for X - f * ( j + 1 3 )(j* + 100) ( j ’ + 8 l) ( s + l) j* + i 3 s * + i o a ! ’ + i3 o a ! ' j* + j* + 8 1 j+ 8 l Differentiate the equation with respect to s .5? (c) Find the values of K that yield closed-loop poles with the damping ratio ^ = 0.6 + 6.54 to find the locus of closed -loop roots with respect to x ■ Since system contains cascaded blocks.5y)’ + 1 0 0 )(-5 .5 ) '][ V ( 7 0 .6)'+(6. Step 8 of 10 (c) To find the value of x ^bat yield closed loop poles with damping ratio ^ s 0. 6 ) ' + (6 .5)' [ V ( ^ . (a) Find the locus of closed-loop roots with respect to K.4)' +(6. The other angles are marked in Figure 2.707.1 is part of the locus. ds y + » * + 8 1 i+ 8 l) ( 5 / + 52s" +300s’ + 2 6 0 0 s)- ( i ’ +13 i‘ +1 OOj ’ +1 300j " ) (3 * ’ + 2 i + 8 1) (s’ + i'+ 8 1 s + 8 l) (j* +»’ +81i+8l)(5j*+52s’ +300J*+2600j ) - =0 ( j ’ +13j * + 100j ’ + 1300j ')(3 j *+2 i +81) 5i’ +57 s‘ +757 j ’ + 7517J* +31112j’ +234900*' + 210600*' =0 -(3*’ +4I*‘ + 4 0 7 / + 5153** + 10700*’ +105300*') 2 * '+ 1 6*‘ +350*’ +2364** + 20412*’ +129600*'+ 210600* = 0 Step 5 of 10 Use MATLAB to solve the equation.5. Where ( j + l ) ( i ’ + 8 l) i( j) = i ’ ( j + l 3 ) ( j ’ +100) Loop gain.den): » rltool(sys) Step 7 of 10 Observe the MATLAB output on the Figure window. The locus of closed loop poles with respect to x shown in Figure 3. RULE 2 : The real-axis segment defined by —13 ^ ^ ] is part of the locus. Graph to find 5 value Step 9 of 10 The intersecting point in Figure 4 is the value of s . Root Locus Edior for Op«n Loop 1(OL1) From the root locus in Figure 4.6+ 6 . . .6 + 6. » sys=tf(num.707) = 45' Draw a line at an angle ^ —45*00 the root locus plot and mark the intersecting point as shown in Figure (4).5 i f ( ( .5 6 '. RULE 4 : Compute the departure angle from the pole at ^ +ylO-The angle at this pole we will define to be ^ . (d) Use Matlab to plot the response of the resulting design to a reference step.2 8 '. ans = 0 -5. roots are of form -^6)^ ± Jo>^ ( complex conjugates).1 3 ^ j S .[1 0 81]). » den=conv([1 13 0 0].5 . RULE 3 : Calculate the centre of asymptotes. (-1+yi.cos"'(0. » den=[1 2 1]. » num=10.i& J luation (1) is of the fonn Equation 1+KL(s) . RULE 2 : There is no part of root locus.5. (s+l)(i +l) step 3 of 6 Procedure to draw root locus: RULE 1 : There are two branches to the locus.299 Therefore.A : r ) L ( j) has two poles. yw GW R(s) 1+G(i)ff(j) 1+ K M The characteristic equation for the feedback system is. Problem 5.732+l)| 10 -2.5 damping ratio to get asymptotes that cut the root locus as shown in Figure 2. ^ I | i( * ) | • (j +l)(i+l) K = (1) 10 To find value of ^ that yields closed loop poles with ^ = 0.5 damping ratio asymptote cuts the root locus at 5=-l±yT.55 in the textbook. find the value of the gain K that results in dominant closed-loop poles with a damping ratio ^ = 0.732. observe that the 0. Figure 1 Step 2 of 6 Calculate the transfer function of the feedback system. Simplify the feedback system {Figure 5.S is . » sys=tf(num. .1-1 < r*- 2-0 -2 The angles of asymptotes are at ^ 9q* .den): » sisotool(sys) Step 5 of 6 MATLAB output: Root Locus EdHorfor Open Loop 1(OL1) Figure 2 Step 6 of 6 From Figure 3.5. one on real axis and other pole value depends on value . To make system stable consider the ^ value as -1. cos$=^ 6 ■cos"'C ..0. the value of that yields closed loop poles with the damping ratio ^ = 0. Substitute value of s in equation (1).5) *60* Draw the root locus using sisotool function in MATLAB and add design parameter as 0.732+l)(-l +. 10 i( s ) . Where 10 ( 5 + l) ( l. both of which approach asymptotes.55) as shown in Figure 1.991 10 I =0.14PP For the feedback system shown in Fig. Step 4 of 6 To find the value of fc that yield closed loop poles with damping ratio ^ = 0. U ( j+ i) J L i. Figure Feedback system Step-by-step solution Step-by-step solution step 1 of 6 Refer to Figure 5.5• Consider the magnitude condition. rvalue is to be found. As Rvalue is known.i. Then. calculate the angle $ . Let Dc(s) = kp at first. 0.5 s+ 6 .V r .6kp.6 6 )(s^-0. Use axes -4 < x < 4.6 6 .15) ' K = 0 points: -0 .25'> = 115.38"+253. A r(5^-0.3) (b) a f s ) = --------.24s+ 0.12±j0.Z ' f e = 70°+86. = -3 3 4 " . Step 6 of 6 .4” = 244° Z f e = 90° If. (c) Verify your answer using Matlab.3 ) a {s ): (s+ 0 . (s+0. (d) Suggest a practical (at least as many poles as zeros) alternative compensation Dc(s) that wil at least result in a stable system. ^ « -64.75" Step 4 of 6 Angles o f arrival: ^ = 180"+<t» ■Where if = S f e . <bi. .75" = 334" And.75" = 64.66)(«2-a24c + 0.8(s’ -0 . (a) Compute the departure and arrival angles at the complex poles and zeros. Use the command axis([-4 4 -3 3]) to get the right scales. -3 < y < 3. S te p -b y -s te p s o lu tio n step 1 of 6 9.15PP A simplified model of the longitudinal motion of a certain helicopter near hover has the transfer function <H*).368 iT s a points: 0.75" And. ^ = -3 3 4 " We can plot the root locus o f the given transfer fimction as shown below.IS> and the characteristic equation 1 + Dc(s)G(s) = 0.5°+87.25±j2. = 64.(«+0. ------------i ^ where K = 9 .61" = 360" ^ = 180"+244.5ff+6. (c) Verify your answer using Matlab. Step 5 of 6 .5 As <ti.8 ^ . Problem 5. (b) Sketch the root locus for this system for parameter K = 9. Use the command axis([-4 4 -3 3]) to get the right scales.245+0.75" ^ = -64.15) Step 2 of 6 w Step 3 of 6 Angles o f departure: = 1 8 0 ° -« t> Where = 2 < b > -Z 'fe 2 4 ^ = 90®+25. = 334° And.66)(ff^-0.25" 2<tir = 106.75" = 424.. Problem 5.16PP (a) For the system given in Fig., plot the root locus of the characteristic equation as the parameter K^ is varied from 0 t o w i t h A = 2. Give the corresponding L(s), a(s), and b(s). (b) Repeat part (a) with /\ = 5. Is there anything special about this value? (c) Repeat part (a) for fixed K^ = 2, with the parameter K = A varying from 0 to Figure Control system Step-by-step solution step 1 of 15 Refer to Figure 5.56 in the textbook. Draw the modified block diagram. Figure 1 Step 2 of 15 Move the summing point of a block as shown in Figure 2. R Figure 2 Step 3 of 15 Draw the reduced block diagram of Figure 2. Figure 3 Step 4 of 15 ^ Move the summing point ahead of a block as shown in Figure 4. Figure 4 Step 5 of 15 Draw the reduced block diagram. Step 6 of 15 ^ (a) Calculate the characteristic equation. I + G (s )ff(j) = 0 f lOiC, Y 0.U(^-fA)-fg,(0.2»-Mn . K, )- j ( j + IO )(j + A) j (j + 10)(5 + A) step 7 of 15 Simplify further. i ’ + ( 1 0 + A ) i’ + 10.1i irgf 1-0 (1) Substitute 2 f o r A in equation (1). ‘ ■ ^ ^ { j ’ + { l l + 2 ) j ‘ + l l ( 2) j ] “ “ From the characteristic equation, the loop transfer function is. r( 2 £ + I2 _ j ’ +13 j ‘ + 22 s o ( i ) = »* + lJ s ’ + 2 2 i And, i ( j ) = 2 i+ 1 0 Step 8 of 15 Write the MATLAB code to draw the root locus. » num=[2 10]: » d e n = [1 13 22 Oj; » sys=tf(num,den) sys = 2 S + 10 s ^ 3+ 13s^2 + 22s Continuous-time transfer function. » rlocus(sys) Step 9 of 15 Draw the root locus plot using MATLAB. Root Locn$ Real Axis (seconds'*) Figure 6 Step 10 of 15 (b) Substitute 5 for >l in equation (1). 2j + 10 1 + Jf,l ^ j’ + ( l l + 5 ) i " + l l( 5 ) ; 1 + Jf,( ' 2*+10 'I 0 k^j’ + 1 6 f* + 5 5 5 j 1+ JC,| ^ j( j+ 5 ) ( j + l l ) j l+ if ,l' * 1 0 From the characteristic equation, the loop transfer function is. And, b (s )^ 2 Step 11 of 15 Write the MATLAB code to draw the root locus. » num=2; » d e n = [1 11 Oj: » sys=tf(num,den) sys = 2 s^2 + 11 s Continuous-time transfer function. » rlocus(sys) Step 12 of 15 Draw the root locus plot using MATLAB. Root Locus ^ 2 - ■n 1 1- I® fr e -1 ^ -2 h -12 -10 Real Axis (seconds'*) Figure 7 step 13 of 15 A (c) From the block diagram, the characteristic equation is, 1+ G ( s ) f f ( s ) = 0 ,.r lOiC, „ [i(j+10)(i+A)Jl, K, ) s{s+ X )+ K ,{2 s*\0) j ( i + 1 0 )(j + 2 ) Substitute 2 fo ritr, 3 (3 ^ 2 )> 2 (2 3 ^ I0 ) i ( s + 1 0 )(s + 2 ) j (j 1-2)+43H-20 j (3 + 1 0 ) ( * + 2 ) j( j- F ll) 1 + .1 - j ’ + lls * + 4 r + 2 0 From the characteristic equation, the loop transfer function is. ' ^ * ’ + lb * + 4 ff + 2 0 fl( 5 )= 5 * + I I j* + 4 s + 2 0 And. i(j)=j(s+ll) Step 14 of 15 Write the MATLAB code to draw the root locus. » num=[1 11 0]; » d e n = [1 11 4 20]; » sys=tf(num,den) sys = s^2 + 11 s s^3+ 11 s'‘2 + 4 s + 20 Continuous-time transfer function. » rlocus(sys) Step 15 of 15 Draw the root locus plot using MATLAB. Root Locus Problem 5 .1 7PP For the system shown in Fig., determine the characteristic equation and sketch the root locus of it with respect to positive values of the parameter c. Give L(s), a(s), and b(s), and be sure to show with arrows the direction in which c increases on the locus. Figure Control system Step-by-step solution step 1 of 4 Step 1 of 4 Refer to Figure 5.57 in the textbook. From the block diagram, the process transfer function is. /'C + I 65 V 9 '' G ( ,) V C+S Calculate the characteristic equation. 1+ G (» )H (» ) = 0 ( c + j) i* + 9 c + I 4 4 5 (c + j)j* cy* + 5* + 9 c + 144s * 0 c (s* + 9) + s ( s * + 144) = 0 l+ c - i- = 0 (1) j(j" + 1 4 4 ) (s*+9) Therefore, the characteristic equation is s(s* +144) Step 2 of 4 From the equation {1), the loop transfer function is. s (s ^ + 1 4 4 ) From the characteristic equation. a(s)=s*+144s And. 6(s)=s*+9 Therefore, the loop transfer function, L { s ) i: ( ) andi(s) e values of 0 5 p(s*+144) are |a(j)sj^+1 44 s and ^(s) = s* Step 3 of 4 Write a MATLAB code to plot the root locus. » num=[1 0 9]; » den=[1 0 144 0]: » sys=tf(num,den) sys = s^2 + 9 s^3 + 144 s Continuous-time transfer function. » rlocus(sys) Step 4 of 4 Draw the root locus for all positive values of c . Problem 5.18PP Suppose you are given a system with the transfer function d+z) where z and p are reai and z > p. Show that the root iocus for 1 + KL(s) = 0 with respect to K \ss circle centered at z with radius given by r= {z -p ). Hint Assume s + z = rej<p and show that L(s) is real and negative for real <punder this assumption. Step-by-step solution Step-by-step solution step 1 of 2 L (s )= -^ let s=a+jflo T / \l _ (cT+p)^V+2jcD(<T+p) Phaseof L ( s ) ^ 1 8 0 “ Step 2 of 2 Imaginary part of L (s) =0 - 2 g j ( o + z ) ( o + p )-I- co^(<H- p ) ^ - o j ^ J = 0 (D^ +CD^+2oz+2pz-p^ ] =0 (or) a^+© ^2pz+z^^^+z^-2oz or o^+©^+2oz+z^^^+z^-2pz (o+z)* +©^=(z-p)^ IWhich is a circle with radius; z-pl Problem 5.19PP The loop transmission of a system has two poles at s = -1 and a zero at s = -2. There is a third real-axis pole p located somewhere to the /e/tof the zero. Several different root loci are possible, depending on the exact location of the third pole. The extreme cases occur when the pole is located at infinity or when it is located at s = -2. Give values for p and sketch the three distinct types of loci. Step-by-step solution step 1 of 4 s+2 T .M s - Step 1 of 4 L ( s ) = ----- ---------- ( = + l) C = ^ ) Step 2 of 4 (i) Let p = -2 Step 3 of 4 (ii) p= ^ Step 4 of 4 (iii) p = -oo For the feedback configuration of Fig., use asymptofes, center of asymptofes, angies of deparfure and arrivai, and fhe Routh array to sketch root ioci for the characteristic equations of the iisted feedback controi systems versus the parameter K. Use Mattab to verity your resuits. (a) C(i) = ;p T n ^ r F I= 57' (») C ( j) = 5 , (c ) = m = '+ 3 * Figure Feedback system m Figure Feedback system S tep-by-step s o lu tio n S le p t of 30 (a) Consider the foiiowing equation. « , ) = (. a (j+ l+ 3 y )( s + l- 3 y ) s+2 i( a ) = a (a + I+ 3 y ) (a + I- 3 y ) ( s + 8 ) Consider the generai torm of characteristics equation. l + * i( a ) = 0 ......( 1) fyfor ^ 5)in Equation (1). i( a + I 0 ) ( a + I+ y ') ( a + I - y ) a+2 1+ i r - = 0 ...... (2) a ( a + l+ 3 y ) ( a + l- 3 y ) ( i+ 8 ) Consider the roots of the generai form of an equation by the root iocus method. l +iC - ^ - O ......(3) ZKs) Where. The roots of ^ ( 5) = 0 are catted the zeros of the probiem. The roots of O(a) = 0 are the potes. Consider the number of potes and zeros from the characteristics equation. Compare the Equation (2) and the Equation (3). To find zeros put numerator fV(s)=0 Thus, the zero is -2. To find potes put denominator D(_s) = 0. j ( j + l+ 3 y ) ( i+ l- 3 y ) ( » + 8 ) = 0 ......(4) The roots of the equation (4) are 0 .-8 . —1—3y and —l+ 3 y . Thus, the tour poies are 0. - 8. —I - 3 y and - l+ 3 y . Step 2 of 30 ^ Consider the formuta for the asymptotes. n -m (sum o f finite poles)-(sum o f finite zeros) (numberof finite poles )-(n i]in b e ro f finite zeros) - l- 3 y - l+ 3 y - 8 - ( - 2 ) 3 3 Thus, the asymptotes is |-2.67|. Consider the formuta for the angte of asymptotes. 180<-f360°(/-l)......,5, n -m Where. Number of potes is n Number of zeros is m / = I,2 ,..ji- m Substitute f for /. 4 for n and 1 for m in equation (5). , 180+3«0(l-l) " “ a 180 “ 3 = 60° Substitute 2 for /. 4 for n and 1 for m in equation (5). , 180+360(2-1) ft “ e 540 3 = I80» Substitute 3 for /. 4 for n and 1 for m in equation (5). , 180 + 360(3-1) ft” + 300 3 = -60« Thus, the angte of asymptotes are |6Q**| . |]go«*|and | - 60**l- Step 3 of 30 Consider the foiiowing equation. 6 (-l8 0 °-ta n[3 |^1 = 108.43“ 6^=90“ [3 6( -tanj^Y = 71.56“ = 23.19“ Consider the departure angies in the transfer function. sum o f angleof vector to' 180“ - the complex pole A angleof depature ) from otherpoles from a complex A j sum o f angleof vectors to the'l 0 j = lS « ° -[0 ,+ e ,+ e ,]+ 0 , {complex pole A from zeros J Substitute ait vatues in equation. 61, = 180“-[108.43“+90“+23.l9“]+71.56“ Thus. the departure angie in the transfer tunction is |6i^ =-29.94“ | at s = —l+ 3 y . Step 4 of 30 Consider the foiiowing transfer function. C (s )W (s ) l+ G ( s ) // ( s ) s ( s + l+ 3 y ) ( s + l- 3 ./) ( _______ ^ (_s(s+ I+3y)(s + 3 y )(s + l-3 y )J (. s + 8 / Consider denominator poiynomiai to find the K vaiue. s(s + l+ 3 y ) ( s + l- 3 y ) ~ s (s + l+ 3 y )(s + l-3 y )(s + 8 )+ A :(s+ 2 ) s ( s + l + 3 y ) ( s + I - 3 y ‘) (s + 8 ) _______________ 8T (s+8)______________ “ s (j+ l+ 3 y )(i+ l-3 y )(j+ 8 )+ 6 r(i+ 2 ) Consider the denominator poiynomiai is equai to zero. s [s ’ + s -3 y s + s + l- 3 y + 3 y s + 3 y -9 y * ](s + 8 )+ 8 r( s + 2 ) = 0 [s ’ + s * -3 y s “ + s *+ s -3 y s + 3 y s *+ 3 y s -9 y *s j(s + 8 )+ H r(s + 2 ) = 0 (s’ + 2s’ + s - 9 y ’s ) (s + 8) + I f ( s + 2 ) = 0 s *+ 2s’ + s’ - 9 y V + 8s’ +16s’ + 8s - 72y’ s + a: (s + 2) = 0 [s ’ + 10s’ + s ’ + 9s’ + 8s’ + 16s’ + 8 s+ 72s]+ K ( s + 2 ) - 0 s‘ +10s’ +26s’ + 8 0 s + K (s + 2 ) - 0 s“+10s’ + 26s’ +(80+AT)s +2A: = 0 ......(6) From equation (6). the characteristics equation is given beiow. A ( s ) = s‘ + 1Os’ + 26s’ + (8 0 + K ) s +28: s’ + l0 s ’ + 26s’ + ( 8 0 + i: ) s + 2 8 : = 0 .......(7) By appiying Routh-Huiwitz criteria to equation (7). 1 26 2K 10 80 + / : 1 8 - - ^ 2K 10 - I f l - lOOAT+14400 1 8 0 - /: 2K Figure 1 step 5 of 30 ^ The sysfem is stabie if fhe equation satisfies the foitowing condition. • Aii the terms in the tirst coiumn of the Routh’s array must be positive sign. From Figure f. the stabie condition is 0 < g f <gQ. Thus, the range of/Cis 0< 8 :< 8 0und it is satistied the stabiiity condition. Substitute 80 for K in equation (7). s’ +10s’ +26s’ +(80 + 80)s+160=0 s“+I0 s’ +26s’ + 160s +160 = 0 W The roots of the equation (8) are ±4 y. —S—y J lS ^ ^ ^ lS —S- Thus, the imaginary axis crossings are and step 6 of 30 ^ Procedure to draw root iocus piot: • Take reai and imaginary iines on X axis and Y axis respectiveiy. • Mark the potes on the reai axis. • Locate the asymptotes on the reai axis, and draw the asymptotes from centroid at an angte of W • Draw the root iocus. The root iocus piot is shown in Figure 2. Hence, the root iocus is ptotted for the given transfer function and it is shown in Figure2. Step 7 of 30 ^ Consider the foiiowing given function. j+ 3 L {s )- x(x + 10)(x+l+ y )(« + l-y ) Write the MATLAB program to obtain root iocus. s=tf('s'); sysL=(s+3)/(s*(s+10)*(s+1+1j)*(s+1-lj)); riocus(sysL) The root iocus piot is shown in Figure 3; Step 8 of 30 ^ Thus, the root iocus is verified from MATLAB output. Step 9 of 30 (b) Consider the foiiowing equation. i( j) = >-m] ■ j ’ (j+ 3) Step 10 of 30 Substitute ) * for £ (j)in Equation (1). i ( j+ 3 ) .(9) i ’ (i+3) Compare the Equation (9) and the Equation (3). To find zeros put numerator N {s^= 0 Thus, the zero is -1. To find poies put denominator D(_s) = 0. The roots of the equation (4) are 0.0 and -3. Thus, the tour poies are 0. 0 and -3. Step 11 of 30 -rv Consider the formuta for the asymptotes. n -m (sum o f finite poles)-(sum o f finite zeros) (numberof finite poles )-(m u n b e ro f finite zeros) 2 ~ 2 Thus, the centre of asymptotes is . Substitute 1 for /. 3 for n and 1 for m in equation (5). 180+ 360(1-1) ft “ + 180 ” 2 =90“ Substitute 2 for /. 3 for n and 1 for m in equation (5). 180+ 360(2-1) 540 2 = 270“ = -9 0 “ Thus, the angle of asymptotes are |90“ |and Step 12 of 30 Consider the following transfer tunction. T -r 0 (s )ff(s ) l+ G (s )W (s ) X (» + 0 ” s’ (s+ 3 ) l+ X s’ (s+ 3 ) Consider denominator polynomial to find the K value. s’ +3s’ + J&+A: Consider the denominator polynomial is equal to zero. s’ +3s’ +&+A: = 0 (10) From equation (10). the characteristics equation is given below. A ( s ) = s ’ +3s’ + A i + j : s’ +3s’ +&+A: = 0 (11) By applying Routh-Huiwitz criteria to equation (11). a : a: 2a : K Figure 4 Step 13 of 30 ^ The sysfem is stable if the equation satisties the following condition. • All the terms in the tirst column of the Routh’s array must be positive sign. From Figure 4. the stable condition is JK > Q. Thus, the value of K is JK = Qand it is satistied the stability condition. Substitute 0 for X in equation (11). *’ +3»’ + k » + a: = o j =0 Thus, the imaginary axis crossing is . Step 14 of 30 Procedure to draw root locus plot: • Take real and imaginary lines on X axis and Y axis respectively. • Mark the poles on the real axis. • Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of 9(r • Draw the root locus. The root locus plot is shown in Figure 5. RootLocus Hence, the root locus is plotted for the given transfer function and it is shown in Figure 5. Step 15 of 30 Consider the following given function. (j+ 1 ) Write the MATLAB program to obtain root locus. s=tf('s'); sysL=(s+1)/(s“ 2*(s+3)); riocus(sysL) The root locus plot is shown in Figure 6: Step 16 of 30 Thus, the root locus is verified from MATLAB output. Step 17 of 30 (0 Consider the following equation. Substitute Ifo r L (j)in Equation (1). U + lA s + 3 j ■•'(Srlef)-".. '■» step 18 of 30 Compare the Equation (12) and the Equation (3). To find zeros put numerator N {s^= 0 Thus, the zero is -5 and -7. To find poles put denominator D(s) = 0. The roots are -1 and -3. Thus, the tour poles are -1 and -3. Step 19 of 30 ^ Consider the formula for the asymptotes. Asymptodes=(numberofUnitepoles)-(numberoffinltezeros) = 2-2 =0 Thus, the asymptotes is . Consider s 0 and simplity equation is given below. ds 8s’ +64s+I04 = 0 From the above equation, the break in and break away points is -2.27 and -5.73. Thus, the break in and break away points is -2.27 and -5.73. Step 20 of 30 Procedure to draw root locus plot: • Take real and imaginary lines on X axis and Y axis respectively. • Mark the poles on the real axis. • Locate the asymptotes on the real axis, and draw the asymptotes from centroid. • Draw the root locus. The root locus plot is shown in Figure 7. Root Locus Hence, the root locus is plotted for the given transfer function and it is shown in Figure 7. Step 21 of 30 Consider the following given function. i( i) = M S I) MATLAB program to obtain root locus: s=tf('s'); sysL=((s+5)*(s+7))/((s+1)*(s+3)): riocus(sysL) The root locus plot is shown in Figure 8. Figure 8 Step 22 of 30 Thus, the root locus is verified from MATLAB output. Step 23 of 30 (d ) Consider the following equation. w A - (< ^ 3 s )( s + 3 + 4 y )( s + 3 - 4 ;) j( s + l+ 2y ) ( j + l - 2y) Substitute ( '• ’• ^ ) ( 4 ’* '3 + y ) ( s + 3 - 4 y ) ^^^ Equation (1). i ( j + l + 2y)(s + l - 2y) *4 f u (l+ 3s)(3+3+ 4y)(s+3- 4y) (13) i ( i + l + 2y ) ( j + l - 2y) Compare the Equation (13) and the Equation (3). To find zeros put numerator ^ ( s ) = 0 Thus, the zero is -0.33. —3 -4 ya n d —3 + 4 y. To find poles put denominator /}(s) = 0 ■ The roots are -1 and -3. Thus, the poles are 0. —l-2 y a n d —l+ 2 y . Step 24 of 30 ^ Consider the formula for the asymptotes. A sym ptodes=(num ber o f Unite poles )-(n u m b e r o f finltezeros) = 3 -3 =0 Thus, the asymptotes is . Step 25 of 30 Consider the following equation. 6 (-I8 0 “-tanj^Y] = 116.56“ = 108.26“ 6!, = 90“ 6)4 = tan [I] =71.56“ 6( = tan | [2 1 = -45“ Consider the departure angles in the transfer function. sum o f angleof vector to' 180“- the complex pole A ang leof depature 1 from otherpoles from a complex A j sum o f angleof vectors to the'l 6>, = 180“-[61 +6l,]+[tf, +6I4 + 6I5] {complex pole A from zeros J Substitute all 0 values in equation. 6>, = 180“-[1 16.56“+90“]+[108.26“+71.56“-45“] Thus, the departure angle in the transfer tunction is \0^ =108.26“| at s = —1+2/. Step 26 of 30 Consider the following equation. d l- lO O - ta n j^ jj = 126.86“ 6 ^ .1 8 0 - ta n [Y ^ ] = 123.62“ 6( =lS0-tan|^Y = 135“ 6)4 = 180-tan = 108.43“ 6 1 ,-9 0“ Consider the arrival angle in the transfer function. sum o f angleof vector to' ang leof arrival 1 sum o f ang leo f vectors to the] from a complex A J = 180“ — the complex zero A 6>. = 180“ - [ 6i + 6) ] + [ 6l + 6) + ^ 4] from other poles zeros {complex zero A from poles J step 27 of 30 Substitute all 0 values in equation. 6!.-180“-[l23.62“+90“]+[l26.86“+135“+108.43“] = 336.67“ Thus, the arrival angle in the transfer tunction is |6t^ = -22.33“| at s = -3 + 4 /. Step 28 of 30 ^ Procedure to draw root locus plot: • Take real and imaginary lines on X axis and Y axis respectively. • Mark the poles on the real axis. • Locate the asymptotes on the real axis, and draw the asymptotes from centroid. • Draw the root locus. The root locus plot is shown in Figure 9. R oot Locus Hence, the root locus is plotted for the given transfer function and it is shown in Figure 9. Step 29 of 30 Consider the following given function. w A - 0 ^ 3 s )(s+ 3 + 4 y )(s+ 3 -4 ;) s(s + l+ 2 /) ( s + l- 2 /) MATLAB program to obtain root locus: s=tf('s'); sysL=((1+(3*s))*(s+3+4i)*(s+3^j)V(s*(s+1+2i)*(s+1-2j)); riocus(sysL) The root locus plot is shown in Figure 10. Root Loon Step 30 of 30 Thus, the root locus is verified from MATLAB output. Problem 5.21 PP Consider the system in Fig.. (a) Using Routh’s stability criterion, determine all values of K for which the system is stable. (b) Use Matlab to draw the root locus versus K and find the values of K at the Imaginary-axis crossings. Figure Feedback system ELh- S te p -b y -s te p s o lu tio n step 1 of 6 Refer to Figure 5.59 in the textbook. Calculate the characteristic equation. 1+ G ( j ) f f ( j ) = 0 ' K (s + 3 ) 1+ m -’ i(j+ l)(s ^ + 4 i+ 5 )+ A :(j+ 3 ) j ( j + l) ( i^ + 4 i + s j + j){ » ^ + 4 » + 5 )+ K s + 3 X = 0 t S s * + 9 j ^ + 5 j + * s +3A: = 0 j*+5j’ + 9 j ^ + ( 5 + A :) s + 3 X = 0 ( 1) Step 2 of 6 ^ Calculate the range of using Routh’s stability criteria. I 9 3K 5 5^K 4 S -5 -K 3AT 4 0 -K 5 3AT From the Routh’s stability criteria, first column elements should be greater than zero. Therefore, S 4 0 -K > 0 K<A0 And. 3^>0 ^>0 Therefore, the values of for which the system is unstable is |0 ^ AT ^ Step 3 of 6 (b) Calculate the loop transfer function. L (s ) = G (» )W (i) ^ ^(^-^3) Y 1 j [ i( i^ + 4 s + 5 ) J li+ l y (» + 3 ) j(j+ l)(j* + 4 s + S ) A T (j+ 3) + & ’ + 9 i* + 5 i Consider that, = Therefore, (^ ^ 3 ) i( s ) . s * + J j’ + 9»"+ 5s Step 4 of 6 ^ Write a MATLAB program to find the root locus of the system. » num=[1 3]: » den=[1 5 9 5 0]; » sys=tf(num,den) sys = s+3 sM + 5 s'^3 + 9 s'^2 + 5 s Continuous-time transfer function. » riocus(sys) Step 5 of 6 Draw the root locus. Figure 1 Step 6 of 6 The root locus is crossing the imaginary axis at ^ = ± J { ,38. Therefore, the value of gain, is I S U Problem 5.22PP C (j) = Using root-locus techniques, find values for the parameters a, b, and K of the compensation Dc(s) that will produce closed-loop poles at s = -1±yforthe system shown in Fig. Figure Unity feedback system Step-by-step solution Step-by-step solution step 1 of 2 Refer to Figure 5.53 in the textbook. From the Figure 5.53, the loop transfer function is, £ (* ) = G (« )i)(5 ) I i(,)= (5 + 2 )( j +3) ______ A '( j+ o ) ( j + 2 ) ( j + 3 )^ 5 + 6 ) Since, the closed loop poles are at S = - l ± J Thus, the characteristic equation of the 2nd order transfer function is, (j+i+ j ) { s + \ - y)»o ( j + l)^ + l = 0 s^+2s+2 = 0 Since, the closed loop poles are poles than the poles of the transfer function G (^ ) • Thus consider that the zero of D ( s ) is at 3 to cancel the pole at 3. That is, a s - 3 . Step 2 of 2 Calculate the closed loop transfer function. r(,) G (^ )g M R (s) i+ G (s )D {s )H {s ) \+ (^ + 2 )(^ + 3 ) K ( i+ 2 ) ( s + i) “ K (s * 2 )(s + b ) _________ K ________ * i'+ ( 2 + f t ) s + 2 * + ^ : From the transfer function, the characteristic equation is, j* + ( 2 + i) j+ 2 4 + K = 0 (2) Compare equation (1) with equation (2). 2+i=2 i =0 And. 24 + A : = 2 2 (0 )+ ^ = 2 K =2 Therefore, the parameters Oy b and K of the compensation V ( s ) to produce closed loop poiesat » = - l ± y are |a = - 3 , t = 0 a n < |g = 2| (3) Where.1 and —l±1. a*+ 4 a ’ +9a’ +10a + 16..5 K tS . Number of poles is n Number of zeros is m / = l.25. | 135<>|. Thus..' . the range of /CIs 0 < AT< 16.58114y • Thus.| ) % 4 ( . Step 4 of 8 Consider the following formula for the departure angle ^ ^ fro m the pole at . Hence. 180»+360»(3-l) 4 -0 «225« = -45» Substitute 4 for /.2 . 4 for n and 0 for m in equation (5). j = . (■s + 2 j + 5 j j+ 2 1 1 + ii: .4°-90°+180° = -90° Thus. (2 ) + 2 j+ 5 j Consider the roots of the general form of an equation by the root locus method.. step 7 of 8 Take differentiate to equation (6) with respect to £ *45’ +12 j * + 18«+10 ds 4 j ’ + 12«*+ I8 j + 1 0 . the departure angle ^ ^ fro m the pole at —l°h2y is |-9Q^|. A(a)=a(a+2)(j’ +2a+5)+X A(a)=a‘ +4a’+9a“+10a+X.25 for K in equation (6 ).225y)-FA: a: = 16. • Locate the asymptotes on the real axis. To find poles put denominator D {s) = 0 • j ( j + 2 )( j '+ 2 j + 5 ) = 0 . there is no zero in the transfer function.. sketch the root locus with respect to K of the characteristic equation for the ciosed-loop system. paying particuiar attention to points that generate multiple roots. and how many multiple roots there are.58114y| and |-l-5 8 1 1 4 y |. Step 5 of 8 From equation (2)..225y • Substitute . w c w = o (1) I K Substitute 7 1 — Z— G( j ) and -------for D^(ff)in Equation (1). —\ + 2 J and —l .l 10 - 13 50 K Figure 1 Step 6 of 8 The system is stable if the equation satisfies the following condition. The root locus plot is shown in Figure 2. To find zeros put numerator JV(j) = 0 Thus. 1 ^ 9 4' ^10 6.2. Compare the Equation (2) and the Equation (3). (4) The roots of the equation (4) are 0 .0A = 45* Substitute 2 for /. .2 2 5 y fo rs in equation (6 ).1 ± 1 .I ) ’ + 9 (-I)*+ 1 0 (-1 )+ A : = 0 a: = 4 Substitute —l+ 1 .58114y and ±1. . ( .ji-m Substitute 1 for /.25 • Thus. and draw the asymptotes from centroid at an angle of 45T • Draw the root locus. • Mark the poles on the real axis.. the location of multiple roots are |_ \\ and | . I80°+360°(1-1) ^ ” > 41.0 (8) The roots of the equation (8 ) are .2 y . (-1 -I-1. = .23PP Suppose that in Fig.135<>|.l+ 2 y ..25^nd it is satisfied the stability condition. Step 8 of 8 Procedure to draw root locus plot: • Take real and imaginary lines on X axis and Y axis respectively.l + 2 y -l-2 y )-(0 ) (4 )-(0 ) =-I Thus. the angie of asymptotes are [45 ^ .6°-63. The roots of J V (j)= 0 are called the zeros of the problem. Find the value of K at that point. the four poles are 0..• a») Step-by-step solution step 1 of 8 Consider the general form of characteristics equation i+ o .ta n -'^ I j . the centre of asymptotes is O Step 3 of 8 Consider the formula for the angle of asymptotes. l8 0 °+ 3 6 0 °(2 -l) 4 -0 = 135» Substitute 3 for /.— ^ and Dc(s) = j (j2 + 2 s +5) s +2 Without using Matlab. state what the location of the mulitple roots is.t a n . (sum of finite poIes)-(sum of finite zeros) (numberof finite poles )-(nunU>er of finite zeros) ( 0 . 2 2 ^ . 4 for n and 0 for m in equation (5). l8 0 °+ 3 6 0 « (/-l) (5) n —m Where. the imaginary axis crossings are |l.225j Y + 4 ( . !.225y) ' + = 0 I0(-I-H. Figure Unity feedback system ..2 . the corresponding value of/C is 4 and 16. 180**+ 3 6 0 » (4 -l) 4 -0 -315« = -135» Thus.25 Thus. Problem 5. Figure2 Hence.2 . .25 = 0 P) The roots of the equation (7) are —2±1. —l + 2y'and —1—2y- Step 2 of 8 Consider the formula for the centre of asymptotes.90°+180° — 116.. From Figure 1. Substitute 16. 4 for n and 0 for m in equation (5). |-45 °|and | . the characteristics equation is given below. The roots of D ( j) = 0 are the poles. • All the terms in the first column of the Routh’s array must be positive sign.225j f + 9 ( .l -t-1. 4 for n and 0 for m in equation (5). the root locus is plotted for the given transfer function and it is shown in Figure2. the stable condition is 0 < AT < 16. . Consider the number of poles and zeros from the characteristics equation. (6) Apply Routh-Hurwitz criteria to equation (6 ).1 for s in equation (6).l +1. . p =6 K ^16 Kz = 16 So. l + G ( j) D c W = 0 „ j f f i ± £ ¥ 'l = o 5* ( 5 + p ) + ( 5 + z ) » 0 s **p s **K s *K z ^ 0 Step 2 of 2 ^ The dominant poles are at s = .2 is.2 ± 2j.24PP Suppose the unity feedback system of Fig. has an open-loop plant given by G(s) = 1/s2.W . Figure Unity feedback system . ( i + 2 ) [ ( i + 2 ) ' + 2’ ] = 0 ( j + 2)( j *+45 + 8) = 0 j ’ + 6 s ‘ + I f a + I6 = 0 Compare the characteristic equation of the system with the desired characteristic equation.A r ( ^ ) ( l) The characteristic equation of the system is. Assume a real pole at j s —2- The characteristic equation of the system with dominant poles at s = —2 ± j 2 and a real pole at 5 = . Design e lead compensation Dc(s) = to be added in set of the closed-loop system are located at s = .• a») Step-by-step solution _ Step 1 of 2 The loop transfer function of the system with lead compensation is. Z»1 s designed lead compensator .2 ± J 2 . G W £ > . Problem 5. the design conditions are met with the selected values. 5+ Z ^(^ + />)(s + 3 )(s + 6 ) The error equation is. j ( s + 3 ) ( i+ 6 ) The lag compensator transfer function is. » step(sys_f) Step 4 of 4 Get the MATLAB output for the step response.den): » sys_^feedback(sys. • The steady-state emor to a unit-ramp input must not exceed 10%. Step-by-step solution step 1 of 4 Refer to Figure 5. Problem 5. Thus. that is.9% { less than 17%). E= 1+ ^ (^ + p )(^ + 3 )(^ + 6 ) Step 2 of 4 Determine the steady state error to unit ramp input. the designed lag compensation is. The open loop plant transfer function is.60 in the text book for the unity feedback system. Observe from the step response that the settling time is 3. Hence.>)(^ + 3)(5+6) + (j + r) J _ l Bp The steady state error to unit ramp input is less than 10%. » z=200*p: » num=25*[1 z]. . Figure Unity feedback system -T . Z z> \Z 0 p Select z = 2 0 0 p Step 3 of 4 ^ Select p and z using MATLAB by trial and error method.1).y(5 + /> )( i + 3 )(^ + 6 ) + ( « + «) (» + p )(n -3 )(j+ 6 ) 1 « l im *. = I im j£ ( 4 I s t im j s-tO + p ) ( 5 + 3 )(5 + 6 ) s lim j 1 1 «-»o . • The step response overshoot is to be less than 17%. e . » den=conv([1 p 0].42 s (less than 5 sec) and the overshoot is 13.25PP Assume that the unity feedback system of Fig. • The step response settling time is to be less than 5 sec.0001.^ —\ 2 > P 5+ P The loop transfer function is. has the open-loop plant ^ J(l + 3)(s + 6)' Design a lag compensation to meet the following specifications. » p=0..1.[1 9 18]). D { s ) .* + . » sys=tf(num. 0. I « E ^- 1+GZ> 1 1+ - s { s+ p ){ s * 3 ) ( j + 6 ) The input is unit ramp. 4 6 Hence. ^ T (s .R\ . 120® Substitute for A = -> /3 i+ i- J ^ ( £ Z £ l= . (-\± jS )(-i± jS + i) ' (-i± jS ){± j^/3 ) ( ± jS ± i) Closed loop poles lie on root locus and m ust satisfy the angle criterion. (b) Design a lead compensator Dc(s) = that will meet the specification. Step 2 of 3 (b) is the lead compensator.4 6 . Take positive sign and check the angle criteria Z r ( 5 ) s . 1 3 (2 ) 3 Substitute — for z ■ P B ) . calculate the value of z . Problem 5.4 6 — 0. 3 (2 ) ..V 3 Substitute 3 for pz. to meet the required specifications .{ s ) = K s + 6 .4 6 Therefore. Write the loop transfer function. — if ] — 30« Take negative sign and check the angle criteria.l g 0 " .46 Therefore. are satisfied if the closed-loop poles are located at :s = —l ± j ^ .5 + 0 .6 . Figure Unity feedback system Figure Unity feedback system * ° —■ Gt*) I o Step-by-step solution step 1 of 3 (a) Assume is the proportional control. Dc(s) = kp.4 6 D . 3 Z-— P 3 . pz=Z Step 3 of 3 Determine the phase angle at ^ s > 1 ± y'-Ts • r. ^ =0 r/7* + 2d ^ = Since the closed loop pole is located at .26PP A numerically controlled machine tool positioning servomechanism has a normalized and scaled transfer function given by I C ( j) = ■ J(5+ !)■ Performance specifications of the system in the unity feedback configuration of Fig. required specifications cannot be achieved using proportional control alone.. ^ T [s ) — 30® Therefore. A (*)= « — ' ' s+p Substitute Ja> for s . r ( « ) = f l. =tan''| —|-tan''| —I (1) Differentiate equation (1) with respect to d*. 1 r n I (1 ^2) ( a»Y l i 'n f j p dw 2^+0^ p ‘ *<^ Equate equal to zero and find & . if D ^(s) is replaced by proportional control alone then the obtained closed loop system is not satisfied the angle criteria. (a) Show that this specification cannot be achieved by choosing proportional control alone.46 5 + 0 . the lead compensator transfer function is fC 5 + 6 . r(«)=*.? * + 6 p -3 * 0 p — 6. W G W 1 Substitute • ] ± for s . D (jeo) Determine the phase angle of lead compensator £ . So. r . 27PP A servomechanism position control has the plant transfer function 10 G (j) = - j(j+ l)(j+ IO ) You are to design a series compensation transfer function Dc(s) in the unity feedback configuration to meet the following closed-loop specifications. (e) Give the Matlab response of your final design to a reference step. G et help from a Chegg subject expert. (b) What is the velocity constant Kv for your design? Does it meet the emor specification? (c) Design a lag compensation to be used in series with the lead you have designed to cause the system to meet the steady-state error specification. specifications. ignoring the error requirement.4 sec. • The steady-state emor to a unit ramp at the reference input must be less than 0. • The response to a reference step input is to have no more than 16% overshoot. (d) Give the Matlab plot of the root locus of your final design. Problem 5. • The response to a reference step input is to have a rise time of no more than 0.05. Step-by-step solution There is no solution to this problem yet. ASK AN EXPERT . (a) Design a lead compensation that will cause the system to meet the dynamic response specifications. ignoring the emor requirement. 02) Step 3 of 3 The dominant poles of the closed loop system is.5.2. 0 2 ) Therefore.l+ y + 0 . = Urnj G(5) = lim ^l —r ^ — r 1 s lim (5 + 1) >1 Assume the lag compensator having transfer function.2 K .47 for in equation (1).> — ’ 0.9 S * j) 1.> 5 The factor by which steady state gain is to be increased is.414)(1)(1. the transfer function of the compensated system is. Write the formula for steady state error to a ramp input for type 1 system.0 . (a + 0 . Step 2 of 3 The required steady state error to a unit ramp input Is less than 0.0 2 ) ( -i+ y + 0 1 ) ' ( .1) D(*) = (*+ 0 .2 K. has a feed forward transfer function Design a lag compensation so that the dominant poles of the closed-loop system are located at s = -1 ± yand the steady-state error to a unit-ramp input is less than 0. Figure Unity feedback system Step-by-step solution Step-by-step solution step 1 of 3 Consider the following feedfonvard transfer function. — < 0 .345K.1 ) ' s(j +1)(5+0.2 -S Place the zero and pole of the lag compensator very close to the origin. 1 <?(*) Calculate the velocity error constant for uncompensated system.47 (*+ 0 . the transfer function of the lag compensator is. ------------ \p j K y o f uncom pensated 0.47(*+0.28PP Problem 5. d ( .1) =1 '* ( * + l) ( * + 0 .0 2 ) Therefore. the transfer function of the compensated system is 1.2.4) X =1. i= 5 P M . ( 1) ' ' '( * + 0 . ( * + 0.47 1. =1 (1. Let zero of compensator at a s .1 Calculate the value of pole using the following factor.5 P p = 0M Therefore.0 2 ) . Apply magnitude condition at this location of dominant closed loop pole.0 2 ) i-O-9+j) '( .i* J ) { J ) ( . ----. (*+ 0 .1 - That is.l + .1 ) D(*) =1. Where the values of z andp are small. ) = a: . K . j s -1 ± y .28PP Assume that the closed-loop system of Fig. z = 0. ) ( _ l + y + l) ( .0 . 02s"-20 s"-1000 Step 5 of 5 D (s)= ^ fc !^ l let 1=31. the voltage e on the photo detector is related to the ball displacement x (in meters) by e = 10Ox. Problem 5.5i+20rH3flg Step 2 of 5 b.5Vg^ng V„=2mg=0.665 D (s)= 0.5 V (s)+ 20X (s) 0. The mass of the ball is 20 g and the gravitational force is 9. (a) Write the equations of motion for this set up. K=0. (c) What is the transfer function from u to e? (c) What is the transfer function from u to e? (d) Suppose that the control input u is given by u = -Ke.02s’ -20] E (s )_ r 50 V (s) 0. and p that yield improved performance over the one proposed in part (d).29PP An elementary magnetic suspension scheme is depicted in Fig. (e) Assume that a lead compensation is available in the form values of K. Sketch the root locus of the closed-loop system as a function of K. For Equilibriiun at X = 0.02s^-20 Step 4 of 5 K50 2500K 0.392V Step 3 of 5 c. 7 ( H .8 N/kg. sm X (s)= 0 .6 ^^ s4p p=50 for ^ . z. (b) Give the value of the bias VO that results in the ball being in equilibrium at x = 0. Figure Elementary magnetic suspension Step-by-step solution step 1 of 5 =0. The upward force (in newtons) on the ball caused by the current / (in amperes) may be approximated by f= 0. 0. For smail motions near the reference position.5/+ 20x.5V (s)= X (s)[0. The power amplifier is a voltage-to-current device with an output (in amperes) of i = u + VO.665^ ^ ^ ' ^ ^ '■ ' (s+50) . . the required step response is plotted. (b) Use Matlab to plot the system’s response to a reference step. y l a b e l ( ’ imag p a r t ’ ).. x l a b e K ’ r e n l p a r t ’ ). . which is the regular positive locus for -G .. r l o c u s (nuihr den^ k) ... Thus. Step 4 of 4 The step response obtained is RwipnnM* 04 ■ A .. t i t l e ( ' R o o t . Thus. W ith the 0 (^ ) as given... < -0 4 m m : : - ■• ■ o ' .. ■. MATLAB needs to plot the negative locus.■ •6 ■ ... Step 2 of 4 ^ (b) The MATLAB program to find the root locus of the given S3rstem is n u m = [0 2 .. figure s t e p (niunr d e n ) . d e n = [l 1 9 ].. Step 3 of 4 ^ The root locus obtained on execution o f the above program is. Problem 5.. ’ x ’ ) .. 1 . [the value of gain for closed loop roots at dangling o f 0.. '2 • . The locus is plotted below. 4 .0 0 5 :1 0 . .30PP A certain plant with the nonminimum phase transfer function is in a unity positive feedback system with the controiler transfer function Dc(s).....707.To get a positive ou ^u t we would use a positive gain in positive feedback....4 ] . . Step-by-step solution Step-by-step solution step 1 of 4 4 -2 s A certain plant with the nonmininmm phase transfer function = -y s^ + s + 9 Is in a unity positive feedback system with the controller transfer function D(s)....l o c u s ’ ). (a) Use Matlab to determine a (negative) value for Dc(s) = K so that the closed-loop system with negative feedback has a damping ratio ^ = 0. 'With all the negatives. s g rid %plot ( r ... the problem statement might be confusing.7 is ^ = -1. k = 0 :0 . 8 ic 12 Tim* {wei • " ■ The final value o f the step response plotted below is -0. A .887 ...04|. ls+ l At K=5. find a value for the gain K that will provide the maximum damping ratio. the lead compensator stabilizes the system.1 sec time constant and unityDCgain. (b) Assume that the sensor transfer function is modeled by a single pole with a 0. Using the root-locus procedure. Damping ratio i s maximum . H . Figure Block diagram for rocket-positioning control system Figure Block diagram for rocket-positioning control system X Step-by-step solution step 1 of 2 Root locus shows that the system is a stable system Step 2 of 2 b.31 PP Consider the rocket-positioning system shown in Fig.( s ) * O . (a) Show that if the sensor that measures x has a unity transfer function. Problem 5. (s) ^ s * + 1 2 7 . (a) Find the locus of closed-loop roots with respect to K.i o o j : R{s) s ' +12s" + 40s" + 1 0 0 r s + 1 0 0 r s (s ’ + 1 2 s^ + 4 0 s+ 1 0 0 r) ^ ” s * + 1 2 ^ + 4 0 s ’ + 1 0 0 ^ s + 1 0 0 i: t „ = I m s S [s) ■When R(s) = - Step 7 of 8 100 m .29.6 6 3 Break away point: s = -5 . . (c) What is the steady-state error {e = r .y) for a step change in r? (d) What is the steady-state error in y fo r a constant disturbance w1? (d) What is the steady-state error in y fo r a constant disturbance w1? (e) What is the steady-state error in y for a constant disturbance w2? (f) If you wished to have more damping. ot. Step 8 of 8 100 s^(s^+ 12s+ 40) l ( i .36. Assume /C= 2 for the remaining parts of this problem.92® Step 6 of 8 100 jg' r(s) s ^ (s“+ 12s+ 40) (c) W ) (1+ 5)100^ 1+ s’ (s^+ 12s+ 40) ^ lOQg_________ s ^ (s“+ 1 2 s+ 4 0 )+ (H -s)1 0 0 i: ^ lOQg__________ 7 + i2 7 + 4 o 7 + T o o F s + Io o Z ^ s*+125^+405^+ i o o j : s + i o o g . s* + 12s* + 4 0 7 + 200s + 1 0 0 „ . 0 .336^ a +336 = 0 I 12 r 12s^+336 = 0 s = ±J5.29 The root loci intersect the imaginary axis at 5 s ± ^5.6 0 7 + 1 0 0 i: s + 1 0 0 g ^ .( s ) l + k ( ' + l ) Ts “(s“+ 7 T 12s+ 40) 100 7 + 1 2 7 + 4 0 7 + 1 0 0 ^ ( s +1) J’(s) _ 100 (r. = 0 I . M 8 0 .= -3 .7 + 1 2 7 + 4 0 7 + io o i:(s + i)-io o .[s )-r(s ) .. a a Asyn^totcs: 60®. -6 ± J 2 K = a points: .2® ^ = 254..9® = -74.36 M 8 0 -1 0 0 ^> 1 a ^ — . K >0 iooi!:(480-iooj!:) > 14400^ 4 8 0 0 0 ^ -1 0 0 0 0 i:^ -1 4 4 0 0 £ ’ > 0 . Problem 5.56®+90® = 413.r ( s ) s*+ 1 2 7 + 4 0 7 + 1 0 0 iT s+lO O ^-lO O s’ W. . . (e) ir . ^^ sV l2s^+ 40s^+2005+200 ^ When Ifa(^) ~ constant *„ = Im s fl(s ) I e.92® And. . ____ ____ s’ + 12s+ 40 100 7+127+40 ________ lOOs^___________ 7 ( 7 + 1 2 s + 4 0 )+ ir(l+ s )(1 0 0 ) l'( s ) 1007 ir.2 j) K = 0 points: 0. . (b) Find the maximum value of K for which the system is stable. Step 4 of 8 Angles o f departure: = 180®-<t> Where = 2 < b > -Z 'fe 2 < t^ = 161.(s) “ 7 + 1 2 7 + 4 0 7 + 1 0 0 i:(s + l) E {s ) = W r.1 .0 0 0 ^ “ +33600^: > 0 10.000A: > 33600 K > 3.1 0 .( s ) 7 + T 2 7 + 4 o 7 + io o ^ 7 fio o Z B (s) = » I ( s ) . what changes would you make to the system? Figure Control system S te p -b y -s te p s o lu tio n Step 1 of 8 iooii:(s+i) (a) D O H {s ) = s ’ (s + 6 + 2 j) ( s + 6 0 . ^ = 74.^— .32PP For the system in Fig.12® 2<t!r = 158. 180®.8 Characteristic equation: s'(s'+ 1 2 a + 4 0 )+ 1 0 0 J!:(s+ l) = 0 s‘ +12s’ + 4 0 s^ + 1 0 0 rs+ 1 0 0 r = 0 Step 2 of 8 7 1 40 100 K 7 12 100 K 7 480-lOOJi: 100 K 12 7 I0 0 ji:(4 8 0 -i0 0 ji:)-i4 4 0 0 i: 480-100^: 7 100 K Step 3 of 8 From the above array.— js ^ + m x = 0 At K = 3. “ 7 + i2 7 + 4 o 7 + io o i:s + io o ^ Substituting K = 2. 300® Centroid = --------.56®+161. s* + 1 2 7 + 4 o7 + iooa : s + iooji: ‘ «„ = Im s £ (s ) When = constant I «„ = 0 I . and let M = 1. and show that K and z can be chosen to meet (b) Repeat part (a) assuming that Dc(s) = K(s + z). adjust the controller parameters so that the specifications are met. In this problem. (d) Now suppose that the small mass m is not negligible. but with a practical controller given by the transfer function Dc{s)=K*- s+ p Pick p so that the values for K and z computed in part (b) remain more or less valid. (c) Repeat part (b). This is the transfer function relating the input force u(t) and the position y(t) of mass M in the noncoilocated sensor and actuator problem.33PP Consider the plant transfer function bs + k C(i) = ' fiimMfi + (Af +m)te + (M + m)k] to be put in the unity feedback loop of Fig. A SK A N EX PERT .1 sec and an overshoot of less than 10%. b = 0. but is given by m = MA 0. (a) Approximate G(s) by assuming that m ^ 0. If not. and show that K and z can be chosen to meet the specifications. Problem 5. Figure Unity feedback system O— — » G (i) I O Step-by-step solution There is no solution to this problem yet.1. G et help from a Chegg subject expert. Can K be chosen to satisfy the performance specifications? Why or why not? (b) Repeat part (a) assuming that Dc(s) = K(s + z). Check to see if the controller you designed in part (c) still meets the given specifications. You may use Matlab for any of the following questions. and Dc(s) = K. we will use root-locus techniques to design a controller Dc(s) so that the closed-loop step response has a rise time of less than 0. k = 1. ASK AN EXPERT . ic f i-ina values of me gams Kpana kuror uc{sj = Kp+ kusmat meet me aesign specincations with at least a 10% margin. Using root-locus techniques. Problem 5. (a) Show that proportional control alone is not adequate.34PP Consider the Type 1 system drawn in Fig.7. G et help from a Chegg subject expert. (b) Show that proportional-derivative control will work. and (2) the damping ratio ^ = 0. (c) Find values of the gains kp and kD for Dc(s) = kp + kDs that meet the design specifications with at least a 10% margin. Figure Control system Step-by-step solution There is no solution to this problem yet. We would like to design the compensation Dc(s) to meet the following requirements: (1) The steady-state value of y due to a constant unit disturbance wshould be less than V s. ASK AN EXPERT . G et help from a Chegg subject expert. After completing each hand sketch. (a) Us) = (b) ^(*) — fZ(r4iO)(yi6s+2S) (c) UA) — 10)^1^4. Evaluate the time response using Simulink.5at www. find the Dc(z) that is the discrete equivalent to your Dc(s) from Problem using the trapezoid rule.35PP Using a sample rate of 10 Hz. Problem 5. {Note: The material to do this problem is covered in the Appendix W 4.com or in Chapters. and determine whether the damping ratio requirement is met with the digital implementation.F(i+ iW ?+ 4l+ 5) Step-by-step solution There is no solution to this problem yet. Turn in your hand sketches and the Matlab results on the same scales.25) iA\ . Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s).) Problem Mixed real and complex poles.FPE7e. verify your results using Matlab. (*+3)(J^+4»4«) . Problem 5. /Co = 10 V/rad. (a) What is the range of the amplifier gain KA for which the system is stabie? Estimate the upper limit graphicaily using a root-locus plot.. km = /Cf= torque constant = 0.1 Vsec Ra = armature resistance = 10Q. va = KA(ei .inn a rnnt-lncii<. G et help from a Chegg subject expert.36PP Consider the positioning servomechanism system shown in Fig.1 N m/A. eo = Kpotdo. A SK A N EX PERT . (a) What is the range of the amplifier gain KA for which the system is stabie? Estimate the upper limit nranhicallv u<.7. Where are all three closedloop root locations fo this value of KA7 Figure Positioning servomechanism Step-by-step solution There is no solution to this problem yet. (b) Choose a gain KA that gives roots at 0. Ke = back emf constant = 0.ef). JL + Jm = total inertia = 10-3 kg m2. Gear ratio = 1:1. nint va = KA(ei .ef). T = motor torque = Kt ia. where ei = Kodi. =1.8) 7 w “ (* + 1)(J^+ ! . t-= ------= = ^ ^ 4 m s >|a^>0.815 = 0. Show on the same plot the region of acceptable pole locations corresponding to the specifications. ( ! 1 ^ S4*-0.05. G (s)D (s) s(s+ l)(s“+ U s + l) M . k0.2 and Kp=0.37PP We wish to design a velocity controi for a tape-drive servomechanism.75x0. and select the best (b) Assume a proportional-integral compensator of the form kp(s + a)/s. Mp < 0. ts < 15 msec.588 for a.58&+:j 0.381rad/msec ii-cos’V § ) ^ . Problem 5. (b) Assume a proportional-integral compensator of the form kp(s + a)/s. indicate the closed-loop poles with a dot (-) and include the boundary of the region of acceptable root locations..^ 15ms . Sketch the root-locus plot of your design.S) a. giving values for kp and a.91 + 0. ! » + 1 ) ' We wish to design a Type 1 feedback system so that the response to a reference step satisfies tr< 4 msec. and select the best possible values of kp and a you can find. and the velocity constant Kv your design achieves.6 . Step-by-step solution step 1 of 3 _ 1 5 k .8 = 0.= e ^ t = ----.9s+0.840rad/m8ec| Step 2 of 3 Step 3 of 3 d (0 = k . (a) Use the integral compensator k l /s to achieve Type 1 behavior.(s “+0.05 = 0. The transfer function from current l(s) to tape velocity QfsJ (in miiiimeters per miilisecond per ampere) is Q (l) I3 (l2 + 0. On your plot. and sketch the root locus with respect to k l . Eq. of mass me holding an inverted uniform pendulum of mass mp and length / with no friction are Eq. (a) Draw a block diagram for the system with V input and both Y and 6 as outputs. Draw the root locus with respect to the gain of De(s). (3). For this problem. These equations can be used to compute the transfer functions Eq. G et help from a Chegg subject expert. and the input is force normalized by the system weight v = . . 2-3 e 1 r V J2CJ2-1)' In this problem you are to design a control for the system by first closing a loop around the pendulum. cart position. y + p0=v. ASK AN EXPERT . (e) Use Matlab to plot the control. (d) Design a controller De(s) for the cart position with the pendulum loop closed. 2-3 e 1 Eq.s) = for the d loop to cancel the pole at s = -1 and place the two remaining poles at -4 ±. let the mass ratio be me = 5mp. Time is measured in terms of r = toot where motion y is measured in units of pendulum length as ^ C(4«c-Hllp) y = | | . The new control is U(s).1 0-9 =-v.38PP The normalized.75. where the force is V(5) = U(s) + De(s)&(s).Draw the root locus of the angle loop. Eq. (2). and then. (c) Compute the transfer function of the new plant from U to Y with De(s) in place. scaled equations of a cart as drawn in Fig. and pendulum position for a unit step change in cart position. Figure Figure of cart pendulum Step-by-step solution There is no solution to this problem yet. with this loop closed.4.. Problem 5. where fi = is a mass ratio bounded by 0 < /3 < 0. (b) Design a lead compensation Dc(. closing a second loop around the cart plus pendulum. 0184*[1 0. Use the following MATLAB code to plot the step response of the function: num = -0. rad/sec.2647)(s+ 0. Gd = tf(num.den). sysci = sysw*feedback (Gd. Take the pole -0.[1 0.) If the steady-state value of the heading due to this wind gust is more than 0.den). Consider that the maximum allowable rudder angle deflection is less than |0« for a 5^ change in heading angle.39PP Consider the 270-ft U.0063]).1111 for the compensator. the required compensator is I j .).0063) * fis ) 0.0063]). Gd = tf(num. Figure 2 Observe that the settling time is less than 50 s. rad/sec.[1 0.1III) Use the following MATLAB code to sketch the step response. Select a value on the root locus to get the values for gain and roots. Note that the maximum deflection in rudder is possible at the initial instant. Hence. rad.1). sysci = feedback(Gd. w ( i) 0. is calculated when = 0 >so the controller <7^ j acts as a feedback. The second-order transfer function that shows the response of the heading angle is. Problem 5. is the negative real part of the pole. Figure USCG cutter Tampa (902) If Vf J f Xf M/ M Step-by-step solution step 1 of 10 Refer to Figure 5. and w is the wind speed. G (5 ) = I. if)r= reference heading angle.2647)(i+0.0184*[1 0. Find the transfer function of r(s)- r -0. Coast Guard cutter Tampa. deni =1.0068 Step 6 of 10 The value of fc is less than the required gain value. Select a point in the graphics window selectedjx)int = 0. w = wind speed.0184*[1 0. the transfer function of the compensator is. there is no change required in the design. den = conv([1 0. r = yaw rate. (c) Check the response of the closed-loop system you designed in part (b) to a wind gust disturbance of 10 m/sec. Step 2 of 10 (a) Consider a step change in rudder angle to calculate the settling time of r .S.018 4 (t+ 0. Substitute 10 for ^ (O ) and $ for io = /:5 5 Consider that the settling time of ^ to a step change of is less than 50 s.1111]. Note that the value selected for gain should be less than 2. Here.2 i -0. step(sys2) Step 7 of 10 The step response of the closed loop system is.2605*-0. Figure 1 Thus. Gc = tf(num1.2647)(j+0. (Model the disturbance as a step input. The settling time of to a step change in ipr is specified to be less than 50 sec.2647)(s+0. .00021 K= 1. step(Gd) Step 3 of 10 ^ The open-loop step response of the function r ( ^ ) is. sysl = tf(num1. r = yaw rate.0063/0. Therefore. ess = 1/Kv The output of the code is.0068) "(j+0. Response for disturbance. design a compensator that uses ifj and the measurement provided by a yaw-rate gyroscope (that is. rad. Substitute 50 for 71 to find the value of <r. numi = 1. sysw =10. step(syscl) Step 9 of 10 ^ The step response of the system for 10 m/s disturbance is. W ( j ) = — • ' s Calculate the response with disturbance. g is the rudder angle.0068) S is ) s is +0.0000064 j(s+0.2605(5-0.0068) |_s(j+0.1111]. Parameter identification based on sea-trials data (Trankle. Coast Guard cutter Tampa (902) shown in Fig. The output of the code is.0063) Here.0063) * where (fj = heading angle.roots] = rlocfind(syscl) The result of the code is the root locus.2647]. numi = 1. 4.0063) Draw the step response of the function and locate the settling time for a single step change.5.den). the steady state error is very much less than 0.[1 0.conv([1 0. modify your design so that it meets this specification as well. The result is that the response of the heading angle of the ship ip to rudder angle 5 and wind changes w can be described by the second-order transfer functions f(5) -0. m/sec.2647]. Here.0000064 G^(s) = ■wW ~ s is + 0.1111 -0. (a) Determine the open-loop settling time of rfo r a step change in & (b) In order to regulate the heading angle ifj.0068].01g4(t+0.0063])). ^ is the heading angle.0068 is present to eliminate the zero in the system and -0.2605*[1 -0. Step 8 of 10 (c) 10 Consider that the disturbance.2647)(5+ 0. Therefore.sys1).den1). 1987) was used to estimate the hydrodynamic coefficients in the equations of motion.0063) The second-order transfer function that shows the wind changes is. if)r= reference heading angle. rad.1111 . Thus. 6 = rudder angle.0063]). ess = 0. 0.3754) in the compensator. Step 10 of 10 Use the following command to find the steady-state error of the system: Kv= 10*1.0068. rlocus(syscl) [K.6 50 *0.0068].Gc).den).0063) j(j+ 0 . sysT=series(sys. den = conv([1 0.0068].[1 0. by ^ = /■. % r(s) is taken in the open-loop to make the overall system equal to second order othenvise take Gd(s) with additional pole (that is 0.0184(j-I-0. rad.0184(1+0.092 Step 5 of 10 Use the following MATLAB code to find the gain and roots on the root locus.0068]. den = conv ([1 0].W G c ( ^ ) Use the following MATLAB code to find the response of the system for 10 m/s disturbance: num = -0. den = conv([1 0.3754 is present to eliminate the pole in the system.2647/0.2647]. deni =1.1111/0.0068) i(s+0.1). r isthe yaw rate in rad/s.0184*[1 0. sys = tf(num.5°. y (4 G A s) » 'W l + G . and for a 5° change in heading.den1).Q .67 in the textbook for the 270 ft U. sys2 = feedback(sysT.0647 Therefore. Observe the roots to note that -0.2647)(i+0.0184*0.3754 0.S. the maximum allowable rudder angle deflection is specified to be less than 10°. num = -0.2605*[1 -0.2647]. -0. Gd = tf(num. The general expression for the step response is.2605 roots = -0. m ! ] - Note that the design of compensators can be done in many ways. num = -0.0068*-0. the open-loop settling time of r is |21Q si • Step 4 of 10 (b) Design a compensator to regulate the heading angle. [1 4 5]).9352 + O. Step 8 of 10 (g) Observe from Figure 3 that the maximum damping factor of the complex roots is |Q. r» 0 i ( j + 10) + 4 s + 5) Enter the following code in MATLAB to draw the root locus. ft Figure 2 Step 6 of 10 (f) The loop transfer function is.7 ' ' * ( s + 10)( j ’ + 4 j + 5) Determine the steady state error of the system. the system is unstable. (a) What value of K is required to keep the steady-state error in 6 to less than 0.5 . Overshoot (%): 0.894 V. „ J+2 ^ 1 + 20 The plant loop transfer function is.-1 3 . Step 10 of 10 (i) Adding lead compensation to the design lowers damping. » num=conv([1 2]. A:(h -2)(» + 3) <?w = ’ i ( i + 10)(s + 2 0 )(j’ + 4 i + 5) Enter the following code in MATLAB to draw root locus.[1 4 5]). y ( j + 3) T (s) j ( i + 1 0 ) ( j ^ + 4 i + 5 ) + 3 r ( i + 3) Substitute 600 for K ■pi^\ _________ 6 00(^+ 3)__________ ' ^ " s ( j + 1 0 ) ( s ’ + 4 j + 5) + 60 0 (i + 3) 600(^ + 3) ' s (j + 1 0 )( s * + 4 j + 5) + 600( j + 3) ________ 600(^4-3)________ ** + 14j ’ + 45 j ’ +650 s + 1800 Enter the following code in MATLAB to find the roots of the system. In order to automatically adjust for the sudden weight shift due to passengers rushing to the bar when it first opens. The transfer function of the system is. Connect the Rate gyro as feedback at the input. 600(»4-3) (s + 1 0 ) ( s '+ 4 i+ 5 ) <?w = N .) (f) For the rate gyro in part (e).[1 3]).9 . the range of K for stability is reduced with the introduction of extra lead network. (e) You are given a black box with rate gyro written on the side and told that. observe that the value of A for which the system roots move to right-half is 150. K (s + 3 ) (s + 1 0 ) ( V + 4 i+ 5 ) <?w = 1.40PP Golden Nugget Airlines has opened a free bar in the tail of their airplanes in an attempt to lure customers.3 3 Thus. (Include transfer functions in boxes. Support your comments using Matlab or with rough root-locus sketches.den): » rlocus(sys) The following is the MATLAB output: Step 3 of 10 (c) From Figure 1. when installed. .5014+ 0.02.2 ± 6 . with a maximum expected value for MO of 0.894|• Step 9 of 10 (h ) Observe from Figure 3 that for maximum damping factor. » sys=tf(num.24 • - •10 -15. .02 rad 1“)? (Assume the system is stable.6 y | • As the complex roots lie to the right-half of the s-plane. » C E = [1 14 45 650 1800].2183+ 6._ M £ ± 3 ) 0 (* + 1 0 )(j* + 4 j + 5) The characteristic equation is.0000i 1.[1 4 5 0]). Show that this value yields an unstable system with roots at s = -2. * 0 .6284i -2. » sys=tf(num. Assume that K = 600 as in part (d) and draw a block diagram indicating how you would incorporate the rate gyro into the autopilot. » num=[1 3]: »den=conv([1 10]. with output K T ' 6. Determine the loop transfer function.5.2183-6. 1+^.2 R eal A xis (seccmds*^) Figure 3 Thus. Step 5 of 10 (e) The value of K is 600. e) (* + 1 0 )(j* + 4 j + 5) Substitute 600 for K 600(»4-3) (s + 1 0 ) ( s '+ 4 i+ 5 ) a { s )- N . Thus. Figure Golden Nugget Airlines autopilot Step-by-step solution step 1 of 10 (a) Refer to the block diagram in Figure 5.6284i 1. '■ Root Locus 10 [System:sy$ Gain: 0 Pole:-2+ li .+ l. V AT(5 + 3) <7(j) » — ------.-13. it provides a perfect measure of 0. (.We will model the passenger moment as a step disturbance Mp(s) = MO/s. » den=conv([1 30 200]. what is the value of K when the system becomes unstable? (d) Suppose the value of K required for acceptable steady-state behavior is 600.) (b) Draw a root locus with respect to K. Figure shows the block diagram of the proposed arrangement. sketch a root locus with respect to K T . the value of is [ ^ .den): » rlocus(sys) The following is the MATLAB output: Observe from Figure 4 that. (c) Based on your root locus.2±6. » num=[1 3]: » den=conv([1 10 0]. » roots(CE) ans = -13.den): » rlocus(sys) Step 7 of 10 The following is the MATLAB output for root locus with respect to K j. £ s l im 5 G ( f ) A:( j + 3 ) s lim j i ( i + 1 0 ) ( j* + 4 5 + 5 ) ^ ( j+ 3 ) s lim -»“ ( j + I0 )(5 “ + 45 + 5) 50 Equate the steady-state error to 0.187 (rad/s):2.68 in the text book.12 *10 -6 .4 . » sys=tf(num.+1.OOOOi The roots of the system are at |5 = -2 . Consider the following lead network. what is the value of K when the system becomes unstable? (c) Based on your root locus.9._ M £ ± 3 ) 0 (* + 1 0 )(j* + 4 j + 5) Draw the block diagram of the system by connecting the block. +3) < + ^ r7 = 0 ’^( j + 1 0 )(5 *+ 4 j + 5) Enter the following code in MATLAB to draw root locus. Dancing: 0. the value of A is h sol- Step 4 of 10 (d ) The value of K is 600. the root locus is shown in Figure 3. K {s + y ) j ( 5 + 1 0 )(i* + + 5) 5 ( j + I0 )( j ^ + 4j + 5) The characteristic equation of the system is. the airline is mechanizing a pitch-attitude autopilot.6).6. Discuss the advantages and disadvantages of adding an integral term and extra lead networks in the control law. (g) What is the maximum damping factor of the complex roots obtainable with the configuration part (e)? (h) What is the value of KT for part (g)? (i) Suppose you are not satisfied with the steady-state errors and damping ratio of the system with a rate gyro in parts (e) through (h).' '----------. the value of K is Step 2 of 10 (b) The transfer function of the system without disturbance is. The plant transfer function of the system without disturbance is. Problem 5. £=6. If you can.5| Step 3 of 4 b.41 PP Consider the instrument servomechanism with the parameters given in Fig. K=55000 D ( s ) = f . (c) Lag network: Let J+1 and Deis) = K- s+ p Using proportional control. let z=7 p=42 . roots are at the samepositioo.4? Select K and p so that the dominant roots correspond to the proportionalcontrol case but with Kv = 100 rather than Kv = 12. Select K T and K so that the dominant roots are in the same location as those of part (a). Figure Control system Step-by-step solution g W =t t s(s“+51s+550) Step 2 of 4 a.5+j26. (b) Output-velocity (tachometer) feedback: Let H(s) = 1 + /Crs and Dc(s) = K. Step 4 of 4 c.04| |K=18000 I.s)=K. and indicate the location of the roots corresponding to your final design: (a) Lead network: Let s+z = De(. Compute Kv.^ lx 5 5 0 0 0 |8a=-11. H (s)=l+KjS for |Kt=0. For each of the foliowing cases. is it possible to obtain a Kv = 12 at 0. draw a root locus with respect to the parameter K. s+ p Select z and K so that the roots nearest the origin (the dominant roots) yield ^2 f> O A (b) Output-velocity (tachometer) feedback: Let H(s) = 1 + /Crs and Dc(s) = K. Then. |K=110000| . Problem 5. give a physical reason explaining the reduction in Kv when output derivative feedback is used. ^^ s4p -= ^ = 1 0 0 ^ K=55000p 550p let p=2. F (i+ iw ? + 4 l+ 5 ) Problem 6 RHP and zeros. Turn in your hand sketches and the Matlab results on the same scales.+i)(.25) (* + 3 )(J^+ 4 » 4 « ) . (c) Us) = ^ (d) l. (b) Us) = ^ 2^ . (b) W = (<=) = (t) = ^ Problem 5 Mixed real and complex poles. ASK AN EXPERT . j 2(j + 1 )( j + 5 ) ■ (a) Draw the real-axis segments of the corresponding root locus.l) t( jW + 3 ] S te p -b y -s te p s o lu tio n There is no solution to this problem yet. verify your results using Matlab.+5)(. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). t h e magnetic levitation system with integral control and lead compensation. Problem 2 Real poles and zeros. After completing each hand sketch. W «*> = ?T 5 T i5 (b) t ( j ) _ Problem 4 Multiple poles at the origin. ■ = 0. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). After completing each hand sketch. | . Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the listed choices for L(s). verify your results using Matlab. verify your results using Matlab. Be sure to give the asymptotes.42PP Plot the loci for the 0* locus or negative K for each of the following: (a) The examples given in Problem 1 (b) The examples given in Problem 2 (c) The examples given in Problem 3 (d) The examples given in Problem 4 (e) The examples given in Problem 5 (e) The examples given in Problem 5 (f) The examples given in Problem 6 Problem 1 For the characteristic equation K 1+ .+io) (b) K s) = (c) Us) — fd) K s) = f+ray Problem 3 Complex poles and zeros. Sketch the root locus with respect to K for the equation 1+ KL(s) = 0 and the listed choices for L(s).j(. Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. verify your results using Matlab. Turn in your hand sketches and the Matlab results on the same scales. and the arrival and departure angles at any complex zero or pole. Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole.( s ) _____ What Is the largest value that can be obtained for the damping 5(j-f-20)^ —2i+2) ratio of the stable complex roots on this locus? (*> . ( j. (b) Sketch the asymptotes of the locus for K - (c) Sketch the locus (d) Verify your sketch with a Matlab plot. Turn in your hand sketches and the Matlab results on the same scales. Turn in your hand sketches and the Matlab results on the same scales. After completing each hand sketch. verify your results using Matlab. Problem 5. (a) L(J) = (b) Ufl) — j2(. G et help from a Chegg subject expert. Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. Be sure to give the asymptotes and the arrival and departure angles at any complex zero or pole. (a) Us) . After completing each hand sketch. After completing each hand sketch.+io)(y^+25) (c) Us) — 10)^1^4. Turn in your hand sketches and the Matlab results on the same scales.the model for a case of magnetic levitation with lead compensation. (a) Us) = ^ io ^ | . 8 6 6 y Step 2 of 7 ^ Consider the formula for the asymptotes.0. 360“(1-1) 2-1 = 0» Substitute 2 for /.5 -0 ...0. ^ l80°+36(y*(/-l) n —m Where.8 6 6 ^ - Step 5 of 7 Consider the formula for the asymptotes.. .5. • Draw the root locus. ji ..866y)-(-I) (2 )-0 ) =0 Thus.0.866y'and -0 .5 -0 . and draw the asymptotes from centroid at an angle of 0»and 360® • Draw the root locus. • Locate the asymptotes on the real axis.5 + 0. • Locate the asymptotes on the real axis. the asymptote is .866j . n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoles)-(nund>eroffinitezeros) (-0..866y’ and -0 .5 + 0.866y'and -0 . 180°+360o(l-l) 2-1 =180» Substitute 2 for /. 5+ 1 1 + o r. + (1 + or) j + ( I + t f ) 1 ^ * + j + l+ o rj+ o r 1 # * + j+ l+ o r ( j+ l) Consider the characteristics equation for the given function. the angle of asymptotes are [ ^ a n d |36Q^|. 3 for n and 1 for m in equation (3).5 + 0. * = « : ! ) . • Mark the poles on the real axis.5 -0 . The root positive locus plot is shown in Figure 1.. The negative root locus plot is shown in Figure 2. the positive root locus is plotted for the given transfer function and it is shown in Figure 1. (5) The roots of the equation (5) are -0 .1 in the above transfer function and rearrange the equation for negative root locus form.5. i+ „ (^ K i± l)= o m S -t-i + l To find zeros put numerator N {s ) = 0 Thus.8 6 6 y - Thus. 2 . Step 4 of 7 ^ Consider the characteristics equation for the given function. the one zero is . the angle of asymptotes are | i 3Qo|and |54Q^|. S te p -b y -s te p s o lu tio n step 1 of 7 Step 1 of 7 Consider the following given function. . the asymptote is . To find poles put denominator D{s) = 0 • »*+S + I = 0 . _ 180^+360^(2-1) 2-1 *540* Thus. Step 6 of 7 Procedure to draw the negative root locus plot: • Take real and imaginary lines on X axis and Y axis respectively.866y-0.5 -0 . Hence. RootLocus step 7 of 7 Thus.. 1 5+ 1 _ l + a -3----. Consider the formula for the angle of asymptotes.5.43PP Suppose you are given the plant ^ ^ a®+ (I + + (I + o)* where a is a system parameter that is subject to variations.5+0.-= 0 Multiply by .. Step 3 of 7 Procedure to draw the positive root locus plot: • Take real and imaginary lines on X axis and Y axis respectively. • Mark the poles on the real axis. To find poles put denominator D(s) = 0 • j '+ J + l. the two poles are -0 . the system is stable for all l a > .. (6) n -m Where.866y)-(-!) (2)-0) =0 Thus. 2 for n and 1 for m in equation (3). the one zero is .866yand -0 .5+0.43PP Problem 5.O (2) The roots of the equation (2) are -0 . 3 for n and 1 for m in equation (3). the two poles are -0 . Use both positive and negative root- locus methods to determine what variations in a can be tolerated before instability occurs. RootLocus Figure 1 Hence. Number of poles is n Number of zeros is m Substitute 1 for /. the negative root locus is plotted for the given transfer function and it is shown in Figure 2. the given system is stable for all ^ . Consider the formula for the angle of asymptotes.l l .m Substitute 1 for /.1 . Number of poles is n Number of zeros is m / = l. 360®(2-l) 2-1 =360® Thus.1 . and draw the asymptotes from centroid at an angle of I go* and 540«. (1) j + i+ i To find zeros put numerator Thus.8 6 6 y - Thus.5 + 0.. n -m (sumoffinitepoIe$)-(sumoffinitezeros) (numberoffinitepoles)-(nunU>eroffinitezeros) (-0. From Figure 1 and Figure 2. 2 for n and 1 for m in equation (3). Problem 5.44PP Consider the system in Fig. (a) Use Routh’s criterion to determine the regions in the /C1, /<2 plane for which the system is (b) Use ritool to verify your answer to part (a). Figure Feedback system . ( . + IX .+ IX 5 ) Step-by-step solution step 1 of 1 SA: 760 SR: 3976 Sketch the given figure. 1 4 ( 4 + l ) ( j + 0 .5 ) We know that = ir^and k^ = and the characteristic equation is s* + 1.5s’ + 0.5s^ + i , s + * ,= 0. The Routh table is / 1 0.5 1.5 0 1.5x0.5-ifc_ 0 ^ 1.5 ' -\.5 k , s' For the Routh table, we get Jt^>0 1.5x0.5-yfc, — 1.5x0.5-A ^ > 0 k^ <0.75 So, \K, <0.75 Again we have, (l.5 x 0 .5 - * ,) t,- 2 .2 5 i,> 0 k ‘, + 2.25k,-0.15k, <0 4 * J - 3 i , + 9 * ,< 0 |4 i!:i^ - 3 i!:i+ 9 j:ig , < o| This represents aparabola in ^k^, ib,Jplane. w The region of stability is the are a under the parabola and above the k^ axis as shown in the figure below. Thus, the regions in the (iT^, plane for which the system is stable is determined. Problem 5.45PP The block diagram of a positioning servomechanism is shown in Fig. (a) Sketch the root locus with respect to K when no tachometer feedback is present KT = 0. (b) Indicate the root locations corresponding to /< = 16 on the locus of part (a). For these locations, estimate the transient-response parameters tr. Mp. and ts. Compare your estimates to measurements obtained using the step command in Matlab. (c) For K = draw the root locus with respect to K T . (d) For K = 16 and with KT set so that Mp = 0.05 = 0.707). estimate fr and fs. Compare your tn th p actual v p Iii p p nf tranrl /.<; nhtpinprl ii.<;inn Mntlph (d) For K = 16 and with KT set so that Mp = 0.05 = 0.707). estimate fr and fs. Compare your estimates to the actual values of tr and ts obtained using Matiab. (e) For the values of K and KT in part (d). what is the velocity constant Kv of this system? Figure Control system Step-by-step solution S te p i of 15 Given block diagram o f a positioning servo mechanism. Step 2 of 15 The root locus with respect to . ^ v ^ n no tachometer feedback is present is Root Loois Step 3 of 15 (b) The root locations corresponding to .^ = 16 on the locus o f part (a) is Step 4 of 15 'With iT s 16 p&om the root locus, the transient response parameters are The overshoot is Af, =44.4% . The dancing ratio is ^ = 0 .2 5 . The frequency is at^=A rad/s. The rise time is 2 .16^+ 0.6 ir = - 2.16x0.25+0.6 = 0.285 8 C = 0.285 s| The settling time is 4.6 =- 4.6 ■0.25x4 . = 4.6 s Step 5 of 15 The MATLAB code to plot the step response of the system » num = 16; » d e n = [ l 2 16] ; » s y s = t f (mun, d e n ) Step 6 of 15 T ra n s fe r fu n c tio n : 16 3 ^2 + 2 s + 16 » t= D : 0. 0 0 0 1 :5 ; » y = s te p ( s y s , t ) ; » p lo t ( t,y , ’b * ) » g rid » x la b e l t » y la b e l y Step 7 of 15 • The step response ofthe system using MATLAB is Step 8 of 15 From the step response. The overshoot is Af^ = (1 .4 4 -l)x l0 0 i / =44.4% The rise time is /^ = 0.4322-0.1173 = 0.3149s The settling time is It. =4.783 s I Hence the parameters almost agree with that obtained from root locus. Step 9 of 15 (c) For ^ = 1 6 th e root locus with respect to ^j<is » num = 16; » d e n = [ l 3 0] ; » s y s = t f (niuttf d e n ) T ra n s fe r fu n c tio n : 16 » rlo c u s (s y s ) » g rid Step 10 of 15 The transfer function with iT = 16 and with tacho feedback is r{ \ ^ ® ^ ° s ^ + (2+JC-..)s+16 Given, ilf, = 0.05 C = 0.707 Step 11 of 15 So, we have, a, = ^ = 4 rad/s ec Also, 2Cat^ = 2 + K j. 2x0.707x4 = 2+£'j. K j. = 3 .6 5 6 K j. « 3.7 Now, the transfer function is 16 T {s) = s" + (2 + 3 .7 )s+ 1 6 16 s" + 5.7s+16 Step 12 of 15 The rise time is _ 2 .1 6 ^+ 0 .6 2.16x0.707+0.6 = 0.532 s J = 0.532 s| The settling time is 4.6 C<On 4.6 '0 .7 0 7 x 4 = 1.63 s| Step 13 of 15 The step response using MATLAB is Step 14 of 15 From the above response. The rise time is i , = 0.6677-0.1265 L = 0.5412 s The settling time is h =1.683 s| Hence the results agree with that obtained &om the estimatioa Step 15 of 15 w The velo city constant o f the system is = lim s (__ l_ s ( s + 2 + i: r ) J K ~2-\-K j. 16 “ 3.7 + 2 = 2.81 Thus, we get |J:, = 2.81| Problem 5.46PP Consider themechanical systemshown in Fig., where g and aO are gains. The feedback path containing gs controls the amount of rate feedback. For a fixed value of aO, adjusting g corresponds to varying the location of a zero in the s-plane. (a) With g = 0 and r = 1. find a value for aO such that the poles are complex. (b) Fix aO at this value, and construct a root locus that demonstrates the effect of varying g. Figure Control system Step-by-step solution Step 1 of 4 L (s) = =>11( g ^ ^ ) s ( ts+1) Step 2 of 4 a. ■ts^+s+a^s+a,=0 s^+s+a^=0 for g=0 and x=l. For getting complex conjuguate roots, l< 4a^ ““4 Step 3 of 4 4s^+4s+l+gs=0 H - ^ = 0 ( 2 ^ Step 4 of 4 Vu Root Locus X > -0 - - 1/2 lu iu iu u e ii. 5 .4 7 P P Step-by-step solution step 1 of 1 Step-by-step solution step 1 of 1 G (s)— 5- s For the above transfer function, as3miptoti c angles are 6 0 '. 180', 300', Two roots will always be in right half plane if pole is not cancelled Hence,plant G (s )= -^ cannot be made unconditionally stable if pole s cancellation is forbidden Problem 5.48PP Prove that the plant G(s) = 1/s3 cannot be made unconditionally stable if pole cancellation is forbidden. luiuiuueii. Step-by-step solution ste p 1 of 1 G (s)— 5- s For the above transfer function, asymptoti c angles are 6 0 '. 180', 300', Two roots will always be in right half plane if pole is not cancelled Hence,plant G (s )= -^ cannot be made unconditionally stable if pole s cancellation is forbidden Problem 5.49PP For the equation 1 + KG(s). where J (J + P )[(I+ 1 )2 + 4 ]’ use Matlab to examine the root locus as a function of/<for p in the range from p = 1 to p = 10, making sure to include the point p = 2. Step-by-step solution step 1 of 2 Step 1 of 2 G (s)= - A t p=2. Two roots are asymptotisc to imaginary axis Step 2 of 2 Problem 6.01 PP (a) Show that oO in Eq. (4), with A = Uo and (jjo = oj, is «o = [ g W — = - V o G ( - jo ) ) ^ , and (b) By assuming the output can be written as y ( 0 = O b*->“ ' Hariwo Pnc = derive Eqs. (1)-(3)- Eqs. 1 y(») = AM cos(aib> + ^), Eqs. 2 M = |G (7 < h ,)| = |C(j ) | ^ = y (Re[GC/<o„)]}2 + {Im[G(yii^)])2, Eqs. 3 Eqs. 4 S -p \ M -p i M -P a S + J te t S -J to , S te p -b y -s te p s o lu tio n There is no solution to this problem yet. G et help from a Chegg subject expert. ASK AN EXPERT 0894 -26. and 100 rad/sec. S te p -b y -s te p s o lu tio n S te p -b y -s te p s o lu tio n step 1 of 2 (a). Problem 6. 5. and compare these with your computed results from part (a).0980 -11.846x10-^ -78.0447 -63. (b) Sketch the asymptotes forGfsj according to the Bode plot rules.7" 7 100 9.0995 -5.10.02PP (a) Calculate the magnitude and phase of 1 G is) = - s+ lO by hand foroj = 1. G (s)= — ^ ^ s+10 M i^ itu d e = |G(jo[))|= Vffi^+ioO Phas e = Z.6" A 10 0.7* 2 2 0. 50.4" 6 50 3. 20.95X10-’ -84.3" 3 5 0.07071 -45" 5 20 0.3" Step 2 of 2 .G (joo) =-tan'^ j s.no ai(rad/s) Magnitude Phase 1 1 0. 2. 1) (f) L(s) = i ( j + l) ( i+ 8 ) ^ (g) L(s) = i ((J-|-5)(J-|-10) j+ l) ( j+ l0 0 ) (h) L(s) = (s + 4I0j (5-|. verily your result using Matlab.I> +100 l)(O J s + l) 1 (c) L ( i ) = 5 (s + T 7 ^0 7 E r+ T y I tdl L(s) = '" ■ (d) L(s) = 1 (T P T p tT P W m +4) (e) L (I) = i( J + l) ( J + l( I0 ) I000(<-1-0. G et help from a Chegg subject expert.03PP Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions.10) 0 )(j+ 3 0 0 ) 5 (i) L(s) = S te p -b y -s te p s o lu tio n There is no solution to this problem yet. ASK AN EXPERT . Turn in your hand sketches and the Matlab results on the same scales. Problem 6. After completing the hand sketches. (a) L(s) = 2000 j(s+ 20 0) (b) L(s) = l(O . 2 0 l o g ^ + I =20log(0.20dPfdfradfto-40dB/decadedue to the (ill) At presence of (ya>+l) in the denominator. -20dB/decade- (N) The Initial low frequency slope due to pole at the ohgin is at=1 lad/sec■ ^Idpe changes from . (iv) At a r = S rad/sec >tde slope changes from -40 dB/decade t o -60 dB /decade due to the presence of (y'ffl+5) in the denominator.56° 1 -125.den). = I rad/sec a>2 -2 rad/sec ai.97 dB. Step 8 of 25 (b ) i(s) = i(j+1X4+5X4+10) O'ar+2) L(jd.13 Step 5 of 25 Draw the magnitude and phase plots as shown in figure 1. den=[1 16 65 50 0]. verify your result using Matlab. sys=tf(num.^ + lj i(ya»)= ^‘^'^'>(f+')(f+') B reak o r c o m e r fiequencies: cs. oXrad/sec) * 0. sys=tf(num. Step 4 of 25 Consider the phase values.24 Step 11 of 25 Draw the magnitude and phase plots as shown in figure 3.97° 5 -132.02 y < » ( / < » + l) [ ^ + l) g + l) Breakorcomerfiequencies: ai| = Irad/sec 0^=5 rad/sec ai^=10lad/sec Step 2 of 25 The magnitude of L(ja>) in dB is.16° 1 -111.1 -92. num=[1 2]. Step 16 of 25 Consider the phase values. .1 0 rad/sec step 21 of 25 A Follow the steps to draw the magnitude plot. the slope changes from -20dB/decadeto-40dB/decadedi presence of (yVu+lO) in the denominator.0 2 |-2 0 Io g |/< ti|-2 0 Io g |y < » + l|-2 0 lo g r ^ + I .=5 rad/sec 0)4 -6 rad/sec 0 . Ibe slope changes from . (iv) At at=2 rad/sec.8° IK -359. Bodsooponi Step 24 of 25 Execute the following MATtAB code. k>g|£. the slope changes from -20 dB/decade to -40 dB/de< presence of (yru+ 10) in the denominator.(yV»)| = 2 0 Io g |0 ./ intersects the 0 dB line at o ) = l ra d /s e c - (ill) At 4u = I ra d /s e c . = 1 rad/sec atj=2 rad/sec CD. (ii) The Initial low frequency slope due to pole at the ohgin is -20dB/dccade - (Hi) At u > = 1 rad/sec >Ibe slope changes from -20 dB/decade to -40 dB /decade due to the presence of ( y ® + i) in the denominator. bode(sys) Step 13 of 25 A Obtain the magnitude and phase plots as shown In figure 4.= S rad/sec a i.02)-20log(<»)-20log(Vl+<»’]-201og|^l+^yj 2 0 > o g [f(^ ] step 3 of 25 Follow the steps to draw the magnitude plot. Problem 6. oXrad/sec) * 0. * 1 rad/sec a >.8° 100 -174.2° 10 -145° 100 -175. Step 14 of 25 (c ) L U ) _____ ' ’ 4(4 + 1X4 + 5X4 + 10) L(ja>) - (/ar+2X/flH6) yaX/oH-1)(/ru+5)(/ru+10) B reak o r c o m e r fiequencies: to.) - yaX/aH-l)0'a>+S)(/ar+10) 0 ./ ^ d this intersects the 0 dB line at o)=l rad/sec- (Hi) At u)=1rad/sec>Ibe slope changes from -20dB/decadeto-40dB/decadedue presence of (yV»+l) in the denominator.94° IK -269. Bodeouopom Step 6 of 25 Execute the following MATtAB code. . the slope changes from .6° 10 -135. = 5 lad /se c. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop transfer functions. (i) The constant term 0.428“ 1 -152“ 5 -240. den=[1 16 65 50 0].46° Step 23 of 25 Draw the magnitude and phase plots as shown in figure 7. Figure6 step 20 of 25 (d) i ( j ) = -----(4 -f 2 )(4 + 4)----- ' ’ 4(4 + 1X4 + 5X4 + 10) (/<ir+2)()ar+4) L (y ® ) = .37° IK -179. (i) The constant term 0. (vi) At ^ = 1 0 rad/sec >lbe slope changes from -40 dB/decade t o -60 dB /decade due to the presence of (yVu+lO) in the denominator.04PP Real poles and zeros.4 dB.1 -97. (ii) The Initial low frequency slope due to pole at the ohgin is -20dB/decade . num=[1 10 24]. the slope changes from 0 dB/decade t o -20 dB/decad< presence of (y a )+ 5 ) in the denominator./a)(/a)+1)(/ru+5)(/a)+10) B reak o r c o m e r fiequencies: ra. Turn in your hand sketches and the Matiab results on the same scales.dteslopechangesfrom -60 dB/decade t o -80 dB /decade due to the presence of (yat+lO) in the denominator. bode(sys) Step 19 of 25 Obtain the magnitude and phase plots as shown In figure 6.24 causes an increase in magnitude of -12. num=1.1 -93. B oeetsogram Step 18 of 25 Execute the following MATtAB code.6° 1 -115.i ----------------- ’ y-fflOaH-lX/ai+SX/ffl+lO) > 0. B o d e D ia g ra m . tbe slope changes from -20 dB/decade t o -40 dB /decade due presence of (ya)+5) in the denominator.T 100 -350.23° IK -179. (vii) At rp s 10 rad/sec.58° Step 17 of 25 Draw the magnitude and phase plots as shown in figure 5. (vi) At 4u = 5 lad/seC'lbe slope changes from -40 dB/decade t o -20 dB /decade due presence of ^yu )+ 6 ) in the numerator. (i) The constant term 0. (vii) At a i= l0 rad/sec.den). (v) At <u= 4 rad /se c . (v ) At u r = 5 rad/sec >Ibe slope changes from -20 dB/decade t o -40 dB /decade due to the presence of (ya>+ 5) in the denominator. tbe slope changes from -20 dB/decade to 0 dB/decade presence of (y ru + 4 ) in the numerator.1 -94.1 0 rad/sec step 15 of 25 Follow the steps to draw the magnitude plot.= 1 0 rad/sec step 9 of 25 ^ Follow the steps to draw the magnitude plot. (a) = iii+ T T § T W + T D 5 " .KJ-l-l)(J+S(a+10) (d) L (s ) = _tJ + 2 )(a + 4 ) Step-by-step solution step 1 of 25 L (a )- a(j +lXa-l-SXa+10) t ( /<») -------------------. num=[1 8 12].4° 5 -120. sys=tf(num. (iv) At ^ = 2 rad/sec >lbe slope changes from -40 dB/decade t o -20 dB /decade due to the presence of (y<u+2) in the numerator.40 dB/decade to -20 dB/deca presence of (y u )+ 2 ) in the numerator.04 causes an increase in magnitude of -28 dB. sys=tf(num. the slope changes from .2 0 dB/decade to -40 dB/deca presence of (yV »+l) in the denominator.0 4 ^ . (i) The constant term 0. (ii) The Initial low frequency slope due to pole at the ohgin is -20dB/decade. <o(rad/sec) * 0. den=[1 16 65 50 0]. den=[1 16 65 50 0]. After completing the hand sketches.44° 5 -172° 10 -204° 100 -261. bode(sys) Step 25 of 25 ^ Obtain the magnitude and phase plots as shown In figure 8.40 dB/dfcadfto-20dB/decadedue presence of (y ru + 2 ) in the numerator.24 causes an increase in magnitude of -12. (iv) At 0) = 2 rad /se c . (v ) At <u= 5 rad/sec. Step 10 of 25 Consider the phase values. <o(rad/sec) * 0. Bodeefoonn Step 12 of 25 Execute the following MATtAB code.4 dB. (vi) At n . (v ) At dt=10rad/sec.26° 10 -282.=2 rad/sec <o. Step 22 of 25 Consider the phase values.den).02 causes an increase in magnitude of -33.den). bode(sys) Step 7 of 25 Obtain the magnitude and phase plots as shown In figure 2.= 4 rad/sec 0)4=5 rad/sec 0 . and approximate the transition at the second-order break point. i (j 2 + L + 1 0 ) (e) L(S) = + 1) Step-by-step solution There is no solution to this problem yet. Turn in your hand sketches and the Matlab results on the same scales. verify your result using Matlab. After completing the hand sketches. based on the value of the damping ratio.05PP Complex poles and zeros. Problem 6. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop transfer functions. ASK AN EXPERT . G et help from a Chegg subject expert. G et help from a Chegg subject expert. ASK AN EXPERT . (a) U s) - Step-by-step solution There is no solution to this problem yet. After completing the hand sketches. Turn in your hand sketches and the Matlab results on the same scales. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop transfer functions.06PP Multiple poles at the origin. Problem 6. verily your result with Matiab. 52° 1 -146.07PP Mixed real and complex poles. Embellish the asymptote plots with a rough estimate of the transitions for each break point. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop transfer functions.24 .5° 5 -24.05J(i>fl lio J [ J Break or comer frequencies: 0 ^ = 2 rad/sec 0)2=8-24 rad/sec 0)3=9.1 -93. and this asymptote intersects the OdB line at CD= Irad/sec (c) A t CD= 2 rad/s e c . U .77° 100 -84.-.-. verify your result with Matlab.1 -179° 1 -173.29° Step 4 of 20 ^ BooeoKQion Step 5 of 20 (s+2) W.58° 2 7.22 -301.2 4 JC D + 1 in ttie denominator. . the slope changes from OdB /decade to -40dB /decade due to presence of + 0 .-. Step 15 of 20 Phase plot: 0(rad/sec) 0.6 (e) At CD= 10 rad/sec.15° 5 -228. . L (s) = s(r+10)(s’+2s-t2) (Jm+2) L ( J cd) = J(D(Jo&t-l 0) (-(D^-t-2J(Brf2) L(Jm) = U J + Jo + l ■ fe)’ Break or comer frequenci es: = ->J2 rad/sec gq> 2=2 rad/sec is^~10 rad/sec Step 2 of 20 Magnitude plot: (a) The constant term '0. the slope changes from -40dB /decade to -20dB /decade /■j® ^ due to presence of I — +11 in the numerator.6 | (e) At CD= 10 rad/sec.1—-20 dB.-.55° 100 -175.707] (d) A t ixi= 2 rad/sec. (b) The initial low frequency slope due to pole at tiie origin is -20dB/decade. -(!)■ Since 2 ^ = 6 and cc^=5 . the slope changes from 0 dB/decade to -20 dB/decade fJot due to tiie presence of I — +11 in the denominator.63 dB. (b) The initial low frequency slope due to the presence o f two poles at the origin is -40 dB/ decade.1' causes an increase in magnitude of 20 log 0.016* causes an increase in magnitude of 20 log 0.1 -179° 1 -161.29° Step 20 of 20 .15° Step 12 of 20 ^ BoMOogum Step 13 of 20 T _ (!+2)( s“ -M s+68) . the slope changes from -40dB /decade to OdB /decade due to presence of +J©+1 in the numerator. the slope changes from -20dB /decade to -60dB /decade { due to presence of "I y j +0.2431 (e) At 0D= 9.1 -180.16'causes an increase in magnitude of 20 log 0. ■ te)' Since 24co^ = 2 and <Og='j2.24J<»I-1 Break or comer frequencies: ® j= 2 rad/sec 0)2=5 rad/sec 0)3=10 rad/sec Step 10 of 20 M s^ itu d e plot: (a) The constant term *0. I. I. the slope changes from -40dB /decade to -20dB /decade {Ja. I. Since 2413^=4 and o^=9.16° 10 -208. (b) The initial low frequency slope due to tiie presence of two poles at the origin is -40dB/decade.^ ( s+ 1 0 )( = M s+85) jf j \ _ CT(d+^(-< d^44J0^-68) flea ^ r ^ mV 1 L ( J cd) = ■ -1 — 1 +0. i (j + 1 oW + 2 j + 2 ) ^ ( j + 1 0 )(J + 6 i+ 2 5 ) (c) r ( x \ = ______ ^ ( j+ 1 0 )(j^ + 6 s + 2 5 ) (c) r ( x \ = ____ f’ + p ^ ______ ^ ( j+ 1 0 )(j^ + 6 s + 2 5 ) ' ' j ^( j + I0 ) ( s^ + 4 j +85 ) (e) U s ) = + ji( l+ 2 M l+ 3 ) S te p -b y -s te p s o lu tio n Step 1 of 20 (s+2) (a).0 0 8 ^ ^ |+ lj L(Jcj) = 24JfiH-l Break or comer frequencies: 0 ^ = 2 rad/sec 0)2=5 rad/sec a>3=10 rad/sec Step 6 of 20 Magnitude plot: (a) The constant term *0.^ . the slope changes from -60dB /decade to -80dB /decade due to the presence of I — +11 in the denominator. and tiii s slope intersects tiie OdB line at 0 = Irad/sec (c) At 0D= -72 rad/sec. = 0.. 4 = 0. (e) At 0D= 3 rad/sec.46° 8.06° 9. At Q0= 5 rad/sec.44° Step 16 of 20 BodedKQKm Step 17 of 20 w.o ^ + 6 J cd+ 2 5 ) 0 . L (0 = s^ (s+10) (s^+6s+25) (-<D^+4JayH) L(Jm) = -<D^(Jco+1 0)(-co^+6Jcd+25) 0. UO ) Step 11 of 20 Phase plot: 0(rad/sec) 0.65° 100 -172° Step 8 of 20 Booe otagnm step 9 of 20 (s+2)^ (c). the slope changesfrom -60dB/decadeto -40dB /decade (Jot \ due to presence of I “^■*'11 erator. the slope change s from . the slope changes from -40dB /decade to -60dB /decade (J<x> ^ due to the presence of I — +11 in the denominator.24JCO+1 in the denominator.24 -111.77° 100 -84.16=-16 dB. the slope changes from -20dB /decade to -40dB /decade {J a due to tile presence of I — +11 in the denominator. At 0 = 8.33'causes an increase in magnitude of 20 log 0.13° 5 -24. I. andttiis slope intersects the OdB line at 0) —Irad/sec (c) A t o = J2 rad/sec. Turn in your hand sketches and the Matlab results on the same scales. M = 0 .008= -42 dB.58° 2 -131° 5 -340.008' causes an increase in m s ^ t u d e of 20 log 0. Step 19 of 20 Phase plot: 0(rad/sec) 0 0. \ due to presence of I — +11 in the numerator.06J1S+1 in the numerator.86° 10 44.36° 10 -107. 4 = 0.40dB/decade to OdB/decade due to presence o f I -I ^ I + Jfl^l in the numerator.22 rad/sec 0)4=10 rad/sec Step 14 of 20 ^ Magnitude plot: (a) The constant term *0. Step 3 of 20 Phase plot: 0(rad/sec) 0. L (s ) = '■ ' s^s+2)(s+3) _ (-a)*+2JGi+2) L ( J cd) ~ -m’ (J(ii+3)(Jai+2) 0.33 r ^V 1 L(Jm) = ■ comer frequencies: 02=2 rad/sec 03=3 rad/sec Step 18 of 20 Magnitude plot: (a) The constant term *0. '[-(fj- Since 2^cc^ = 4 and cc^=2 (<Q At ®= 5 rad/sec. the slope changes from -60dB /decade to -80dB /decade n® due to the presence o f -----1*1 in &e denominator.70^ At 0 = 2 rad/sec. Problem 6.016 = -3 6 d B (b) The initial low frequency slope due to the presence of two poles at the origin is -40dB/decade.56° >12 -150. the slope changes from -20 dB/decade to -40 dB/decade (Jot due to tiie presence of I — +11 in the denominator.42° 1 -132. Step 7 of 20 Phase plot: 0(rad/sec) 0. the slope changes from -20dB /decade to +20dB /decade due to presence of -f—T +0.24 rad/sec. (e) At iX)= 10 rad/sec. ■teJ Since 2^gd^ = 2 and cd^ = . 24 J Since 24o^ = 4 and 13^=8. 4 = 0.1 -177. L (0 = s^(s+10)(s^+6s+25) (Jarf2) L(Jco) = -<D^(Jco+1 0 ) ( .64° 1 -158. After completing the hand sketches.2171 (e) At 0D= 10 rad/sec. 4 = 0.33=-9.22 rad/sec. the slope changes from +20dB /decade to -20dB /decade due to presence of 05JO+1 in dte denominator. And this asyrnptote intersects the OdB line at ® = Irad/sec (c) At CD= 2 rad/sec.and this ass3nnptote intersects the OdB line at o = Irad/sec (c) At CD= 2 rad/sec.97° 100 -263.22 . the slope changes from -20dB /decade to -60dB /decade due to presence of +Ja5rl-1 in die denominator. (b) The initial low frequency slope due to tiie prasence of two poles at the origin is •40dB/decade.016 -(f) L(Jco) = • ■H). Since 2 1 ^ = 6 and cc^=5 .86° 10 44. 1(2))) = -236.1(6))) = -140® rad Therefore.25<»*-l)(l+0. Term slope changeinslope(<l%^) -40 ( i) ' -1+y® 20 -40+20=-20 Step 17 of 27 rad Calculate the magnitude of . 4+ 2 £ ( 4) = - 4(4-1){4+6)= Substitute y® for 4 . s+2 1 i( s ) = (s + 1 0 )(s = -4 ) j+ 2 s’ + 1 0 s = -4 s -4 0 Draw the MATtAB code to plot the response of £ ( s ) . ^ .tan"' (0. ® 2 X (y ® ) i0"= -270 10“ -217. num=[1 2 1] den=]1 38 322 -720 800 0 0] sys=tf(num.2 Start with the low frequency asymptote (J< s) The magnitude plot of this temr has a slope of -2 0 dB per decade. Substitute y® for s . Turn in your hand sketches and the Matlab results on the same scales.2 0 (y ® )= + i2y ® + 6= -4 0 -2 0 .5 (1 0 ))-tan "' (0. sketch the asymptotes of the bode plot magnitude and phase for the open loop transfer function £ ( s ). the slope shifts to -4 0 dB per decade - 1 Consider the fourth low frequency asymptote (1+o. num=[1] den=]1 3 3-7] sys=tf(num.iy®) The magnitude and phase of the function are.ly®) The magnitude plot of this temr has a slope of -2 0 dB per decade. Step 14 of 27 Draw the MATtAB code to plot the response of £ ( s ) .5y® ) y®((y®)’ -i)(i+o. the slope shifts to -6 0 dB per decade. -0.1 rad = 10 Step 10 of 27 0. The magnitude plot of this temr has a slope of -4 0 dB per decade.iu — (-0. Problem 6. the phase curve rises to —236. verity your result with Matlab.1(6))) = -230® rad Therefore. ^(l=+(0.1(500))) ^ rad The phase curve again falls to —lg0« at at = 500---. sketch the asymptotes of the bode plot magnitude and phase for the open loop transfer function £ ( s ). rad Frequency — sec F ig u re s ste p 18 of 27 Draw the MATtAB code to plot the response of £ ( s ) .5 -2 — sec _1_ °0.- sec ste p 13 of 27 Determine the phase Z £(y® ) at sec Z £ ( y 2 ) .) = - s ( s + 1 0 ) ( s = -l) Substitute j a for s . Consider the third low frequency asymptote (l+0.2( y ® ) + 2) The break point frequencies are pp s l i ^ . At the break point frequency of the pole.— ^ in dB at nr= 10' ( » sec A.5y<g) ” (-0.05 start with the low frequency asymptote -0.1)) -1 ° -28.2 (y® )+ 2 ) ■irS) ste p 20 of 27 ^ Tabulate the terms of £ (y ® ) in the increasing order of their frequencies. sketch the asymptotes of the bode plot magnitude and phase for the open loop transfer function £ ( s ).25(y(0.i-(0.tan"' (0.2 0 + 40 = 20 1 -4 0 20 .2 The phase curve starts at ^ = -270® corresponding to the low frequency asymptotes .3® at at .den) bode(sys) The Bode plot for the function £ ( 4 ) by using Matlab is shown in Figure 10.- ^rad sec Combine all the asymptotes to draw a composite curve. Step 23 of 27 (e) Write the expression for £^ 5 ).2 10= -9 0 Combine all the asymptotes to draw a composite curve. I<») Step 2 of 27 Determine the breakpoint frequencies.2(l+0. rad Therefore. I Consider the third low frequency asymptote (1+O. et = 2 0 ^ ^ . = 2 0 log (7(0.25(y(0.028 dB Step 4 of 27 Consider the second low frequency asymptote (l+0. Teim slope change in s l o p e ( < l % ^ J _1_ -2 0 i<o (y ® )= + 2 ( y ® )+ i 40 .2 s + 2 ) Substitute y® for s .2 ^ ^ - sec ZL (J2) . At the break point frequency of the zero. (a) =. Step 27 of 27 Write the MATtAB code to plot the response of £^ 4 ).4 1 4 ! ^ .1 = 102^ sec Step 3 of 27 0.- sec rad Determine the phase ZL (y® ) at at = 500---- Z £ (y 5 0 0 ) = (-1 8 0® + tan"'(0.Sm) .08PP Right half-plane poles and zeros. in dB at at = 10"' (y®) sec 0. F igure? ste p 22 of 27 Draw the MATtAB code to plot the response of £ ( s ) .5 (6 ))-ta n "' (0.- sec ste p 6 of 27 Determine the phase Z 6(y® ) at a t .l 0. __1_ *■' " 0.4 0 = -6 0 Determine the phase at various values of ® . + 2)X+ 3] Step-by-step solution s te p ! of27 (a) Write the expression for .(yi0) -(-2 7 0 ® + tan"' (0.1(10))) — 2363® . rad The phase curve again falls to —236. £ (y ® )= — (y®)‘ + 2 (y®)+i y®(y®+ 20 )= ((y®)= . the phase curve rises to —230® at n» = 6 ---- rad Determine the phase Z £(y® ) at nr=10 Z£.- sec Determine the phase Z £(y® ) at nr = 500---- rad Z£.6 ---- sec Determine the phase Z £.3® at n> = 2 ---.(-180®+tan"' (0 . Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop transfer functions.5®)=) |£ (y ® ) |= - ® (-< t^ -l)^ (l= + (0 . ) -------------------------------- y ® ( y ® .5y® ) The magnitude plot of this temr has a slope of 20 dB per decade. sketch the asymptotes of the bode plot magnitude and phase for the open loop transfer function £ ( 4 ).i) ( y ® + 6 ) 2(i+yo.( . The magnitude plot of this temr has a slope of -4 0 dB per decade. . Step 26 of 27 (f) Write the expression for £^ 4 ).4 0 — 60 ( ( y ® ) = .5(500))-tan"'(0. . Make sure that the phase asymptotes properly take the RHP singularity into account by sketching the complex plane to see how thexLfsJ changes as s goes from 0 to + After completing the hand sketches.3® at ® = 1 0 ---.2 0 ( y ® + 20)= 1 -4 0 . a-f2 I L(a) ( i + I O ) ( j" .3® .l ) 0.1 ® )') Z L { ja ) = -90® -180® + tan"'(0.05 x . Calculate the magnitude of .2 0 log 0.tan"' j Tabulate the terms of L{jip) in the increasing order of their frequencies.25(yV»)’ . L i j a . Term slope change in s l o p e ( « l % ^ J -2 0 y® y ® -i -2 0 . Step 15 of 27 (c) Write the expression for £ ^ s ) . num=[1 2] den=[1 10-1 -10 0] sys=tf(num. the slope shifts to -6 0 dB per decade - Step 12 of 27 0. the slope shifts to -2 0 dB per decade. sketch the asymptotes of the bode plot magnitude and phase for the open loop transfer function L{s).tan"' ( .3® at a r= 1 0 ---.6 ^ ^ - sec Z £ ( y 2 ) .25<»’ -l)^(l*+ (0. _ |) [ ( .den) bode(sys) Step 8 of 27 The Bode plot for the function L (s ) by using Matlab is shown in Figure 2. Step 19 of 27 (d) Write the expression for £ ^ j ) . s= + 2j + 1 £ (s) = - i( i+ 2 0 ) = ( s = .5 (2 )).1(2))) — 146.5 = 2— sec _1_ °0. the phase curve rises to —140® at ar . = 2 0 log (-/(«•>)) = 20 log A = 6.(a = -l) Determine the phase Z £ (/® ) at ^ s 2 ^ ^ - sec Z £ (y 2 ) .4 ) Substitute Jio for s . _ l rad sec at = - “ 1 _ l rad sec 0. At the break point frequency of the pole. Therefore. So.den) bode(sys) The Bode plot for the function £ ( s ) by using Matlab is shown in Figure 6.5 ® ) 20 -4 0 + 2 0 = . Therefore.2 ---.7 10= -180® Therefore.2 A.05 = 20log 0. ® 2X(y®) 10-= 0 10“ -4 5 10 -84.5® )-lan"'(0. At the break point frequency of the pole.2 0 .5®) y m ( y ® -i) (( y ® ) = + i2y ® + 6=) ^ { j a ) = -9 0 ® + tan"' (0. (js>) |£(y“^^(y®)=[(^i+tan-'(-®) ste p 16 of 27 ^ Tabulate the terms of L{jfp) in the increasing order of their frequencies.25<»’ . the approximate hand sketch of magnitude of asymptotes is as shown in Figure 7.(-270®+tan"' (0 . 0.2 0 . the slope shifts to -4 0 dB per decade - Step 5 of 27 The phase curve starts at ^ = -180® corresponding to the low frequency asymptote i= -4 Determine the phase Z i ( / ® ) at n t . num=[1 2] den=[1 11 24 -36 0] sys=tf(num.sy«i) 10(l+0.5 i»)M |l O'<u) | = -----------.tad The phase curve again falls to —146.05(l+0.5 (6 ))-ta n "' (0.1yV») 4(o.1(10))) = -1463® .1))’ .5y® ) The magnitude plot of this temr has a slope of 20 dB per decade - At the break point frequency of the zero.5 (2 )). Step 9 of 27 (b) Write the expression for L{s) - s+2 1 £ ( . “ sec sec sec Since the break point frequencies are very closer.5(500))-tan"'(0. Step 21 of 27 Combine all the asymptotes to draw a composite curve. Term slope change in s l o p e ( < l % ^ J -2 0 fe ) 1 -4 0 -2 0 -4 0 = -6 0 (y ® + 2 )= + 3 Determine the phase at various values of ® .1yV») 0.02dB Step 11 of 27 1 Start with the second low frequency asymptote This is a second-order pole with ^ = 0.05 A .3® .den) bode(sys) The Bode plot for the function £ ( s ) by using Matlab is shown in Figure 8.20 = -4 0 (i+ y o . ® z £ (y ® ) i0"= -180° 10“ 123.(y500) = (-2 7 0® + tan"'(0. B o d e d iig iM i Step 7 of 27 The transfer function is.- sec Combine all the asymptotes to draw a composite curve.iy®)((ya>)’ -i) 0. Therefore. Figixc9 Step 25 of 27 Write the MATtAB code to plot the response of £^ 4 ).r This is a second-order pole with ^ = 0. cannot observe the accurate change in slope on the scale of logarithm.01))= = 201og!10®| = 80 dB Determine the phase at various values of ® . ^ ^ ( y ® -l) (( 7 ® + 2 )= + 3 ) Z £ (y ® ) = -90® -tan"' Tabulate the terms of £ (y ® ) in the increasing order of their frequencies.(-180®+tan"' (0 . y® (y ® + 1 0)((ya))= -l) 2(l+ 0.(/® ) at a t = 1 0 ^ ^ - sec Z £ (y i0 ) -(-1 8 0 ® + tan"' (0 . .40 = . 1 ± 2 1 (The model for a case of magnetic levitation with lead compensation) C-* U s ) = ^ (The magnetic levitation system with integral control and lead compensation) (c) L(s) = (d) L (f) _ (f) L(s) . £ ( 4) = - ^ 4 -1 )((4 + 2 )= + 3 ) Substitute y® for 4 . num=[1 2] den=[1 10-4 -40] sys=tf(num. e r = 1 .25® =-1 0. ( y « )+ io )((y v » )* _ 4 ) 2(i+o.43 10' -193 10= -270 Step 24 of 27 Therefore.5®) . sketch the asymptotes of the bode plot magnitude and phase for the open loop transfer function £ ( 4 )./(P .1(500))) = -270® The phase curve again falls to —270® at nr = 500---. rad Therefore.(-270®+tan"' (0 .1® ) Determine the breakpoint frequencies. Therefore.iy®) The magnitude plot of this terni has a slope of -2 0 dB per decade.l< »)') ZL^Ja)= -180®+tan"' (O. the phase curve rises to —146.5y® ) ■ioy®(i+o.® ) .den) bode(sys) The Bode plot for the function £ ( 4 ) by using Matlab is shown in Figure 12.5(10))-tan"' (0.05 Calculate the magnitude of indB at n) = iO "'— .tan"' (0. num=[1 -1] den=]1 0 0] sys=tf(num.den) bode(sys) The Bode plot for the function £ ( s ) by using Matlab is shown in Figure 4. » y=1+10. C ( . Figure Magnitude portion of Bode plot Step-by-step solution step 1 of 5 Refer to asymptotic Bode diagram in Figure 6. £ = 10 Step 3 of 5 The response to unit step input is.05 a: =10'“ = 10 Step 2 of 5 The transfer function <?{*) is. 201ogX'-201ogl0=l 20IogAT-20=l . Apply Laplace transform.1:10. > A (f)s tf(/)+ 1 0 fu (/) =(l+10f)i/(/) Therefore. to a unit step input is [(l+ 1 0 /) i/(f)|. a + 1 0 _ i< ^ s s+ \0 = As + B (2) Equate s coefficients. ^ =1 Equate constant terms.y) Step 5 of 5 The response to unit step input is shown in Figure 1. » plot(t. Figure 1 Therefore. s And as 10 . The initial slope represents the term —.*t. 21 ■ o g A T -- k>gA: = 1. Find and sketch the response of this system to a unit-step input (assuming zero initial conditions). I loj From the asymptotic bode plot. » t=0:0.85 in the textbook. it represents the presence of zero. the slope has been changed to OdB/dec. the response to unit step input is shown in Figure 1.09PP A certain system is represented by the asymptotic Bode diagram shown in Fig. the response of the system. Step 4 of 5 MATLAB code to plot output response to unit step input. That is 11 4 — 1 in the numerator. .) (s + lO ) (1) s Calculate the response to unit step input. y ( j) = c ( j) c /( i) J410 Apply partial transforms to simplify the equation. Problem 6. Problem 6.) |= t 201og]QK=201ogK.E-201ogjp(D Which has solpe of -20dB per decade Also if = 2c0i then itis called as Octave since -201og2 = -6dB . the slope could also be expressed as-6dB/octave .1 OPP Prove that a magnitude slope of -1 in a Bode plot corresponds to -20 db per decade or -6 db per octave. Step-by-step solution step 1 of 1 Step 1 of 1 The m ^ i t u d e of G(jo)) in dB is obtained by multiplying the logarithm to the base 10 of |G (-j0)| by 20 |G(jcD)|dB=2Caog|G(j(!))| If |G ( .j.ffi |G (jco) IdB=201 ogu.. the resonant peak M .15 + 1 5*+5 + l Write the MATLAB code for the step response of the transfer function.30 10 1.1 1]. ________1 J14.15 + 1 5*+5 + l Write the MATLAB code for the frequency response of the transfer function. resonant peak for various values of a has been determined. » num=[0. This leads to expect extra peak overshoot M p in transient response.1 • 5*+S + l 105 + 1 5*+5 + l Write the MATLAB code for the step response of the transfer function.den): » step(sys) Step 4 of 15 Consider the seconder order system for a s 1.01 5*+5 + l 1005 + 1 5*+5 + l Write the MATLAB code for the frequency response of the transfer function. No Significant change in the frequency response is observed from a = 10 onwards.den): » bode(sys): Step 13 of 15 Consider the seconder order system for a s 100 • *7 + 1 G { S ): 100 5*+5 + l 0.16 As a is Increased. _L 5*+5 + l 5+ 1 5*+5 + l Write the MATLAB code for the step response of the transfer function.01 • 7+ 1 G(s)> 0.01 98.16 0. Step 8 of 15 The maximum value of the frequency-response magnitude is referred to as resonant peak .1. Similarly significant change In the transient response is observed at a = 0.den): » step(sys). » num=[1 1]: » d e n = [1 1 1]: » sys=tf(num.5 ) ^ l .1 and a = l- .5. _L 5*+5 + l 5+ 1 5*+5 + l Write the MATLAB code for the frequency response of the transfer function.0. For ^ = 0.den): » step(sys).1 ■ 5*+5 + l 105 + 1 5*+5 + l Write the MATLAB code for the frequency response of the transfer function. 0.01 5 *+ J+ l 1005+1 5*+S + l Write the MATLAB code for the step response of the transfer function. and 100 with the Mrfrom the frequency response of each case. Peak overshoot K ) is generally related to the damping of the system expressed as.den): » bode(sys) » hold on Step 10 of 15 Consider the seconder order system for j s 0.16 100 1. » d e n = [1 1 1]: » sys=tf(num. Step 2 of 15 Consider the seconder order system for a s 0.5 ) * = 1.11 PP A normalized second-order system with a damping ratio ^ = 0.015 + 1 5*+5 + l Write the MATLAB code for the frequency response of the transfer function.1 0. » d e n = [1 1 1]: » sys=tf(num. » num=[0.0. » hold on Step 3 of 15 Consider the seconder order system for a s 0. From the Table 1. peak overshoots for various values of a has been determined.den): » bode(sys): Step 11 of 15 Consider the seconder order system for a s 1.1 and a -\ Therefore.. » num=[100 1]: » d e n = [1 1 1]: » sys=tf(num. » num=[1 1]: » d e n = [1 1 1]: » sys=tf(num. Is there a correlation between M r and Mp? Step-by-step solution step 1 of 15 Step 1 of 15 Peak overshoot K ) is the maximum peak value of the transient response.01 1].01 1].« 5*+5 + l 0. » num=[0.den): » bode(sys): Step 12 of 15 Consider the seconder order system for a s 10 • 0 {s). » d e n = [1 1 1]: » sys=tf(num.01. the resonant peak in frequency response Is decreases.1.5 and an additional zero is given by s /a + \ G(s) = j^ + j+ r Use Matlab to compare the Mpfrom the step response of the system fo ra = 0.01 • *7 + 1 G {s).01.10.93 4.015 + 1 5*+5 + l Write the MATLAB code for the step response of the transfer function. Problem 6.46 0.1 1].den): » step(sys) Step 5 of 15 Consider the seconder order system for a s 10 • 5*+5 + l 0. Step 7 of 15 Step response for various values of a is. Resonant-peak magnitude is generally related to the damping of the system expressed as. Frequency response resonant peak is related to the closed loop frequency response. in frequency response and peak overshoot response M p in transient response are correlated at a = 0. Therefore.den): » bode(sys): Step 14 of 15 Bode plot response of for various values of a is. Step 6 of 15 Consider the seconder order system for a s 100 • *7 + 1 inn u ^ 5 . » num=[10 1]: » d e n = [1 1 1]: » sys=tf(num. the resonant peak magnitude is.01. » num=[100 1]: » d e n = [1 1 1]: » sys=tf(num. Tab le 1 a Mr "r 0.den): » step(sys). i+ 1 G(j).94 1 1. To*' 5*+5 + l 0. » d e n = [1 1 1]: » sys=tf(num.1547 Step 9 of 15 Consider the seconder order system for q s 0.( 0 . Step 15 of 15 Tabulate the peak overshoot and resonant peak value for various values of a .8 54. i+ 1 G(j). » num=[10 1]: » d e n = [1 1 1]: » sys=tf(num. Therefore. » num=[0.1 9.15 0. = - 2 ( 0 .0. Step 4 of 6 For p=10.6 . .01.01. 0.5 and an additional pole is given by ^ -------------.1.The bandwidth is increased due to the introduction of extra pole. and 100. Step 6 of 6 ^ Conclusions: (1) .10. Step 5 of 6 For p=100.1 2PP Problem 6 . (2) . What conclusions can you draw about the effect of an extra pole on the bandwidth compared with the bandwidth for the second-order system with no extra pole? Step-by-step solution step 1 of 6 Step 1 of 6 Forp=0. [(■s/p) + 1 ](*^ + 1 + 1) Draw Bode plots with p = 0.1. as the p value goes on increasing.The bandwidth value goes on increasing. Step 3 of 6 For p = l.1.1 2PP A normalized second-order system with ^ = 0. Step 2 of 6 Forp=0.. |= u^ere M s|T(ja>)| (’■ M ) " '’M Step 2 of 2 solving for -^5” ^ get ^ = l.13PP For the closed-loop transfer function j'/ g \ — ________5______ s^ + 2((OnS + < i^ ’ derive the following expression for the bandwidth ojB W of T(s) in tenns of u)n and ^ : = a > n y jl\ .order system can be found by Bnding that frequency for which M = . Oil <d^ = o>. Step-by-step solution Step-by-step solution step 1 of 2 T ( s ) = t -----. 8 + 25oa^s+cc^ The Bandwidfti of the standard second .--------------J. .2 ( '^ + ^ 2 + 4 t * ^ ^ ^ .^(1-2?= )+ V 45'^?“ +2 Hence proved. plot ojSlVfor 0 < ^ < 1. Assuming that oj/7 = 1.2 4 = ± j4 ? 4 ? t i since ^5SL>Q. Problem 6. . in equation (3). From bode plot.+ e » f^ _ 3 . (b) Define the bandwidth as the distance between the frequencies on either side of cuO where th magnitude drops to 3 db below its vaiue at uX).1 to 10 for magnitude.1 and 2 for Q.|— . .1 to 10 for phase margin. V* Substitute 0.. we get phase with normalized frequency — varying from 0. BW -(5) At 3db point which corresponding to magnitude is given below. Step 3 of 4 Where magnitude at 0 ^ and 0 ^ are equation which may be frequency which also include e frequency at which magnitude drops to 3db.. substitute Q value in equation (4).V W OH) W o^J o>^) =l (6) 0 ). A (2) j0 Take modulus for Equation (2).i io « 10' Figure 1 Similarly. » ' + + «H. Bandwidth at 3db point from centre frequency from bode plot.(ja)+<a^Q A„aja> -o ^ Q + a J a + C D ^ Q S] G (j0 ).| (4) 1. (8) s^+ 2C a ^+ ef Compare the equation (7) and equation (8). the frequency relationship is given below. The bode plot is shown in Figure 1. We get phase with normalized frequency — varying from 0.5. and show that the bandwidth is given by BW (c) What is the relation between Q and ^ Step-by-step solution step 1 of 4 (a) Consider the magnitude and phase of the transfer function. Hence |G(M)|-|C(M)| From bode plot.^1 «>) Consider the following formula for the normalized frequency. C (* ) = 0 s * + < V + « ( . Compare quality factor Q and damping factor ^ Consider the given transfer function. the bandwidth is BW : 1 ( e ''' Step 4 of 4 (c). the centre frequency is 0 ^ > 0 ^0 ^. (a) Compute the magnitude and phase of the transfer function analytically. BW ^ Thus.3. And equa magnitude from symmetry. Substitute the value in equation (4). and 5 as a function of the normalized frequency oj/wO. Magnitude at frequency 0 ^ s magnitudeatfrequeocyoiji Where a)[<d)^and 0 ^ > 0 ^.0 t 2x =aL.:2 .. " ^ = la n .’ = » '+ Consider s scentreFrequency Thus. from mean value Betweenat3dbpoints uppercutoff-lowercutoff 2x ^ 0 j.5. (7) Consider the standard second order transfer function.14PP Consider the system whose transfer function is i4o<wo^ G (s )^ + <3 This is a model of a tuned circuit with quality factor Q. Problem 6.J Compare the equation (5) and equation (6).1. (1) Q s ^ + a ^ + o ii’ Q Substitute Ja>tor s in Equation (1). there is even symmetry from centre frequency 0 ^. ^ ■ 3 1 I 2 "F F F Take square on both sides 1 "I2 . Step 2 of 4 (b).’ g A ^ G (4 . and plot them for Q = 0. we have From the above equation. Gijosi)— Q (ja ) +a. 1 ^ V ..2. and show that the bandwidth is given by magnitude drops to 3 db below its vaiue at ciX). the quality factor is .. if r . ^ --------------- (10) + {2 C a .(11) When d \ n j < 4 0 . the amplitude and phase angle are 0. In (0 .. Figure Voltmeter schematic /= 4 0 X I( r« k g m 2 J t= 4 x l( r « k g iii2 /s e c 2 T = input torque = Kg^v v = input voltage j r . the transfers function is given c e (s ) (2) Consider the standard second order transfer function.1 » e Take logarithm on both sides. <9= 0.= 4 x H T ® N i ii / V Step-by-step solution Step 1 of 7 (a) The DC voltmeter is shown in Figure 1.15PP A DC voltmeter schematic is shown in Fig. the damping frequency is 0.. d& <»= 0.59)(0.316 for <»^in equation (12).37288 0.316)(y<»)+(0.. the amplitude and phase angle are |0.316 rad/sec....m ' * = 4 0 x l 0 '‘ k g .549 x 0.25 rad/sec ■ Step 4 of 7 (c) Consider the Magnitude and phase response of the given function. pftpr in itip l trp n p ip n tp h p v p HIp H n u t ? W h p t i. the Magnitude equation is given as. the natural frequency is 0.0256rad|and |-169°I .... s 0. deo * .1 ( > ) ' + 0. BodeDiagnm Step 6 of 7 5 peak frequency is [Q.1 for A/^in equation (6).s+<». Substitute 0..‘ k g . .0256radand —1< Thus. / = 4 0 x l0 . F (4 ) g (» ) K JI (8) K(s) 4’ +2^<». rearrange the Magnitude equation.316 rad/sec • Step 3 of 7 (b) Consider the following formula for the peak overshoot.. Thus. Use the Isim command in Matlab to verily your answer in part (d).59 Consider the formula for the damping frequency. Substitute 0. Problem 6.549<».0998 Write the Matlab program and draw the bode plot. ... th p nhp<..316for <9^.. . ffl. The pointer is damped so that its maximum overshoot to a step input is 10%.p \an W h flt a m n litiiH p w ill th p m p t p r in riin p tp (d) Suppose this meter is now used to measure a 1-V AC input with a frequency of 2 rad/sec. .316 =0.. (4) " I (5) Substitute the given values in equation (4) and equation (5).37288y®+0.0998])) The output of the Matlab program is given in Figure 2...173 rad/sec Step 5 of 7 Substitute0.m /v Consider the equation of motion. = 4 0 x l 0 '‘ N . (1) dt dt From equation (1).(12) Substitute 0.. 0. 4x10"*^°'^ 4 0 x 1 0 ' * ^ ° ' ^ ^ ® * ^ * ^ ^inequation (9)- 8(j<o ) (4 xl0 -‘ /4 0 x l0 -‘ ) V(J<») (y®)^+2(0.m ’ K .3 lW l-0 .[1 0.<. 0. bode(tf([0. (a) What is the undamped natural frequency of the system? (b) What is the damped natural frequency of the system? (c) Plot the frequency response using Matlab to determine what input frequency will produce the largest magnitude output? (d) Suppose this meter is now used to measure a 1-V AC input with a frequency of 2 rad/sec.a > y J Take differentiate with respect to a>. What amplitude will the meter indicate after initial transients have died out? What is the phase lag of the output with respect to the input? Use a Bode plot analysis to answer these questions. ( j a ) + a ’ From equation (9).5 9 ’ =0.1 )= I f = 0.316 for ajj^and 0.173rad/se^ Step 7 of 7 (d) With 2rad/secfrom Figure 2.173for <»...59 for ^ in equation (7).= 0 . J _ (3) Compare the equation (2) and equation (3). Step 2 of 7 Consider the given data.316)’ 0.1].’ Consider the value of s is »(J<») IC J l (9) V (M ) { j( u ) ‘ + 2 i a .25 rad/sec Thus...m V se ' r = 4 0 x l 0 ‘‘ kg. Substitute _1_^____ 1 ATG(j) = lo r ^ T T v . ' ’ ■•fe-') ii:<. • Mark the poles on the real axis. the Bode diagram is plotted for the given transfer function and it is shown in Figure 3. x e |. 4 fOr rx and 2 for m in equation (9). To find zeros put numerator JV(j) = 0 Thus. n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoles)-(niunberoffinitezeros) -3 0 -(-3 ) 1-1 = co Thus. Step 12 of 31 (b) Consider the transfer function. 3r(l-F4/r». the root locus is plotted for the given transfer function and it is shown in Figure2. ^ ( 4 + 10X4 + 11 .16PP Determine the range of K for which the ciosed-ioop systems (see Fig) are stabie for each of the cases beiow by making a Bode piot for K = 1 and imagining the magnitude plot sliding up or down until instability results. the system is stable for |jc > Q |.+ 5 )1 Fiaure Block diaoram for YfsVRfst = KGfst/tl + KGtstl r#-r v r> t-\ A T fa-F I 0 ) ( J .5 10 54. The Bode diagram is shown in Figure 5. =30 (5-) -20 0 Calculate gain A using table 1. 1 ^ ^ and higher frequency ax = 1 0 0 ^ ^ - sec sec Calculate gain (A) at a —a . Step 20 of 31 The root locus plot is shown in Figure 4. Assign lower frequency ax = 0 .( . the asymptotes is a step 18 of 31 Consider the formula for the angle of asymptotes. The root locus plot is shown in Figure 6. • Draw the root locus. Table 2 Corner Frequency Slope Change in slope Term (S) m m (y-AX+l) "'■=T -20 - -1 (y-AX+l) AX„=1 -20 ■40 (if*') AX„=10 -20 -60 Calculate gain A using table 2. Assign lower frequency = 1 ^ ^ and higher frequency =1000^^- ' sec sec Calculate gain A. =|G(yAx)| = -20db Calculate gain (A) at ax= ax.3 100 15. (a) js:g ( s) = ' ^ (b) K G (s) - v r * t ^ \ _ ^(a-F IO)(j-F 1) (0 * G ( * ) .(a) . 3 for rx and 0 for m in equation (9).. To find zeros put numerator JV(j) = 0 Thus. the phase angle is never crosses —180®.58 The Bode diagram is shown in Figure 1.. 1 ^ ^ ond higher frequency ax = 1 0 0 0 ^ ^ - sec sec Calculate gain (A) at a —a ..(12) Find the magnitude of K G [ ja ) In dB using Equation (12) as listed In table (3). Step 21 of 31 A Thus. =|G(y<»)| = -9.. • Mark the poles on the real axis. v <6> Plot the straight Bode diagram for the transfer function KG{^s) ■ Consider the standard form of transfer function up to third order system. the system is stable for |jc > Q |.2.^=-S6db Calculate gain (A) at ax = AX. 3 for rx and 0 for m in equation (9).' ) ■(2) ■ ( & • ') Step 2 of 31 Plot the straight Bode diagram for the transfer function K G {s ) ■ Consider the standard form of transfer function.(10) ' ^ (4 + 1 0 0 )(4 + 5 ) Rewrite equation (10)..F I ) (0 « G (** ) . 2 180°+360«(/-l) -(9) w— n —m Where. r . Verity your answers by using a very rough sketch of a root-locus plot. the results are verified from root locus plot. and draw the asymptotes.^ . the asymptotes is □ u Step 29 of 31 Substitute 1 for /.- A. From Figure 5.8 30 39.. the zeros are -10 and -1.10. Step 8 of 31 'v Consider the formula for the asymptotes.. Number of zeros present in the system is 0 Number of poles present in the system is 3. .-L . the angle of asymptotes are |60v|. the pole is . + . the asymptotes is . 180+360(2-1) 3 540 3 =180* Substitute 3 for /. the system is stable for |jc > Q |. oo)(.60**l- Step19of31 Procedure to draw root locus plot. From Figure 6. .5. g ( lF a M ) ■(3) (1 + a M ) Compare Equations (2). . the results are verified from root locus plot.(.1 2 0 dB Find the phase plot for equation (8). Number of zeros present in the system is 1 Number of poles present in the system is 1 The corner frequencies are =30aodo)^ =30- Substitute j a for s in Equation (2). Table 3 Corner Frequency Slope Change in slope Term (y-AX+l) AX„=1 -20 - (f-) ax„ = 5 -20 ■40 -20 (if-) AX„=10 -60 -20 -80 (£*■) AX„ = 100 Calculate gain A using table 3. n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoles)-(numberoffinitezeros) .4 ( f * ') a v . and draw the asymptotes. Problem 6. + . ^ . 4 fOr rx and 2 for m in equation (9). - Step 25 of 31 Calculate gain (A) at ax= ax^. Root Locus Step 31 of 31 Thus. the system is stable for |j(f < 242 |ond the system unstable for |j(f > 2421 ■ Step 17 of 31 ^ Consider the number of poles and zeros from Equation (5). Step 11 of 31 Hence. Step 16 of 31 Hence. Step 27 of 31 Hence. • Take real and imaginary lines on X axis and Y axis respectively. the results are verified from root locus plot. ... Step 10 of 31 A Thus. * de g 0. A ^ s^sk x p c fitx m A x ^jto A jjiX lo g -^ j+ A ^ = -6 0 1 o g ^ ^ ^ j-6 0 d b = . Number of poles is n Number of zeros is m / =l. . • Mark the poles on the real axis.+ 5 )3 Figure Block diagram for Y{s)/R(s) = KG{s)/[1 + KG(s)] —■ act*) I or Step-by-step solution step 1 of 31 (a) Consider the given transfers function. K (4 + 1 0 )(4 + l) ( 4 + l) Substitute 1 for K in above equation. Step 26 of 31 Thus.. the root locus is plotted for the given transfer function and it is shown in Figure 6. . Step 28 of 31 ^ Consider the number of poles and zeros from Equation (1).=30 20 - ax. 180+360(1-1) 3 180 s ----- 3 -60» Substitute 2 for /. the system is stable for |j(f < 242 |und the system unstable for |jc > 2 4 2 l ■ Step 23 of 31 ^ Hence.)| = -2 0 d b Calculate gain at At = a>^ ■ A ^ . To find poles put denominator D (s) = 0■ Thus. y -(4) Step 3 of 31 'v Find the magnitude of K G [jiii) in dB using Equation (4) as listed in table (1). Table 1 Corner Frequency Slope Change in slope Term (S) © © ®. the zero is -3. and draw the asymptotes. A ^ = -8 3 d B Calculate gain (A) at a —a.(11) Plot the straight Bode diagram for the transfer function KG{^s) ■ Substitute J a fOr s in Equation (11).1 -12.1 and .=|G (y<i. toAx^jxlcxg^T. Step 7 of 31 'v Consider the number of poles and zeros from Equation (1).( .30.).S42db Calculate gain (A) at At=At^2 - A=-^=|G(y<»)| = 0db Calculate gain (A) at A „ 2 = |C (H = 0db Step 4 of 31 Find the phase plot for equation (4). • Take real and imaginary lines on X axis and Y axis respectively. • Locate the asymptotes on the real axis. wIslopefiomAX^. \ ^ ----. Step 14 of 31 A Thus. the pole is . Step 9 of 31 Procedure to draw root locus plot. • Take real and imaginary lines on X axis and Y axis respectively. • Draw the root locus. .1 1 -95. the bode diagram is plotted for the given transfer function and it is shown in Figure 1. K K G (s ) = -^ ■(5) ( j + 10)(s + l) Rewrite equation equ (5). From Figure 2. ■ =-62db Calculate gain A at ax = a ^ ■ A^=-59db Calculate gain (A) at ax= ax^. (1) ' ’ a+30 Rewrite equation (1). The root locus plot is shown in Figure 2. the gain is equal to the 242 when phase angle crosses —IgCP.4 0 1 o g ^ Y j. +A „ ^ I = . the root locus is plotted for the given transfer function and it is shown in Figure 4.ji-m Substitute 1 for /. the system is stable for |J(f >Q|. the pole is .9 10 -213 100 -263 Step 13 of 31 The Bode diagram is shown in Figure 3. Step 15 of 31 From Figure 3. From Figure 1. To find zeros put numerator JV(4) = 0 Thus. To find poles put denominator D (s) = 0 ■ Thus. Assign lower frequency ax = 0 . the gain is increased or decreasing on the bode diagram and the phase angle is not less than -18CP Step 6 of 31 Hence. Step 5 of 31 Thus. Hence. the Bode diagram is plotted for the given transfer function and it is shown in Figure 5.) G (z) -(7) 0 + 4 M )0 + 4 M )0 + V < » > ) Compare Equations (6). at A „. ..2 0 d b = -« 0 d b Calculate gain (A) at ax= A^. .5. A. 180+360(2-1) 2 540 B----- 2 =270® Thus. To find poles put denominator ^ ( 4) = 0- Thus. n -m (sumoffinitepoles)-(sumoffinitezeros) (numberoffinitepoles)-(numberoffinitezeros) . K JCG(4)=7 ( j + 1 0 )(j+ l)(s + l) Substitute 1 for K in above equation. there is no zero. the angle of asymptotes are |9Q®|and |270®l- Step 30 of 31 Procedure to draw root locus plot. Consider the formula for the asymptotes. and (7). ..5 and . Step 22 of 31 A From Figure 4. oo. | 180°|and | .z lizM 3-0 =-4 Thus.. The corner frequencies are =1.- A ^ = -1 2 0 d B Find the phase plot for equation (4). 7 -----------. 180+360(3-1) 3 300 3 —60* Thus. Step 24 of 31 ^ (c) Consider the transfer function. z iiH im 4 -2 =-52 Thus.<»^2 andfiVa -1 0 - Substitute j a for s in Equation (6). • Locate the asymptotes on the real axis. • Locate the asymptotes on the real axis. * de © g 1 16. K G {J a )= .1.100. • Draw the root locus. and (3). - 34= |G ( » | = -20db Calculate gain A at ax = a ^ ■ A„^. ( f ^ ') K G (jo> ) = 0 .1 1000 1. -(8) Find the magnitude of K G [ ja ) In dB using Equation (8) as listed in table (2). 180+360(1-1) "■ 2 ^ 180 2 =90® Substitute 2 for /.. f e * " . Consider the formula for the asymptotes. 3 fOr rx and 0 for m in equation (9). a ( j* 3 i K [ ? * '] Substitute 1 for K in Equation (1). Problem 6 . G et help from a Chegg subject expert. ( .1 7PP Determine the range of K for which each of the iisterJ systems is stabie by making a Bode piot for K = 1 and imagining the magnitude piot siiding up or down untii instabiiity resuits. ASK AN EXPERT . ^ 2 K j2 + 9 ) tr4 'i)t» ? H Step-by-step solution There is no solution to this problem yet. Verity your answers by using a very rough sketch of a root-iocus plot. (c) K G {s) . sketch s= C x where C1 is a contour enclosing the entire RHP. Problem 6. {Hint Assume Cl takes s small detour around the poles at s = 0. 2) (b) Repeat part (a) for an open-loop system whose transfer function is Figure 1 An s-piane plot of a contourCI that encircles the entire RHP Figure 1 An s-piane plot of a contourCI that encircles the entire RHP c. that is. 1.18PP (a) Sketch the Nyquist plot for an open-loop system with transfer function 1/s2. as shown in Fig. / V Figure 2C1 contour enclosing the RHP for the system Step-by-step solution step 1 of 3 M o p in g imaginadty axis ( 0 < ( 9 < od) idiich lies on the negative real axis Step 2 of 3 Step 3 of 3 . as shown in Fig. 5 + 2 )a (d) Using your plots. Don’t be concerned with the details of exactly where the curve goes. Problem 6. (a) X G ( S ) = (b) K C ? ( s y = i s -+. and then compare your result with that obtained by using the Matlab command nyquist._ j_ io o ) ( .1 0 )C ^ + 2> 2 + 1 0 ) ( j + 1) (C ) K C ^ i^ s y — if '\ + 1 0 ) ( j + 1) (C ) K C ^ i^ s y — ( . but do make sure it crosses the real axis at the right spot.19PP Sketch the Nyquist plot based on the Bode plots for each of the following systems. and qualitatively verify your result by using a rough sketch of a root-locus plot. ASK AN EXPERT . Step-by-step solution There is no solution to this problem yet. G et help from a Chegg subject expert. has the correct number o f -1 encirclements and goes off to infinity in the correct direction. estimate the range of K for which each system is stable. Step 8 of 10 To choose the contour to the left of the singularity on the imaginary axis. there are no characteristic equation roots in the RHP. So the system is stable for positive values of k ■ Thus. in equation (1) must be chosen negative. » system=tf(numerator. Z = A ^+I = . The Nyquist plot for the feedback system.20PP Draw a Nyquist plot for KG(s) = Ks{s(s+ 1) + 3) * choosing the contour to be to the right of the singularity on the yw-axis. denoted by p a re . Thus. Next. denoted by p are zero. the range of for which system is stable when contour is to the right of the singularity on the ja> axis is |j^ > Q |. » nyquist(system) Step 5 of 10 The Nyquist plot for the feedback system is shown in Figure 1. it is clear that the range of x for the system to be stable when the contour is to the right of singularity on the imaginary axis is same as that of range of x for the system to be stable when the contour is to the left of singularity on the imaginary axis.denominator). » numerator=[1.l The number of unstable (RHP) poles of ATG(yfl>). choosing contour to the left of the singularity on the imaginary axis is as shown in Figure 3. Step 4 of 10 To choose contour to the right of the singularity on the axis. check the range of K for which the system is stable. the plot is as shown in Figure 2. The results should be same in either way. 1+1 =0 Hence. iV = . .1].ta n -' j (3) Step 2 of 10 Magnitude of ATG(y<9) at ^ s O is. Z = JV+P = 0+0 =0 Hence. Number of unstable (RHP) poles of KG[^jG>). J . Phase of JSrG(yfl») at = \KG{jo>) = tan"' f y l ” tan"' f j * -9 0 * Magnitude of ATG(y<9) at n>sco i Step 3 of 10 Take ^ c o m m o n from the above equation. using the Nyquist criterion. » denominator=conv([1. Step 6 of 10 Choosing the contour to the right of the singularity on the imaginary axis. the value of in the equation (1) must be positive. the way of choosing the contour around singularity on the j & axis does not affect the system stability criterion. P =\ Numbers of unstable closed loop roots Z are. this time choosing the contour to be to the left of the singularity on the imaginary axis. Step 10 of 10 Hence. Numbers of unstable closed loop roots Z are. j =ta n -' . determine the range of K for which the system is stable. FtgDre2 Step 7 of 10 The numbers of clockwise encirclements of -1 {denoted by ).3]). n if n . Enter the following code in MATLAB to draw the Nyquist plot with contour to be to the right of the singularity of y<naxis. Then redo the Nyquist plot. So the system is stable for negative values of . Step 9 of 10 The numbers of counter-clockwise encirclements of _ j by the contour in the Nyquist plot of Figure 4 is.0]. Thus. Problem 6. the range of x for which system is stable when contour is to the left of the singularity on the y<»axisis |jj^ > Q |.[1. by observing the Nyquist plot of Figure 2 are zero. there is no characteristic equation root in the RHP. Magnitude of ATG(y<9) ■' ■(2) Phase of JirG(y(9)is. Again.V + 1 l^ c o < » )L = M f— =0 Phase of ATG(y(9) at <n>oois |^G(ya>)| =tan-' [ j] = 9 0 --9 tr-9 (r = -9 0 ‘ Follow the above procedure and calculate magnitude and phase of KG[^jG>) tor different values of I The Nyquist plot will always be symmetric with respect to the real axis. Are the answers the same? Should they be? Step-by-step solution Step-by-step solution step 1 of 10 Consider the following open loop transfer function of the system: (1) ' ' j(j+ 3 ) Nyquist plot represents frequency response of the system. value of x. using the Nyquist criterion.9<r. to n . Z . Step 7 of 12 Enter the following code in MATLAB to draw the Nyquist plot. Nyquist plot doesn’t encircle _ j. Number of clockwise encirclements of -1 . Consider both positive and negative values of K. by observing the Nyquist plot of Figure 2 are zero. ^ ( 2 . y (t) KG{s) R {s )° l + KG(s)H{s) ^ I ( i ’ + 2 i + 2) I l+Ky (s ' + 2 i+ 2 ) ( i+ l) From the closed loop transfer function.G (ja)H ( J a ) = O . Hence. The number of clockwise encirclements have to be now determined depending on value of . Consider positive value of k ■ V The maximum gain of G (7 ® )« (y < » ) isobtainedat ^ s Q ^ n d the value is — .' j" [7 ] = . Z = JV+P = 0+0 =0 For > —2 . The plot is normally created by the NYQUIST MATLAB m-file. Step 11 of 12 Consider value as negative. So the system is stable for > —2 • Step 12 of 12 ^ When j f < _ 2 . Problem 6. Using the Nyquist stability criterion. The Nyquist plot will always be symmetric with respect to the real axis. Z = 0 : that is there are no characteristic equation roots in the RHP. \G (ja \H {jo ii ■ ^ ( 2 . [f] Step 5 of 12 Magnitude of G ( ja ) H { ja ) at (psQ is.[1 1]). it is clear from the obtained results that the range of for which system is stable is [Z H 5 I . denoted by /» are zero.a a -‘ [ 3 ^ ] . When j f > _ 2 . Figure Control system Step-by-step solution Step-by-step solution step 1 of 12 Consider the feedback system shown in Figure 1.tan-' j. Z = JV+P = 0+0 =0 For positive value of . ^ ( 2 .9 0 '. Number of clockwise encirclements of -1 . Step 2 of 12 The closed loop transfer function for the above feedback system is. » sys=tf(num. (0) = (T Step 6 of 12 Magnitude of G { j a ) H ( j a ) at d>»oo is. A ^=0 Number of unstable (RHP) poles of G { i a ) H ( j a \ denoted by /> are zero. ZG{j(o)H(ja>) = 0 .e ^ f + A e ^ y ll + a ^ Phase of G {ja > )H {ja) is. JV = 1 Number of unstable (RHP) poles of G { i a ) H ( j a \ denoted by /» are zero.den): » nyquist(sys) Step 8 of 12 The Nyquist plot for the feedback system is as shown in Figure 2. determine the range of K for which the system is stable.W = -180' Follow the above procedure and calculate magnitude and phase of G { j o \ H { j a ) for different values of to. FigDre2 Step 9 of 12 Determining stability of closed loop system based on frequency response of system’s open loop transfer function is Nyquist stability criterion.00)* + 4 ( 00)^ V l +«>“ =0 Phase of G (j< o )H (ja) at <2>sco is.21 PP Draw the Nyquist plot for the system in Fig. The numbers of clockwise encirclements of -1 {denoted by U ). So the system is unstable for < —2 • Thus. ZG(ya))W(yffl)=0 .0 ) ’ + 4 ( 0 ) ’ V i+ 0 £ ' 2 Phase of G {ja > )H {ja) at (psQ is. Z = J\7+P = 1+0 =1 For fC < ~ 2 ’ Z = l^ that is there are characteristicequation roots in the RHP. Number of unstable (RHP) poles of G (7 ® )« (y < » ). »den=conv([1 2 2]. the open loop transfer function is obtained G (s)H (s) = (s“ + 2 j + 2 ) ( i + l) Step 3 of 12 Nyquist plot represents frequency response of the system. Nyquist plot encircles _ j. So the system is stable for positive values of j^ . » num=1. Z = 0 : that is there are no characteristic equation roots in the RHP. K 1 (2 -ffl’ + J2to) (y'ffl+l) Step 4 of 12 Magnitude of G { ja ) H ( ja ) is. Step 10 of 12 ^ The numbers of unstable closed loop roots is. is stable for all values of K ^ ^ s+10 For G (S) = 0 < K < 1 for stabili^ s+10 s+1 F o r G ( s ) = . sketch the phase of the minimum-phase system and the nonminimum-phase system G(s) = --y . (1)? (b) Does an RHP zero affect the relationship between the -1 encirciements on a poiar piot and the number of unstable closed-loop roots in Eq.2 8 ) Step 3 of 4 Step 4 of 4 s+1 d.. (1)? Eq.— r 0 < K < 1 forstability -(s-l) .22PP (a) For w = 0.1 I noting that A{joj . Problem 6. F o r G f s ) —-----. Step-by-step solution step 1 of 4 G (= )= — G(S): ■ .lO . (1) Z ~ N + P.M ) '■ ' s+10 S+10 Step 2 of 4 ^ b. (b) Does an RHP zero affect the relationship between the -1 encirciements on a poiar piot and the number of unstable closed-loop roots in Eq. (c) Sketch the phase of the following unstable system for w = 0. (d) Check the stability of the systems in (a) and (c) using the Nyquist criterion on KG(s).1 to 100 rad/sec: J+ 1I G{s) = j. it does not have any eSect on £Q (6 . No.1 to 100 rad/sec. Determine the range of K for which the closed-loop system is stable.1) decreases with oj rather than increasing. and check your results qualitatively by using a rough root-locus sketch. V ^’ 4 (4 -1 )’ Substitute Ja for s. 1 K G (ja ) = K - { J a .4 j + i ) KG (s) = K (4 -1 )’ Substitute Ja for s. 4 ( j. 1 KG(s) = K j (» -1 )(4 + 2 ) Substitute J a for s.l ) ( j1. Substitute jo> for s. ( y o .^ j . for the systems given below.denG). The magnitude of the beginning point is oo and phase is 0® and hence has the correct magnitude and phase. Step 6 of 14 (c) Consider the open loop transfer function of the classical cunre called Folium of Kepler. K G (ja )-K — (y V » -l)(y « > + l)’ ^ ( l + a i‘ ) . » sysG=tf(numG. » sysG=tf(numG.denG). _____ Step-by-step solution step 1 of 14 (a) Consider the open loop transfer function of the classical cunre called Cayley’s Sextic. r V (^’ + 0 KG (s) = K-^ -----.t a i T ' ^ . » denG=conv(conv([1 -1].[1 -1]).[1 2 1]). » denG=conv([1 -1]. Hence the magnitude and phase of the beginning point are verified. Nyaebl Dtoaraw Hence the magnitude and phase of the beginning point are verified.[1 1]). ' '4 + a JCG(0) = i = 0. The magnitude of the beginning point is 1 and phase is Q* and hence has the correct magnitude and phase. Freeth. » nyquist(sysG) Step 8 of 14 The Nyquist plot for the cunre is shown in Figure 3 Hence the magnitude and phase of the beginning point are verified.[1 -1]). » denG=[1 1 0]. (y»-l) ■ (v /n v )’ Determine themagnitudeat a> = 0 b n d Jtf = l ( V T iV ) a:g ( o) = i Determine the phase at a> = 0 b n d Jtf = l ZKG(ja) ■ -3tan"' (-<») =0 Enter the following code in MATtAB to plot the Nyquist plot for s | » numG=[1 0 1]. » denG=4*conv{conv([1 -1]. » sysG=tf(numG.[1 -1]). » nyquist(sysG) The Nyquist plot for the cunre is shown in Figure 7. 2 ( j + 1)(4’ . » nyquist(sysG) Step 12 of 14 The Nyquist plot for the cunre is shown in Figure 5. 1 KG (s) = K - (4 -l)(4 + l)' Substitute j a for s. with K = and verity that the beginning point and end point for the jw > 0 portion have the correct magnitude and phase. Hence the magnitude and phase of the beginning point are verified. named after the English mathematician T J.t a n " ' j Enter the following code in MATtAB to plot the Nyquist plot for Jtf = i » numG=1. Step 3 of 14 (b) Consider the open loop transfer function of the classical cunre called Cissoid. 1 CG(a) = K ( a . Step 11 of 14 (e) Consider the open loop transfer function of the classical cunre called Nephroid. » sysG=tf(numG.denG). 1 KG(s) = K- (» + !)’ ■ (b) The classical cunre called the Cissoid. 1 JTG(a) = K i ( i + 1) (c) The classical cunre called the Folium of Kepler. 1 KG{s) = K - Substitute j a for s. . » nyquist(sysG) Step 10 of 14 The Nyquist plot for the cunre is shown in Figure 4.« .l) ( J a * 2 ) 1 =K - Determine themagnitudeat ^ s Q ^ n d Jtf = l K G ( ja ) = . K G (ja > ) = K — L ^ [ ja + l) Determinethemagnitudeat ^ s Q a n d Jtf = l 1 K G (jta)= - ■ ( i/T iv ) ’ X G (0 ) = I Determine the phase at ^ s Q a n d Jtf = l z a :g ( o) = o° Enter the following code in MATtAB to plot the Nyquist plot for =i » numG=1.S Determine the phase at ^ s Q ^ n d Jtf = l ■ . meaning ivy­ shaped. » nyquist(sysG) The Nyquist plot for the cunre is shown in Figure 2. » denG=conv(conv([1 -1].a ‘ )^ + 4 a ‘ step 7 of 14 Determine themagnitudeat ^ s Q ^ n d Jtf = l 1 K G {ja) = - ^ { l+ a ’ ) ^ j{ l. ^ ( l .a ? f+ 4 a ‘ K G {0 )-\ Determine the phase at ^ s Q ^ n d Jtf = l Z K G (ja )= -lS iy‘ Enter the following code in MATtAB to plot the Nyquist plot for =i » numG=1.[1 2]).F l) ( j3 . meaning kidney-shaped.[1 1]).l) 3 (g) A shitted Nephroid of Freeth.denG). Determine the Nyquist plot.[1 -1]).[1 0 3]). » nyquist(sysG) Step 13 of 14 The Nyquist plot for the cunre is shown in Figure 6. Hence the magnitude and phase of the beginning point are verified. » denG=conv([1 -1].5 Determine the phase at a> = 0 b n d Jtf = l Z K G (ja ) ■tan"' (ar)-3tan"' (-®) =0 Enter the following code in MATtAB to plot the Nyquist plot for Jtf = i > numG=2*conv([1 1]. » denG=conv(conv([1 1]. » nyquist(sysG) Step 2 of 14 The Nyquist plot for the cunre is shown in Figure 1. » sysG=tf(numG.tan " '(r» ) . (a) The classical cunre called Cayley’s Sextic.F 2 ) (e) The classical cunre called the Nephroid. meaning kidney shaped.l) ( s .denG). Step 9 of 14 (b) Consider the open loop transfer function of the classical cunre called Folium.H ) 2 ' (d) The classical cunre called the Folium (not Kepler's). Jta(ja>+\) 1 = AT -a + ja step 4 of 14 Determine themagnitudeat ^ s Q a n d Jtf = l 1 KG{Ja>) ■<lar+a X G (0 )= « > Determine the phase at ^ s Q a n d Jtf = l Z K G O v») = .denG). » sysG=tf(numG. Problem 6.( i ) ZATG(0) = 0° Step 5 of 14 Enter the following code in MATtAB to plot the Nyquist plot for =i » numG=1. meaning Ivy-shaped. » sysG=tf(numG.[1 -1]). studied by Kepler in 1609.23PP Nyquist plots and the classical plane cun/es. 4 (y < » -l) ^ 2 7 ( l4 ‘t»’ ) ( 3 .l) KG(0) = 2 Determine the phase at at = 0 b n d Jtf = l ^ G [ja ) ■tan"' (nr)+tan"' ^|-^^j-3tan"' {. ( i.[1 -4 1]). Step 14 of 14 (g) Consider the open loop transfer function of the classical cunre called shifted Nephroid of Freeth. ’ ) 4 ( V IT ^ )’ Determine themagnitudeat nt = 0 b n d Jtf = l 4 ( v T iV ] XG(0) = j = 1.a ) = 0» Enter the following code in MATtAB to plot the Nyquist plot for s | » numG=2*conv([1 1].F 3 ) KG(s) = K 4 ( r . KG{s) = K - (1 -1 )3 (f) The classical cunre called Nephroid of Freeth. (f) Consider the open loop transfer function of the classical cunre called Nephroid of Freeth.[1 -1]). discovered by Maclaurin in 1718. KG(s) = K( j .denG).l) Substitute Ja for s. using Matlab. Sketch what the corresponding root locus must look like for such a system.= — = 2.24PP The Nyquist plot for some actual control systems resembles the one shown in Fig.77 2 0. given that a = 0.5 system is stable If gain > 2.4. What are the gain and phase margin(s) for the system of Fig.4 If gain lies between 0. Describe what happens to the stability of the system as the gain goes from zero to a very large value.= 0..5 a n f 2 = .5 system becomes unstable Root locus will look like Step 2 of 2 Re(s) . Problem 6.77 and 2. and 0 = 40*. /5 = 1.3.. 1 /n / 1 / 1 / 1 X 1 I / 1 I / 1 1 \ / 1 -1 t /W tf/ ll Re[G(s)] V S te p -b y -s te p s o lu tio n Step 1 of 2 P M = q ) = Aiy G a fl = . Also. sketch what the corresponding Bode plots would look like for the system. Figure Nyquist plot Im [C (j)] ^ U -a. +0 100 =0 Step 5 of 8 Phase of at a> = ools |G ( ja > ) g ( y a > ) = 0 ° .01 1]). Numbers of unstable closed loop roots Z are. It is clear the gain of the system equals 1 if the gain is decreased by a factor of 100 . »nyquist{sys) Step 6 of 8 The Nyquist plot for the system of equation (1) is as shown in Figure 1.»den=conv ([1 -1 0].91 from the text book.t a n .( 1 8 0 ° .37® »270® Step 2 of 8 It is clear that the pole at 5 s l in equation (1) contributes to a phase of —igo* d> = 0.01 >pole at 5 s -10 0 io equation ( 1) contributes to a phase of 0* for ^ s 0. that is gain becomes 1.01< pole at 5 s 0 io equation (1) contributes to a phase of _90* for ^ s 0.1 ] [ ( V 1 0 0 ) + 1] is shown in Fig.9 0 '> ) . +1 I/--/ ur/ M Phase of G{ja>)H {ja) Is.t a i r ' ^ ^ j = -2 7 0 « Magnitude of G { ja ) H ( ja ) at a> = oo I 100 "vU . So the system is stable for any value of f c . Figure Bode plot S t e p . Enter the following code in MATLAB to draw the Nyquist plot: » n u m =[10 100].' ( ^ j Step 4 of 8 Magnitude of G ( ja ) H { ja ) at (2> a 0 is 100 7(0)’ +1 |g ( H w ( / < » ) L = = 00 Phase of G{ja>)H {ja) at a^sOis | G ( ^ f f l ) £ ^ = 0 '> + t a i r '^ ^ j.5 7 2 ® )-0 .01 • Calculate phase of the system at <ps0. Thus.91 started from the angle of —270®- Step 3 of 8 (b) Nyquisf plof represents frequency response of the system. | £ ( y ® ) £ ^ = 0 » + t a n .9 0 » = -1 8 0 » Follow the above procedure and calculate magnitude and phase of G{ja>)H {ja) for different values of a>.' [ y ] ] . stable? (c) Is the closed-loop system for the Bode plot shown in Fig. total phase=-180®-90®+0®-0® = -270® Thus.' ^ ^ j . The Nyquist plot will always be symmetric with respect to the real axis. Problem 6. Refer to the Figure 6.[ ^ 1 8 0 » . it is clear from the total phase obtained that the phase at low frequencies is —270®.f a n .25PP The Bode plot for 1 0 0 [(i/1 0 ) + l ] G (J) = 4 W D . The plot is normally created by the NYQUIST MATLAB m-file. stable? (d) Will the system be stable if the gain is lowered by a factor of 100? Make a rough sketch of a root locus for the system. So the system becomes unstable if the gain condition is < 1 .01 ^od zero at 5 = -1 0 contributes to a phase of O' fo r® = 0. (a) Why does the phase start at -270° at the low frequencies? (b) Sketch the Nyquist piot for G(s).s t e p s o lu t io n step 1 of 8 Consider the transfer function of the system.) (1) j[i-l] (a) Calculate the phase at low frequencies.b y .0 5 7 2 ® -9 0 ® -(l8 0 * -0 . 100 fe-') 100 G (j< o )H (ja )= - UOO ) Magnitude of G { ja ) H ( ja ) Is.'[ ^ j . looj^ G (.t a n . = 0 ® + 0 .^ 1 8 0 ° .9 0 ‘> . It is clear from the Figure that the phase of —27CP is obtained at ^ s 0.' [ ^ j j .9 0 ° . Thus. »sys=tf(num. and qualitatively confirm your answer. Step 8 of 8 (d) It is clear from the equation (1) that the gain of the system is 100. so the phase graph in Figure 6.! ” ' ' [ ^ ] = 9 0 ° . It is clear from the equation (1) that there is one right hand side pole at $ s 1.t a n .[0. so jy = 1.0 0 5 7 9 * — 269.01 The phase at low frequencies is obtained by calculating the total phase.^ 1 8 0 » . (c) Is the closed-loop system for the Bode plot shown in Fig. for all values of ^ the system is stable. the system becomes unstable when the gain decreases by a fector of 100.den).t > i r ' ^ ^ j j .9 0 » .01 for the system shown in equation (1).9 0 » . . Figure 1 Step 7 of 8 (c) It is clear from the Figure that there is a counter clock-wise encirclement. Z = JV+/* = 1-1 =0 For any value of Z = 0 : that is there are no characteristic equation roots in the RHP. » margin(sys) Step 2 of 4 Draw the bode plot. Problem 6. Step 4 of 4 The phase margin is more commonly used to specify control system performance it is most closely related to the damping ratio of the system.. num=[25 2 5 ]. conclude which margin would provide more useful information to the control designer for this system. Figure Control system G(s) -O Y Step-by-step solution step 1 of 4 The loop transfer function is. Bode Diagram Frequency (rad/s) Figure 1 Step 3 of 4 From Figure 1. G M |3. s 101^ Therefore.» den=conv{[1 2 0]. the gain margin.91dB The phase margin is.9! dBl phase margin.den)sys = 25 s + 25 — -----------. .sM + 4 s^'3 + 20 s^'2 + 32 s Continuous-time transfer function. 2 5 (s+ l) G{s) = - s(s + 2 ) ( j 2 + 2 r + 1 6 ) Use Matlab’s margin to calculate the PM and GM for G(s) and.26PP Suppose that in Fig. 2 S (j-t-l) C(s)= s (5+ 2)( s^+2 s + 6) Write the MATLAB code to calculate the gain margin and phase margin. on the basis of the Bode plots. The gain margin is.» sys=tf{num. <7A /s3.[1 2 1 6 ]). /> A /Is | | q | o|. Figure 3 Step 10 of 15 From the characteristic equation.2 + 2 A : = 0 Apply Routh-Hurwitz criterion to find the intersection of the RL with the imaginary axis. (c) Use hocus to determine the values of K at the stability boundaries. L ( » ) = A rC ( * ) W ( j) •)+ l ( i .2).5 s + 3 + s^2 ..» den=conv([1 -1]. p = \ . (d) Sketch the Nyquist plot of the system.l ) ( i ’ + 2 j+ 2 ) Step 2 of 15 (a) Substitute i for in loop transfer function.5 is Step 7 of 15 (c) Calculate the characteristic equation. s’ 1 K s’ 1 -2+2iC s' K -2 + 2 K 0 s' -2 + 2 K 0 For stable system. l + J C G (j)W ( i) = 0 s*2 \ + K\ = 0 . Thus.den)sys = s+2 - .93 in the textbook. the range of x for the stable system is ||< jj^< 2| • Step 4 of 15 (b) Calculate the phase margin for = 1. From the block diagram.42rad/s. iW = 7 ( j . Draw the bode diagram. the system has two unstable closed loop roots. (b) Verify the stable range of K by using margin to determine PM for selected values of K. Therefore. and use it to venfy the number of unstable roots for the unstable ranges of K. K -2 -2 K > 0 K<2 And.[1 2 2 ]).i) ( ( * + i) ^ + i) Step 8 of 15 Write the MATLAB code to plot the root locus. Z = N+P = 0+1 =1 The system has one unstable closed loop root.1+1 =0 Thus the system is stabie.sys=tf{num. »’ + i ’ + K t . (a) Use Matlab to obtain Bode plots for K = 1. Z = N+P = 1+ 1 =2 Thus. » margin{sys) Step 5 of 15 Draw the bode plot. Therefore..» sys=tf{num. and use it to venfy the number of unstable roots for the unstable ranges of K. » num=[1 2 ] : » den=conv([1 -1].1 ) ( s * + 2j + 2 ) ( j . The number of encirclements are. the range of x for the stable system is | | < jj^ < 2 | • . the range of x for the stable system is | | < jj^ < 2 | • Step 15 of 15 (e) Recall the characteristic equation. Figure Control system Step-by-step solution step 1 of 15 Refer to Figure 6. » num=[1 2].2 Continuous-time transfer function.den)bode(sys)sys = s-2 s^2 . the phase margin at A T s l. Problem 6. Bode Diagram Freqieacy (rad/s) Figure 2 Step 6 of 15 From the plot. ( j . the phase margin is. » num=[1.2 Continuous-time transfer function..5 • Write the MATLAB code to plot the bode diagram. the loop transfer function is. (ii) For 1 < a: < 2 . The number of encirclements are.66^^^ Al =1. Bode D iig ram Freqaeaqr (rtd/s) Figure 1 Step 3 of 15 From the bode plot.» sys=tf{num.den)sys = 1. and use the plots to estimate the range of K for which the system will be stable. p = l . determine the ranges of K for closed-loop stability of this system. PM = 6. all first column elements should be positive. » num=[1 2 ] : » den=conv([1 -1].den)sys = s+2 • . JV *1 Pole. (e) Using Routh’s criterion.den=conv([1 -1]. P ^ o is t Diagram Step 14 of 15 (i) For 0 < K < \ . (Ill) For 2 < K . K -2 -2 K > 0 K<2 And. the range of x fo'’ fbe stable system is |} < j^ < 2 | • Step 12 of 15 (d) Write the MATLAB code to plot the Nyquist plot.w) Step 13 of 15 Draw the Nyquist plot. -2 + 2 J ^ :> 0 ^>1 Therefore. all first column elements should be positive. The number of encirclements are. Thus.[1 2 2]).l ) ( j ' + 2 i+ j)+ /:( j + 2)«0 j ’ + « ^ . P a l .27PP Consider the system given in Fig. Pole. (1) ( * .2 Continuous-time transfer function.5 3 ] . Therefore. Therefore. where phase plot crosses the —]80^ line. 1 K 1 -2 + 2 K K -2 + 2 K 0 -2 + 2 K 0 For stable system. (d) Sketch the Nyquist plot of the system.2 Continuous-time transfer function.» sys=tf(num. But this can make the closed system stable with positive gain margin by increasing the gain K up to the crossover frequency reaches at ^ s 1.l ) ( s ’ + 2 i+ 2 ) Write the MATLAB code to plot the bode diagram.2 + / r ( i+ 2 ) = 0 »’ + * ’ + K t-2 + 2 A : = 0 (2) Step 11 of 15 Apply Routh-Hurwitz criterion to find the intersection of the RL with the imaginary axis. -2 + 2 K > 0 K>\ Therefore. Therefore.. Z = N+P = .» rlocus(sys) Step 9 of 15 Draw the root locus plot.5- Therefore. the closed loop system is unstable for ATs 1.. JV = -1 Pole.[1 2 2 ]).[1 2 2 ]).» nyquist(sys.» w=logspace(-2. 3 .2 16. . it is clear that the gain margin is. 0 2 rad/s The phase margin is. sys=tf(num.4i*+32* Step 2 of 4 The MATLAB code to obtain gain and phase margins is.4 32 0].90 in the textbook. That is the roots of the system are real. .28PP Suppose that in Fig. Problem 6.8* ■ damping will be 1. and comment on whether you think this system will have well-damped closed-loop roots. G M = l d B a t ^ = 4 . bodeplot(sys) margin(sys) grid Step 3 of 4 The Bode plot is shown in Figure 1. it is clear that a very small change gain would cause instability to believe that the damping of roots is very small.2 (1 + 1 ) <Hs) = - j (i + 2)( i 2 + 0.2(i + l) / + 2 . The open loop transfer function is. Bode Diagram Freqaeaqr (rad/s) Figure 1 step 4 of 4 From Figure 1.2 j+ 1 6 ) 3.ten ^ nliitinn Step-by-step solution step 1 of 4 Refer to the control system in Figure 6. 3 .2 3.2 ( i + l) g (5 )= ■ ^ V ' ' j ( i + 2 ) ( s ’ + 0 .2]. den=[1 2. However from the Figure. 8 'at ffl = 0 .den). PM = 9 2 .2 s + 1 6 ) ' Use Matlab’s margin to calculate the PM and GM for GfsJ. num=[3. Figure Control system |cw|—p o r Sten-hv-<. 2 i ’ +16.1 n u j/s Since p |^ * 92. Problem 6.29PP For a given system. show that the ultimate period Pu and the corresponding ultimate gain Ku for the Ziegler-Nichois method can be found by using the following: (a) Nyquist diagram (b) Bode plot (c) Root locus Step-by-step solution Step-by-step solution ste p 1 of 4 K g and Tg are related to gain andintegtal dehvative times Tr and Tn Step 2 of 4 ^ a. At that point Pg = — Tiriiere a is phase cros sover frequency and gain is Step 3 of 4 ^ In Bode plot the point at vidiich plot crosses 0 axis is the desired point to find K g andPg Step 4 of 4 In root locus plot the point at which plot crosses imaginary axis is the desired point to find R g and Pg . In Nvquist diagram the point at plot crosses real axis 2fr is the desired point. 36 .7 Sor ^ = 0.f 2x0.31 is 2^ PM = tan" 2x0. PM = 170.7. ^ " --------y.4 Jbr ^ = 0.1 Jbr 4 = 0. then the closed-loop transfer function is given by Q)i T (s) = .5- s +25oa^s+cc^ Step 2 of 2 PM as given by equation 6.0 .4 .7 .) A nd closed loop transfer function is given by T ( s) = t -----^ -----.V V l+2x0.4.0. Problem 6. P i f = tan = 110. PM=taa~^ =1 ^ .\ .4" -2 X -0 .2 x .7^ All these values matche closely with fig: 6.1.33*1 _> /V n-2x0. + 2f0)nS + Verify the values of the PM shown in Fig.42*1 V > f t + 2 x 0 . Figure Damping ratio versus PM Step-by-step solution step 1 of 2 G (0 = - s(s+2?<ii.7*-2x-0. f o r^ = 0.4 V 2x0. and 0.f .30PP If a system has the open-loop transfer function G (s) = - s(s + 2(o>n) with unity feedback. p h a s e . n u m G = [ 1 ] . Bode Diagnzn Step 6 of 10 The magnitude in dB for a phase margin of 45 ® is. \ 2 is 14.) ( |G ( y < » L . the gain margin for K . 0 4 0 . 0 4 0 .08 j + 1) ^ __________ K __________ 0. the velocity constant. for phase margin equal to 45® is m step 10 of 10 (d) Draw the root locus of the system using MATLAB and indicate the roots for a PM of 45®.892 = 1.12_________ ^ ' " j (0. d e n G ) . the gain K for a phase margin of 45® is m step 7 of 10 ^ Write the MATLAB code to determine the gain margin for fC = 1. Thus. d e n G = [ 0 . . R o o tL o c u Observe from the plot that the roots are -0. G (4 = K iR iH +1 ” f [Q.971. 08(y<») + .9dB . [ m a g . s y s G = t f ( n u m G .12 -U s j (0. ^ i * o. 1 2 1 . K .0 4 j’ +0.892 Determine the gain K for a phase margin of 45® using the condition that the magnitude is 1 at the phase margin. 2 0 lo g |C ( > ) | = -0.12 j ’ + I.045*+0.12 Thus.12 is. 1 + 1 . and indicate the roots for a PM of 45*. Problem 6. 1 2 1 . n u m G = [ 1 ] . 1 2 1 .r ) |= ' ___ 1 _ * 0.08*+ l] ____________ K ___________ “ j (0 .468± y 0 .9 7 l|.CMi"+0.31 PP Consider the unity-feedback system with the open-ioop transfer function K GW = - ■ s(s + l) [ ( P / 2 5 ) + 0 .«> Step 2 of 10 (a) Write the MATLAB code to draw the Bode plot for with J ^ s l.04j* + 0 . Step 9 of 10 ^ (c) The open loop transfer function of the system for Jir = 1. s q u e e z e ( m a g ) ) s e m i l o g x ( w . K.12 • n u m G = [ 1 .12 j * + 1.468± y0. i 2 ( y « ) ) + i .\ .0K s + 1) = 1. s y s G = t f ( n u m G .08j + 1) =U " (0.996 |G(7<»)| = 0.4 ( i/S ) + 1] ■ (a) Use Matlab to draw the Bode plots for G( jo)). (d) Create a root locus with respect to K. l o g l o g ( w . m a r g i n ( s y s G ) Step 5 of 10 iviai i\ u ic pun II ui I u ic pi laoc piui wi icic ii ic pi lasc lo " luiuaico a pi laoc iiiaigii i ' 45» Mark the corresponding point on the magnitude plot to get the gain in dB. assuming that K = (b) What gain K is required for a PM of 45°? What is the GM for this value of K7 (c) What is Kv when the gain K is set for PM = 45“? (d) Create a root locus with respect to K. 1 2 ] . when the gain K \s^ . 0 4 0 . 0 8 1 0 ] . the roots for a PM of 45® are |-Q . Step-by-step solution step 1 of 10 The open-ioop transfer function of the system is.0 4 j ’ + 0 .12. and indicate the roots for a PM of 45*. s q u e e z e ( p h a s e ) ) Get the MATLAB output for the Bode Magnitude plot.12 Thus. that is. d e n G = [ 0 .\2s^ +\. 0 8 1 0 ] . Step 3 of 10 Get the MATLAB output for the Bode phase plot. gf _ 1. 0 8 1 0 ] . |( ^ L « . d e n G ) .04j * + 0. d e n G = [ 0 .1 2 i’ + 1. = Lt sG (s) 1.08i’ + j G ijoA — . BodeD U grw Thus.04 s ’ + 0.04(y<») + o . s y s G = t f ( n u m G j d e n G ) .08s’ + i+ 0 . w ] = b o d e ( s y s G ) .08 j + 1) Determine the velocity constant. . Step 4 of 10 (b) Write the MATLAB code to get the gain K for a phase margin of 45®. m a r g i n ( s y s G ) Step 8 of 10 Get the MATLAB output for the Bode plot. den ) Step 3 of 25 Transfer fimction: 1 8^4 + 9 s^3 + 28 s^2 + 36 s + 16 » rlocus(x) » rloc£ind(x) Select a point in the grs^hics window selected_point = 0.6db shouldbe added.[]) » gtext('\2eta*0. Problem 6. » den*conv(c. » <^onv(a.b). and determine the GM that results for PM = 65®.6db 201og£'=29. The magnitude correction is independent of frequency.den) Step 6 of 25 Transfer function: 1 sM + 9s^3 + 28 s^2 + 36 s + 16 » rlocus(x) » v=[. » num=[l]. » ^f(n u m .S6 K = -[0 ^ K = 2.^. the transfer-function blocks are defined by 1 1 G (s ) = 'H ( s ) = (5 -1 -2 )2 (5 -I-4 ) 5-1-1 (a) Using hocus and riocfind. (c) What is the GM of the system if the gain is set to the value determined in part (b)? Answer this question without using any frequency-response methods. how does the transfer function V2(s)/R(s) differ from Y1 {s)^(s)7 Would you expect the step response to r(t) to be different for the two cases? Figure Block diagram.d ). (b) Using hocus and riocfind. » num=[30.6db is contributed by the term K.3—^^------------ ^ ' ( s + 2 ) '( s + 4 ) ( s + l ) Bode plot for open loop transfisr for iT = 1 a ^ l 4 4]. » sgrid([0. What damping ratio would you expect for this PM? (e) Sketch a root locus for the system shown in Fig (b).b).707 Step 9 of 25 c) Gain margin is g = -? L 5.den) Transfer function: 30.928 G = 13. The value is calculated by equating 2 0 1 o ^ to 29. » b = [1 4 ]. » ^£(n um .b). » c=[l 1].9288 Step 8 of 25 Root locus is K = 5. = -115® Step 13 of 25 ^ With 1 the db gain at 135® is -2 9 . » dei^o n v (c.12 open loop transfer function is a { s ) H [ s ) ----------------------------- ^ ' ( s + 2 )’ ( s + 4 ) ( s + l) Now the bode plot for above system is » a = [ 1 4 4]. » rlocfmd(x) Select a point in the graphics window Step 7 of 25 selected_point = -0. determine the value of K that will produce roots with damping corresponding to 0.6 294 A T =10» ^ = 30.32PP For the system depicted in Fig{a). » n u m = [ l 1]. (b) H(s) in feedback S te p -b y -s te p s o lu tio n Step 1 of 25 (=0 Given o ( s ) = ------------2------------ ^ ' ( s + 2 ) \s + A ) H (s ) — ! - ^ ' s+1 ( s + 2 ) ^ ( s + 4 ) ( s + l) Step 2 of 25 MATLAB program for root locus » b = [1 4 ].b). determine the value of K at the stability boundary.(s) K a (s )H {s ) R [s )~ l+ K O {s )H {s ) 1 .d ).56db Gain margin in linear scale 201ogK=8.12 sM + 9 s^3 + 28 s^2 + 36 s + 1 6 » bode(x) » grid Step 15 of 25 Step 16 of 25 From the above bode plot Gain margin = 8. (d) Create the Bode plots for the system.6db iqswards.0179 + 2.6 d b .( s ) i:(s + i) 5 (s) J i:+ (s + 2 )’ ( s + 4 ) ( s + l) step 23 of 25 M AILAB program for step response » a = [ 1 4 4]. (a) and (b).12].66 Step 10 of 25 <0 Given 0 ( s ) « ( s ) = -------.12 Step 14 of 25 With K = 30.9598 Step 4 of 25 From the matlab code for . » num=[l].AT= 81 root locus crosses imaginary a For ^ > 8 1 system is unstable Step 5 of 25 b) ^F=[14 4].707') » axis(v).6 5 Step 18 of 25 Fig (a) Step 19 of 25 Fig(b) Step 20 of 25 Root locus is drawn for 0(^b)H(^ s) For figures 0 { s ) H { s ) = K O { s )H (s ) So Root locus for both the figures is same And root locus is Step 21 of 25 £) For figure (a) Transfer function is T. » b = [1 4 ]. How does it differ from the one in part (a)? (f) For the systems in Figs. (c) What is the GM of the system if the gain is set to the value determined in part (b)? Answer this question without using any frequency-response methods. » c=[l 1].56 8.9092 + 0.707. » ^conv(a. » b = [1 4 ]. 1 yi{s) (s + 2 ) ^ ( s + 4 ) s+ 1 (s+ 2 ) (s+ 4 ) s+1 1 J^(s) ( s + 2 ) ^ ( s + 4 ) ( s + l) (ff+2) (ff+ 4)(ff+ l) W __________ f R{s) J5 T + (s+ 2 )^(s+ 4 )(s+ l) Step 22 of 25 For figure (b) Transfer function is r.00821 ans = 81.6B So gain margin is 2. » c=[l 1]. » ^conv(a. Hence the magnitude of 29.b). (a) unity feedback.90861 ans = 5. Hence to every point o f magnitude plot a db gain of 26.den) Transfer function: 1 sM + 9 s^3 + 28 s^2 + 36 s + 1 6 » bode(x) » grid Step 11 of 25 Bode plot is Step 12 of 25 ^ Phase margin ^ = 1 8 0 ° + ^ where is the pahse o f G(j(zr) of a = v4ien y = 65® 65=180®+^^ d>. » ^£(num .<5. » ^conv(a. » b = [1 4 ]. » }^f(num. This gain shouldbe made zero to have to PM of 65®.d en ) Step 24 of 25 Step 25 of 25 . » dei^o n v (c.7a7]. .9288 Produces roots with damping ratio corresponds to 0.6 0 -3 3]. » ^tf(num .<5. The corrected magnitude plot is obtained by shifting the plot with R=1 by 29. » den=conv(c. » c=[l 1]. » c=[l l]i » ^conv(a.( s ) _ K a js ) R (s )~ \+ K O [s )H [s ) jv 1 r^ (s) (s + 2 )^ (a + 4 ) K 1 ^ W " l+ - (s+ 2 ) (s+ 4 ) s+1 jv ] ^ (s + 2 )^ (s + 4 ) ^ ------------ ( s + 2 ) ( s + 4 ) ( s + l) r . » num=[l].68 Step 17 of 25 Approximate danq^ing ratio PM 100 65 c=- 100 ^ '* 0 . » den=conv(c. use Bode and root-locus plots to determine the gain and frequency at which instability occurs.. instability occurs A !soalffl= 034rad/sec .4 d B at PM = 20“ .96 and Gain —S 2 |. |K =4.33PP For the system shown in Fig. What gain (or gains) gives a PM of 20*? What is the GM when PM = 20*7 Figure Control system Step-by-step solution Step-by-step solution step 1 of 1 Open loop transfer function of the unity feedback system is given by T (s) = s^ (s+3) [s^+2s+25j From Madab we get &at at ^d= 4.17 to get PM = 20°| Also G M = 2 2 .96 instability occurs A t \tD— 4. Problem 6. approximately the value of K is. the transfer function of the system for the Figure 6.34PP A magnetic tape-drive speed-control system is shown in Fig. T {s ). 1 Equate this expression to equation (2). the value of reel time constant r^is 4 seconds. where b = the output shaft damping constant = 1 N m.- 13(0. I G (*) I = 0.94 is as follows. If the gain is less.5 * + 1 4 Step 6 of 10 The transfer function of the fonvard loop is.m. Thus.5 j + 1) '■ ' ( 0 . 0. I G M «t 1 0.sec The steady state error of the system is. for the emor less than 7% of the reference.^ p and motor I ^ drive Step-by-step solution Step 1 of 10 Refer to the Figure 6.2 8 -!£ a: 13. Step 5 of 10 (b) Recall equation (1). the steady state error e„ will be high.1 with 180^. the phase margin PM is 0 Step 10 of 10 Since GM is low. this is not a good design and needs a compensator.m. the gain should be low. Find the PM. the value of output shaft damping constants is 1 N. 1 <?(*) ( j + l)(4 i + 1 ) Simplif Simplify the expression further. (a) Determine the gain K required to keep the steady-state speed error to less than 7% of the reference-speed setting. Thus. i s l second. . ( * ) U .5 * '+ 5 .nLsecand the value of motor time constant r . the damping ratio is low and the overshoot is high.f 180® -7 ® Thus. For the system to be more stable. + l)( y s + i) + iT Step 2 of 10 (a) It is clear from equation (1) that the system is of Type 1. 1 _ 1 1 + /:= !+ * . 1 N. The steady state error of the system is. Step 8 of 10 Hence.79 at ZG(s) = -180’ . PM = [ ^ C ( * ) | | ^ . Obtain the transfer function of the system.94 from the text book. (b) Determine the gain and phase margins of the system.5 secfor . | . . ------------!--------------- ^ ( l + 4® ’ ) ' + (5® )’ Find |G (* )|a tZ G (* ) = -180' The phase shift is.5 sec.0 7 " I4 .5 i + l ) ( j + l ) ( 4 j + l)+ 1 3 1 3 (0 . the gain must be . f— If—1 1 -1- ( 1) ( s r .5*’ + 3 . e^£7% 1 7 ^ 0 . Step 4 of 10 Calculate the value of gain such that the steady state error is less than . Is this a good system design? Figure Magnetic tape-drive speed control I Amplifier [ 7 ™ .2 7 Thus. we get Z G ( f ) = -I73®- The phase margin PM is obtained by adding Z G ( 4 ) when | G(iS)| . + ! ) ( « .79 -1 .sec for b.5 i + l) 0.5 seconds. + l ) ( J s + 4 ) + ^ T i ( s T . Thus. the gain margin GM is ||.28£JC Hence. The GM can be calculated as the inverse of the magnitude of G(s) at Z G (4 ) » -180*- Find the GM. the system is very close to instability and since PM is low.0 7 1 + ^ :" 1 z^\+K 0 . It is clear that it^in equation (2) represents a positional error constant.27l- Step 9 of 10 Find the Z G (4 ) when | G ( j ) | .2 8 S 1 + /: 1 4 . it is clear that the value of K is equal to k^ . and the motor time constant is r1 = 1 sec.+ + 1j + X Substitute 13 for/C.sec ^ in steady state emor. + 1) r(j)= { « ’. Problem 6. I Substitute 1 N. 1 C (*) = (4i* + j + 4 j + l) I + J5a> + 1 I ( l + 4<P*) + JS49 step 7 of 10 Find the magnitude of G(s) | c . It is clear that the value of speed measurement time constant is 0. + l)(*r. It is clear that for Type 1 system. sR{s) l + G{s)H{s) = lim- 1+ n i+ ( i) ( jr ) Step 3 of 10 It is clear that the value of b is 1 N. ] + 180» ■ -I73 ® . the reel time constant \sTr= J/b = A sec.1 • Thus. . + l ) ( i r . there exists positional error constant . The speed sensor is slow enough that its dynamics must be included.sec. Thus. A r(tr.m. The speedmeasurement time constant is rm = 0. and 1 sec for r. 1 (2 ) 1+*.. Calculate the steady state error. it is clear from the root locus plot shown in Figure 2 that the number of roots in RHP for those values of x for which system is unstable is zero. ZK ° 4 Consider positive value of k ■ 3 gain of G { ja ) H ( ja ) at = is y . » den=[1 4 3 0]. by observing the nyquist plot of Figure 1 are zero.» ) « ( H -----.3 5 P P For the system in Fig. » sys=tf(num.3 of root locus or —l < ^ < 0 > s o ^ d o e s n ’t represent break away point. Number of unstable (RHP) poles of G { ja ) H ( ja ) .' = -18<r . Procedure to draw root locus: RULE 1 : Number of poles of the feedback system shown in Figure 1 are three. The number of clockwise encirclements have to be now determined depending on value of . Enter the following code in MATLAB to draw the root locus: » num=[3]. So the system is stable for >-4. 2 1 Step 12 of 14 ^ The root —2.4 5 j. [ j ( j + l)(*+3)J The number of roots in RHP is zero. Thus. Nyquist plot encircles Number of clockwise encirclements of .300*- Step 11 of 14 RULE 4: calculate break away point.2 ’ that is there are no characteristic equation roots in the RHP.0.l l “ 3 = -1.1. gain of the system when the Nyquist plot touches the real axis is to be known. Consider transfer function shown in equation (1).. » sys=tf(num.ff(y®)=0-90* .(M l. So. Find ® value by equating phase of the system to —igQ*- 0 . A^ = 0 Number of unstable (RHP) poles of G ( ja ) H ( ja ) .(3 » * + 8 s + 3 ) ' = 0 3 3s* + 8 j + 3 = 0 The roots obtained for the above equation are 5. Z^N +P = 0+0 = 0 For positive value of K . Figure Control system S te p -b y -s te p s o lu tio n step 1 of 14 (a) Refer to the feedback system in Figure 6. The root . JV = 1 Step 9 of 14 Number of unstable (RHP) poles of denoted by p are zero.A ’ Z = I ’ that is. all three poles approach asymptotes.0 . Numbers of unstable closed loop roots Z are. Check your answer by using a rough rootlocus sketch. ''^prosents break away point. » den=[1 4 3 0].45 lies in the range —] < 5 < 0 of root locus.4 .tan"'^yj = -90*-90*-90* = -270* Follow the above procedure and calculate magnitude and phase of G ( j o ) f / ( j a ) for different values of ® for drawing the Nyquist plot. Z = N -¥ P = 1+0 =1 For K < . When Nyquist plot doesn’t encircle » j. Draw the root locus for equation (1). To determine the range of for the stability. RULE 2 : The real-axis segment defined bv j< .3 a n d —1 < < 0 is part of the locus. The closed loop transfer function for the above feedback system is: R I+ G (s)N (s) 3 j( j+ I) ( j+ 3 ) 1 + a: 3 |i « ( j+ l) ( « + 3 ) J 3R 5 ( j + 1) (j + 3 ) + 3 ^ From the above transfer < 7 ( ^ )//( j) c a n be obtained as shown below. Thus. the system is stable fo r positive values o f . Step 6 of 14 ^ Determining stability of closed loop system based on frequency response of system’s open loop transfer function Is Nyquist stability criterion. ^ -1 _ j ( j + l)( j+ 3 ) Perform — = 0- ds j(j+ l)(j+ 3 ) 3 Simplify the above equation. G (s)ff(s)^/C 5(« + I)(5 + 3 ) i( i * + 4 s + 3 ) +3s] Enter the following code in MATLAB to draw the Nyquist plot.: — ^ = 9 < r -T 4ffl 3 -a ^ b O e )= J i Step 7 of 14 Substitute the above obtained value of ® in equation (2).I a n " ' t a n " ' Step 3 of 14 Magnitude of G ( j a ) f / ( j a ) at (psQ is *00 Phase of G ( j ( » ) f / ( j a ) at ® a Q is |G (y '® )tf(y ® ) = 0 -9 0 * . 0-1-3 3-0 .t a « . Z = JV+/* = 0+0 = 0 For AT > . there are characteristic equation roots in the RHP.i. there are three branches to the locus. denoted by p are zero.33 The angles of asymptotes are at 60*. Step 8 of 14 Consider k value as negative.den): » rlocus(sys) Step 13 of 14 ^ The locus of closed loop poles with respect to x shown in Figure 2: F ^o re 2 Step 14 of 14 Thus.. the system is unstable for K < . . The plot is normally created by the NYQUIST MATLAB m-file.2 . So.den): » nyquist(sys) The Nyquist plot for the feedback system is shown in Figure 2.A - Thus..t»d-‘ j .Z .95 in the textbook. (y®) = 0 -90* .1 (2) + V 9+ < ir The phase of is. denoted by p are zero.1. = . When j ^ < . The numbers of clockwise encirclements of -1 (denoted by ^ ). it is clear from Figure 2 that the roots in RHP are zero. Step 5 of 14 The Nyquist plot will always be symmetric with respect to the real axis.21 doesn’t lie in the range j < .tan"' j . so 5. » num=[3]. it Is clear from the obtained results that the range of x. The magnitude of is.tan"' j “ ^j a -9 0 * Step 4 of 14 Magnitude of G ( j a ) f f ( j a ) at ® a 00 Is =0 Phase of G ( M ) f / ( j a ) at ® so o ls |G(y®).4 . . Procedure to determine gain of the system when the Nyquist plot touches the real axis is explained as follows. RULE 3 : Calculate the centre of asymptotes. tor which system is stable is S > 3 Step 10 of 14 ^ (b) The number of roots in the RHP for those values of x tor which system is unstable can be determined from the rough root locus sketch. Z = 0 : that is there are no characteristic equation roots in the RHP. Number of clockwise encirclements of . 1(5 + l)(*+3)] Step 2 of 14 ^ Nyquist plot represents frequency response of the system. t a n '. P ro b le m 6 . « .180*. determine the Nyquist plot and apply the Nyquist criterion (a) to determine the range of values of K (positive and negative) for which the system will be (b) to determine the number of roots in the RHP for those values of K for which the system is unstable. | G ( y .9 0 . 36PP For the system shown in Fig. OM = 2 K<2 Also for negative K . In both regions of instability. two roots lies on right half plane . G M =1. determine the Nyquist plot and apply the Nyquist criterion (a) to determine the range of values of K (positive and negative) for which the system will be (b) to determine the number of roots in the RHP for those values of K for which the system is unstable. R > 1 Hence 1-1 < K < ^ Step 3 of 3 b.. Check your answer by using a rough rootlocus sketch. Figure Control system Step-by-step solution step 1 of 3 K G ( 0 = ^ Step 2 of 3 a. Problem 6. Z —V that is there are characteristic equation roots in the RHP. Figure Control system Step-by-step solution step 1 of 9 Consider the feedback system shown in Figure 1. N ^ \ Number of unstable (RHP) poles of G ( j t t . one of which approach finite zero and other of which approach asymptotes. So the system is unstable for AT < -1 • It is clear from the above results that there is no range of x tor which system is stable. ) H ( ja ) .tan"' I 1 U -H J «l80® -9 0 ® -9 0 ® = 0® Follow the above procedure and calculate magnitude and phase of G ( j a ) H ( j a ) for different values of of. the expression for G ( r ) f f ( r ) is. AT=0 Number of unstable (RHP) poles of G ( j a ) H ( j a ) . Step 9 of 9 Enter the following code in MATLAB to draw the root locus: » num=[1 .§ . ■ (^+ 1)’ ic O v » -i C { ja ) H { ja ) K ( j a . Step 5 of 9 Determining stability of closed loop system based on frequency response of system’s open loop transfer function is Nyquist stability criterion. so ^ doesn’t represent break ii point.= 3 5j«-l The root 3 doesn’t lie in the range —1 < 5 < 1of root locus. Consider positive value of . r CM R 1 + G (r )ff(*) s-\ 4 -1 \+ K . The number of clockwise encirclements have to be now determined depending on value of x ■ When Nyquist plot doesn’t encircle _ j.'^ j ^ j Step 3 of 9 Magnitude of G { j a ) H ( j a ) at ^ s O ■ Phase of G (y«l))H 0V ») at ® = 0 is ^ G ( J a ) H (J a ) = 1 8 0 ° .i ’ .la n .t a n .1]: » den=[1 2 1] » sys=tf(num.' ( 0 ) . Z = I . 2( ^ . So the system is unstable for positive values of k ■ Step 6 of 9 Consider x value negative. the numbers of roots in the RHP are checked from the root locus as shown in Figure 3. Step 8 of 9 (b) Number of roots in the RHP should be determined from Nyquist plot first and then verified using root locus. The numbers of clockwise encirclements of -1 (denoted by f j ). Number of clockwise encirclements of -1 . Figure 1 Step 2 of 9 (a) Write the closed loop transfer function for the feedback system shown in Figure 1. Check your answer by using a rough rootlocus sketch.l ) ( j + l ) . it is clear that root at $ s 1 is in the RHP which results in system to become unstable. 1]: » sys=tf(num.t a n " '( < » ) . represents break in point. Z = 0 : that is there are no characteristic equation roots in the RHP. ) H ( ja ) isobtainedat ^ s Q a n d the value is jff.. Z = N+P = 1+0 = 1 For A T <-1 . The Nyquist plot will always be symmetric with respect to the real axis. ( 1) (s + \) Nyquist plot represents frequency response of the system. The plot is normally created by the NYQUIST MATLAB m-file.4 0J • 0.2 .tan"' («>) . K ^\ +0 ‘ |g ( » « ( H = - +Aa? K y l \ + e i‘ K Phase of G (ja > )H {ja ) is.den): » nyquist(sys) The Nyquist plot for the feedback system of Figure 1 is as shown in Figure 2. RULE 3 : Calculate the center of asymptotes.1 ■ Ifr » .I • Step 7 of 9 When at < -1 >Nyquist plot encircles _ ].( i + l)‘ ( * .2 i = 0 (* + •)’ j^ -2 j -3 = 0 The roots obtained for the above equation are s .') ’ 2j ' .37PP For the system shown in Fig.1 ■ • • ■“ • I i -5 -4 J -2 -I 0 1 2 R e d AzbCsccoMb'*) Figures Thus. that is there are characteristic equation roots in the RHP. Draw the root locus for equation (1).den): » rlocus(sys) The locus of closed loop poles with respect to x shown in Figure 3: R oo tL o o u 0.') Magnitude of G ( j a ) H { j a ) is. The maximum gain of G ( j t t . Z G ( ja i) H {ja f) = 0 + tan"' = 1 8 0 » . Procedure to draw root locus: RULE 1 : There are two branches to the locus. Z = N -¥ P = 1+0 =1 For positive value of K . So the system is stable for AT > . so 5.0 . RULE 4: calculate break in point.l . Numbers of unstable closed loop roots Z are. denoted by p are zero Z = JV+/* = 0+0 = 0 For AT > . Enter the following code in MATLAB to draw the Nyquist plot. Problem 6.1 . H z IH !) 2-1 -3 S ---- 1 = -3 The angles of asymptotes are at 180®. determine the Nyquist plot and apply the Nyquist criterion (a) to determine the range of values of K (positive and negative) for which the system will be (b) to determine the number of roots in the RHP for those values of K for which the system is unstable. Number of clockwise encirclements of -1 .tan"' = 180“ Magnitude of G ( j a ) H ( j a ) at is step 4 of 9 Phase of G ( j< o ) H ( ja ) at n>soois Z C O 'o ) H (j& i) = 180®. . Number of unstable (RHP) poles of denoted by p are zero.2 • 0.-1]: » den=[1 2 . The root -1 lies in the range $ < —lo f root locus. by observing the nyquist plot of Figure 2 are one. K =- (» -•) Perform — = 0 ds Simplify the above equation. Thus.1 [ ( « + !)■ From the transfer function. RULE 2 : The real-axis segment defined bv 5 < —l a n d —1 < ^ < 1 is part of the locus. » num=[1. denoted by p are zero. (d) System type (0. 1. Figure Nyquist plots \ 1 -1 1 j V H (I) (b> Step-by-step solution step 1 of 4 For first sjrstem phase margin is very low while for second system Phase margin is high.1 since semicircle is enclosed . The proposed operating gain is indicated as KO.2 since full circle is Enclosed Second system is Type . In each case give s rough estimate of the foilowing quantities for the closed-loop (unity feedback) system. Step 3 of 4 For first sjrstem Gain > Ko For second system Gain < Ko Step 4 of 4 ^ First system is Type . or 2).38PP The Nyquist diagrams for two stable. 1. and arrows indicate increasing frequency. or 2). (d) System type (0. Problem 6. (c) Range of gain for stability (if any). open-loop systems are sketched in Fig. Step 2 of 4 b. (b) Damping ratio. First system has low damping ratio while second has high damping ratio. (a) Phase margin. 0362 ) Step 2 of 4 b. (c) Is the ship-steering system stable with K = 0. U .a jg 1 at m = 0 . Problem 6.0 3 6 2 + l ) ’ where V is the ship’s iaterai veiocity in meters per second.2? (c) Is the ship-steering system stable with K = 0.f —^ l + l l G (s)= - . PM. (b) On your plot.2? (d) What value of K would yield a PM of 30*.0 5 7 6 ra i/ sec \P M = .3 2 5 + l)(5 /0 .2.1 4 2 J J UO. No . |g M = .0318raA /s»c |Aad K = 0.087 n jrf/sw Step 3 of 4 c. (a) Use the Matlab command bode to plot the log magnitude and phase of G( jo)) for K = 0. ship steeling system is not stable Step 4 of 4 A A t PAf = 30* ® = 0. and 5r is the rudder angle in radians.042| . and what would the crossover frequency be? Step-by-step solution step 1 of 4 K .5 . r \ at » = 0.2 3 . indicate the crossover frequency. and GM.( j/0 .39PP The steering dynamics of a ship are represented by the transfer function V js ) ^ ^ n .1 4 2 ) + l] S f( s ) f ( j / 0 .325 A o. PM = 1 8 0 » + Z G ( y a > ) L .577(80fli-<i>’ )-19< !> *-100 = 0 -0 3 7 7 < »’ .108 Therefore.860y r a d /s e c Step 4 of 4 ^ Find the magnitude of the open loop system.c)=tan"'^j^^j i'-© J Simplify further. Z G 0 « ) = U n .a > H . 1 + G (j)« (i) = 0 i^ ( j+ lo p ' » * { i+ io ) * + A : ( i+ i) = o »^(** + 100+20s)+AS+i(r = 0 j ‘ + 2 0 j ’ + 1 0 0 i* + K j + A : = 0 Apply Routh-Hurwitz criterion. From ^2 row. I __________ = .. . 30®=tan"* j . « . ’ s ^ ( ja + \ o f Phase margin Is.00141822 Therefore to get crossover at that frequency x become..g ) -I50° =tan-‘ .-(i)-. n r -8 0 ® ^ 3 0 . ^ 1 9 ^ 80® 0.3 1 )* ) 35.00141822 -70S.40PP For the open-loop system K (s + \) K C (S ): determine the value for K at the stability boundary and the values of K at the points where PM = 30“ . ^ ^ .3 1 ))V ( i 0 0 -(-3 5 . the fc value at which PM s 30® i® 1705.1 9 ® * + 4 6 ..44)-1146.324 (35.( l) . m — 2 m -K -m > o \€ 0 0 > K Range of value at the stability boundary is |0 < /r < l6 0 0 l- Step 2 of 4 Consider the open-loop system.2019±1.80--2. jc = — !— 0. the value becomes.1 8 0 " . 30° = 1 8 0 °+ Z G (. the value x becomes.»-g) -150'=tan €) a> 180'-tan since 2tan"'(.f f l* ) ^ Find the magnitude at frequency ®s— 35^3lad/sec- ------------- a/(-3 5 -3 1 )'^ ((2 0 )(-3 5 .1081 .> » ” Step 3 of 4 ^ Find the phase frequency at which angle becomes 150® of the transfer function.. Problem 6. 1 100 K 20 K 0 2 00 0 -X 20 2 0 0 0 .-- ^ / ^ ^ ( 2 0 ® ) * + ( l0 0 . Step-by-step solution ste p 1 of 4 Step 1 of 4 Write the characteristic equation.18< »-100 = 0 Solve the equation.387) = 0.3 1 rad/sec» ® = 1. ----------.324 ~ (35. Substitute 30® for PM in the equation.2 . s * ( j + 10)^ Convert the function to the bode fonn. 2000^^0 20 20o o > a : From j t row.a: 20 K For system to be stable there should not be sign change in the first column. |G(y®)|---------.7961 35.31)(705. ® = -3 5 .3I)^(498718.tan"' <a i'-© J ( i ) ' 30* tan"' i'-O J since tan" X-tan" tan' ‘ — fe) <11 -a [-© j A x-20 tan30«»- F IR -I9 a )^ -I0 0 tan30®» «i’ -l0 0 < » + 2 0 i» -1 9 « » * -1 0 0 lan w » . ) (d) What is the PM of the system as drawn? (Estimate to within ±10“ . ASK AN EXPERT . Problem 6. (a) What is the velocity constant Kv for the system as drawn? (b) What is the damping ratio of the complex poles at w = 100? (c) Approximately what is the system error in tracking (following) a sinusoidal input of cu = 3 rad/sec? (d) What is the PM of the system as drawn? (Estimate to within ±10“ .) Figure Magnitude frequency response Step-by-step solution There is no solution to this problem yet. Assume that the plant is open-loop stable and minimum-phase. G et help from a Chegg subject expert.41 PP The frequency response of a plant in a unity-feedback configuration is sketched in Fig. 42PP For the system 1 0 0 (j/a + l) G(i) = - '5 ( j+ l) ( j/ f r + l) * where b = 10a.63 T_____ .8 Va cV = . is alead compensatorwith a=0. find the approximate value of a that will yield the best PM by sketching only candidate values of the frequency-response magnitude.8 7>l2 T=0. Problem 6.l O d B occurss at cd^=17.5 j==17.1776 =-^=5.1 Gain o f .2 0 l o g . Step-by-step solution step 1 of 1 Step 1 of 1 G(s)=- ^ e r e b=10a. 2 . PM = 180® +^^ The phase of the system is.6 3 6 .2 t a n " '— 20 Step 2 of 10 Hence.2 1 -0 . K G ( s ) = .den).0 3 1 4 = 20.21)^ ((2 5 . the maximum possible bandwidth is 120.178 rad/s Thus.(f < e++ 4< 0o 0 ) Step 7 of 10 Substitute 25.+ 2 0 S 0 step 5 of 10 The roots are.178 rad/sl • Step 9 of 10 The MATLAB code to k=26085. ) ® ^ ] ^ '^ 656.' 20 ■ .l a n " ' ^ . j= (s + 2 0 ) The phase margin of the system is. bodeplot(sys) margin(sys) grid Step 10 of 10 ^ Thus. PM ^30® > 30“ S t a n " ' .3 7 Thus.2 1 )* + 400) 7 (2 5 .21 The bandwidth is.t a n " ' ^ 2 30“ 1^20+n. the phase margin of the system is. num=[k k]: den=[1 40 400 0 0].0 3 1 4 > 2 5. a: ( > / ® '+ i ) |A:c(a)|= i*(V ® “ + 4 0 0 ) Since the gain of the system at 0 ^ is unity or 0 dB.371- Step 8 of 10 ^ Find the maximum possible bandwidth.£ • ” 20 tan"'«i .t a n " '. L +400 T (fflj. Use Matlab to find the bandwidth. 379a.1 8 0 ® + t a n " '® . 1 9 ® . 379® . 0 ^ ^ 2 5 .t a n " ' ^ a 3 0 “ 20 20 a tan"' . B o d e D ia g ra m F re q a c n c y (ra d /s) .'® . Problem 6. ® ? .a 30“ 20 ” 20 J t a n " ' .21. * . Step-by-step solution step 1 of 10 Step 1 of 10 The open loop system is.2 1 )’ + ! 658133.4 4 7 ® .-^ 20 230“ 20.24 <9^=0.J. Hence.2 1 rad/s Step 6 of 10 The gain of the system is.21 rad/s for 0 ^ . 0 ^ — 25.37. sys=tf(num. 1 2 0 + ( 2 0 + « ^ . . the value o f /< is 126085.^ ' ' j ‘ (s + 20) AT(a+ l) “ j* + 4 0 i’ +400s' ®l = 0 .0314 = 25. = 2 5 . I = tan *0 . (25.447® ^ 2 20 + 20® ^ + aij.2 tan — 20 Since the phase margin is calculated at gain cross over frequency PM * Ian"' -2 ta n " ‘ — 20 Step 3 of 10 Since the phase margin.94 25. 20 . PM = l8 0 * .+ 4 0 0 ) = a: ( ^ ® .21 The suitable value for 0 ^ is 25.23 > 2 6 0 8 5 .J 20 Step 4 of 10 Further simplify.2 ta n .43PP For the open-loop system determine the value for K that will yield PM > 30° and the maximum possible closed-loop bandwidth. 2 tan 30“ 2 0 + ( 2 0 + ® i) » .2 ta n " '.1 8 0 » + t ii n . the plot of MATLAB is sown in Figure 1. a a ta n ^ 4 h ~h ] (13) 1+ (1I 4) step 9 of 9 Let ^ 5 2 ^ = 0.a ) [ [ 2 ( l+ a * 2 S ® * + r „ V ( l+ a ^ ) ) ] .. 4 = A . ja > + p The phase angle equation is. the value of frequency at maximum phase is. the required equation is derived and shown below.. ^ a V + l Substitute J(p for s. A (*)= « — ' ' s+ p Substitute j a for s. (12) Consider the trigonometric relationship equation.[ 4 o ’ 7 j ® '+ 2 rJ ® ’ ( l + o ' ) ] = 0 [2]-[2o'J2®'] = 0 2 ‘2aX ___l _ * a«2r< Tn 1 n>_ = .f r Consider the equation of sin ^^- . . d ta n ^ [ '* ( w r ] ] ( A |p |] [ S I r ' r ) da 2 1+ (ii)](H ‘ R]“(iSi)(R‘ R]°® ( R 'R j't e lf e 'R ) ( R 'R )t e ) ° ( R 'R ) ® '. TJa>+l D . r r „ « . B and A—B.. ( l. 4 = A . J l+ a ’ 2 .(4) Consider the trigonometric relationship equation. ( i. ' ^ l+tan(i4)tan(£) Substitute equations (2). * .g ) . log«m „ = 5 ( l o g ^ + l o g j ^ ) . Differentiate equation (13] with respect to a . (b) Show that the frequency where the phase is maximum is given by I Td ->A* and that the maximum phase corresponds to o«nu = = —7= and that the maximum phase corresponds to 1 —a sin^nm = +o (c) Rewrite your expression for ojmax to show that the maximum-phase frequency occurs at the geometric mean of the two corner frequencies on a logarithmic scale.<») Thus. 1+ tan*# Substitute equation (5) in (6). g ® ..B ..« ) T 1+ Ti+«r„vJ [r„ a » (l-g )]* { l+ o T jf f l’ ) + ( r o < » ( l. = tan*' (Tjyoi) .a ) ) ' l + o ’7^<»* + 2 « r > = +72<»’ ( l . 4 = tan*' {Tj^0) . Thus the phase equation is. > 4 a t a n . the value of phase is. (11) and (12) for A. r r „ « > ( i. ____________ l+ a * .a ) ^ l + o ’ 7 2 ® * + 7 -„ W (l+ a ’ ) ( 4 a ' r „ V + 2 r X l + a ’ )) . B and A-B. (a) Show that the phase of the lead compensator is given by <p= tan-1 {TDw) . the value of phase is.o g ( ^ ) + i|o g ( ^ ) Thus. = (2) B = tan"' { a T ^ ) (3) Substitute equations (2) and (3) in (1).g ) ' 1 + (l + a* ) Take square root on both sides.. k .B .. (d) To derive the same results in terms of the pole-zero locations. T. .. ' ' l + ta n ( i4 ) ta n ( £ ) Now substitute equations (10).. ( jt a } = a T g j6 f+ l l+ a T i. 1 **® s -7 = r+ *o « -r= r v*o >1^*0 Io g « . — B E f L ^ 1 + o ^ 7 j® * T 7 2 ^ ( l+ a ^ T M l..j —— . (3) and (4) for A. tan*d^ S in V = : ------.« ) T ....(10) ■“ "(fi) B ( 11) ■ " " fe ) Now substitute equations (10) and (11) in (9). = i. \-a s in ^ ^ l+ a step 6 of 9 (c) The maximum frequency is occurs midway between the two break frequencies on a logarithmic scale... 2..= lo g ^ .'f ® l. [ i+ a r > ' J ™ ^ . ( 8) daTo Thus the frequency value at maximum phase is verified and shown below.a ^ 7 S ® ‘ + r „ V ( l + « ' ) ) ’ 7 i® ( l . = 0 (4a=72®‘ + 2 r ji» ’ ( l + a ' ) ) r . s in tL .44PP For the lead compensator Tds + 1 Dc(s) = - ’ aToS+ r where a < 1.^an-^{aTDu)). the maximum phase value is verified and shown below. da Differentiate equation (7) with respect to 7 i® ( l .tan*' (grp<p)| (1) Step 3 of 9 (b) Consider following equations... rewrite Dc(s) as Dc(s) = s+ p and then show that the phase is given by such that .R R F llf l Thus.. 1 5 ^ Step 5 of 9 Substitute equation (8) in (7).. (9) Step 8 of 9 Consider following equations. log 9 M 2 fe w a step 7 of 9 (d) Consider the compensator transfer function. 10gtf> a .[ 4 a 'r J ® V 2 r „ W ( l+ a ^ ) ] ] = 0 [ 2 ( 1+ a ’ r > * + r ja i" ( 1+ o ' ) ) ] .2 + I 2 0 + 1+ 0 ' 1 = -7 ^ 1 2o + l+ o 0 -° ) V20+1+0' 0 -°) Vo+«)’ 1 -0 1 -fa Thus. t ^ U ) ___ ij.6 ) j From this equation.tan*' (or7i. ® * + r > ’ ( l + o ' ) ( 4 o ^ r „ V + 2 r > ( i+ o * ) ) 2 ^1+ o*72 ® ‘ + r > ' ( l + o ^ ) =0 ( ^ l + o ^ J S ® * + 7 i V ( l+ a ^ ) ) ’ ' ( 4 a ^ r „ V + 2 r > ( l+ a * ) ) ' = 0 2 ^ \ + a ’ T iei)*+ T ia/‘ { l + a ‘ ) ■ 2 ( i + a ' r „ V + r > ’ ( i + a ') ) ' =0 (4 o ’ 7 ji» ’ + 2 r j ® ( l + a ’ )) [2(l+a’r>*+7->’(l+o'))(r„(l-o))].. Problem 6. = v'UTR- Hence the frequency at which the phase is maximum is the square root of the product of the pole and zero locations..(r« 0 -« )) </(sinj>) 2 ^ l+ g = r > ‘ + r > ^ ( l + a ‘ ) da (^U .. Step-by-step solution step 1 of 9 (a) Step 2 of 9 Consider the transfer function of lead compensator. J.2 a + o ' ) r „ V ( i....o ) (7) ^ l + a ’ 72®‘ + r ^ ® ’ ( l + g ’ ^ Step 4 of 9 g f(s in jt)^ Let ^7 s 0 frequency value at maximum phase value.T T . 2 0 .084S+1 D(s) = 0.003955+1) But still this system cannot have phase margin greater than 50" .5" at (1r 28.6 dB at cc^=58rad/sec cc^ = .084| 0. Step-by-step solution step 1 of 1 Step 1 of 1 50.00395S+1 50.000(0.5" Gain of ( .45PP For the third-order servo system 50.5 ')+ y = 70.0845+1) ■■ ^'( j + io )(5+50)(0. Then verily and refine your design by using Matlab.^ = 5 8 rad/s y/cLt |t=0.6 rad/sec let e=5" Adds onal phase 1ead = 50 .2 0 1 o * ^ ) = .000 s(s+10)(s+50) PM—10.5' l+sin^^ l+sin70.(-1 0 .000 (K S ) = « (i+ IO )(f + 50)* use Bode plot sketches to design a lead compensator so that PM > 50° and cuBW > 20 rad/sec. Problem 6.. 035 Step 4 of 7 Determine the transfer function of lead compensation.7 * Therefore.2 8 3 ) = 0. ± =8 a .53) 5 ( 5 + 1 ) (5 + 5 ) ( 5 + 2 8 .04 rad/s.den2).)« .denG).0 . »den1=conv{[1 1 0]. » margin(sys) Step 6 of 7 Get the MATLAB output for the Bode plot.[1 5]). 40® -I-5*+ 3. the phase margin to be compensated by the lead compensator is. the value of — is.53) 5 + 28.2835+1 CM 0. PM s 3.6 Determine the transfer function of the system with lead compensation. C M - The transfer function of the iead compensator is. the approximate bandwidth of the system is 0..54 in the text book for the maximum phase for lead compensation.94*.53) j ( j + l ) f | + l ] ( i + 28. » numG=200*[1 3. Step 7 of 7 Observe from the MATLAB output that the phase margin of the system is 4 |. As the required phase margin is P A /^ 4 0 * . Then verify and refine your design by using Matlab.7 dB at 0. Select the maximum gain cross­ over frequency as 10 rad/s.125for a . lO V O lM = 0. » sysG=tf(numG. » denG=conv(den1. » numG=5.6].1).035J + 1 8 (^ + 3.denG).46PP For the system shown in Fig. » denG=conv([1 1 0]. Band width is defined as the maximum frequency at which the output of the system is attenuated to a factor of 0. What is the approximate bandwidth of the system? Figure Control system S te n -h v -< .707 times the input. » margin(sysG) Step 2 of 7 Get the MATLAB output for the bode plot showing the gain and phase margins. D (s) is 8(5 + 3 . Bode D iofron Step 3 of 7 The phase margin of the open loop system. » sysG=tf(numG. 1 =10 1 r= - " lo V a Substitute0. 0. 4 0 ( 1 + 3. » den2=[1 28.6 Thus. aT s*\ Write MATLAB code to plot the Bode plot using margin to get the phase margin.[0.or < 1 and with unity DC gain.53]. Observe from the MATLAB output that the magnitude is 0.te n ^ n liitin n Step-by-step solution step 1 of 7 The open loop transfer function is.I2 5 )(0 . allowing a margin of 5 * . » sys=feedback(sysG.67 rad/$ • .2 1]). the lead compensation with unity DC gain. Ts+l £ )(. Thus.283 tfT = (0 .94* = 49* Refer to Figure 6. Problem 6.1 2 5 The new gain cross-over frequency is greater than 2. the design requirements are met with the designed lead compensation.6) J 200(5 + 3 .669 rad/s.S3) i + 28. suppose that c ( j) = - ■ s(i + 1 ) ( 5 /5 + I) ‘ Use Bode plot sketches to design a lead compensation Dc(s) with unity DC gain so that PM > 40°. For maximum phase lead equal to 4 9 *.6 ) Step 5 of 7 Write MATLAB code to plot the Bode plot of the feedback system using m argin to verify the phase margin with the designed values. Then apply the Final Value Theorem (assuming Td = constant) to detennine whether 0C“ ) is nonzero for the following two cases.(s) *-»® (s+2 ) + 1 .9 '! s+ 2 V 09(s+2) ” !"(ft-2 )+ 1 . (a) When Dc(s) has no integral term: constant.9 1d(s) 1+1 2 ^ 0 .9K(s+2) 0 (oo) = Imt———^7 . Problem 6.8 D(s) Step 2 of 3 0 (oo)=lmts0 (s) 0.9=(=+2)T.8D(0) D(0) Step 3 of 3 b. constant. Figure Block diagram of spacecraft control using PID design Figure Block diagram of spacecraft control using PID design Step-by-step solution step 1 of 3 0. (b) When Dc(s) has an integral tenri: In this case. 0. T4 (s)=— 8 .47PP Derive the transfer function from Td to 0 for the system in Fig.8 D(s) TT As T4 =constant=K.=0 [Hence 9 (00) isZeroj .^ ^ ^ 1. li^tD^(s)s=coiistant=4t^ 0 (00)=lmt . Problem 6. Then verify and refine your design by using Matlab..m jP )s ^ . ASK AN EXPERT . which is similar to (a) Use Bode plot sketches to design a lead compensator to achieve a PM of 30®.mpgl{mt + nip) * Step-by-step solution There is no solution to this problem yet. yuu uuieiii i m e iiev^uetiuy ledpui u iis sy aiei 11 eApei iiiiei iieiiiy r Eq. G et help from a Chegg subject expert. (c) Could you obtain the frequency response of this system experimentally? Eq... (b) Sketch a root locus and correlate it with the Bode plot of the system. 0'{j) _ Mpl U (s) ~ ( ( / + m pp)(m t + nip) .48PP The inverted pendulum has a transfer function given by Eq. num=100*[1.2)(3.3 1].47.4®at a cross-over frequency of 3. .784 rad/s 0.92) = 0.4 deg (at 3.1])).784 T r = i. margin(sys) Get the MATLAB output for the Bode plot. Hence. (i) The steady-state error to a unit-ramp reference input is less than 0. den=conv([1/5 1 0].conv([1/50 1]. 0.den). D {s) = a — — . the design is verified and cannot be refined. (ii) PM > 40“ . den=conv([1/5 1 0]. (b) Verify and refine your design by using Matlab. that is.S n d h ) . C M - Determine the open-loop gain K that gives a phase margin of 40«. “i-i-K S ri) Thus. 3 Step 4 of 5 Determine the other corner frequency. Choose the comer frequency * — to be one octave to one decade below the cross-over frequency.den).20. Step-by-step solution step 1 of 5 (a) The open loop transfer function of a unity feedback system is.0 0 2 6 j + U Step 5 of 5 (b) Verify the design using MATLAB.3 ) _J^ °2 6 The desired compensator is. BadeMagraM Increasing the gain does not result in further increase of phase margin. 1 aT 1 ■ (2 0 )(1 . (0. sys=tf(num. The steady state error to a unit ramp input is. sys=tf(num.5 0 0 jU .49PP The open-loop transfer function of a unity-feedback system is C(*) = - j(«/5-M )(j/50-H )‘ (a) Use Bode plot sketches to design a lag compensator for G(s) so that the closed-loop system satisfies the following specifications. Pm .[26. the desired compensator is l.01 /:2 io o Take X = 100 Step 2 of 5 Draw the Bode plot of the uncompensated system using MATLAB for K —5- num=5.[1/50 1]). Problem 6. so the lag compensation should have a equal to 20.S dB (it IS. margin(sys) BodeDiafmn Gm .« > 1 ' ' a r f+ 1 The low frequency gain must be raised by a factor of 20.01.92 n d ^ ) Step 3 of 5 The value of K equal to 5 gives a phase margin of 47. s lim j- i F ) F ) 1 1 .92rad/s The compensator transfer function is. 4.^ = 3 2 .46' l-s in ^ l .4| . Including a direct computation of the damping of the dominant closed-loop poles. PM£1004»40" P M o f G(s) = 6.4 Vott r = 0.4rad/s a ^ = . (ii) For the dominant closed-loop poles.01 Now ?0. (i) The steady-state error to a unit-ramp reference input is less than 0.6 . Step-by-step solution step 1 of 2 G (0 = - 1 Now.6 dB occxirs at CD^=32.50PP The open-loop transfer function of a unity-feedback system is G(s) = .s i n 39.-.54' Let € = 6 ' Additional phase added = 40^ . (b) Verify and refine your design using Matlab.4 .5 4 '+ 6 ' = 39.46' Step 2 of 2 Gain lin of ^-2 0 1 og j =-6 .01.D(s) = 100(0.066 0 . the dominant closed-loop poles.46* l+sin<^ 1+sin 39. a (« /5 + I)(5 /2 0 0 + l)‘ (a) Use Bode plot sketches to design a lead compensator for G(s) so that the closed-loop system satisfies the following specifications. the damping ratio ^ > 0. Problem 6.066s+l) Hence D (s)G (s)*- 5 )5 3 ^ ^ And its |^=0.0 6 6 S + 1 . K y= — = m ^ 0. 56" ■1-sincp.6 Vat |t=0. (i) The steady-state error to a unit-ramp input is less than 1^00.14dB atca^=47. Its open-loop transfer function is given by SO C(s) = ■ «(s/5 + l ) ' (a) Use Bode plot sketches to design a compensator for the motor so that the closed-loop system satisfies the following specifications.04“ Step 2 of 2 let £= 6 “ additional phase =45.0478| 4(0.=0. (iii) The bandwidth of the compensated system is no less than that of the uncompensated system.S"-9.56 l+sin(p^ l+sin42.0478s^1) “ [5 ^]('> '> '> S 7 3 s+1) PM of componsated system = | 47. (ii) The unit-step response has an overshoot of less than 20%.04'’+6"=42.0478s+l) D(S): (0. (b) Verify and/or refine your design using Matlab. including a direct computation of the step- response overshoot.456| PM a 1005=^5.00973S+1) . Problem 6. _ l-sin42. MP=20%=100e =>|.56 gain of [-2 0 1 o g -^ |—7.6rad/sec 1 now. D (s)G (s)^ 200(0. cd^= -= -= 4 7 . Step-by-step solution step 1 of 2 50 G(s)=- (F^ = in n « ^ K ya200.6 ’ original PM=9.51 PP A DC motor with negligible armature inductance is to be used in a position control system.5” | . (i) The steady-state error to a unit-ramp input is less than 0. Step-by-step solution step 1 of 3 G (.2dB| 201ogp=30.)= - Step 2 of 3 K = K k . (iii) The steady-state error for sinusoidal inputs with a)< 0.^50. including a computation of the closed-loop frequency response to verily (iv).52PP The open-loop transfer function of a unity-feedback system is C (J ) = ■ j (H-5/5K1+»/20)* (a) Sketch the system block diagram.2 rad/sec is less than 1.76 rad/sec is |30.^50.01.i. (ii) PM > 45“ . 1 = ^ z 8 3s+l D (0 = 97s+l 100(3s+l) H e n c e D ( s) G ( s) = and its PM is 47° .36| Step 3 of 3 Placing upper corner frequency 3 octaves below <9 . (b) Use Bode plot sketches to design a compensator for G(s) so that the closed-loop system satisfies the following specifications.2 |p=32. (iv) Noise components Introduced with the sensor signal at frequencies greater than 200 rad/sec are to be attenuated at the output by at least a factor of 100. 0^=2.= 1 0 0 ’ 0. including input reference commands and sensor noise.01 Required PM ^45" . Initial PM =-28.2 rad/sec is less than 1.76rad/sec And Gain at 2. Problem 6. (iv) Noise components Introduced with the sensor signal at frequencies greater than 200 rad/sec (iii) The steady-state error for sinusoidal inputs with a)< 0. (c) Verify and/or refine your design using Matlab.? I Lag compensator should be i^ lie d I phase angle = -180*+45^+9=-126' At -126^ phase angle. margin(sys) Step 2 of 7 Bode plot with phase and gain margin is shown in Figure 1.08945 + 1 3.263) 5 + 11. 292 ] «(-20)(0. Pm ^ I2.den1).426(5+3. K <?(*) s{s+ \) Substit Substitute 20 for K .2638 1 Tn=Z 3.8 ” +6“ -33.04rad/s That is. den2=[1 11. Bode Diagram Frequency ( n d /s ) Figure 2 Step 7 of 7 From the Bode plot shown in Figure 2.1) ‘ Use Bode plot sketches to design a lead compensator so that Kv = 20 sec-1 and PM 40*. 3.53PP Consider a Type 1 unity-feedback system with C(J) = J (J -I.3063)a + l 0.292 for a . the phase margin of the system is 12. l+ s in A m l-sin(33. sys=sys1*sys2.4 0 « .426(5+3. sys=tf(num.2* Calculate the value of a . -6>04rad/s The zero at — is. 20 <?(*) j( j+ l) MATLAB code to calculate the phase margin.426*[1 3.den).2») 1-0.267) = -5.18 MATLAB code to plot bode plot with phase margin: num1=[20]: den1=[1 1 0].3063 Step 5 of 7 The lead transfer function is. Problem 6. Bode Diagram Om«bifdB(atbifrad/s).3063)j+l (0. margin(sys) Step 6 of 7 The Bode plot is shown in Figure 2.2638 *0.1 2 . Use Matlab to verily and/or refine your design so that it meets the specifications. the gain —5.547 = 0.547 1+ 0.3063 for Tp and 0.263) Therefore.8deg(at4. sys2=tf(num2.42rad/s) step 3 of 7 ^ From the bode plot.18].18 .34dB From the bode plot shown in Figure 1.30635+1 0. 1 — « (6 .2*) ■ l+sin(33.263].292 for a .6®- Therefore. the designed lead compensator is D{s) 5 + 11. the phase margin is 42.8* • Let. ^ . dB=-20log| ii] =-20log! ( v o . ^ (0.den2). num=[20]: den=[1 1 0]. To Substitute d.0 4 )V O !^ = 3. the lead compensator meets the required specifications.292)(0. num2=3. ' ' aV +1 Substitute 0.34 dB occurs at 6. sys1=tf(num1 . e=e> The maximum phase contributed by the lead is. Step-by-step solution step 1 of 7 Step 1 of 7 The open loop transfer function is.292 Step 4 of 7 Calculate the gain.04 rad/s for and 0. 54PP Step-by-step solution Network error loading content <ErrorCode. Problem 6.2> RELOA D PAG E . (i) Use Matlab to plot the root locus with respect to K for the system.01) ft/s e c = 77:7 = - S{s) f( i2 + 0 . Begin by considering e proportional control law Dc (s) =K.0 0 2 S ) deg The autopilot receives from the altimeter an electrical signal proportional to altitude. S te p -b y -s te p s o lu tio n There is no solution to this problem yet. (ii) Modiiy the G(s) stated above for the case where the variable to be controlled is the rate of altitude change. This signal Is compared with a command signal (proportional to the altitude selected by the pilot). and the result is used to command the elevator actuators.e. G et help from a Chegg subject expert. where |G | = 1) of 0.0. (g) What steady-state error would result if the command was a step change In altitude of 1000 ft For parts (h) and (i). F in u r e C o n tr o l s v s t o m fo r P r o b le m fi FtFt You have been given the task of designing the compensation. ASK AN EXPERT . including the compensator you designed in part (h). only the long-period airplane dynamics are important. Locate the roots for your value of K from part (h).. (f) Use Matlab to plot the root locus with respect to K.0 1 i-l'0 . (iii) Design Dc (s) so that the system has the same crossover frequency as the altitude hold mode and the PM is greater than 50°. the autopilot of a Jet transport is used to control altitude.16 rad/sec and the PM is greater than 50°. (i) Sketch the block diagram for this mode. Figure Control system for Problem 6. The linearized relationship between altitude and elevator angle for the long-period dynamics is k(s) 20( 1 -1. Begin by considering a proportional control law Dc (s) =K.16 rad/sec? (c) For this value of K. and locate the roots for your value of K from part (b). Problem 6. would the system be stable If the loop were closed? (d) What is the PM for this value of K7 (e) Sketch the Nyquist plot of the system. (j) Altitude autopilots also have a mode in which the rate of climb is sensed directly and commanded by the pilot. The error signal is processed through compensation. assume a compensator of the fonn T p s+ l Deis)- a ro S -h l (h) Choose the parameters K. and the difference provides an error signal.55 Hm (a) Use Matlab to draw a Bode plot of the open-loop system for Dc (s) =K = (b) What value of K would provide a crossover frequency (i. and locate carefully any points where the phase angle is 180° or the magnitude is unity. and a so that the crossover frequency is 0.55PP In one mode of operation. You have been given the task of designing the compensation. TD. For the purpose of designing the altitude portion of the autopilot loop. and measure the PM directly. Verify your design by superimposing a Bode plot of Dc(s)G(s)/K on top of the Bode plot you obtained for part (a). A block diagram of this system Is shown in Fig. 10 G (s ).45*+4.= ^ 0.8 — ® sec Gainmaigins 12.4 )( 5 + 3 ) 42 +4. the lag compensator needs to lower the gain at from 10. rad Therefore.4 to 1.rad at <pa2. the approximate bandwidth of the system is .9 dB (at 2. What is the approximate bandwidth of this system? Step-by-step solution step 1 of 10 Step 1 of 10 Write the open loop transfer function. to avoid influencing the phase at 1 r " 20 7. 1 At < » » — .4 j *+4. „ = 0.04 3. let the new cross over frequency is ® .0S — ■ 0.0 — “ 4. .4 for a and 25 for j* in the compensator £>^(5 ) a a T s* \ 25s+ \ 2605 <fl 7+ 1 -----0.- 10505 + 42 2605^+11 4 5 5 ’ + 1096.81 >25 Step 6 of 10 Choose the value of a . the lag compensator is 0.44 sec Step 4 of 10 To get the phase margin of 40® the lag compensation needs to lower the cross over frequency.846x10“’ *-+l ^ -Q .S )!^ sec -.2 j U 6 0 5 + l j5 ’ +4.25 Step 8 of 10 Draw the bode plot for compensated system G(s) Bode D ig ra m G m « 12.7 .04 r+ 1 0.01 rad/s). Pm » 42.806 rad/s) F ig u re 2 step 9 of 10 The stability margins of compensated system are. 1 . r(5 )-A (5 )G (5 ) _ ^ 2 5 5 + n _____ 42 +4. .8 — ® sec a „= 2 a .Jl(s/1. .81— is. rad phase margin s 4 2 ^ at ^ 0 .5 deg (at 0.45*+4. . From the uncompensated bode plot.56PP For a system with open-loop transfer function 10 .«rad at A»a2. |G ( M ) | = 10-4 Therefore.4 sec Step 10 of 10 Determine the approximate bandwidth. rad phase margin s 4 2 ^ at ^ sO. Step 5 of 10 Choose the zero breakpoint of the lag a factor of 20 below the cross over.4 10.«.. 42 j ( j + 1 .2 j Assume lag compensator of unity gain £>^(5) s Ts+\ a T s* \ step 2 of 10 Draw the bode plot for uncompensated system <7(s) Bode Diagram Step 3 of 10 From the bode plot.8I — rad The gain at 0.9dB . i3 d B ^ «. D ^ { ji o ) s — T ct 1 1 = 10 .2 { 0 .0038 Step 7 of 10 Determine the loop transfer function. rad phase margin a -2 0 ^ at a3 G a in i n a i ^ s .4)+ ll[|(l/3) + l l ’ design a lag compensator with unity DC gain so that PM > 40°.0038 Therefore.04- r+ 1 0.6 ! ^ sec .«. Problem 6. (s ) i (j /0J25-f 1)(j /0.0362-H)’ where V is the ship’s lateral velocity in meters per second. PM. (b) On your plot.0154. and 5r is the rudder angle in radians. m. ‘J |PM=54.-. (i) Velocity constant Kv = 2.0139 201ogp=42. and what would the crossover frequency be? S te p -b y -s te p s o lu tio n step 1 of 2 [l.U 2i+ \] S . Problem The steering dynamics of a ship are represented by the transfer function Problem The steering dynamics of a ship are represented by the transfer function m ^ ^ K {-is/0.0. (b) For your final design.2.D(s)G(s)= s\( * V lll ' V l l L U . Phase angle at crossover frequency = -1 8 0 ' + 8 *+ 50= -122' .9 S + 1 ) .0.33 .H I] Placing upper comer frequency 3 octave below atg 1 at i = -i= > T = 583. (a) Use the Matlab command bode to plot the log magnitude and phase of G( jo)) for K = 0.=0. (iii) Unconditional stability (PM 0 for all oj < ojc.0143±j0. (a) Design a compensator that meets the following specifications. initialPM =-llf let €= S' and Lag compensation since PM of uncompensated system is very low.325j J[l.y I for compensated S3rstem Step 2 of 2 Roots are at-0. and GM. Problem 6.8 ■ . the crossover frequency). and indicate the location of the closed-loop poles. 0 3 6 2 .57PP For the ship-steering system in Problem. indicate the crossover frequency.0362j J K ^=K=2. -0.0139 at [G(jcj)=-122® and gain =|42. (c) Is the ship-steering system stable with K = 0.8dB ] at oa^=0.9 ^ ' (80584S+1) ( 5 8 3 .3 2 5 J J [ < 0 . (ii) PM > 50“ . draw a root locus with respect to K.2? (d) What value of K would yield a PM of 30*.-. 2| K y = K = 3 0 .= 31.2' Step 4 of 5 c. (d) Determine the PM of the compensated design. Dole of tfie laa comoensator at 3.31 '------. How must the gain be changed to obtain crossover at cue = 31.6 rad/sec.05s+1) (O.6dB -201og K=-29. n ew zero = ^ ^ = -^ ^ = |0 . Gain —29.Ols+l) At G^—31.2 Step 3 of 5 100 _ b.6. and what is the resulting value of Kv7 (b) With the lead compensator in place. and determine the zero location that (c) Place the pole of the lag compensator at 3. what is the required value of K for a lag compensator that will readjust the gain to a /Cv value of 100? fcl Place the.58PP Consider a unity-feedback system with 1 G (s) = s (i/20 + 1) (i2/1002 + O.16 rad/sec To maintam <d.16 rad/sec.2^ . and determine the zero location that will maintain the crossover frequency at ojc = 31. K=Ky KforLag = 30. PM of compensated design = 61.' Step 5 of 5 d. pole=3.16 rad/sec. Problem 6. Plot the compensated frequency response on the same graph.6 rad/sec.9 5 4 | ‘ K 3.6 => |K=30.6.Sj /100 + 1) ■ (a) A lead compensator is introduced with a = 1/5 and a zero at 1/T = 20. Step-by-step solution step 1 of 5 G(s)=- f -5— +l l UOO 100 ) Step 2 of 5 K(0. 4 s -f 1) ( 3 ^ + l) ( 4 ’ + 0. On the other hand. » d en= conv(fl. Most swept-wing airplanes have a “yaw damper.7 deg (a l U.2 4 1] .75(4 s ^ + 0 . the frequency of the lightly damped mode is at =-10 rad/s .8 55. one at a time. where Dc(s) = s /r p -p 1 (iv) Adding a notch filter as described in Section 5.7° 97.75*[4 0 . 0 1 1] . I’m . thelowpassBlteris relatively robust to where to place its break point Evaluation of the margins with the bending mode frequency lowered by lO^will show a drastic reduction in the margins for the notch Biter and very httle reduction for the low Biter.24s+ Where.f 0 . and qualitatively describe the differences in fhe step response. From the Bode plot of the do sed-loop system.97A deg (a l D. » f 4 = [ l 1 ]. Step 2 of 21 The MATLAB program to draw the bode plot for the given open loop function ii » num =8. den) » m a rg in (sy s) » g rid Step 12 of 21 A The Bode plot obtained onexecutionofthe program is Kudc IN agram Step 13 of 21 The gain and phase margins from the Bode plot are |gM =7. » f l = [ l / 0 . increasing the frequency is e^ensive and makes the structure heavier.02 The MATLAB code to plot the Bode plot for the revised open loop system IS » num =0. » den=cx>nv ( f l .34. Step 20 of 21 W Generally. which caused the revelers to spill their drinks.085 ra d ^ Step 5 of 21 The revised system is r(s ) 8.PM.2 4 s + 1) + 2{s/o)i. Step-by-step solution S le p t o f21 The transfer function is given by r(s) _ 8 . you must have a good estimation ofthelocaBonofthe bending mode poles and the poles must remain at that location for all aircraft conditions.17dB ata»= 2 0 rad/s I \PM = 97. What is the frequency of the lightly damped mode that is causing the difficulty? (b) Investigate remedies to quiet down the oscillations. » s y s = t f (num. 0 1 1 ]. » f 3 = [ l / 4 0 0 0 .6 Hz or 10 rad/s . » f 2 = [ l 0 .0 1 + l ) ( s 2 + 0 .0 0 /1 0 1 ].04.75(4 ^2 + 0 . .02 to 0. and resonant-peak amplitude. den) » m a rg in (sy s) » g rid Step 15 of 21 A The Bode plot obtained on execution of the program is B odcD iagiaB i Gm . Pm . they discovered that there was some flexibility in the fuseiage that caused a iot of unpleasant yawing motion at the rear of the airplane when in turbulence.4 deg (a l D.5 9 P P Golden Nugget Airlines had great success with their free bar near the tail of the airplane. + 1) ’ where rub is the frequency of the bending mode (= 10 rad/sec) and ^ is the bending mode damping ratio ( =0. the notch Biter is very sensitive to where to place the notch zeroes in order to reduce the hghtly damped resonant peak So if you want to use to notch Biter. invesfigate each of the following. it is a better design to use a low pass Biter because of its reduced sensitivity to mismatches in the bending mode frequency.0 4 /2 0 1 ) . the proportional feedback gain /( = 1. f 4 ) ) ) . » f 3 = [ l / 1 0 0 0 . c o n v (f2 . and resonant peak anq/litude is Low pass Biter Notch Biter GM 34. but maintain the same low-frequency gain in order not to affect the quality of the Dutch roll damping provided by the yawrate feedback.08 0. c o n v ( f 2 . den) » m a r g in (sys) » g rid Step 6 of 21 -o.2 4 1 ].4° (ai = 0. (1) with KDc (s). » s y s = t f (num. closed-loop bending mode damping ratio. » f l = [ l / 0 . The revised transfer function is found that the transfer function needed additional terms which reflected the fuselage lateral bending that occurred due to excitation from the rudder and turbulence.13 iIB (a l ID ra d /s e c ).c o n v (f4 . Problem 6 .2 4 1 ].3. Step 21 of 21 The table comparing the GM. Step 17 of 21 The MATLAB code to draw the Bode plot with notch Biter with the transfer function.) Problem For the open-loop system AT(»-f 1) KG(s) = j2(j -F10)2’ determine the value for K af fhe stability boundary and the values of K af fhe points where PM = 30” .8 dB at a = 8. it might be difficult and e^ensive to carry out. Step 7 of 21 » s y s = tfr (num. Specifically. 1 Uoo J » num =0. they found that the transfer function needed additional terms which reflected the fuselage lateral bending that occurred due to excitation from the rudder and turbulence. well below magnitudel.4 1 ]: » f l = [ l / D . Likewise. The Bode plot obtained onexecutionofthe program is R ede Diagram G m . conv(f2.5° at a>= 0. The approximate transfer function for the rigid body roll/yawl motion.7S*[4 0 .04 is » f l = [ l / 0 . (i) Increasing the damping of the bending mode from ^ = 0. » f 2 = [ l 0 .3. » s y s = tf(n u m .2 4 1 ]. » s y s = tf(n u m . 2 4 1]1 » f3 = [ l/1 0 0 0 . » s y s = tf ( n u m . keeping the DC gain of the filter = 1.7° at a>= 0. 7 5 ( 4 s ^ . with a damping of ^ = 0. where Eq.02).1) is r(s) 8 . closed-loop bending mode dancing ratio.24a + l ) ’ ■Where. Therefore the best recommendation would betousethelow pass Biter. d en) » bode (sy s) » g rid Step 3 of 21 -e.0 4 /1 0 1 ].4 1 ]. f 4 ) ) ) .0847 r a ^ which are healthy margins and the resonant peak is again..8 d ll (a l 8.7 5 (4 j ^ + 0 .0847 rad/s) (ai = 0. not a good idea in an aircraft." which essentially feeds back yaw rate measured by a rate gyro to the rudder with a simple proportional control law. Of the remaining two options. » d en= conv(fl.04 (would require adding energy-absorbing material in the fuselage structure). thus the system will be much b etter behaved Step 11 of 21 The MATLAB codetodrawtheBodeplotofthe system with the frequency of the bending modeincreasedfrom <z^=1 0 rad/sec to = 2 0 rad/sec is » num =8. conv(f2. = 10 rad/sec ^■=0.085rad^ Step 10 of 21 We see that the (jM has increased considerably because theresonantpeslc is well below magnitude 1 .1 ) » m a rg in ( s y s l ) » g rid The Bode plot obtained onexecutionofthe program is Bmle liia g n u n The low GM is c aused by the resonance b eing close to is stabihty.085 r a ^ The MATLAB codetodrawtheBode magnitude plot of the do sed loop system IS » s y s l= fe e d b a c k (s y s . Step 14 of 21 @ii) By picking up = 1the MATLAB code to draw the Bode plot is » num =0. f 3 ) ) . » f3 = [l/1 0 0 0. » f 2 = [ l 0 .4° at ai = 0 085 r a ^ which are healthiest margins of all the designs since the notch Biter has essentially canceled the bending mode resonant peak. and pick the denominator poles to be ( s /100 + 1) 2. PM. Pick the frequency of the notch zero to be at tub.083 lad A e c) Step 19 of 21 The gain and phase margins from the Bode plot are lGA/ = 55. D l 1] . For the new Golden Nugget airplane.4 s+ 1) M s) (s/0.0 4 /1 0 1 ] .97.92. » d e n = c o n v ( f l. (ii) Increasing the frequency of the bending mode from rub = 10 rad/sec to rub = 20 rad/sec (would require stronger and heavier structural elements). (c) Investigate the sensitivity of the preceding two compensated designs (iii and iv) by determining the effect of a reduction in the bending mode frequency of-10%.1 5r(s) = -Kr(s). dan) » m a rg in (sy s) » g rid Step 8 of 21 -a.4 s -F l) Ifs’ -F 0. f 3 ) ) . » f 2 = [ l 0. » d e n = c o n v ( f l.4 s + 1) M s ) ^ ( s /0 . 2 4 1] .5°at ai=0.0 2 = 0. co n v ( f 2 .02 to ^ = 0.7S*[4 0. 4 1 ] . The revised transfer function is r(s) 8.75*[4 0. Specifically.62 la d /H v ) . (See Problem) However. (a) Make a Bode plot of the open-loop system.01 + I)( s2 + 0 . The Bode plot obtained onexecutionofthe program is B e d e D iag ram Step 4 of 21 The gain and phase margins from the Bode plot are |gj|/= oo| \PM = ?7.068 The magnitude plot of the closed-loop system for Low-pass Biter is B ede liia g n u n The magnitude plot of the closed-loop system for Notch Biter is B ode D ia g ra m (d) ■While increasing the natural damping of the sj/stem would be the best solution.2 (ar = 8. 0 1 1 ). c o n v ( f 3 . (iii) Adding a low-pass filter in the feedback—that is. 4 1 ] .7. reexamine the two designs by tabulating the GM. The Bode plot obtained onexecutionofthe program is H ade D iagram The gain and phase margins from the Bode plot are lOilf = U d B a t 4 >= 1 0 rad/s~l \PM = 97.62 rad/s) (ai = lOOrad/s) PM 92. In performing a finite element analysis (FEA) of the fuselage stmcture and adding those dynamics to the Dutch roll motion.085 l a ^ And again the GM is much improved and the resonant peak is signifrcantly reduced from magnitude 1 . 0 1 1].DII5 rad/sec) Step 9 of 21 The gain and phase margins fromtheBodeplotare |g J / = 7 1 3 d B a t a i= 1 0 r ^ ^ lfM = 97.5° at a>= 0.2dB atai = 100 r ^ \PM = 97. » f 4 = [ 1 / 1 0 0 1] . » f l = [ l / 0 . B ode D ia g ra in G m . » f 2 = [ l 0 .5° at ai = 0. when they purchased a much larger airplane to handle the passenger demand. replacing A in Eq.62 rad/s I \PM = 92. (d) What do you recommend to Golden Nugget to help their customers quit spilling their drinks? (Telling them to get back in their seats is not an acceptable answer for this problem! Make the recommendation in terms of improvemenfs to the yaw damper. 4 1 ] . » d en= conv(fl. called the "Dutch roll” mode (Section 10.4. 2 4 S + 1 ) ’ where r is the airplane’s yaw rate and dr is the rudder angle. determine the PM and GM for the nominal design and plot the step response and Bode magnitude of the closed-loop system.085 rad/s) Qosed loop bending = 0 .S U dB (a l I M r a d /s e c ) . f 3 ) ). den) » m argin(sys) » g rid Step 18 of 21 A The Bodeplotofthe system with the given notch Biter is shown below. f 2 ) . (b) The MATLAB codetodrawtheBodeplotofthe system with the bending mode damping increased from ^ = 0. » f l = [ l / 0 .U847 nid/icc) Step 16 of 21 The gain and phase margins from the Bode plot are |gM = 34. And thisisborneoutbythestep response that shows a hghtly damped o scillation at 1. r is the airplane’s yaw rate and is the rudder angle.04 Mode Damping ratio Resonant peak 0.75*[4 0 . P m . 0 1 1]. » f 2 = [ l 0 . Problem 6. Step-by-step solution step 1 of 5 G(s) = Step 2 of 5 a. (c) Compute the vector margin (VM).0. VM=Shortest distance firom (-1. and find GM and PM.60PP Consider a system with the open-loop transfer function (loop gain) 1 G (i) = - i ( i + l) ( j/ 1 0 + l ) ' (a) Create the Bode plot for the system.769.1 6 9 f +(1.68-4^0.689)’ =0.0) touches l^quist plot at -0.0) The smallest possible circle of center (-1. 1+ - S(j<D) = step 4 of 5 Step 5 of 5 c. (b) Compute the sensitivity function and plot its magnitude frequency response. GM=20.829 .8dB PM=45* Step 3 of 5 b. V M = ^ { 0 . What does this imply about the shape of the Nyquist plot if closed- loop control is to outperform open-loop control at all frequencies? Step-by-step solution step 1 of 2 Step 2 of 2 S(s) = l+D(s)G(s) 1 |l+D(jcD)G(j(D)| Now distance of any point inside the circle &om its centre is alwsQrs less then its radius If any point of D (s)G (s) lies inside the circle then its distance from centre = |l+D(jco)G(jo)| ■ |l+D(jo))G(j(i))|<l 1 |S(jo)|= >1 Hence proved |l+D(jm)G(j<D)| .61 PP Prove that the sensitivity function S(s) has magnitude greater than 1 inside a circle with a radius of 1 centered at the -1 point. Problem 6. (iii) Sinusoidal inputs of up to 1 rad/sec to be reproduced with < 2% error. (e) Explain why a lag network alone cannot meet the design specifications. 1 with the plant transfer function ^ i(f/1 0 + l)‘ (a) We wish to design a compensator Dc(s) that satisfies the following design specifications: {\)K v = 100. | 0. (b) Create the Bode plot of G(s). (b) Create the Bode plot of G(s). 2.0 7 8 s + l) (1. (c) Show that a sufficient condition for meeting the specification on sinusoidal inputs is that the magnitude plot lies outside the shaded regions in Fig.0 7 8 S + 1 And lead network = 0 .l) ( 0 . Problem 6. Lead network will not affect the lower frequency Step 6 of 7 ^ e.0 2 1 S + 1 10° ( 0 .358s+l)(0.021s+l) . K»=100 KG(s) = Step 3 of 7 Step 4 of 7 c.4 6 5 n . (iv) Sinusoidal inputs with a frequency of greater than 100 rad/sec to be attenuated at the output to < 5% of their input value. Lag network will not affect the higher frequency Step 7 of 7 f F irrt sq>ply Ls^ compensation to Gain P M ^ 5 ^ 0 .62PP Consider the system in Fig. 05 Step 5 of 7 ^ d. forcD<Irad/sec and error ^2% . Recall that Y KG ^ E 1 and —= ■ R l + KG R l+ K G ‘ (d) Explain why introducing a lead network alone cannot meet the design specifications. Figure 1 Control system Figure 2 Control system constraints S te p -b y -s te p s o lu tio n step 1 of 7 G(s)=- Step 2 of 7 a. (ii) PAf>45“. |H(jcD)|=50 For (3>>100rad/sec and attenuation ^5% . choosing the open-loop gain so that Kv = 100. choosing the ooen-looD aain so that Kv = 100.4 6 5 S + 1 Lag networl^ 1 .3 5 8 S + 1 0 . (f) Develop a full design using a lead-lag compensator that meets all the design specifications without altering the previously chosen low frequency open-loop gain. redo the design by selecting 1/TD = 0. 3. n d th e sensitivity fu n c tio n .-4 0 d b p erd eca d e)a o d —3 (—60 db per decade) show that the system would be unstable for any value o f i t if no derivwive feecl»dc were used. 3. This can be accomplished primarily by adpisting To. (6. Problem 6. 7^ 4 -Tf. Determine waw andmona. 5. 2. most equal zoo. 4. derivative cootrol is required to bring the slope to —1 at the crossover frequency that was shown in Secdon 6.H)eBlopescrf—2 (tfa aiis. S — . a PM o f 6S”.47). 2. Hgure 6£1 defines the qrtfeni. The frequency response o f the spacecraft and sensor. G et help from a Chegg subject expert. which shows that the phase would be —iS irf w th e —2 s k ^ and —270" for die —3 slope. mwI the ■em iti wtv reasonaUy possible. 5 ''j L 0 5 at -2D 5 sV \ ac101 a01 a1 2 5^ . then Te — —7a.5 to be a requiremeiU for s t^ lity . esTor to a Gonstant- distnrfaance torque.05 and then determining the highest possible value 1/T/ that will meet the PM requirement. in the response to a step disturbance torque.68 Compensation fo r PID 11 1 \ to design in Example 6 . wUch would correqiood to a PM of (P or —90". Hieadbre. where 0. Design a PID controller to have zero steady. H ie only way diis can be true with no error (a s 0) is for Dr(s) to contain an integiml term. Therefore.40) Rgura 6.20 111 \ too T ntnT V to 20 - 10 20 T . Pint the elnMwUlnnn f t r r r n n f leainM e. respectively. and 7/—that will satisfy the speci6catioiis. This is clear o f Bode’s gain-^>hase relationship. including in t^ ra l control in the compensation will meet the steady*state requtremenL This could also be verified mathematically by use of the Final Vdne Theorem (see Problem 6. ASK AN EXPERT . Plot the cloBed-loop frequency response . The eariest apiMooch is to worii first on the phase so that PM = 65" is achieved at a leasmiably high frequency. n otii^ that 7 / has a minor eSect if sufficiently larger than Rgura $.9 C(a) = \H(s) (6. GH.20 is shown ioH g. oonunent on the usefirlness o f this controller to SolutioiL R iit.2 0 PID Compensation DesignforSpacecraji Attitude Control A nmplified design for ipnoeciif t Mtitiide control was presented ni Section 6 J . 6. Plot the step response rewus a command input and the step reqwnae to a constant disturbance torque.f tocf) 4 < - aoDi ooi ai 1 2 5 10 loo t»(ndtae) Step-by-step solution There is no solution to this problem yet.t>1“ (g-1)1 L. here we hare a more realistic sHaation that jnchides a senior lag and a dittuibing torque. Plot the step response versus a coounand input and the slq» reqMOse to a constant disnubance torque.67 Block diagram o f spacecraft co n tro l using FID design. Hence. For a torque distuihtmce from sUar pressure that acts as a sinusoid at the orbital fate (o» = 0.68. Then examine the improvement. To. 1.001 rad/secorn> lOO-minote period). For die spacecraft to be at a steady final value. V 1 s s. if any. 0( m=1 // —D 3(1).^-at \ -IM* I t/ / -210" . and as a bandwidth as is reasonaUy possible. Bcample 6.63PP For Example.(»tacS) 1111111___ A. the total input torque. The problem now is to iMck values for the three parameters in Eq. Example EXAMPLE 6 . if 7a 9^ 0.^ . however. let us talm case o f the steady-stme error.48)— K. Step-by-step solution step 1 of 4 •UM G(s ) = . While maintaining a phase margin > 40°. (a) One lead-compensator section Dc(s) = Kr f.j l f s + b* where b/a = 100. (c) Comment on the statement in the text about the limitations on the bandwidth imposed by a delay.1 Let K=1000 and take cc^=2 rad/sec D ( s ) = 1 0 0 o i^ ! ^ L ^' (s+2 0 )' |PM=6 8 “ | Step 4 of 4 B 'W in bodi cas es is reduced.. 0 7 -J2 D ( s ) = 1 0 o f l^ ^ l '■ ^ {s ¥ 7 m } |PM=55°| Step 3 of 4 0 ^ .64PP Assume that the system has a 0.2-sec time delay {Td = 0.2 sec). find the maximum possible bandwidth by using the following. .— ^ ^ s+ 1 0 Step 2 of 4 Magnitude—201o g . (b) Two lead-compensator sections (b) Two lead-compensator sections whereb/a =10.= ^ = 7 .= —20dB at cd<10 Vet L et CD. Problem 6. |G(jo)|=201ogK-7.65PP Determine the range of K for which the following systems are stable.5dB |K<2.5dB 20logK<7. (a) G(s) = (b) G(s) - Step-by-step solution step 1 of 2 Step 1 of 2 Ke-* a.37| . Problem 6.05rad/sec Ato>=cc^. Intersection with real axis at 4oc^— 2 G M =- _____ 8 K<- Step 2 of 2 Ke- GW = f(s+2) \ ^ I cosco-isinffl I G n © )= K — ^ — r GQ^ = 1. The temperature o f the water in the pipe will v a y contuniously along the p ^ as the heat flows from the steam to the water.OSrad/sec. Example 2.86) Wv = mass flow rate o f the water. ccu=— = ^ . being downstream from the exit temperature in the pipe. (2.87) A. Tgg = temperature of the inflow steam.? ( 1 0 .85) Cg = iHgCggis the thermal capacity o f the steam in the chamber R = the thermal resistance o f the heat flow averaged over the eikiie exchangee U tew ise. The resulting equatioita are then c. 0 8 T v /i |T=15. Steam eaters the chamber tbiDugh the controUable valve at the top.16| _6(15.(i) Ke-V ' ' = ------------------------. (2. (2. 2 J 7 . For many cmkrol appUcatkms it b not necessary to have great accuracy because the feedback will correct for a considerable amount o f error in the modeL Therefore.85) b nonlinear because the quantity Tg b multiplied 1^ the control input At. Problem 6. This net flow determines the rate of temperature change of the steam according to Eq. + i r . Kg = Bow coefhcieat o f the inlet valve. 7 ^ = temperteure erf' the inenming wale^ Tw = temperature o f the outflowing water: l b complete the dynamics. There b also a flow of beat into the chamber from the inlet steam that depends on the steam Bow rate and its temperature according to Eq. (2. = CwTw = . the time delay between the measurement and the exit flow is described by the relation vdiere 7 ^ b the measured downstream temperature of the water and b the time delqr. In o rd n to eliminate the Twi term in Eq. cw = specific beat the steam. Equation (2. respectively.)G (0 = i(10s+l)(60s+l) PM=44“ . the difierential equatioo describing the water temperature is 1 (2.3 S + 1 ) Step 3 of 3 Let us bring co^ at co=0.66PP Consider the heat exchanger of Example 2. Cew= specific heal of the water.16 EXAMPLE 2 . vdiich would be modeled in the same manner. Tg = temperature of the outflow steam.02(272s+l)e* D (. lags the temperature by ts sec. A ltiiou^ the time delay b not a nonlineariQr.86). (2. as given by Eq. Ta = Therefore.) PM= 4 3 . The sensor that measures the water outflow temperature.04rad/sec Placing comer frequency at 0.t. which we will define as ATg.82X Cgtg=AgKgCgg(rgi-Tg)-^(Tg-T.Find the diffemnial equatioas that describe the dynamics o f the measured water outflow temper­ ature as a functioo o f the area Aj o f the steam-inlet cootrol valve when open. L cte= 6 ® additional phase = l r _ l-sin(p^ 0 ^ -------- 1 +sintp^ 1 cou^. Heat exchanger and Tw for the outflow steam and water temperatures.1 6 Equations fo r M odeling a H eat Exchanger A beat exchanger b shown in Hg. . the transfer function o f the heat excfamiger has the form T . The equation can be about Tgo (a specific value <rf Tg) so that Tgi —Tf b assumed constant for purposes of ^tproximating the tenn.81).(») (T is-H )(n * -H ) ' * S te p -b y -s te p s o lu tio n step 1 of 3 G (.) = (10s+l)(60s+l) Step 2 of 3 Now KnotmettionedsotakeKw 6 NowPM=40*' to get PM = 45®. Wg = KgAg. Solntion. There may also be a rlehy in the measurement of the steam temperature Tg. and cooler steam leaves at the bottom. There b a coostant Bow of water th ro n g the pipe that winds U u o u ^ the middle of the chamber so that k picks up beat from the «team.16 with the open-loop transfer function -5 i G (i) = - (10s + 1 )(6 0 5 + 1 )‘ (a) Design a lead compensator that yields PM > A5“ and the maximum possible closed-loop (b) Design a PI compensator that yields PM > 45° and the maximum possible closed-loop bandwidth. A( = area o f the steam inlet valve. The net beat flow into the chamber b the difference between the beat from the hot incoming steam and the heat flawing out to the water.(2. We flwn MSMine that the heat transfer from steam to water b proportional to the dif­ ference in these temperatures.gy.004rad/sec. m an flow rtee o f the w—m. An accurate tbennal model o f tb b process b therefore quite involved because the actual heal transfer from the steam to the water will be propor­ tional to the local temperatures o f each fluid.6s+l) D (. T. 2J37. The temperature the steam vrill also reduce in the chamber as it passes over the mare of pipes. 0. it b conveniem to measure all temperatures in terms of deviation in degrees from Twi. it makes sense to combine the spatially varying temperatures into single temperatures Tg Rgura 2J7 X. Steam enters the chamber through EXAMPLE 2 . we will see in Chapter 3 tlu t operationally.84X qu = wjCw(7W —Tt).1 6 Equations fo r M odeling a H eat Exchanger A heat excfanger b shown in R g. \M^ = 1. (b) From Bode plots of the system. Compare that with the approximate value obtained in part (a). (c) Plot the data from the Bode plots [adjusted by the K obtained in part (b)] on a copy of the Nichols chart in Fig. |. 2. 3. The closed-loop system is specified to have an overshoot of less than 30% to a step input. (d) Use the Nichols chart to determine the resonant-peak frequency turand the closed-loop bandwidth.6 7 P P A feedback control system is shown in Fig. 1.=0. Problem 6 . / ’Jf£ 1 0 0 { = 35. determine the maximum value of K that satisfies the PM specification.8| Step 4 of 5 c. | g = 7.8" Step 3 of 5 b. and determine the resonant peak magnitude Mr. Figure 1 Control system R (a) Determine the corresponding PM specification in the frequency domain and the corresponding closed-loop resonant-peak value Mr. (See Fig.358| Step 2 of 5 a. Figure 2 Transient-response overshoot {Mp) and frequency-response resonant peak {Mr) versus PM for f(s) _ a»S Fhae mfgifl Figure 3 Nichols chart Step-by-step solution step 1 of 5 M .) (b) From Bode plots of the system.=30% =100e’^ % .5 by Mchols chart also Step 5 of 5 d BandwidA = |4.36 rad/sec| . determine the maximum value of K that satisfies the PM specification. £andwidtli(Uncompensated)=70radf/f Bandwidth (Compensated) = 3Qrad f s Step 4 of 4 d. . (a) What are the resonance peaks of each system? (b) What are the PM and GM of each system? (c) What are the bandwidths of each system? (d) What type of compensation is used? Figure Nichols plots Figure Nichols plots Step-by-step solution step 1 of 4 a.68PP The Nichols plots of an uncompensated and a compensated system are shown in Fig. GM (Uncompensated) = 5 GM(Compensated) =10 PM (Uncomp ensated)=40* PM (Compensated) ^63" Step 3 of 4 c. Mg (Uncompensated)—l. Problem 6.S Mg (Compensated)=1.0S Step 2 of 4 b. Lagcompensationisused. GM=0dBPM=0" K*2.02dB . and identify corresponding points in the two plots.9. ^ forK= 2 .3" K=1. 5=0. (c) For K = A .) (b) Show how the value of K for neutral stability can be read directly from the inverse Nyquist plot. To what damping ratios ^do the GM and PM of part (c) correspond? Figure Control system Step-by-step solution step 1 of 4 3K a. Problem 6.1" Step 4 of 4 d for K=4. To what damping ratios ^do the GM and PM of part (c) correspond? (d) Construct a root-locus plot for the system. 1 8 3 for K=l. (a) Construct an inverse Nyquist plot of [Y(jo))/E(yojj]-1. and 1. K is intersection with real axis Step 3 of 4 c. (See Appendix W6.381 .PM=18. determine the gain and phase margins. KG(s)=- (s+l)(s+3) Step 2 of 4 b. K=4.2.GM=12dB J>M=38.2 . (d) Construct a root-locus plot for the system. and identify corresponding points in the two plots.GM=6.69PP Consider the system shown in Fig. ^ . 9. At K < 0 system is unstable Step 4 of 4 Root locus plot of Y/F At PM = 4 ? .2.7 0 P P An unstable plant has the transfer function Y(s) _ ^+1 F(s) ~ A simple control loop is to be closed around it. Problem 6 .45 . (See Appendix W6. in the same manner as in the biock diagram in Fig. ^ = 0.) (b) Choose a value of K to provide a PM of 45®. In this case. to what value of ( does PM = 45° correspond? Figure Control system Step-by-step solution step 1 of 4 Step 2 of 4 b. (a) Construct an inverse Nyquist plot of Y/F. and identify corresponding points in the two plots. O M = -5. K = 3.1SdB Step 3 of 4 c.S9 Jbr PM =A ^ . What is the corresponding GM? (c) What can you infer from your plot about the stability of the system when K < 0 7 (c) What can you infer from your plot about the stability of the system when K < 0 7 (d) Construct a root-locus plot for the system. 02dB.12dB.4 .1* K=2. PM=44.) (c) Consider closing a control loop around G(s). To what value of ^ does each pair of PM/GM values correspond? Compare ^ versus PM with the rough approximation in Fig.7" K=1. 2. read from your Bode plot the values of GM and PM when K = 0. The s^roximation of ---. (b) Use your Bode plot to sketch an inverse Nyquist plot. Using the inverse Nyquist plot as a guide.7.1. What value of K yields PM = 30“? (d) Construct a root-locus plot. and label the same values of K on the locus. GM=15. PM=54.7 . and 2.0 .0. (See Appendix W6.4" AtK=1.is valid in all die cases 100 . and label the same values of K on the locus.0 .4.1® K=1.1.54. GM=9.71 PP Consider the system shown in Fig. as shown in Fig. 1{a).9.1dB. PM=30“ Step 4 of 4 PM d. PM=33. Figure 1 Control system Figure 2 Damping ratio versus PM Phue niMgtn Step-by-step solution step 1 of 4 Step 2 of 4 Step 3 of 4 K=0. (a) Construct a Bode plot for the system.2. 1(b). To what value of ^ Ho p s p a n h n a ir nf PM /f^M v a lu p a m rrpR nnnri? C n m n a rp (^vpraiiR P M with th p rniinh (d) Construct a root-locus plot. GM=12dB . Problem 6. GM=6. PM=21. t) . we get Step 5 of 10 ilie transfer function of the given sjrstem is r(s)= Step 6 of 10 To verify the solution using MATLAB. Problem 7.y ) . y { l) + 2 y [t) + y \ t ) = u (t) Taking L^lace transform on both sides. The MATLAB ou^ut is Response nsing l.aplace Transfarm Step 10 of 10 The response obtained using i n i t i a l command and that obtained using Lt^lace Transform are identicaL Hence the solution is verified.^ [ h ) . s ai^ + bu ^ = [’ % = tX v&ere X = [y y f Step 7 of 10 ^ The MALAB program to verify the answer is » a = [ 0 .Verify your answer using the initial command in Matiab.0 ] . .) + r ( s ) = U{s) Substituting the initial conditions shown in the figure. * e x p ( .u ( t ) + L — + S i( t) +. y(g=o >(U=lV. l} .01 PP Write the dynamic equations describing the circuit in Fig. » s y s = s s (ar h.. » b = [ 0 .+ ') ' Equating the coefficients we get A =\ ^+5=2 5 =1 So. » g rid .. (3) dt L L dt U ^ ' Step 3 of 10 Given that £ = 1H C = 1F Substitute the given values of £ . we get ^ (0 1 .= 1H Jt= 2 n y(0 ?(U= 1V. x ] * i n i t i a l ( s y s . » t i t l e (* I n i t i a l c o n d i tio n r e s p o n s e * ) . 0 0 0 1 . re-write the differential equation in state space form.s .2 + r (s) = 0 (s“ + 2 s + l) r ( s ) = s + 2 r(s)= ' 7 + 2 s +1 r(s )= _ £ ± ij Step 4 of 10 Using partial fi^ction e^ansion method.2 ] .2 ) \j.l. Step 8 of 10 The MATLAB ou^ut is Initial ceadilion nKuiHise Step 9 of 10 The MATLAB code to plot the response obtained using Ls^lace transform is » syas t » t= 0 .l. » x l a b e l ( ^Tinie ( s e c ) * ) . 0 . tY l ^ s ) . Write the equations as a second-order differential equation in y(t).y '[ h ) + 2 s r ( s ) . Figure Circuit £.y(i) = 0 dt Therefore. 1 ^ dt Rearranging the equation. Assuming a zero input. » y l a b e l ( *y ( t ) * ) .2 7 ( i .y ) . =f1. . The current through the Cc^acitor is iW = ( 1) ^' dt DifTerentiating equation (1) we get ( 2) dt The voltage across the inductor is -di V= L — dt Writing mesh equation to the first loop.x o ) . solve the differential equation foryffj using Laplace transform methods for the parameter values and initial conditions shown in the figure.). sV (s) . » p lo t (t. s+ 2 _ A ^ B (s + l)’ j4( s + 1 ) + J ( . » x o = [1 .3Kg=0 Step-by-step solution Step 1 of 10 - The given circuit is L=1H R = 2 ii ----------— WV— -------------. » c = [ l. 8 . d) .A dt L L dt L L\ d t) dt L L dt Now substitute equation (2) in the above equation. » d = [0 ]. » [ y / t . » y = e x p (-t)+ t.0 ]. i? and C in equation (3).0 u(t) m C =1F “ 0 o y ( 'o ) = iv /(/„ ) = o Step 2 of 10 From the figure. we have g+ 2 1 ^ 1 (s + l)“ s + 1 (s + 1 ) ' Taking inverse l^ la c e transform So.-1 “ £ .0 + 2 s r (s) . » p lo t ( t. Assume that the mass of the spacecraft plus helium tank. (11) Cohsider the mass value of probe.. . . z2 = 106y2. p=eig(A) z=tzero(A.B. -(3.0 0 0 1. and v= lOOOw.. -(4. -y .. (13) Consider the viscos damping value.. (12) Cohsider the spring constant value.0 0 0 1..... enter the equations of motion into Matlab in the form x = Ax-l-Bv y = OL+Dv. (a) Write the dynamic equations of motion for the system consisting of masses ml and m2 using the Inertial position variables.. -* 2)+ ij^ v j " (2.C..) 4 ^ v -(3. aIicicivjic...( “ «X 10'’)(Z2 .-z. the poles and zero locations are obtained by using MATLAB..( 3 .) .. = l.) (16) Hence.0 0 0 te . y1 and y2. 6 ( i a ^ Step 10 of 14 (c) Consider the value of z . = ..z . C and D.6xl0’)(l0-i..4 .6xl0’)(l0-‘)(i. (17) Consider the value of x^- .3 1.) (2. . the differential equation for the mechanical system is h j > i = .. Step 3 of 14 Refer Figure 7. (c) Put the equations in state-variable form using the state X = (z| i | Z2 z2.> ’i) = 0 * y ’2 = .) ‘ (1.(7) >.2 x 106..z ...OOOOi -0.. 2 0 0 ( z a .z.) + 0 ... Step 11 of 14 (b) Write the MATLAB program from equations (27) and (29) for plot the response of y caused by the impulse. B.3200 4. y= . .2xI0‘ .-z.> ’2 ) ^ 3nd ) .2xl0‘)(l0'*zj -10‘‘z.2004 is sketched in Fig..=10*.. .OOOOi z = -695..)+0.v..5:0... Step 13 of 14 (e) Write the MATLAB program from equations (27) and (29) to identify the location of poles and location of zero of the system.0]: c =[0 01 0]: D=[0].) 10* h . This is the signal the rotor must follow. (b) The actual disturbance u is a micrometeorite.3 1.-z.. clc. The viscous damping b is 4.000 Consider the mass value of spacecraft plus helium tank.6 -3200 -4.600 2.2xl0‘)(l0-‘z. B=[0:0. F is the vector sum of all forces applied to each body in a system a is the vector acceleration of each body with respect to an inertial reference frame.Z j) .( 2 .V 2 ...B.> l) .6 -3.y 2 ) .D) Step 14 of 14 The output of the MATLAB program is given below.3 0.C.200(zj-z.. u= 10-36ffj N sec on mass ml.6]....* U . rewrite your equations with the scaled variables z1 = 1063^1.lO-z..z .-z. m is the mass of the body.* ( > ’2 .) 'l) (3) Thus.) .. . “ ij’i - Write the equation of motion for the mass .r..-i.3(z.t): plot(t.200 -4.* ( > ’| . = . m1. = -(4.3 0.. Figure Schematic diagram of the GP-B satellite and probe i /•“ Step-by-step solution ste p 1 of 14 (a) Step 2 of 14 The coordinates of the two m asses are jtj and y is the displacements of the masses from their equilibrium conditions.2(z. 10-42..6xl0-’)(z.y = [0 0 1 0 ]* + 0 (29) 0 1 0 0 0 -1.3 1..)' (1. Step 5 of 14 The free body diagram 1s shown in Figure 1. the equation with the scaled variables are s —l. F = m a .6 -3. A= [0 1 0 0.. m2... (6) Consider the value of Z. z2 = 106y2.z . -(4.-1600 -2. .3 ( i..5 Hence. Step 12 of 14 Hence.6xl0’)(l0-)(z. y=impulse(sysGPB.. lev*I lie yuui ei^uauui lo win i u k si/aieu V z1 = 1063^1. . D) to find the zeros of the system.J'i )+M.2 .D).600 -2.) (2.6xl0’)(l0-‘i. (22). Refer Figure 1 and write the equation of motion.. -(4. (24) Substitute equations (17) to (24) in equations (15) and (16).600 -2. (d) Using the numerical values.-lO-i..600 -2. (22) Consider the value of x^- * 4 = i j .. (19) Consider the value of = . (23) Consider the value of ^4=^2 ...5 i i (25) X.6].6 ( z . yj. t=0:0. the state variable form is irs x+ U 0 0 0 1 0 3.* ( ^ 2 ... Therefore.(8) j!. Step 4 of 14 Consider and fixed and increase from 0.3200 4.0]: c =[0 01 0]: D=[0]: sysGPB=ss(A.* | ) .-10-‘z.. 2 x 10‘ ) (1 0 -4 )(z. (25) and equatioh (26).. Problem 7. i ..5 x+ (27) 0 0 0 1 0 0 Consider the output equation.4 ..3 ) ( z ..3 1600 2.19611 -0.) (26) Write the state matrix form from equatiohs (18).-lO-z.C.600 2.1961 i -3. (28) Modify equation (28) by equation (21).6 0 0 ( z ....6 x l0 ’ ( z .0000 + O.( 3 ..6JL*4J 0 1 0 0 o' -1..6(z. * = 3 ..... Consider the output equation.)-2.6522 Hence...)-4. — and |z.)-(4. -z..-1600 -2.. = . p = -3. " W j + *(> ’2 ..( 2 0 ) Consider the value of Zj. ) .) ^ ) .i | ) ] zj =-3.02PP A schematic for the satellite and scientific probe for the Gravity Probe-B (GP-B) experiment that was launched on April 30.3 .5:0.. 0 1 0 0 -1.4500 -69. A rotor will float inside the probe and will be forced to follow the probe with a capacitive forcing mechanism. (18) Consider the value of x^.... The spring constant of the coupling k is 3...001:1. is 2000 kg and the mass of the probe.000) 10-‘i:..) +^ v 10* ^-3.. The it spring will be compressed producing its spring force..000)10-‘ 2.6 x 1 0 ’ (14) Step 9 of 14 Substitute equations (4) to (14) in equations (2) and (3). and the resulting motion is very small.B..)] ste p 8 of 14 (b) Consider the value of z. = 10*.5v.600 2.200 4.y) The output of the MATLAB program is shown in figure.3 1600 2. ( 21 ) Consider the value of x^- * 4 = ^ .200 -4.0 0 0 /ig .0000 . = ..* ( > .. (e) Use the Matlab commands p=eig(F) to find the poles (or roots) of the system and z =tzero(A.. the response of the output y caused by the Impulse is by using numerical value of A.(9) Consider the value of v=U000ir 1 ( 10 ) 1.000) 10-‘ z. y.6 -3200 -4.6 0 |y = [0 0 1 0 ]x + 0 |.(1) Where. .2xl0’)(l0-‘)(z... is 1000 kg.1 . A= [0 1 0 0. I».C... B=[0:0.-10-‘z. * . (5) #.000)10-*z.-i.200 4....6 x 1 03..-10-2.3.B. Plot the response of y caused by the impulse with the Matlab command impulse(sysGPB). =-l. and define the Matlab system: sysGPB = ss(A.4500 +69. . and v= lOOOw. » H = 2 ...3 0 0 0 1 3.000) z... (4) y.. .. (15) -(3.D).O..84 in textbook and draw the free body diagram to identify the direction of the forces on the object...= [0 0 1 0] +0 .3. and the input an impulse.600(z. . 2 0 0 ) ( z ..=10-4*.4 . Step 6 of 14 b ih r h ) Figure 1 Step 7 of 14 Consider the Newton’s law of equations of motion for any mechanical system..-10-‘i.... 5) 0.. = [-1 4 0 ] D =20 A . = 8 1] 0] Step 6 of 8 (d) Consider the gainof G (4) ...5] Hence.. -3 -21 1 oj pj 8 1] D . =0 Step 3 of 8 (b) Consider the gainof G ( 4) G(«) = _ £ a I 10 ) =2 0 i^ 4 + 10 201+60 G (4 ) 4 + 10 204 + 200-140 4 + 10 20(4+10)-140 4 + 10 G (4) = 2 0 . the value of state description matrices in control canonical fonn is B .5] B.. the value of state description matrices in control canonical fonn is B.. = [l] C.. (7). 1 0 ..= 0 1 0 0 (5) 0 1 ••• 0 0 . (8) and (12). = [l] C . -a . the value of state description matrices in control canonical fonn is B . (6).. *.5] D. = [-140] D... (6).. A . 1 0 B .03PP Give the state description matrices in control-canonical fonri for the following transfer functions: (a) G(J) = "■> « ' ) = wmTTT- "F+3+2" (d) C(j) = (d) C(i) = Step-by-step solution ste p 1 of 8 (a) Consider the numerator part of the gain. ( 12 ) Write the state descriptioh matrices in controi canonicai form from equations (4).. (8) and (11).^ .0 1 0 r 0 0 c . the value of state description matrices In control canonical fonn is B.. =[0.. =20 Step 5 of 8 (c) Consider the gainof G ( f 84 + 1 G (4 ) ( 11) 4*+34 + 2 Write the state description matrices in control canonical form from equations (4). the value of state description matrices in control canonical fonn is .. (7). -3 -38 0 0 0 0 A . (5). (7).. = [l] C . {5).= 0 1 7] »] -2 -2 O' A ...5] D. (6)..= 0 0 0 0 1 0 C. ’ -2 -2 O' 1 0 0 . Problem 7.= [A b. =0 A ...$'^ +ajS*~^ +. (8) and (10)..= 1 0 0 0 1 oJ Hence. =[0. A . S +fljJ +O2S +*** + fl.i c . ste p 2 of 8 Write the state description matrices in control canonical form. = ( 1) Consider the denominator part of the gain.... (6). (5). (5). cM = (3) a ( j) Substitute equations (1) and (2) in equation (3).= 0 0 Q = 0 1 7] 0] Step 7 of 8 (e) Consider the gainof G(4) (4 + 10)(4^+4 + 25) ^ ^ '4 ^ ( 4 + 2 ) ( 4 '+ 4 + 36) . = [l] C..= Hence...= Hence.5 G(s) = (9 ) j+ 0 .] (7) £)^=Seperate valuefromnumeratoraiiddeiiominatorelse 0 (8) Consider the gainof G (j)- 1 G (*) = 2s+\ 1 2{s+0... ^ 4*+ 114^ + 354 + 250 (13) ' * ' ° 4 ’ +34*+384"+724^ ste p 8 of 8 Write the state description matrices in control canonical form from equations (4).0] A . (8) and (13)... - .... a {s)= s "+ 0. ( 10 ) 4 + 10 ste p 4 of 8 Write the state description matrices in control canonical form from equations (4). = [-1 0 ] Hence. -<h . ..+a^ (2) Consider the general canonical form of the equation. 0 A .= (6) C .. = [-1 0 ] B.. (7).= I 11 35 250] o .5 Write the state descriptioh matrices in controi canonicai foim from equations (4) to (9).. =[0. = [0. = 0 1 11 35 250] D.5 C ( .D]=tf2ss(num. den = [13 2]. . num=[0. (d) Consider the value of gain G(5)- 5+7 G (4 5 ( 5 ^+ 25 + 2) 5+ 7 G (« ) = ■(4) 5* + 25*+25 Write the MATLAB program from equation (4) to find the state description matrices. A= -3 -2 10 B= 1 0 C= 81 D= 0 Hence.5000 D= 0 Hence.den) Step 11 of 11 The output of the MATLAB program is given below.D]=tf2ss(num. A= -2 -2 0 1 00 0 10 B= 1 0 0 C= 0 17 D= 0 Hence. den = conv([1 2 0 0]. [A.B.5 ) 0. A= -0. -3 -3 8 -7 2 0 O' 0 0 0 K = 0 0 0 1 0 0 I 0 0 0 0 p C. the value of state description matrices is obtained by MATLAB. (e) Consider the value of gain G(5)- ( s + 1 0 ) ( i’ + j+ 2 S ) G(. (b) Consider the value of gain G (j)- G (« ) = .D]=tf2ss(num. A= -10 B= 1 C= -140 D= 20 Hence.B.B.B.den) Step 8 of 11 ^ The output of the MATLAB program is given below.C.) = ( 1) j+ 0 . den = [1 2 2 0]. [A. the value of state description matrices is obtained by MATLAB. den = [1 10]. [A.D]=tf2ss(num.den) Step 4 of 11 The output of the MATLAB program is given below.[1 1 25]). the value of state description matrices is obtained by MATLAB. Step 5 of 11 (c) Consider the value of gain G(5)- 85 + 1 C (5 ) (3) »’ + 3 i + 2 Write the MATLAB program from equation (3) to find the state description matrices. = 0] Hence.5].m I 10 J f +10 20f + 60 C (5 ) = (2) 5 + 10 Write the MATLAB program from equation (2) to find the state description matrices. num=[1 7].5000 B= 1 C= 0.C. num=[20 60]. den = [1 0. (s + 1 0 ) ( j’ + s + 2 5 ) Step 10 Of 11 Write the MATLAB program from equation (5) to find the state description matrices.den) Step 6 of 11 The output of the MATLAB program is given below. . [A.C.B. A= -3 -38 -72 0 0 10000 01 0 0 0 00 100 0001 0 B= 1 0 0 0 0 C= 0 1 11 35 250 D= 0 Rewrite the output with appropriate notations. the value of state description matrices is obtained by MATLAB. Give the state description matrices in control-canonical fonn for the following transfer functions: Problem ~ 2l+l* "■> « ' ) = wmTTT- "F+3+2" IC| = (d) C(i) = Step-by-step solution (a) Consider the value of gain <7(^) ' ’ 2^+1 2 ( i+ 0 .den) Step 2 of 11 ^ The output of the MATLAB program is given below.D]=tf2ss(num. j ^ ( s + 2 ) ( j '+ i + 3 6 ) .5 Write the MATLAB program from equation (1) to find the state description matrices. num=conv{[1 10]. the value of state description matrices is obtained by MATLAB.04PP Use the Matlab function tf2ss to obtain the state matrices cailed for in Problem.C. Problem 7.C. [A. num=[8 1].5].[1 1 36]). 5 ) 0...= [A 6.2 0 0 0 0 0 -1 -36 0 0 0 1 0 0 B . = [-1 0 ] B..... (35). l l = 38(-1.. C = l-7 C = .472)+36(1.. 1 0 1 0 B .. =20 Hence. 0 0 0 0 -2 -2 0 1 0 1 B. Figure 2 ste p 6 of 10 ..5] 0=0 Step 2 of 10 (b) Consider the value of G (j) G (*) = f e .... 2 (2) Write the general canonical form of the equation. .472 1.+ a . . .288] +0 y = [-1.[ Z -z .346 . 85 + 1 -7 15 7+7 .m (3) a(s) Substitute equations (1) and (2) in equation (3). = [l) C. = [l] C..1 0 ] ....... the value of state description matrices in control canonical form is A .. 0 1 0 .346 3.288 ... 5* ( 5 + 2) ( 5 * + 5+36] '" i (5+ 10)(5^+5 + 25)1 5 ^( 5 “ + 5 + 3 6 ) L (-2 + 1 0 )(4 -2 + 2 5 ) 4 (4 -2 + 3 6 ) . the value of state description matrices in control canonical form is B . "4l 0 0 0 0 0 4j 1 42 1 0 0 0 0 4i 0 = 0 0 -2 0 0 4i + 1 4.... = -7 15] 0] Step 5 of 10 (d) Consider the value of G (5) 5+7 G (5 ).] c .... the value of state description matrices in control canonical form is: Step 8 of 10 (e) Consider the value of G (5) (4 + I0 )(5 ^+ 5 + 25) (31) ' ^~5^(5 + 2)(5’ +5 + 36) Apply partial fraction in equation (31).... (14) Substitute equations (13) and (14) in equation (12)..i" ''+ < i » ''’ + ... 0 0 0 0 0 ■ 1 0 0 0 0 A .... = [-140] D. j+ 7 a (5’ +25 + 2)+(B 5+C )5 5 (5 ' + 2 s + 2 ) 5 ( 5' + 2 5 + 2 ) s + 7 —s^A+ 2sA+2A+Bs^+ C s .^ . 8s + l (s + l)(» + 2 ) '" i 8 s+ I -8 + 1 -1 + 2 A = -7 .. ] (7) £)^=Seperate value&omnumetatoraiiddeiiominatorelse 0 (8) Consider the value of G (j).= 0 0 ...07 0..346 3. (16) Write the state equation in control canonical form from figure 1.45. o ( i) = i" + o .421)+2£ £ = 0. 35 = 72X +38B (36) Substitute equation (35) in equation (36). (27) Substitute equations (24). 1= 3 ( . A + B = 0 . (39) and (41) in equation (32).. (17).....346 3. = [o -2 j Hence.. .(1) Consider the denominator part of the gain.. 1 . +a5’ + «s"+36B5+2fl5’ + 2a5+72B+C5’ + 0 ’ + 36C 5'+ D 5 * + 2 C s ’ + £ 5 ’ + 2 £ s ' 'A5*+ 3 .B .. (37) Write the coefficient of from equation (34).346)+3...+■ Write the state equation in control canonical form from figure 2. (18) Write the output equation in control canonical form from figure 1. ] +0 . Give the state description matrices in control-canonical form for the following transfer functions. (33) Modity equation (32)..07 0.. 35 = 72X+38(3. (17). =[l] C..6 .472) X = -1. l = 3 X + B + C + 2 f l+ £ (40) Substitute equation (33). (35).’h. -« j -<h .l 1 5]*+ 0 (19) Write the state tate description de: matrices in control canonical form from equations (16). (43) and (44).. (5 + 10)(5’ +5 + 25) As+ B ^ C ^ Ds+E (32) 5'(5 + 2)(5’ +5+36) 5' ^ { s + 2 )^ ^ + s + 3 6 Determine A value from equation (32). i .. = C. (29) and (30). - ...l ...45’ +38^5*+72X5 5’ +ll5’ +355 + 250 == +B5’ +3B5’ +38B5+72B+C5* (34) +Cs’ +36Cs* + Q s' +2Ds' + Es'+2Es^ Write the coefficient of constant from equation (34).. Write the state description matrices in control canonical form.(15) (5 + 1)(5 + 2) (5 + 1) (5 + 2 ) Draw the block diagram from equation (15). *1 0 0 0 4) 1 = 0 -2 -2 42 + 1 0 1 0 .. i ] :] -7 15] D.. Figure 1 ste p 4 of 10 Write the general form of state space equation... 0 .5 G(s) = (9) j+ 0 . j + B =0 (25) . (13) Find B from equation (12).. 0 0 0 -1 -36 4.. y = C x+ D .jx l ■ (H ?) s + lO 201+60 G (*) i + 10 20 1 + 2 0 0 -1 40 j+ IO 2 0 (j+ 1 0 )-1 4 0 i + lO G (5) = 2 0 .. (37) and (39) in equation (40).. (25) and (27) in equation (21)...421+2£>+0..[Z -Z . Problem "■> « ' ) = wmTTT- "F+3+2" "F+3+2" (d) C(i) = Step-by-step solution ste p 1 of 10 (a) Consider the numerator part of the gain. .... and realize them as a separate subblock in control canonical form. ll = 38X +3B +36C +2£ (38) Substitute equation (33)... (^^) Write the coefficient of from equation (23). 0 0 0 1 0 .... = -1...0 7 (41) Substitute equations (33). 1 .. the value of state description matrices in control canonical form is .5 Write the state description matrices in control canonical form from equations (1) to (9)..288 D = -0 .= 0 .. (35) and (37) in equation (38). .... (20) 5 ( 5 ’ + 25 + 2 ) Apply partial fraction in equation (20).. 0] -1 01 A. (17).. 2 _ _ 2 . Make sure that all entries in the state matrices are real valued by keeping any pairs of complex conjugate poles together.. *»+l A B ( i + l ) ( * + 2 ) ° ( i + l ) '^ ( j + 2 ) * ' Find A from equation (12).288 5 '( 5 + 2 ) ( 5 ’ + 5 + 3 6 ) s‘ ( 4 + 2) 5*+ 5+ 36 ste p 9 of 10 Draw the block diagram from equation (42).= 1 1 1 p c . A .. Problem 7..= [ ..421 -0. C (5 + 10)(5^+ 5+ 25) ..') . 250 = 72B B = 3.I =J 2 2 J A = Hence.. y = [-1....472 ..45.5 ] B.=20 Step 3 of 10 (c) Consider the value of G W 8>+l G(5) »’ + 3 i + 2 8s + I GW = .. (5 + 10)(5^+5 + 25) -1.. 5+7 A Bs + C 5(5* + 25 + 2) s ^ 5 ’ +25 + 2 Determine A value from equation (21). (35) Write the coefficient of s from equation (34). Figure 3 ste p 10 of 10 A Write the state equation in control canonical form from figure 3....... 0 (5) 0 1 .....] x + 0 (30) Step 7 of 10 ^ Write the state description matrices in control canonical form from equations (16)..421 2>c = 0] Hence..421 -0.I Write the coefficient of s from equation (23). (28) 5(5’ + 25 + 2) 5 5^+25 + 2 Draw the block diagram from equation (28).5] D=0 A .472 1..(11) ( i + l ) ( j + 2) Apply partial fraction in equation (11). (37)... = [I] C.472 1. . „ 8s+l (s + l) ( j + 2 ) < " 1 85 + 1 ■ ( s + 1) -1 6 + 1 -2 + 1 B = 1 5 . 7 7 5+7 -5+6 _____________ .288]x+0 (44) Write the state description matrices in control canonical form from equations (16)..472 1..(10) s+ 1 0 Write the state description matrices in control canonical form from equations (1) to (8) and equation (10) A .421 -0..421 . GW =— ' ’ 2s+\ 1 2 ( j+ 0 .. (6) 0 C . (39) Write the coefficient of 5* from equation (34).. (22) Determine the value 6 and C from equation (21).2 x+ 1 If . (29) 0 1 0 0 Write the output equation in control canonical form from figure 2. = [-0.s”"' + —+a..472+1.. (4|5 + B )(5 + 2)[5* + 5 + 36) (5 + 10)(5’ +5 + 25) +c(5= )(5= +5 + 36) + (fl5 + £)(5* )(5 + 2) *(5 + 2)(5*+ 5 + 36) 5®(5 + 2)(5*+ 5 + 36) '(A5 + B )(5 + 2 )(5 '+ 5 + 36) ( s + 10)(5*+5+25) = + 0 ( 5 ^)(5^ + 5 + 36)+(Z)5 + £ )(5 ^)(5 + 2) A s' + As’ +36As^ + 2As^+2As’ +T2As 5’ +ll5'+355 + 250-.. ' Write the general form of output equation. (24) Substitute equation (22) in equation (24).. (8)(27) 4(38) C = 1.. (18) and (19). the value of state description matrices in control canonical form is B . C (s) = (4) s" +a. 5+ 7 A= 5 ( 5’ + 2 5 + 2 ) 5+7 ' ( 5' + 25 + 2) A = j .3465 + 3.. =[0..0.2 0 0 x+ 1 tt (43) 0 0 0 -1 -36 1 0 0 0 1 0 0 Write the output equation in control canonical form from figure 3..... x = A x+B tr ..5] Hence.... k= [-7 1 5 ][^ ]+ 0 y = [ ....05PP Give the state description matrices in modal carionical form for the transfer functions of Problem... 0 0 0 0 0 0 r 1 0 0 0 0 0 JC= 0 0 ... 0 0 0 0 r x= 0 -2 .. l = 2A + C ... 1 0 . (26) Substitute equation (22) in equation (26). =[0.346)+3(3. = [-140]and£). = [-0 .075 + 0. .. Calculate transformation matrix.(15) Substitute equations (1).. X is transformation matrix... C ■ C =[b AB . (19) Step 2 of 2 Substitute eauations (3) and (15) in eauation (19).. the value of new matrices are B = i] c = > 3] 0 . x ftoid equation (14).A t T"' = . D ^ D . k ..(9) Substitute equations (2). -[I t] Write the general formula for t.[ ? I] Calculate entire transformation matrix. . Compute the new matrices A.. and D..... Problem 7. "i[! 3 C = [l 3] .= t j A . the state matrices describing the dynamics of z are in controi canonical form.. I .(20) Write the general formula for ^ ... f . A . (1^) Substitute equations (2) and (14) in equation (17)... C = C T . C.. j .(6) Consider the general formula for inverse of the transformation matrix... Find the transformation T so that if x = Tz... Write the general formula for entire transformation matrix.- t .. t.. (11) and (13). Step-by-step solution ste p 1 of 2 Step 1 of 2 (a) Consider the state description matrix... oil 3 .... D =0.] !l. B. A = T 'A T .. (5) Where.. d 1 K ® :i (16) Consider the general formula for q . ■[:..(4) Write the general formula for ^matrix. and (11) in equation (12).....(12) Calculate t|from equations (1)..1 oJ Hence. ^^=[3 J Calculate t^from equations (7) and (10). (1) B C = [l 0] (3) D = 0 .... 3" 5 5 (14) 3 -1 L5 5. (9) in equation (8)..... = [0 0 . l]C -‘ (7) Consider the general formula for controllability matrix... '■[I tK ....’ b ] . (8) Calculate AB^rom equations (1) and (2).. (21) Substitute equations (4) in equation (21)..i (13) .. T =L J ......t from equation (6).[ 'J AB = | .06PP A certain system with state x is described by the state matrices -[= -]■ -=[1]- C= [ 1 0 ]. B s T ' B . ll I Bs 5 5 3 ^ 5 5J (18) Write the general formula for q . D ^O -2 -2 ] A= . (14) and (15) in equation (5).. 0 .F y ''a + J = a{s) I =o{s) I . X = PX-\-Ou y = H X -\-Ju a(s) = H(si-py'a+j Assume a change of state X to Z using the non singular transformation T. X=TZ The new system matrices are. y = n ^ + ju 0 (s ) = H ( S l .T .{s)= h t Is t t ^-t ' ftY't -'o+j = H { S I .07PP Show that the transfer function is not changed by a linear transformation of state.( S ) = C { S l-A y 'B + D = H T [ S I .'F T y ^ ’r ^ a + J a. A= T^F T B = C = HT D = J Step 2 of 2 The transfer function is.F y 'o + J Assume a change of state X to Z using the non singular transformation T. . Problem 7. Step-by-step solution ste p 1 of 2 Assume the original system. £3 . are the loop gains. Step 3 of 3 From the figure.(Sum of Products of three Non Touching Loop Gains) + -----Aj^ = Nodiing but A .^»Gainof Jt* forward path A = l-(Sum of IndiTidual Loop Gains)+(Sum of Products of two non Touching Loop Gains) .08PP Use block-diagram reduction or Mason’s rule to find the transfer function for the system in observer canonical form depicted by Fig. s s where . i j + AjS + +a^s +OiS^ . not touching the ^ forward path. Problem 7. Figure Observer canonical form of a third-order system Step-by-step solution Step-by-step solution ste p 1 of 3 Step 2 of 3 Mason’s gain formula: TrMsfer Function = . ^ 4 = V U {s) tl A Where K = Number of Forward Paths C/(s) " A where J/. = [ 0 0 1]’ 0~' . (12) Substitute equations (4) and (7) in equation 7.. Step-by-step solution Step-by-step solution ste p 1 of 4 (a) Write the observer canonical form of the state space equations. [1 0 0] = [t. |. (13) ste p 3 of 4 Write the matrix equation from second and third columns from equation (13).....T “ 'AT..T-i ^ = AT TA = A T ..... and (12) in RHS of the equation (13).C . |t ...=AAt.. t. 0 0 [“ *1^ *1 ^*2 ^ * 2] .. (11) Substitute equation (10) in equation (11). y ..22 in the textbook. (14) t.. (10) ... (5) Refer equation 7. Eqs.C = I ..A t . (16) Substitute equation (12) in equation (16)... (1) A ... t j C A ^ O .. -o.] [1 0 0] = [t.. B. under Eqs.. = A t .] -a.. t.c r . t..'A T .C ] [1 0 0] = t. from the results the transformatioh matrix values are |t .o t.. the new state description matrices will be in observer canonical form..= 1 0 0 0] (4) D . <®) Consider the new state description matrix is in observer canonical fonn....] (7) Modify equation (6).. Problem 7..c tjC tjC ] . t. (18) Substitutes equations (11).. C (D = 0 in this case).= 0 0 0 (3) 0 0 0 1 0 0 c .CA t.. Step 2 of 4 Write the general equation for transformation matrix. t...(19) ste p 4 of 4 c CA CA’ Hence. t. . (2) C .. Eqs. = [0 0 i f o . (1) and (2).. Write the matrix equation from second and third columns from equation (9).. I O' [t. 1 0 0 -Oi 0 1 0 0 A .... t j . |a t .t. [1 0 0]=C[t. Find the transformation T so that.' .... * .C A * t... . t..09PP Suppose we are given a system with state matrices A.21 in the textbook and write the equation for ^ matrix.] -a .C = 0 . 0 1 =A[t. . (6) Where. = A t . . T = [t. tjCA’ .= ... . fi-D ......... l . Substitute equations (3) and (7) in equation (8).... (15) t. CA’ CA C] c CA CA’ t. (17) Substitute equation (10) in equation (15)... X is transformation matrix. (1) Write the output equation.. (2) Write the state description matrices in observer canonical form... t j C ^ O ... A = T .. Hie equation for Bn in terms o f Br is TB* » Br.37b) p ^ + lp + l l. (7...1] IC.= A ns. we need fint to find die eigenveclois and eigenvalues o f the Af maliu.9 Tran^ormation o f Themud Systemfrom Control to Modal Form Rod thee s ia Eq. These computations can be carried out by using the following Matiab T -[4 -3 . = C.l 2 U i = p * * 2 l.I 2 J -4 o' 4 = Ans.14). = [ 2 -1]| A?is. (7 J 4 ) and (7. D»=Dc. to explicitly multiply out the equations at the end of Example.3 . Step-by-step solution Step 1 of 5 According to equation (7. Bm«1nv(T)*Bc Cm-CcT. = CfT. the matrices o f Eqs.3 and. Am-inv(T)*Ac*T.14) are related as follows: Am^T~'A€T. Aoconfing to Eqi. Dm-Dc.c ^ r ■" 4 .3 +2] = [2 .38) "■[-t 1].1 OPP Use the transformation matrix in Eq.4 .t ? ] ’ ']• Example EXAM PLE 7. Problem 7. aJ7 a) ^ t i i + 7 ^ 1 + 12f2i = 0. Step 4 of 5 c .T D . and its solution b I2i = ^1 and l22 — !• Therefore.l o J 1 3‘ •16 9 ] 1 4 4 ..36c) into Eq. = D. Step 2 of 5 and A = 0' 1 3 •7 -12 1 4 1 0 ][-. We arant to select the two scale focton such tha both elements o f Bg| in Eq. Step 5 of 5 = 0| Ans . T = [ . C . = T^B C . 0 -3 Step 3 of 5 aT% 1 3 1 4 'l+ O ' a 1+0 B .3J -16+12 9. V] = [4 . (7. a And die equations are A .36b) results in . Wb take the eigenvectofi to be [S] aaUL'Wei es oi me Af maanL >ve nice me eigenvecton lo He i5I [s] aJ6 a) -7 f|i -1 2 ^1 —pliu 036b) 'll (7J6c) Suhstiniting Eq.30) Transformation Matrix -= [. (7.35). fartfaennore.1 1 3' ■28+12 21-121 1 4 4+10 . = r% T B. (7. -.12) and (7.O . -11]. ] where til and tn a e arbitrary nonzero scale focton. (7.""-[I *]■ Elementary matrix muhqiUcation abows that. (7. (738). a» ) *1b fied the iavMse e f a 2 X 2 ■ M l '22. (7.14a) are mhy.4 . (7.7 p ll i. .37d) Wb have found (again!) that the eigenvalues (poles) a e . B in = T “ *Bc.'* d w ^ die s ^ o f * e **i r H d d ie ‘7 P I [slisE4. (737c) . (7.2 . using T as defined fay Eq. Eq. StdvUon. the traMforinatioo matrix and its iovene^ are a.(7JB)l.1 ^ into the modal fonn of Eq. Eq.36c) tdls os that the two eigeuvecton are r-^ i L ft.9-1 -16+16 9 .3 . *• = [ . a .. C c= s. ».1 2 J].. 0 ’ kJ Co = [1 0 0 . ■Where -1^ 1 0 0 .. = r ‘ 4 . = T . 1 (6) 1 0 0 .= [ 2 ] - Co = [ I 0 ]. = b.. (7) £)£. o' A] -a. c. .. 0 1 0 . Do = 0.. Step-by-step solution ste p 1 of 4 Step 1 of 4 As 4 . to the modal canonical form.= 0 0 ■■■ 1 0 (5) Step 3 of 4 ^ If 4) > Qi > ^ are in observer Canonical form.=A . Substituting the above equations into equations B .= A ..T CD (2) (3) (4) ■Where and are model matrices and are in control Canonical form and Step 2 of 4 -O j - 1 0 4 . 0 4 = A= .11 PP Find the state transformation that takes the observer canonical form of Eq..'c . Problem 7. 0] A=o Step 4 of 4 Comparing Equations (5) & (6) We can say that. 2 .7071 0.00 -0.1236 -0.7377 2.9133 Cm= C« = [ 1.0000 2.0969 -0.10 Using Matlab to Fhui Poles and Zeros ofTape-Drrve System Find the dgeDvahies of die qrttem mMiix described below for the tape-drive conind (see Fig.1214 2.0 Oil Oil] Servomotor ontpm.3162' -0.5075 -0.0001 -0. C3 s [OJ 0.000 0.00 0.9239 0.0000 0.0017 0.323 0.17221 -1.7377 -2. = T N = Tdtagin^n^ .1502 0.0161 -3.S0). [Am. whidi results in Donnormalized values in Bm and € « .0000 -0.0376 0.3627 -0.ooi0i a o o 4 0 -o x» io i aoooo 0. B« 0 a.B. [sysGm. This means that a stq> iiqiut will result in a ramp output.4785 -0.0000 -0. And.7962 -9.Cm.0] Ibnsion output.0000 -0.9227 -2. S.0000 1.0000 0.3439 0..0000 -0.0000 0.000 -0. 3.7291 -0.0000 0. and the real poles fall on the main diagonal o f this matrix.2850 0.10 in modal canonical form but will convert each element of the input matrix Bm to unity.6371 -0.t. = 1^ 'a = [1 1 1 1 i f Ti is full rank matrix.2466 0.B.5023 -0.8379 0. So. Dm= />* = 0.059 0. The inverse is computed from T = ln v a i) and results in 03805 0.58SI -0.3883 0. The rest o f the conqNitatkxis from canon are 0. which results in -0.6274 Bm= B«: -1. %.4708 -1.0000 0.0612 _ B.1847 0.75 . H = T 'a And.1.1053 -1.5022 -0.2282 11334 -3.0257 -0.2077 0.8797 -0.00 0.0817 -2.19251 -a il6 8 -a i9 2 S i -0. B= 0 0 ’ 0 2 0 0 0 * 'o' -ai -035 0.0000 0.6669 0.8233 0.8234 2. = T N B .0017 0.. If we found it desirable to do sex we could readily find further transformatioiis to make each efement o f Bm equal 1 or to mterchange die order in adiich the poles appear: Step-by-step solution Step 1 of 4 (a) For transformation matrix Ti.. The system sare ■0 2 0 ®1 'O'0 -ai -(U5 ai 0ai 0.7376 -5. so we need to invert our Matlab results.1692 -0.9016 -0.Tl]-canon(sysG/modal').00 0.0207 01160 Eigenvalues The eigenvecton computed with [V.6371+0.Bm.0075 01697 Notoce tbit the first two cohunm o f the real tmsfonnation T are com­ posed of the real mid the imaginay ports o f the first mgeiivector in the first cohium of V.8697 -18284 1. The vectors in V are norraalired to unit ler^th.11684-0.4903 ].0561 4.9683 .0330 -18284 2.7741 0.2 0. To transform to modal fmin.6371 0.6371 0.000 -0. » = [»i «i .3406 a4714 -a4112 -0.9227 -6.0449 3.Dm]<sdata(sysGm) The result o f this ctMnputatitMi is -0. compute the tramformatioa o f the equations of the t^ie drive in their giwea fm n to modal canonical fonn.0150 0.1663 0. The sM e vector b defined as xi (tape positiao at capstan) « i (speed o f the drive wheel) JK3 (posttioii of the tape at the head) » 2 (ouQNit speed) The matrix C3 corresponds to malrii^ X3 (die position o f the tape over the lead/write head) the output.00 0.8708 -8.0000 0. = T N = [«ili .P)i^io(F) are -0.9314 0.0000 -0. Example 7.3353 -2.9683 Notice that the system has all poles in the left half-plane (LHP) except for one pole at the origin.5075 0.0.0133 0.4(» 0.7071 0..0129 0.5432 -3.1 ai a?5 0 A» 0 0 0 2 0 .6669i -0.000 -0.0000 0.4133 0. C r .8933 -0.1003/ -a02704-O. D=Qja.3060 03093 03317 1. -0.I003i -0 .A 0 0 2 . B. so we conclude that the system has Type 1 behavior.0000 0.0000 0.0000 0.0014 0.00 -0.28024-0.9341 T g N = diag{n) Step 3 of 4 : T*N ■-0.0014 0. Hence.6371 0..2970 2..00 ■ -17.0130 -a0114 -0.G.00 -0.1847 -0.. = 1 1 1 if 1 = '^ ste p 4 of 4 ■70.2941 -0. 1.4785 0.2933i -aOOOO -01126 -03079 aoo40 4-o. To compute the eigenvalues using MMhb.2239 -1.0897 -0.9133 It happens that canon was written to compute the inverse o f the transforma­ tion we are w<xidng whh (as you can see from TI in the ixevious equaticm).6247 TI = T-* = -0.3533 -4.2 0.0000 -0.0000 2.6360 -0.00 -0. Problem 7.10 EXAMPLE 7.D).0150 -3.6669i Ps 0.[ -0 .3353 -1.0000 Am = A* = 0. Abo.0000 -0.2569 -1.00 -0.8797 0.8708 -4.000 0.5075 -0.5980 -0.1282 -0.12PP (a) Find the transformation T that will keep the description of the tape-drive system of Example 7.0000 0. 0 = T ^ B .3439 -0.29331 -01802 ..000 -0..0000 -0.4 a4 -a4 .4871 03887 -0.6669 -0. we write P-eig(A).9683 Notice that the cmnplex p d es ^^>ear in the 2 x 2 block in the uf^ier-left comer o f Am.00 0. and the matrix Ct corresponds to making tension the output Sclution.3533 -0.0745 1.4 0 0 0 -04)3 0 0 -1 1 C2 * [OD OlO 1.3264 0.O J 0.000 00.9037 -0.0 OJ Oil Oil] Posiboii M leaiVwfite head as oatpal.0000 0.. (b) Use Matlab to verify that your transformation does the job.00 0.4708 -0.0000 0.0000 0.0270-0. a = T.6776 0.3402 -0. Step 2 of 4 (b) Using Matlab.2115 0. we use the Matlab ftmctitm canon: sysG-ss(A.5980 4. It n dib step that causes the complex roots to appear in the 2 X2 Mock in die upper left the A* matrix. = T.8284 1.0247 2.3383 -1.6371 0.0000 0.6767 -1.1844 -1. [Am.0.8697 .1 8 4 7 0. compute the tiansfonnatioa o f die equations of the t^ie drive in their given fnm to modal canmical fbnn.5593 0.0000 ' -0 .0000 0.0000 1.5 0 0].1925/ -0 1 1 6 8 -0 1 9 2 5 / -0.3406 0 .0 Oil Oil] Servomotor ontpm.2 .0040 0.2077 0. This means that a stq> iiqNit will result in a ramp output.9133 = [ 1. Arrange the eigen values in increasing order. To compute the eigenvalues using Mathb.0000 Am = A « = 0.4 1 1 2 -0 .0 4 1 1 2 -0.3 5 3 3 -0 .3060 -2 . [sysGm.6 6 6 9 -0 .indices]=sort(abs(P)).6 6 6 9 /) > Pj (-0 .0002 -0. It b dib step that causes the complex roots to appear in the 2 X2 Mock in die upper left the A* matrix.2 4 6 6 0.0000 -0.0000 -0.0000 0.0000 0.3 3 8 3 -1 . The system ' 0 2 0 0 0 * '0' -a i -0 3 5 0.Cm.0561 4.6 6 6 9 / P * -0. so we ctmclude that the system has Type 1 behavior.0000 + 0.6371 + 0.0000 0.4903 ].0330 -1 8 2 8 4 2. ■0.1236 -0.2850 V = V= 0.0.0001 -0.0000 -0.3060 0J093 03317 1.5093 0. Abo..0 1 6 1 -3 .1 6 9 2 . .(j.1 0.2 0. The Eigen values of matrix are displayed.0000 0. and . -0 .3 . whidi results in normoimalized values in Bm and € « .B.5851 -0 . -0.8697 1. C2 * [OD OlO 1.10 in modal canonical form but will cause the poles to be displayed in Am in order of increasing magnitude. compute the tramformatioa o f the equations of the t^ie drive in their giwea fm n to modal canonical fionn.03 0 0 -1].0 . The sMe vector b defined as xi (tape positiao at capstan) « i (speed o f the drive wheel) JK3 (posttioii of the tape at the head) (ouqwt speed) The matrix C3 corresponds to malrii^ X3 (die position o f the tape over the lead/write head) the output. To transform to modal fm n .5 9 8 0 4. 0.0000 Cm2 = 2. and the real poles ftdl on the main diagcmal o f this matrix.3805 0.2 O.0817 .OOOOi Step 4 of 4 The new set of state matrices are shown below.0 2 7 0 -0 .3264 0. Consider the eigen vector of T.1 .2 0.0000 -0.0000 0.0000 0.6 2 7 4 Bm= B«: .3533 -0.3439 0.1. (b) Use Matlab to verify your result in part (a).4120 dm2 = 0 Hence the results are verified new set of state matrices are obtained using matlab.0000 -0 .2569 .10 in the textbook and find the state transformation of the matrix.8707 -0.0001 0. The vectcrs in V are normalired to unit ler^th.9 6 8 3 Notice that the system has all poles in the left half-plane (LHP) e x c ^ for one pole at the origin.1847 -0 .0000 -0.00101 00040 -OOOlOi ooooo 0.13PP (a) Find the state transformation that will keep the description of the tapedrive system of Example 7.3 2 6 4 0.1847 -0.TI]=canon(sysG. and the matrix CTConespoods to making tension the output.6 6 6 9 /) > /J (-0 .4 -1.0017 0.0.0000 1.9 6 8 3 Notice that the cmnplex p d es ^^>ear in the 2 x 2 block in the uf^ier-! comer <^A«. The rest o f the conqwtatkMis from canon are 0. /i. Am2 = 0.1. Problem 7.indices): n=T2\B.0000 + O.6 3 7 1 + 0 . and give the complete new set of state matrices e .0000 -0.D): [sysGm. Abo.Bm.0000 0.5075 + O.1 8 4 4 -1 .3439 -0.2802 + 0.0017 0.5 9 8 0 ./j.0000/ -0 .0150.3353 -1.6776 0.1334 -3 .2 0.6776 0.0 0 1 1 4 .1 0.4785 0.10 EXAMPLE 7.4714 0.1 -0.0 .0:0.1692 -0.0 .2933/ -0.2 .5 0 7 5 + 0.2466 0. we write P««ig(A).4 1 1 2 .2077 0.P]=eig{P) [f.Oj.0 0 0 0 0.6360 -0.0000+0.6247.3 4 3 9 -0 .58SI -0.c3.0000 0.0000 -0.3 6 2 7 -0 . 3.2802 . -0.2 .2 2 8 2 -0 .1 5 0 2 0. If we found it desirable to do sex we could readily find further transformations to make each efemeM o f Bm equal 1 or to interchange die order in adiich the poles appear: Step-by-step solution Step 1 of 4 (a) Refer Example 7.0000 0. Example 7.P)i4 ig(F) me ' -0.5 0 7 5 -0 . T=inv(TI): % [V.1844 -1.Cm.0002 0.5980.0 .0 .2. To display the poles of Am in the order of increasing magnitude.0 2 9 3 3 / -O O O O O -0.0207 0.0 .6371 .6 3 7 1 .6371 -0.2 -0. Sclution. D=0.0 .3 3 5 3 -1 .7071 0.4785 .6669 0.0000 0.0 1 1 4 .9 6 8 3 + O. T3=T2*diag{n).2282 T= T = 11334 --3. Tj is the rearranged eigen vector T in the order of increasing magnitude.0 . = 0i ■ .8797 08797 -0.0 0 1 7 0.6669/^ .D.7741 0.8 2 8 4 -0 .0 .Dm]<sdata(sysGm) The result o f this ctMnputatitm is ' -0.8 2 8 4 1.3353 -2.66691 0.1502 0.0 0 0 0 0.C. >5 ( .40) 0.0 OJ Oil Oil] Posidoii MleaiVwfite head as ootpal. [P]=eig(A).0000 0. rearrange the eigen vector T in the order of increasing magnitude.0130 -0 .6776 -0. P= -0.0 0 0 0 0.0 .0207 .9133].0000 -0 .2 2 8 2 2.0.8233 0.6014 0.7741 0.4 0.6 371+ 0.0000 -0.0 .6 3 7 1 + 0 .0 -0.Dm]=ssdata{sysGm): TI=[-0. T2=T(:.6 3 7 1 .1168 + 0. which results in .0161 -3.Tl]-canon(sysG/modal').Bm.2160 0.0 0 0 0 Step 2 of 4 (b) Write the matlab program to verify the result and to obtain the new set of state matrices.1334 .9133 It haf^ieiis that canon was writtra to ctmqMite the inverse o f the transfcxma- timi we are w<xidng with (as you can see from TI in the inevious equaticm). the matrix must be arranged as shown below.B.5075 -0.9 6 8 3 + 0.3439 0.D).8 2 8 4 ' -0 . cic A=[0 2 0 0 0.6 6 6 9 i .0] Ibnsion output.3 4 0 2 0.2 4 6 6 0.0002 -0./j.4 a4 -a 4 .4 0 0 0 -0 4 B 0 0 -1 . Therefore the rearranged eigen vector T in the order of increasing magnitude is given below.4 -0.75.3317 1.OOOOi) > P.3402 -0.2 0. B.0017 0.8 2 8 4 2.0 1 5 0 0.0000/ The row values of P gives the corresponding column values of the matrix T.0 3 3 0 2.0130 .0000 -0.1 ai a75 0 A» 0 0 0 2 0 0 a.0000 0.3 .3406 -2 .6 6 6 9 i Ps 0.3 4 0 2 -0 .8705 -6.1 0 0 3 / -0 0 2 7 0 + 0 1 0 0 3 / .3406 a4714 ■ .OOOOi -0.6 2 4 7 TI = _ -0 .7 2 9 1 -0 .9683 Bm2 = 1.0000 -0.0.3317 1. C r . C3 s [(L5 0. .0561 4.4 0.1 . r2= ['5 '2 U Where.5 0 0.6 3 7 1 0.0 .3060 0.6 3 7 1 -0 . 0.0969 -0.0130 .5851 0.S0).'modal'): [Am.0000 -0 .0000 -0. we use the Matlab fimctitm canon: sysG-ss(A.0 1 5 0 . Am2=T3\A*T3 Bm2=T3\B Cm2=c3*T3 dm2=0 Step 3 of 4 The output obtained on executing the code is given below.OOOOi -0.0000 0.a n d /ja re the respective column of matrix T.1]: c2=[0 0 1 0 0]: c3=[0.3805 0.0000 0.0 0 0 2 0.0969 -0 .0 . 3.0. (-0 .3383 -1.0 .1847 0.2126 -0 J 0 7 9 00040 + 0.3353 -2 .66691 -0.0000 0.0 .0000/) Hence.0000 0. so we need to invert our Matlab results.0207 01160 Eigenvalues The eigenvecton compuled with [V. The system contnd (see Hg.0000 0.6371 0.9683 + O.35 0.4 7 1 4 ' -0 .5 0 7 5 + 0.0000 -0.0 .0000 1.5093 2.1 5 0 2 -0 .5108-0.S0).4871 0J887 -0 .(-0.0000 .0000 1.4785.0 3 3 0 .-0.0000/ -0 .000 0 /) > P .7291 -0. The inverse is computed from T = lnv(TI) and results in 03805 08697 -1 8 2 8 4 1. A .7071 aS379 0.5 0 7 5 0.O J 0. cT=[-0.6371 0.0000 0.0016 -0.G. sysG=ss{A.0000 0.[ . B=[0:0.0 1 1 4 0.2160 Let the eigen values of the matrix T is given below. D = 0 .10 Using M atlab to Find Poles and Zeros ofTape-Drive System Find the dgeDvahies of die qrttem mMiix described below for the tape-drive conind (see Fig.8284 1.6755 2.3627 -0.0075 01697 Notice dmt the first two cohimm o f the real tmsfonnadon T are com­ posed of the real mid the imaginary ports o f the first mgenvector in the first cohirrm of V. (2). Problem 7.14PP Find the characteristic equation for the modal-form matrix Am of Eq. det(/r(l —A) = 0. (1a) using Eq. Eq.4 ] = 0 h :-J s 0 0 rs+ 4 0 ] [ 0 S + 3J Characteristic Equation is: (s+ 4) (s+3) = 0 + 7 5 + 12 ^ 0] .. Step-by-step solution ste p 1 of 1 ste p 1 of 1 ■ i r j Characteristic Equation is: det ^ 7 . (la) Eq. (2). (4) Consider the value of matrix a • "=L-2 -..15PP Given the system * = [_ 2 with zero initial conditions..m . (3) and (4) in equation (1).. (7) Substitute equations (5) and (6) in equation (7).J Consider the value of matrix b • ... (6) Substitute equations (2).... ± = A x + B u ..X ... 0 = A x .. -B A -B A -‘ . Write the general form of state space equation... (3) Consider the value of x at steady state condition. Problem 7. .. (2) Consider the value of wat unit step input condition.j' Hence. the steady state value of x is .+ B -B = A x.. Step-by-step solution ste p 1 of 1 Write the general form of state space equation... a = I ... find the steady-state value of x for a step input u.... (1) Consider the value of t a t steady state condition with zero initial condition. .... x= .... i = 0 . Figure A block diagram Step-by-step solution ste p 1 of 6 Step 2 of 6 U [s) =z where Afg = Gain of 1 ^ Forward path A s 1.(Sum of Individual Loop Gains) + (Sum of Products of two Non Touching Loop Gains) ..X 3 + u (1) I3 = Et3(s) = I j( s ) + I .JSfj (s) = . 1 i/ « _L 2s (s +4) ’ ^ s+ 4 -1 A* 2s*(s+4) ’ ^ s(s+ 4 ) 1 1 2s^(s+4) s(s+ 4 ) A. s+ 4 sJf* (s) + 4 ^ 4 (s) = U ...16PP Consider the system shown in Fig.Sjc. (4) ^ = + (5) Step 6 of 6 A '0 0 0 0. Problem 7. = 1 Step 3 of 6 F (s) 2s(s+4) s-l-4 ■■■ U { ^ 1 2 s^ (s+ 4 ) "*"3 (5 + 4 ) y (s) _ 2s+l 2s^(s+4) U [s) 2s(s+4) 2s^(s+4)+l+2s^ s* (2s+l) . (b) Write state equations for the system using the state-variables indicated.( s ) Xj = X2 +X4 (2) Step 5 of 6 sX3(s) = jr ..A....5 A 1 0 0 0 xa A 0 1 0 1 .( s ) ^ = *1 (3) ‘ 2s = O. = 1. m s “(2s+l) 2s'+ 8 s’ + 2s=+l ste p 4 of 6 b) From the Figure.and A. A.(Sum of Products of Three Non Touching Loop Gains) +. 0 0 .4 X2 y = [l 1 0 1] .1 . 4*4 .^= Nodiingbut 'A' not touching the forward path.. (a) Find the transfer function from U to Y. G(*) = ^ = C ( i I .3 . II 4 I^ I 1 I'l (b) Eq. = .2 0 = 1 « = [1 0 1]. Problem 7.10 x4 +tt 0) jr5 = (A .x^ + U .2 1 0 0 0 -10 3-= [1 0 . (2) 4 a^= -X 4 X j=-3xj +X2 (4) y = x i-x ^ (5) S tep 1 3 o f1 4 ^1 ■-3 1 0 O' ^1 xa 0 0 4 0 Xa 0 . Step-by-step solution ste p 1 of 14 Step 2 of 14 BLOCK DIAGRAM REDUCTTON: The above block diagram can be redrawn as: Step 3 of 14 Step 4 of 14 2(s+2.4 x . 1 .j^ ) — S +2 X + 2 x3 . 0 0 -4 1 (5) y = [1 0 1] Step 8 of 14 The above matrix equations are in the form of X = iTx+Gb y= H x+ Ju where 7 = 0 -5 r ‘0* where F = 0 -3 .p r >'2 1 i +10 ^+ 2 5+3 4 1 2 ste p 10 of 14 Step 11 of 14 Step 12 of 14 r(s ) _ s(s+ l)(s’ +3s+4) U{s) ” s(s+3)(s+10)(s^ + 2s+4) (s+ ll)(s’ + 3s+4) (s+3)(s+ 10)[s^+2s+4j (s) ^ s+10 '■ ' SX4+IOX4 =u x^ = .i^+ X 3+2 x3 = -3x2 A-u + 2j^ Xj = a^+X j-S xj (3) SX3+X.1 0 ] j =0 r(s) _ (s+ n )(s^ + 3 s+ 4 ) U{s) ~ ( s + 3 )(s + 1 0 )(s ’ + 2 s+ 4 ) .l.+ U .X 2 6 3 X3 = 2X3+ X4 . write the state equations for each of the systems shown in Fig.17PP Using the indicated state-variables.^3. +u (1) Step 7 of 14 ^ s+3 ’ aJTj ( s ) + 3JSTj (s) = sJTj ( s ) + 2 JTj (s) * . Find the transfer function for each system using both block-diagram manipulation and matrix algebra [as in Eq.].1 0] and £W H (s i-F y ^ a + j where -3 1 0 0 ■ o' 0 0 4 0 0 F * 0 = 0 -1 -2 1 1 0 0 0 -10 1 Step 14 of 14 «= [1 0 . J=0 0 0 -A Y[s) s“+10s+20 U(s) (s+ 3 )(s+ 4 )(s+ 5 ) Step 9 of 14 1 1 . Figure Block diagrams I I .A r ' B + D. (4) ^1' -5 1 1 0 Xj = 0 -3 -2 X2 -f 1 u .5) s+ 4 (J+3X-V+5) Step 5 of 14 Step 6 of 14 r (s) ff^-l-lCte + 20 ■y (s ) (s+ 3 )(s+ 4 )(s+ 5 ) Matrix Algebra: From the given block diagram: U s+A EXi[s)+ AX^{s)^U [s) X. .. jc = A x + B u . G (*) = (4) s" + 02S"~* + Write the state description matrices of controller canonical form. (14).= (6) C .... Draw the block diagram for controller canonical form from equations (23) and (24)...= .. Write the general form of output equation. the state equations of observer canonical form are x ..... (16).. Figure 4.... (1) Consider the denominator part of the gain.r=[i o]j(.. Step 4 of 10 ^ Write the observer canonical form of the state space equations.. 0 B .. y = C x* D .... the block diagram for observer canonical form is shown in figure 2. = . (18) J J .. (15).. (14).. a ( j ) = y + < i ^ " " '+ a j j ^ ^ + .. (13) d 0 0 1 0 0 0 0 Hence... (18).. (16). =Sqwrate valuefromnumeratoraiiddeiiominatorelse 0 (8) ste p 2 of 10 Consider the value of gain.. (2) Write the general canonical form of the equation..] (7) £). . 0 0 1 0 0 1 0 0 0 (11) 0 1 0 0 0 0 1 0 Write the general form of output equation..... 0 1 0 0 A ... (17)... Problem 7. and C..= [1 0 0 0 O j*....+ b ..= = 0 0 1 0 (16) ■ 0 0 0 1 -a .. (24) Step 7 of 10 -2 -2 1 Hence. . (14) Write the output equation. . ’■*1 0 1 0 0 *1 I _ 0 0 0 + 0 •*3 0 1 0 0 ’h 0 0 0 I 0 *4...=[A b.. ... 1 0 . Figure 2: Block diagram for observer canonical form Hence. the state equations of controller canonical fonn are: X S X+ u 1 0 0 b =[3 4]x|.. (15). -2^ 0 1 0 0 o' 1 0 1 0 1 (20) 0 0 0 1 0 0 0 0 0 -2 .. .. draw a block diagram and give the appropriate expressions for A. (19) Write the state equatiohs ih observer canonical form from equations (4). ^. (15) Write the state description matrices in observer canonical form....... Figure 1: Block digram for controller canonical form Hence.. the block diagram for observer canonical form is shown in figure 4. the state equations of observer canonical form are: -2 0 jj-li OKI Step 9 of 10 ^ Draw the block diagram for observer canonical form from equations (25) and (26).y=[3 4 ] * . (18) and (19).2 ] I |.■*4.. (b) (b) = Step-by-step solution diep^by-^iep* S olliiion Step 1 of 10 (a) Consider the numerator part of the gain. (25) "•=[-2 J-=[1 0] . -<h .. 1 0 B.. ■(3) Substitute equations (1) and (2) in equation (3). the state equations of controller canonical fonn are X s x+ U 0 0 0 0 0 0 0 0 |..= [0 1 0 ....= [0 1 0 . Block diagram for controller canonical form Hence... the block diagram for controller canonical form is shown in figure 1... y = C x+ D .. 0 0 1 0 0 (5) 0 1 ••• 0 0 .(17) C .. (12) .+ o . (9)... 0 1 0 O' o' _10 •'2 + 1 1 0 0 0 0 1 *1 0 ..+ 0 0 0 1 0 0 0 0 0 -2 |y = [l 0 0 O " ^ Step 5 of 10 ^ Draw the block diagram for observer canonical form from equations (20) and (21)... -a .. 0 0 0 0.........> = [1 0 0 0 0] .18PP For each of the listed transfer functions. C (* ) = 4 ^ (9) s -s Write the general form of state space equation.2 ] x + 0 . Step 3 of 10 ^ Draw the block diagram for controller canonical form from equations (11) and (13). B. Block diagram for observer canonical form ste p 10 of 10 ^ Hence... the block diagram for controller canonical form is shown in figure 3.f 0 . write the state equations in both control and observer canonical form. 1 0 ••• 0 -a. In each case.. (26 ) -2 1 Hence.. b {s)= b ^^+ b2S^^+ ....*4.. (17)....= 0 . (a) C(s) = (control of an inverted pendulum by a force on the cart).= [ l 0 0 ••• 0 ] .. -o ..■*4. (21) 0 0 0 o' 0 0 1 Hence. (19) and (22). *. Step 6 of 10 (b) Consider the gain value. 0 .. (10) Write the state equation in control canonical form from equations (4) to (10). 3s+4 (22) s’ *2s+ 2 Write the state equation in control canonical form from equations (4) to (9) and equation (22).. ?]'•[:>.... Step 8 of 10 Write the state equations in observer canonical form from equations (4). Figure 3. .. -1 -3 Step 3 of 6 (b) Parallel realization of Apply partial fraction in equation (2).. = .= [-3 4 ]* (22) Hence.... (28) 0 C . + « . . (23) Consider the denominator part of the gain.t.. (34) O . (16) Hence... ■(25) Substitute equations (23) and (24) in equation (25). the state description matrix in series realization of C{4) at cascade connection are -2 0 and |y = [0 l]jc]. F ig u re s Step 6 of 6 Hence. Step-by-step solution ste p 1 of 6 (a) Series realization of <?(#): Consider the value of gain G(... .. s+6 -3 4 r+ 7 (20) ( j+ 3 )(j+ 2 ) ( 5 + 3) ( j + 2 ) Draw the block diagram from equation (20).. (35) and (38). 1 0 .. . „ 5+6 (s + 3 ) ( ^ + 2 ) -3 + 6 <"’L “ (-3 + 2 ) A ...3 jC 2 + « .. (21) Write the output equation in control canonical form from figure 2.. (b) Using a partial-fraction expansion of GfsJ.... (19) Substitute equations (18) and (19) in equation (17). (32) C ... Problem 7.... (38) Write the general form of output equation from equations (34).. in the form m s} s + 3 \s + 2) find a series realization of G(s) as a cascade of two first-order systems. (c) Realize G(s) in control canonical form. * .. iinu a paraiiei reaiizaiion ui u{sj. find a parallel realization of G(s).r)- (1) Write the series realization of G{«)from equation (1).|J f + l ... (12) Substitute equation (7) in equation (3)... Write the general form of output equation of canonical form......9 ..... . (2) Draw the diagram of series realization from equation (2)..(10) Substitute equation (5) in equation (10)....... (15) Write the output matrix from equation (15).. Figure 2 Step 4 of 6 Write the state equation in control canonical form from figure 2. 0 .. + U j .[ : :E 1*[:1- "[• +H 1" =I . + u .. (3) i.] (29) D^sSeperatevaluefromnumeratoranddeiiominatorelse 0 ....... the state description matrix in parallel realization of C{4) are -3 0 and |y = [-3 4 ]x | 0 -2 Step 5 of 6 (c) Consider the numerator part of the gain..= [A ^ »...3 X 2 + U ....19PP Consider the transfer function yw »+6 + + (a) By rewriting Eq. 1 y= y2 „ M = M io - s+ 2 5+ 3 Figurel Step 2 of 6 Write the state description matrix from figure 1. (9) Substitute equation (9) in equation (4)...(17) (j+ 3)(j+ 2 ) (5 + 3) (j+ 2 ) Find A from equation (17)... the state description matrix in parallel realization of C{4) = -5 -6 ' I + u and 3’ = [« '] +0 1 0 .. 0 1 0 0 (27) 0 1 •• 0 c . -<h ...... (11) Substitute equation (7) in equation (11).... j+6 yi = u2.. |w . (5) ......... (4) y . (36) Write the state equation in control canonical form from equations (32). x ^ — 3 x ^ + y t .. (33) and (36)..3 x .. . = ..i[^ ] y=[0 l].. x = A x + B ti ..... (24) Write the general canonical form of the equation. -a.. i. y^[6 IJI *1+0 (39) Draw the block diagram of canonical form from equations (37) and (39)..= [0 ] (35) Write the general form of state space equation of canonical form.... 1 0 I 0 0 ... 3... = . K ] .* 1.. *2 — X .. = y ...c .. (7) y = y ... (13) Write the state description matrix from equations (12) and (13). s+6 ... (o) Osrrlg~a p^n'iar-iractiuh expartsiijifijnj(:>....2 j i + i/.. s+6 A B .... KHrim i..2 j c . . +b.. (30) Consider the value of G (4). ... ..x .(8) » .. i l = ... [a-Kaa*[ii- « ] ■ - Substitute equation (6) in equation (8). y =Cx+D .(s+2) (s+3){s+2y B = 4 ..... 5+ 6 C (5) (31) 5*+55 + 6 Write the state description matrices in control canonical form from equations (23) to (31)..... Find B from equation (17). b{s) = b^^+b2S^^+ ... . y= .... (6) « = « . +b2s"~^+’” +b^ C (s) = (26) Write the state description matrices in control canonical form..=[6 1] .. +Cje*<'-'*Bii(r)</r+D»(/) h Assume x ^ sQ h y { l) = ^ b ( l ..) = C e*<'''’x. the impulse response of the system is |*(<) = Ce"B-H?^(t)| .+|e*<'‘'>Bii(r)</r + Du(t) j-(/) -Ce'^"*>x. C. A(/) = Ce*'B+/M(/) Thus.T 'p u ( i ) d i h Where.+|e*<'-''Bii(r)</r h y (l)= C x (t)+ D u {t) y { ..r)B(r)i<(r)</r+B(/)ii(/) For the time-invariant case. Where.20PP Show that the impulse response of the system {A. D) is given by h(t) = CeAB + D5(t).r)B(r)ii(r)</r (2) Differentiate equation (2) with respect to dt 4 * (0 = + A(/) J®((.»„)i„+|®(/. AV \ ” A‘l* Step-by-step solution ste p 1 of 1 Step 1 of 1 ^ Consider the state variable equations for the inhomogeneous case with a forcing function input » (/) i(0 = A (/)x (0 + B (r)« (/) (1) The solution of {1) is x(») = ®(/. B. where eAf is the matrix exponential defined by -Al /.] The state solution is x(0 = e*<'-'>x. e " = (l+ A /+ ^ + . Problem 7. > K And.+ 4 )n -(6 A | + 7A r. (2) The output equation Is. y=x^+3xj (3) ste p 2 of 9 From equations {1). +7AT.] i Calculate the matrix ^5! —A + B K ) • 5 0] .l+ 3 3 : ) ( . I I-7+2A . Calculate the characteristic equation. -l+ ^ T j 1 .21 PP Consider the plant described by * = [? ..[ a:.( .( . u = .7 ) = 0 Therefore.:h.. .7 _ j+ 6 + 6 j+ 2 1 j* + 4 s-7 I s +21 i'+ 4 * -7 75+27 Therefore..7 + 2JST. the transfer function G { . :.]'* j 0] ro 1 ] r 3 /:] p s j ” [7 -4J'^[2A: 6A:J s+ K -1 + 3 A ] -7 + 2 A s+ 4 + 6 ^ :J Step 9 of 9 To calculate the characteristic equation equate determinant of matrix (A I-A + B K ) to zero. calculate the transfer function of the plant.i] . + 2 A . (ii) u = -Ky. Calculate the characteristic equation.-h. F^ve 1 Therefore.i+ A : . 5 + 4 + 2 /:J Step 7 of 9 To calculate the characteristic equation equate determinant of matrix ( > U .iH The state equations are. >>= C x + D tf Hence From equations (1). -1 + AT..7 +0 7 f i+ 4 + 2 ' 7 + 2i *+6l “ T r ir y C ^1 2j + 7 j+ 6 + 3 (2 i+ 7 ) j *+ 4j .-7) = o|. = 0 »’ + (/:. d e t[A I-A + B K ] = 0 |A I-A -l-B K |> 0 I s+K -1+3AT I-7+2AC *+ 4 + 6A :|" (j+ X ) ( i+ 4 + 6 X ) . (b) Find the transfer function using matrix algebra. draw the block diagram for the plant. +2K.! 4 ] * + [ 2 ] “- y = [ 1 3 ]x. Step 8 of 9 (ii) The state feedback is. Problem 7.l + 2K + 2 \K -6 K ^ = 0 s ^ + {lK + 4 )s + {2 1 K -l) = 0 Therefore. x = Ax+Bu The output description is. (c) Find the closed-loop characteristic equation if the feedback Is rn iu [ii^'‘ur6^bu-iuup urieiiauieiisuu equciuuii ii m e leeuu<du^ is {\)u=-[K1 K2]x.) . the block diagram of the plant is shown in Figure 1.A + B K | = 0 I i+AT.7 + 2 ^ r . (2). : i fj-o 0 .Kj . (2). the characteristic equation is |^’ + (7 A + 4 ) j + ( 2 7 A : .. G(j ) = C( sI-A )"'B + D «+4 I ■[> % i ’ + 4 j .7 + 2 A r ) = 0 s^+4s + 6Ks+Ks+4K + 6K‘ .A + B K » 5 [.) = o s^+4s+2K^ + K ^+4K .. The block diagram for the plant is shown in Figure 1.... u= .A ) :h . y i _ ««ljointof(rf-A ) ^ ~ |iI-A | 1 p+4 1] -7 i+ 4 | ________ 1 p+4 1] 1 p+4 1] “ j ’ + 4 s -7 [ 7 jJ Now. y= [i 3]» ■[.7 ) = ^ . (a) Draw a block diagram for the plant with one integrator for each state-variable.0 _ ■ 5+JC. |/ U . and (3).) of the plant Is 5 ^ + 4 5 -7 Step 6 of 9 (C) (i) The state feedback is. and (3).-[A 3iT] Calculate the matrix ^5! —A + B K ) • 5 l.A:. A= B= C= Step 4 of 9 Calculate the matrix ( 5 l . the characteristic equation is | i‘ +(A:. Consider the state description. + 4 ) i +(6A:. Step 3 of 9 (b) Write the expression for the transfer function.1] “ [o -7 i+ 4 j =[* -7 s -+'1 4] Step 5 of 9 .) ( j+ 4 + 2 /r . The matrices are. .4 j ^ + 2 tt. i+ 4 + 2 A . + 2iT. -2AT.|“ (j+ ir . .. Step-by-step solution ste p 1 of 9 (a) The plant is described by.A + B K ) to zero. + 7JC.. a:.) ( .7 + 2AT. (1) = 7 jc .A> =-js:[i 3] . .(2) Compare equations (1) and (2). K is the state feedback gain K = [JC. AT.5 . it is verified using MATLAB. Problem 7. Hence.707)(1. G =| Write the expression for peak time.707. Steo-bv-steo solution Step-by-step solution Step 1 of 5 The system is. t From the unit step response the peak time Is. s' + ( 5 + a:.707 for ^ in the equation to calculate the natural frequency..707for ^ .) = o .14 sec.707' ___ 1 _ “ 0. F . » plot(t.414)i+( 1.. ‘ ■[i Here.14s« — * 3.y).. = 3 .G K ] 0 -6 K.K.m -4 The state feedback gain is.den): » t=0:0. .] Calculate the value of the matrix.0001:6. = 2 K .14 s for in the equation to calculate the damping frequency. K =H -3] The state feedback controller.j r J * The closed-loop characteristic equation is.414 lad/s Write the general second order characteristic equation.14 s. ± XS F . S +K^ = 2 K jm -3 And. . » xiabel t » ylabel y(t) Step 5 of 5 The following is the unit step response.414rad/s for »'+2(0. I4 s The peak time obtained is at 3.. tt = -K x Here. » grid. u Is F H -3 1 4 Write the MATLAB code to plot the step response.atJ Determine the matrix.i 0] . 6+K. ^4 ■ Substitute 1 rad/s for and 0.5 . F= [ i: ] And. Verify your design with Matlab. » sys=tf(num. s* + ®0 Substitute0.707 ■ 1.a n d 1.G K 0 1 F -G K = -6 -S _r ® ■ [-6 _r 0 ' -5 JC 0 . Substitute 3. Step 4 of 5 » num=1.t). 3.) s + (6 + a: .414)’ = 0 s '+ 2 t + 2 = 0 (1) Step 3 of 5 The linear state feedback expression is. /. (b) Step-response peak time is under 3.22PP For the system design a state feedback controller that satisfies the following specifications: (a) Closed-loop poles have a damping coefficient ^ = 0. » den=[1 2 2]. -0.14 s «1 rad/s Step 2 of 5 Write the expression for damping frequency. » y=step{sys. k = p l a c e ( F . s y s C l= s s (F-G *k. t s = 0 . Write the ouQiut obtained on execution of tiie program. the d e s ^ is verified usit^ MATLAB.1 ] .6 'Hierefbre.1 1 5 .p ) .0 ] . p = r o o t s ( [ l . 2 * ze ta* w n .01. we get ^ 's O . Step-by-step solution Step-by-step solution ste p 1 of 3 (a) Given tliat X = i:. 6 / ( t s ^ z e t a ) . Problem 7. % Tweak v a l u e s s l i g h t l y s o t h a t s p e c s a r e m e t. 4.4 . P = -1 0 ]. If it does not. we get ^ s 0.115 sec: *=[o -10 ]* + [? ]■ ■ • (b) Use the step command in Matlab to verify that your design meets the specifications. we use = 0. Step 3 of 3 Sketch the step response plot obtained. ® * [ 0 . For tile overshoot specification. Hence. J) s te p ( s y s C L ) . we get = ft 46 Thus. For tile lYo setting time specifications. 4 . . H.4.G . Z e ta = 0 .23PP (a) Design a state feedback controller for the following system so that the closed-loop step response has an overshoot of less than 25% and a 1% settling time under 0. G. w n * ^ ]). modify your feedback gains accordingly. and<JL = — ft Step 2 of 3 (b) Write the MATLAB program to verify the d e s ^ . we have = e'^ 1 Thus. wn = 4 . J = 0. H = [1 . xl x2 xl 0 1 x2 -9 8 0 3 -80 ul xl 0 xl x2 yl 1 0 ul yl 0 C o n tin u o u s -tim e m odel. K.6 s 0.994 *’ +3. K2 s + l + K^ Thus. r* I-F + G * : -G l -n H .50 The values of *. 0 -1 1 . 6 f ' . .i .73 2 0 5+ 1 -1 0 Thus 0 ( 5 ) is obtained as -1. » y=step{sys1.8 6 9 f ' 12.6 Substitute the value of 7^equal to 4. det * I .O lV l-f’ ) ' ( “>. thus j / = [! 0 0]x 5+0. (3) 4.+ * .r)’ 2 .31* + 10.269 0. 6 . modify your feedback gains accordingly. In (o . 0 -1 I .t).2 K j det 0 5 0 0 -1 I 0 0 5 l-K . » grid.0001:6. 3+ 2 * i+ A i= 3 . % O S= e^ Substitute the value of ovemhoot 5% in equation (2). j C (* ) d e t[ * I -F + G K ] Assume that y . ln5 (1.24PP Consider the system t (a) Design a state feedback controller for the system so that the closed-loop step response has an overshoot of less than 5% and a 1% settling time under 4.+ 4 l^ = 10.l + K.31 5 -2 * . » t=0:0. »d en = [1 3.83*’ + 10.462 2.19)»+(2.269.456)(2.6 4.368 1 1 0 0 0 Simplify the matrix and obtain G (5) • G (*) = 2*’ +0.50 = 0 Step 3 of 8 It is clear from the equation (1) that.375 5+2. Figure 1 Step 8 of 8 It is clear from the Figure 1 that the design meets the required specifications of overshoot less than 5% and the settling time of 1% under 4.1.50 Step 6 of 8 Obtain the step response for the system with transfer function r* (5 ) • Write the following code in MATLAB to obtain the step response. Steo-bv-steo solution Step-by-step solution Step 1 of 8 Consider the system shown in equation (1).79 s 0 Use ITAE criterion fro step input and calculate the third order characteristic equation. » xiabel t » ylabel y(t) Step 7 of 8 • The plot of the step response is as shown in Figure 1.2 . » sys1=tf{num.( F .6 4. det 0 5+1 -1 =0 [ . Consider the maximum overshoot 5% and calculate the value of damping coefficient ^ . + ffl. the determinant of the matrix is r + 5 ’ ( 3 + 2 i^ + ^ ) + 5 ( 5 + 2 ^ + i^ + 4 ^ ) + ( 5 . c 0 0 =0 0 0 5 I 0 -I K1 1^2 5 0 0 -1-2AT.50].50 = 0 It is clear that.456 in equation (3) and calculate the value of . » plot(t.+ 3 * 2 +3Ai = 10.2 ^ + 3 ic 2 + 3 ^ ) = 0 Step 4 of 8 Compare the determinant of the matrix with the characteristic equation of the system.6 seconds. 4V.31*+ 10.8 3 5 + 2 * .9 .375 1.0.994 Calculate the transfer function of the system. **+3.’ ■ 0 Ths.6 f = 0. s’ +3.006 -1.den).2 -2 ' '2 x= 0 -1 1 jr+ 0 -(1) 1 0 -1 1 (a) It is clear that the overshoot is less than 5%.19)* = 0 Thus.368]| - Step 5 of 8 (b) Calculate G (*) - . ■-1 .83 10. 5-M + 2A:.19 Write the second order characteristic equation for the system using the values of ^ and ■¥ »0 + 2(0. the characteristic equation of the system is + 2 f +4.456 Step 2 of 8 It is clear that the response of the closed loop system has 1% settling time under 4.3 1*+10.375.2 K 2 -2 .6 seconds. the state feedback controller for the system is [[-0.46^” . 2K.269 0.G K ) ] = 0 5 0 0 ■_1 _2 -2 '2 det' 0 5 0 . 4. 2K2T\ det 0 5 0 . » num=[2 0. .7.x ^ .*J.609)’ ( l .’ ) = (i-.368 Thus. 2+2^2 2+2K.y).75 6.2 .456 a>. si* +2 . If it does not.31 10. 0 1^2 ^ 3] =0 0 0 5 I 0 -1 1 5 0 o' -I -2 -2 ‘ [2*:.83*’ + 10.*^ are -0.006»-1. (b) Use the step command in Matlab to verify that your design meets the specifications.6 sec.994]. r '( * ) = G (4 1+ G (*)J9(*) 2*’ +0.006*-1. -1 -2 -2 F= 0 -1 1 1 0 -1 Determine the characteristic equation of the system with state feedback controller..24PP Problem 7.83*’ + 10.2 .6 seconds and the value of ^ equal to 0. the third order characteristic equation is. Write the formula of settling time equal to 1%. = 2. . (10) Write the state equation in control canonical form from equations (4) to (10). G(5)- ^ (4 = 7 ^ (9) Write the general form of state space equation.. jv = Ax+B » ... d e t(5 l-A + B K )= (5 + A :.. the state equations of controller canonical fonn are: 1 0 b = [i o]4 ste p 3 of 4 (b) Write the matrix A from equations (10) and (11). . « = -[K .. JC...... 1 0 I 0 0 ..= [A 6* .A + B K ) = j * + ^ ^ + 4 + ^ j . C (s) = (4) Write the state description matrices of controller canonical form....25PP Consider the system in Fig.= [1 0 ] * (12) Step 2 of 4 ^ 0 -4 Hence....2 y ) ( s + 2 + 2 y ) = 3 ’ + 4 s + 8 .. M (7) D^sSeperatevaluefromnumeratoranddenominatorelse 0 .....(6) 0 C .. »=[o] Consider the law of the form... Find desired characteristics polynomial of the system The two poles are placed at -2+ 2J and —2—2J- ( s + 2 ...si—A + B K ) from Equations (13) and (14) s 0 0 s s 0 0 -4 ] fjC. 1 0 .»+<!. Problem 7.. which will place the closed-loop poles at s = -2 ± 2j. from Equations (19)....... (20) The design of state-feedback gain matrix.. (8) Consider the value of gain.] (15) Find (. = s ' + 4 j + 8 if.. = . = +.)s -(-I){ 4 + /:j) d e t ( s I .. (1) Consider the denominator part of the gain. .......4 A : . -a . Write the general form of output equation...... = 4 .... AT. (17) Step 4 of 4 ^ Consider the closed loop systems satisfy the det(sI-A*FBK) =0 condition..j + 4 . (2) Write the general canonical form of the equation.... (18) Compare the coefficients of s from equations (17) and (18) 3 * + /f . (19) Compare the coefficients of constant from equations (17) and (18) 4 + JC j= 8 ^ j . R]-[: *■[! ^ ' * [ j ... -<h .... Figure (a) Write a set of equations that describes this system in the standard canonical control form as j = Ax + Bt/ and y = Cx.. s 4 |a n d |AT2 = 4 |. (b) Design a control law of the form « = -[* • which will place the closed-loop poles at s = -2 ± 2j.] 0 s 1 o j [o oj (16) Find d e t(s I-A + B K ) = 0 from equation (16)... m (3) a(s) Substitute equations (1) and (2) in equation (3)..... y = C x* D ^. ^=[> Write the matrix B from equations (10) and (11). (20) are [AT.... 0 1 0 0 (5) 0 1 ••• 0 c ...4 ....... Step-by-step solution ste p 1 of 4 (a) Consider the numerator part of the gain. Refer equation (3) and write the controllability equation at the output state.) ] = C C ji‘ (4) Where. (2) Where.j ( 0 ... y(0*f) andy(0-)a> 'e the output state variables at various interval.. A system is said to be output controllable if at any time you are able to transfer the output from zero to any desired output y* in a finite time using an appropriate control signal u*.( 0 + ) = C C j <‘ .. Assume the initial state to reauired final state has finite Step 1 of 3 Consider the linear controlling system. “ '= [* 1 Where. it can be concluded that state controllability express the o u ^ u t controllability but output controllability does not express die value of state controllability. SitSv'SM scalars up to n*order. x (0 + ) a n d x (0 -)a re the input state variables at various interval... [ y ( 0 + ) .. C) to be output controllable. C [ x ( 0 + ) . Consider the continuous system (A ... the state is measurably small while using impulsive inputs. Therefore.. how? Step-by-step solution ste p 1 of 3 Consider the linear controllina svstem. Write the input equation at the initial state for the control signal n*.. Are output and state controllability related? If so. II* is the control signal at infinite time. Assign x is the input at initial state and y Is the output at the final state. C is the inverse of controllability matrix. x ( 0 + ) . In many situations a control engineer may be interested in controlling the output y rather than the state x.... Where.) = C jf* . (3) Where. the system is controllable in the output only if the condition ^CB C A B . it is required to consider the inverse of the controllability matrix C in the equation.C ) . Assume the initial state to required final state has finite time and with finite input.... Write the control signal (u) equation with respect to scalars as considered. For this case.B ...) = 0 . Therefore. Problem 7. CA**'B Jis in full rank. If there is no loss then y ( 0 ... Step 2 of 3 Now.. C is the controllability matrix. y ( 0 + ) = [ C B C A B . ( 6 ) Here. In the equation (2). . assume to achieve the controllability of the state at the output. and write the controllable equation for this continuous system at the output state. B. The controllability matrix C is necessary to achieve the controllability of the state. the equation for the controllability at the output state is y ( 0 + ) = [ C B C A B ..26PP Output Controllability. Hence.. C A ’ ' B ] ii‘ | Step 3 of 3 Equation (6) is always true for SISO (Single Input Single Output) system if the transfer function greater than zero.. (5) Equation (5) gives the controllability equation at the output state. the control signal n*will move the system to arbitrary state when the controllability matrix C is non-singular.> ’( 0 . . equation (3) gives the design to achieve the controllability at the Input state.C A " ^ B ] « ' . Write the mathematical equation for the control signal {u) with respect to time.... is the derivatives of the delta function up to order.i ( 0 .. Derive necessary and sufficient conditions for a continuous system (A. Assume there Is no loss In the output and rewrite equation (4).) ] = C C ji ‘ .. Therefore. 27PP Consider the system 0 4 0 0 ■ ■0 ■ -I -4 0 0 0 s 7 15 x+ 0 0 3 -3 0 _ (a) Find the eigenvalues of this system. and -2. (c) For each of the uncontrollable modes. find a vector v such that vTB = 0. G et help from a C hegg subject expert. {Hint Note the block-triangular structure of A . (c) For each of the uncontrollable modes. Step-by-step solution There is no solution to this problem yet. ) (b) Find the controllable and uncontrollable modes of this system. -2. v T A = /\v 7 . (d) Show that there are an infinite number of feedback gains K that will relocate the modes of th system to -5. find a vector v such that vTB = 0. (e) Find the unique matrix K that achieves these pole locations and prevents Initial conditions on the uncontrollable part of the system from ever affecting the controllable part. v T A = /\v 7 . ASK AN EXPERT . -3. Problem 7. e j 0 1 0 0 ■ 0■ -1 0 0 ml^ ml x+ 0 0 0 1 0 Ka^ (K<^ a\ 1 0 . = —ka^{92 —0i) —mgl92 + Ih< Figure Coupled pendulums (a) Show that the system is uncontrollable. Can you associate a physical meaning with the controllable and uncontrollable modes? (b) Is there any way that the system can be made controllable? Step-by-step solution ste p 1 of 3 (A) Using the state vector. coupled by a spring. 0 ml_ Step 2 of 3 The controllability matriK is determined as. Step 3 of 3 (B) YES. Problem 7. which are applied to the pendulum bobs as shown In Fig. are to be controlled by two equal and opposite forces u.9 i ) . The equations of motion are H^9i = -k^(6i .28PP Two pendulums. III. X > 01 0.mgl9\ . FO F ^a -1 1 ( K a '^ a \ K a ‘ 0 — 0 ml ml 1) nFF -1 „ 1 (K a ‘ o'! r a ’ 0 ml C = 1 „ 0 — 0 ml HFF 1 1 a 'l 0 ml fn t m l \ ml^ 1J a cc p . . (b)A student has computed det C = 2.282xlQ-’ | .= [l 0 0 0 0] +0 ^ = [1 0 0 0 0 ] * + 0 .0195 -0.174 0 0 0 0* -0.645 0 0 0 -0. 0.0195 A*B = -0.... value of |det|C| = -2.09x10’’ -0.035724...174 0 0 0 O' ■-0.(17) CA’ CA’ Step 7 of 7 Calcuiate CAftoni equations (3) and (5). C CA CA’ . -0.174 0 0 0 0 0. 0.174 0 0 0 0 0.0287 -0.005 0 1 0 0 0 JC4 0 U ■■■■ ■(7) 0 0 0 0 0 0 0 0 1 0 0 Write the output equation in control canonical form from equations (2).005 -0... (15) Substitute equations (3) and (14) in equation (15).645 0 0 0 -0. ( 22) 5.005 -0.174 0 0 0 0] 0 1 0 0 0 0 0 1 0 0 0 0 CA’ = [0..035724 -0.157 0..3x io -7 and claims that the system is uncontrollable.030276 0 0 0 0 ..0195 -0.645 0 0 0 CA’ =[5.005 -0.267x10’’ 0..0195 . 0..035724 -0..174 0 0 0 0.035724 ABs -0.2828e-07 Hence.8966*10''-4. 0.09x10’’ 0.035724 0 0 I 0 0 -0. .645 0 0 0 -0.174 0 0 0 0 0.035724 -0.005 ( 10 ) 0 0 Step 5 of 7 Calculate equations (3) and (10).174 0 0 0 0 -6.005 0 0 0 I 0 0 -1. The student computation is wrong..0287 A*B = 0 1 0 0 0 -0.( 8) ste p 2 of 7 ^ Draw the block diagram for controller canonical form from equations (7) and (8). 0.. 0 0 0 -0.(11) Substitute equations (3) and (10) in equation (11). (19).0195 A*B = 0 1 0 0 0 -0.174 0 0 0 O' -1..157 0.005 Bs 0 (4) 0 0 Consider the matrix C. det|O|=0 Hence.645 0 0 0 -0.267x10’ -0..207' 0.645 0 0 0 -0. (18).174 0 0 0 o' ■-0.005 ABs 0 1 0 0 0 0 0 1 0 -0.0287 .645 0 0 0 -0.645 0 0 0 -0.005 -0.0287 0 0 1 0 0 -0.267x10’’ -1.01275 A^B> -0. This value is not equal to zero..035724 -0...157 0.036018 -6. ’0.036018 0. Is the student right or wrong? Why? (c) Is the realization observable? Step-by-step solution ste p 1 of 7 (a) Write the general form of state space equation.267*1 -1.157 0. 1 0 0 0 0 0. A’B = A‘B xA . Figure 1 ste p 3 of 7 Hence. Is the student right or wrong? Why? (b)A student has computed det C = 2..(16) -0. value of det|O| = 0. Step 6 of 7 (C) Determine the system observabiiity O- Write the observabie condition from description matrices. 0. (5) and (6). So this is controllable system..005 -0..’ b ] .... -0..3x 10-7 and claims that the system is uncontrollable.030276 0 0 0 0] 0 1 0 0 0 0 0 0 CA’ =[5. '0.0195 0 0 -0. x = .005 Calculate a ^B ^^^ equations (3) and (14).157 0... 0 1 0 0 0 -0.174 0 0 0 0 0 = 0.09*10^-3 -1..035724 0 0 0 1 0 -0. ans = -2.. *0.035724 -0.( 21 ) Substitute equations (5). A 'B = A B x A .268x10” 0 0 0 O] 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 CA’ = [9.166x10’ * 0 0 0 o ]..01275. (a) Draw a block diagram of the realization with an integrator for each statevariable..0287 -0.157 0.036018 -0.035724 Write the MATLAB program to find detennination of Matrix C- A=[-. (13) Substitute equations (3) and (12) in equation (13).207* 0.. Problem 7. A .0287 -0.157 0. det(A) The output of the MATLAB program is given below.035724 -0..(20) Calculate C A ^^°^ equations (5) and (20). (1) y = C x + D .207 -0.So this is (not observable system)..8% 6xl0-* -0.0287 -0..174 0 0 0 0 -0..207' ■*1 0. (20) and (21) in equation (17).645 0 0 0 1 0 0 ■(3) 0 0 10 0 0 0 1 Consider the matrix B. (3) and (4).005 -0.005 -0. *0... 0 -0.005 0 1 0 0 0 4 B= 0 0 0 1 0 0 0 0 0 0 1 0 0 C = [ 1 0 0 0 0 ]...207 0.174 0 0 0 0 ■-0. 0.0287 A 'B = -0.157 0.09x10’’ -1.01275 0 -0.268x10’’ 0 0 0 0 O.lOOxlO” 0 0 0 0 Determine the determination ^matrix from equation (22)..005 0 Calcuiate a ^BIiiiiii equations (3) and (12).035724 Substitute equations (4). (2) Consider the matrix A.207* -0. 0 0 -0.005 -*» = 0 1 0 0 0 + 0 ■«4 0 0 1 0 0 0 0 0 0 1 0^ 0 0.035724 -0. (9) Calculate AB^rom equations (3) and (4).8966x10"* -0. (12). (14) and (16) in equation (9)..035724 ( 12 ) -0. (10).0287 0 0 0 -0. the block diagram of realization with an integrator for each state variable is IdrawnI - Step 4 of 7 (b) The general fonnula for controllability matrix. .(14) -0.005 -1.157 0..0195 ...29PP The state-space model for a certain application has been given to us with the following state description matrices.268x10"’ 0 0 0 o ] .645 0 0 0 CA’ =[0.036018 -6.645 0 0 0 C A -[1 0 0 0 0] 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 CA = [0. C is. A’B = A’B xA .157 0.645 0 0 CA’ =[0...005 -0. C = [l 0 0 0 0] (5) Consider the matrix D.157 0.0287.174 0 0 0 0..0287 -0.005 I 0 0 0 0 1 0 0 -6. (18) Calculate equations (5) and (18).1S7 0. O = [0] (6) Write the state equation in control canonical form from equations (1).035724 A ’B . C =[b AB .].207 -0.174 0 0 0 0 ] . ’-0..030276 0 0 0 O] (19) Calculate equations (5) and (19). (a) Find the state matrices Ac.. B. Step-by-step solution There is no solution to this problem yet..an-1 ^ 0 for some /. Problem 7. G et help from a C hegg subject expert.. «l . The name of the algorithm comes from the multi-input version in which the ai are the blocks that make A resemble a staircase. where A is an upper Hessenberg matrix (having one nonzero diagonal above the main diagonal) given by ' * ai • O ' ■0■ a = t ’'at = * * '. as in Problem? This algorithm can also be used to design a numerically stable algorithm for pole placement [see Minimis and Paige (1982)]. and C = CT = [ c . (d) Give the state matrices In observer canonical form and repeat parts (b) and (c) in terms of controllability instead of observability. T U T r Orthogonai transformations correspond to a rotation of the vectors (represented by the matrix columns) being transformed with no change in length.. Refer to ctrbf.. fat Prove that if ai =0 and ai+1. c... and Cc in control canonical form that correspond to the given differential equation.B. then the controllable and uncontrollable modes of the system can be identified after this transformation has been done. (b) Sketch the eigenvectors of Ac in the (x1. an-^ # 0 for some /. B = T’'B = * ♦ Oj. (b) How would you use this technique to identify the observable and unobservable modes of (A._l 0 ..C) can be transformed by an orthogonal similarity transformation to (A. Problem Consider the system y+ 3y+ 2y= u+ u. obsvf commands in Matlab. C). (c) Express xO and xCin terms of the observability matrix O. 1978):Any realization (A. 0 . and draw vectors that correspond to the completely observable (xO) and the completely unobservable (xO) state-variables. x2) plane. Cl . 1. (a) Prove that if ai =0 and ai+1.C)? (c) What advantage does this approach for determining the controllable and uncontrollable modes have over transforming the system to any other form? (d) How can we use this approach to determine a basis for the controllable and uncontrollable subspaces. where g1 # 0. then the controllable and uncontrollable Orthogonal transformations correspond to a rotation of the vectors (represented by the matrix columns) being transformed with no change in length. Be.30PP Staircase Algorithm (Var) Dooren et al. B. ASK AN EXPERT . * = [13.You may wish to use the initial command in Matlab. (d) Compute Nx and Nu for zero steady-state error to a constant command input on the cart position.O].= 1 -p ^ = [12 16 12] Step 2 of 4 B) /f = [0010] % Symmetric root locus i= [ O . assume that ^ = 0.You may wish to use the initial command in Matlab. For parts (b) through (d).31 PP The normalized equations of motion for an inverted pendulum at angle 0 on a cart are where x is the cart position. and the controi input u is a force acting on the cart. Step-by-step solution ste p 1 of 4 A) iie t[ S I . Problem 7. (c) Compare the responses of the closed-loop systems in parts (a) and (b) to an initial condition of 0 = 10°.5. c = [0*/?. (b) Use the SRL to select poles with a bandwidth as close as possible to those of part (a).____________________________ (c) Compare the responses of the closed-loop systems in parts (a) and (b) to an initial condition of 0 = 10°.9 13. b. and find the control law that will place the closed-loop poles at the points you selected. d= 0 vlocus (a.0 ’G]. = 0 .5 18.{ S ) Con^aring coefficients yields _ 10-8P 1 -p 1-P 6 A ..98] .36 3. c. (a) With the state defined as X—[ ^ ^ ^ ^ jT find the feedback gain K that places the closed-loop poles at s = -1. d).-1.-1±iy.F + a K ) = a . 0) Step 3 of 4 ^ C) The initial condition response to 6 (0)s10°for both control design in (a) and ( i) Step 4 of 4 D) i Yields W i W^=[0 0 1 o f AT. and compare the step responses of each of the two closed-loop systems. = [ c a -* b ] Step-by-step solution ste p 1 of 13 Refer equation 7. Since /> 0 apply Taylor’s series and rewrite the equation.. Step 10 of 13 Apply final value theorem and write the equation. (6) Step 9 of 13 Consider ramp Input is given to the system for infinite time Interval. . e .... Show that the velocity error coefficient is given by i..] ^ 7 j .) = 1 (j I . D = 0).j^ jj «-»0 Step 12 of 13 The velocity error coefficient is defined as the ratio of the input rate to the steady state error... K = w (") Step 13 of 13 Substitute the value of e ^ (oo) in the equation. ] ^ ^ jj £ ( j ) = [ l + CA-'B+ sCA-*B+ i'C A -’B + . r(j)=C(iI-A)'B+D ste p 2 of 13 Substitute rewrite the equation..jA-* . ] ^ ^ j *lim j .A ' .. ] ^ ^ r-*0 j slim ^ j+ ^ C A -'B + a C A -’ B +o!C A-’B + . (3) Step 6 of 13 Consider r ...'.'B . -.i'C A .45 in the textbook and write the closed loop transfer function of the system. £ ( j)= [i-r (j)] « (j) (5) Step 8 of 13 Apply equation (3) and (4) In equation (5).32PP An asymptotically stable Type I system with input rand output y is described by the closed-loop system matrices (A.. -------^ oCA-'B 1 "C A -’B -[C A -’B]"' Hence the condition I f .)]b r ( j) = [. . C. . Suppose the Input is given by the ramp r = at. . . (2) Step 5 of 13 Apply equation (2) In equation (1).] .s^A-’ . K . .. A is Invertible.. ) . -[C A -’B]"' is proved .A ) “'B (1) ste p 3 of 13 The system is asymptotic stable. for f > 0.. So in equation (1). Mathematically it is expressed as. [ l + CA-'B+ iC A -’B +«’CA-‘B + ..t .. B.j A .j'A-* -.A ) '' ■ (.» C A .'. S te p i of 13 Refer equation 7. Step 4 of 13 ^ Consider the term (rf-A )-' ..A .. Problem 7. £ ( j ) = ^ l-[-C A -'B -jC A -“B .a J and convert the equation to s-domain. ...45 in the textbook and write the closed loop transfer function of the system.’B .. *(4= 7 w Step 7 O f 13 Consider the general emor equation for the transfer function..A ) ( . (j I . [ l + CA-'B+ iC A -’B + i’CA-’B + ... r( j) =c[(-A-' . Step 11 of 13 Substitute equation (6) in the equation and apply limit to the equation.= B .j'C A ^ B . r(s) = C ( j I .C A .. foo) = lim j . write the loop gain as sum of its real and imaginary parts L ( ja ) = K ( ja l.w e get = | D ( > ) f [ l + p |0 ( > ) f ] But |i+ i< (> B i-if)‘‘m|&i Let us re . K. Hence the LQR gain matrix.(if .GJi:]| = dct (51.p y 'o = Re (£ [ j a ) ) + j l ^ (£ [jd>)) Finally {[Re (£iy (b)] + l} ’ + [in [ l (jffi))f a 1 . can be multiplied by a large scalar or reduced by half with guaranteed closed-loop system stability. Problem 7. (-5) = i3 (5 )fl(-5 )[l+ ^ (5 1 -^ )‘‘o] x [l+ i:(-5 1 -if)‘ ‘a] = l+ C 0 .(-5 ) Step 2 of 2 SettingS = jQ .if ) dtt | l + (51 .F .33PP Prove that the Nyquist plot for LQR design avoids a circle of radius one centered at the -1 point. (S) = det [5 1 .(5 )0 .if [l + (51 . Show that this implies that < GM < the “upward” gain margin is GM = and there is a “downward” GM = Vz.if )■' os:]) = I3(5)[l+ i:(51-if)‘"]a a . as shown in Fig. Figure Nyquist plot for an optimal regulator Step-by-step solution step 1 of 2 We have a . (S)a.Gi*:)] = det 151. and the phase margin is at least PM= ±60°. the value of l . Write the observable condition from description matrices. ..f t f t ] .. + 2 J : . .:]-[? .. Hence.[ i] [ A .( 12 ) Find from equations (11) and (12). C) is unobservable.]•[.[I 2 ]([-’ j] ..A ) ‘' B (14) Substitute equations (1). (13) Hence. G „ (j ) = C ( j I .B K .] C A -[0 1] (7) Substitute equations (3) and (7) in equation (6)..1 AT. C rf(i) = C ( f I . °=[^l Find C A froni equation (3) and (1). < . l . (c) Compute a K of the fonn K = [1. TH i ■ t' ^ li'+ 2 i. (2) and (3) in equation (16).ll0 C „ ( s ) = [l 2] *"+ 2 5 -1 i+2 Write the general form of closed loop transfer function of the system. that is.B K ) from Equations (1). (5) Determine the system observability Q. n =>■ 4 : :]-[t . [ . ...] (9) Substitute equations (3) and (9) in equation (6). = 0 .]-P '■]) -t' C " ' ".. — 2 .[ ? . f t ] .4 '^ . C = [l 2] (3) Consider the law of the form.]i. . the O is nonsingular. *-[? . (2) and (3).A : . the pole is [cancelled w ith o n e 2ero| in unobservable system.]. Problem 7.. that is. Hence. C A -[...34PP Consider the system *= [r' ]•«'=“ and assume that you are using feedback of the form u = -K x+ r.iC . the closed loop mode is unobservable from the output function.J ™ Determine the det(0 )fro m equation (10). (d) Compare the open-loop transfer function with the transfer function of the closed-loop system of part (c). where r is a reference input signal. find H2 so that the closed-loop system is not observable.i Write the matrix B.4"’ ft] [t-t-2 -2 + 2 5 + 6 ] f"+ 3 s + 2 fj + 2 2 i+ 4 ir i' j ’ +3j + 2 [o j+ 2 j* + 3 i + 2 j+2 ‘ (4 + 1 )(4 + 2 ) (18) (4 + 1) Compare equations (15) and (18).A + B K ) ’ ' B (16) Substitute equations (1)..'■]) c (A -B K )= [-^ r. “ -[o] Write the matrix C.BK. the value of Jf.( 10) Consider the value of d e t(O ) is zero. 4 . d e t(0 ) = I . ] . * 2 K .. find H2 so that the closed-loop system is not observable.th e n (A ..]-[? 4 . 1 -^ :. and 3 for in equation (17).... C) is lunobservabtel Step 3 of 4 ^ (c) Consider the value of Jf.C) is lobservabiel- Step 2 of 4 ^ (b) Find C ( A . K2] that will make the system unobservable as in part (b).>-[. + 2 X .[.. (b) Show that there exists a K such that (A . What is the unobservability due to? Step-by-step solution step 1 of 4 (a) Write the matrix A.at unobservable system is [ ^ . (a) Show that (A. » -[. (11) Hence.i C ..h . (A.. • . (17) »-<. C) is observable.K . Modify equation (4).] . = 0 ..... AT.. (2) and (3) in equation (14). Step 4 of 4 ^ (d) Write the general form of open loop transfer function of the system.3 . 1-A T .. ] Substitute 1 for Jf.+ 2 (1 ) = 0 .]) . C (A -H K ). 2 ]p . 4 [ . (c) Compute a K of the fonn K = [1.]i* . » = -[«.:] »> From equation (8). :]-P 31.. <•[. -U . K2] that will make the system unobservable as in part (b). .. = ( i ■ ^ 9 ^ :. (35) [-9AT... K ) = ( j. -818T. 3 c .3 —3y- (i-F 3-3y)(s-F3-F3y)=i’ . (e) Write the MATLAB program to verify the value of full order estimator L at —12±12y poles.|.....^ ^ ( 21 ) Find det(5l-A -l-B K )sO from equation (21).-F9A :. ( A ...B . .. det|0| = I .9 Write the state description matrices in observer canonical form from equations (4). det|0|=l Hence.. Bo..A .. L" S-I-9J:. 1^ . value from equations (9) and (29)..a. Bo) controllable? (c) Compute K so that the closed-loop poles are assigned to s = -3 ± 3/.. —B^Kfrom equations (30) and (31)..9]: C=[1 Oj: D=0. [numt.. Po=10. ■l■9A:.^-^9^:. (6) Step 2 of 9 Write the state description matrices in observer canonical form.»-f9A :. m Write the observer canonical form of the state space equations.. Co) for this system in observer canonicai fonri..K x + r. ‘’• [ c l ] " Calculate C^A^from equations (14) and (16). Consider the value of det|0| is zero.z): r=roots(dent): poles=[-12■^12*i-12-12*1]. Po=10.l + 9AT.. .( ^ + 9 A .H 8 ... d e t ( s l . a { s ) = s " + 0.9]: C=[1 0]: D=0.9 A : j . ) ( * -I-9AT.b .J a Calculate the value of C . R ]-[: •fo n r®i [9 « h?+ « w . =Seperate valuefiomnumeratoraiiddeiiominatorelse 0 (10 Consider the value of gain G {s)- (11) Write the state equation in observer canonical fonn from equations (1)to (11). Step 5 of 9 (c) Consider the law of the form.. from Equations (14). . Problem 7.’B . i .. Ts=1. y = C x *D ^. (36) If detjO jis zero. y = C ^ x ^ + D . ] ( .. ‘ ’ = [0 “] Determine the determination ^m atrix from equation (28).A .K )fro m equations (30)and (31).. .= = 0 0 1 0 (7) ■ 0 0 0 1 ..A .9 0]: B=[0..dent]=ord2(wn. C .9 A r . [Again assume an observer canonical realization for G1 (s)..B. L=acker(A'. the reason for the unobservable system is | . the value of A„ = |C .dent]=ord2(wn.. ) ( ..] (34) Modify observable condition in equation (26).poles)' The output of MATLAB program: L= 24 297 Thus the values of full order estimator at —12±12ypolesare =24|and \L^ s 2 9 7 |.. .. C ■ C = [b. ) .. ( a .B K ) = [_ .i . value of l«‘ » t|o |= o | .] Step-by-step solution step 1 of 9 (a) Consider the numerator part of the gain. So this is [obsCTvable system|. z=(-log(Po/100))/(sqrt(pi'‘2-^log(Po/100)'‘2)): wn=10*wn. (d) Is the closed-loop system of part (c) observable? (d) Is the closed-loop system of part (c) observable? (e) Design a full-order estimator with estimator error poles at s = -12±12/ (f) Suppose the system is modified to have a zero. = ..f B .H ) . ’ (4 + 1) (29) c(*)= i '.(9) £)....K ) = 0 s 9 0 9K..+ B . (23) Compare the coefficients of s from equations (22) and (23) 9 A j= 6 A .f 6 j..(30) . « o ■(37) Determine d e t(5 l-A ^ ■ !‘ B ....35PP Consider a system with the transfer function W = (a) Find (Ao. K ) = j ’ -f9/:.K ) = p s 9 0 s 0 0 r 0 0 (rf-A . ^^=[9 3 Determine the value of det|C(from equation (19). A . .u n .K ) from Equations (14) and (15) ( .+ b . So this is controllable system..C'. 1-91C J Calculate det|0|from equation (35). Write the value from equations (5) and (12). (39) Step 9 of 9 From equations (39). .(i) = j2 -9 ■ Prove that if u = . (8) and (29).. • ... ) = * ’ ■l■9A:. = « ’ + *-F 9 ir..9 + 8 l | + 9 i r .. = 3 and A K . 9K. C . (1) Consider the denominator part of the gain. (5). value of |det|C| = -^ I| ...poles) The output of MATLAB program: K= 3..= (31) Write the C .[j ®]-[® 5+9AT. Substitute equations (15) and (18) in equation (17). 1-9AT. *•“ •= [9 0. fo l ■(15) Write the C .. s 0 0 I 0 (rf-A .= [ l 0 0 ••• 0 ] .^-t-SlA.. -.K ) = . Find desired characteristics polynomial of the system The two poles are placed at -3 + 3 y ’ and .-B . Oj [9 j Write the output equation in observer canonical form from equations (1) to (11).z): r=roots(dent): poles=[-3■^3*i -3-3*1]. [numt. 0 1 0 0 A ..[ 0 1] (27) Substitute equations (16) and (27) in equation (26).= [1 0 ] * (13) Step 3 of 9 Write the value from equations (5) and (12). -o .+ B .» + 9 X .[ : .. K ) = « '+ 9 ^ 5 + 9 8 :. (5). b { s ) = b ^ ^ + b 2 S ^ ^ + . A .A . ) ( * .. k ) = [ ^ a :.i..i -j -i-98:.l+ 9 A j =det[' L-9-f9AT.A . K=acker(A. d e t ( jI ..i.. .i Write the state description matrices in observer canonical form from equations (4).0000 0. (25) are AT..s-9-F 9A :.This value is not equal to zero...l v a lu e o f pole is cancelled b y zero o f th eg ain G {s)\- . 1 0 ••• 0 -a .... Ts=1. = 3 .. c ....K) =det[.8r..K . ..• “ ■ K '.6667 Step 7 of 9 (d) Determine the system observability Q- Write the observable condition from description matrices.B . d e t ( j I . 9(5-H ) G.-1 8 A. (38) Substitute equation (37) in equation (38). +r (20) Find -f B ... det(*I-A. . ....j ..i.] (17) Calculate A.. = | . (24) Compare the coefficients of constant from equations (22) and (23) -9 + 9 A .... A= [0 1.. oJ ' “ 1 .= [1 0]| 9 0 j| 9 step 4 of 9 (b) Write the general formula for controllability matrix. ^ ( 5 )- ... .$ '^ + .^.B ..K )] Write the observability matric from equations (34) and (35).... (5) Write the output equation... -9+818T.9 a: . (7).= [ l 0] (16) 0 Hence.9 Oj: B=[0.+B... ] C .+ a ^ (2) Write the general canonical form of the equation.3I Step 6 of 9 Write the MATLAB program to verify the value of K: A= [0 1.= [l 0] (32) Calculate the value of A .B^K)from equations (32) and (33).. . Step 8 of 9 (f) Consider the value of gain. (b) Is (Ao.. 0 . value from equations (9) and (13). B . (25) The design of state-feedback gain matrix. + B . a = -[Ar.[. = j ' .. the system is unobservable.(2 2 ) Consider the closed loop systems satisfy the det(5 l-A * F B K ) = 0 condition.9 ..i-9-F81A r. d e t(jI-A -fB K )=»*-i-9A :.. c A -(i • ( .B^from equations (14) and (15). Substitute equations (1) and (2) in equation (3).= (8) C . and (29).i . .. det|C|»-81 Hence. there is a feedback gain K that makes the closed-loop system unobservable... z=(-log(Po/100))/(sqrt(pi'‘2-^log(Po/100)'‘2)): wn=4/(z*Ts)...( A .i-9-f 9A:. 0 B .A . (35) ’ [c . If the knowledge of the output' Y’ and the ii^u t u (t) over a time interval o f time (O. A system is said to be completely controllable. Step 2 of 2 For the Linear system given by X { t) = A X {t) + £ U ( t) 0) y ( t) = C X {t)+ D U {t) Ifthere exists an input u (0 . Otherwise. the system is aid to be unobservable. These properties p l ^ an in:portant role in modern control system apoic ano a zero m uie system tiansier [unction. I f all the initial states are controllable (or) simply controllable.^ ). if every state can be affected or controlled to reach a desired state in finite time by some unconstrained control u (t). observable.^) suffices to determine the state X ( 0 ) . . Step-by-step solution step 1 of 2 The concepts o f controllability and observability are closely related to the cancellation of a pole and a zero in the S3rstem transfer function.AT(O) to the statX ’ in a finite time the state X (0) is said to be controllable. Problem 7. This subsystem is called an observer whose design based on observability property o f the controlled system The platn is said to be conq^letely observable if all the state variables in X (<) can be observed from the measurements o f the output d(^). need a subsystem that performs the estimation of state variables based on the information received fi'om the input u(<) and output . Otherwise.36PP Explain how the controllability. The Ii^ u t and output o f a system are always physical quantities. The state X (O) is said to be ob servable. These properties play an important role in modern control system design theory. And the ii ^ u tu ( ± ) . and are normally easily accessible to measurement we therefore. I f all initial states are observable. which transfers the initial state .y(<). it is said to be uncontrollable. the system is said to be con^letely observable or singly. observability. Controllability and observability are the properties which describe structural features of a d3mamic system. For the Linear System given by equation (1). Controllability and observability are the properties which describe structural features of a dniamic system. and stability properties of a linear system are related. e0 I?C ~ C LC -R. +/? [a .c . (b) What condition(s) on R.i >0 step 7 of 7 ■-2^ -V L c H’ = 1 0 . L. c . -R ^ C -L -¥ 2 R ^ C ^ . H = [-R 0] j =R -2 R r R' L L L FO-. + V c + ic R = 0 where ic = “ -*s Writing the above eq i»tioa In terms of state variables. R }> \ L . and the output y is a voltage.x ^ R . Cdvc U = tr + ------- ‘ dt u = X i+ C i^ :. and C will guarantee that the system is observable? Figure Electric circuit Step-by-step solution step ^ of 7 ^ I Step 2 of 7 As there are two energy storage elements. L 2R^' L -R L .XiR = 0 . 2/? r L _ R .X2 = — + — (1) ^ C C Step 3 of 7 Applying RVL. Step 5 of 7 -2R r 'R ' L L L F= a= 1 =1 0 Lc J . LC Step 6 of 7 i T = [a jio ] ] 1 LC LC As for the system to IV be VC completely controllable.L x i +X 2 + R u . (2 )& (3 ) Xi . The input u(t) is a current. L. at least two state variables are required to describe the system Let ^ i= h X2=V(j at node. c C. (a) Write the internal (state) equations for the circuit. -2i? _^1 (?) ‘ L ' L ' L and y = ic ^ = u R . Problem 7. 1 2^ ^ K — K + A ^ 0 rC . and C will guarantee that the system is controllable? (c) What condition(s) on R. Let x1 = iL and x2 = vc.37PP Consider the electric circuit shown in Fig. -R ^ 1 .22 r R' L L L Xj — s. 1 0 .iiR = u R ~ XiR (3) Step 4 of 7 From Equ(l).c .i ^ ] = 0 . F 2R} L L d e tr= — > 0 . -1 — ^ 0 UJ 1 (4) . f = 1 X 1. y = 1 x1.. 'A -B K ir B l [lc A -L C -B K J L B AB = f " » .LC .. (21) in equation (20).L l'c J . S tep-by-step solution step 1 of 5 (a) Refer figure 7... (6) and (7)..“ .( A . [b a b .. s A z -B K x ^+ B r i.. ] . Where. I is unit matrix...2BK) j + A '. •(.. (10) Write the general formula for ^m atrix. (5) and (8) in equation (11). (15) Where.ALC .L C . (23) Therefore the value of Mc\=o\ Hence..A + L C ) .(17) (17) Write frite the new coordins coordinate system from equations (12). (21. the state equations of the system are LC A -L C . -_ j_ ri o ifA -BK iri oi . r(j) = C [x I.... Substitute equations (5) and (9) in equation (13)... C = C T ...y=[C 0]z (19) The general fonnula for controllability matrix.. Write the general state space equations. . where T= ...---------. i] C:=[c = [C 0]] ..[ ! ?] ■ Show that the system is not controllable..... The system state is -[?]• and the dimensions of the matrices are as follows. if* ® (14) = i. c = fB .38 (a) Write state equations for the system..L i -I Consider nsider the value of |3 is equal to one.B K 0] Step 2 of 5 (b) Consider the system state x . C'') Step 3 of 5 Substitute equations (4)..38PP The block diagram of a feedback system is shown in Fig... B = T‘‘B . Problem 7..... (15) and (17) in equation (24).l c ..r c m I ”0 (x -A -^ 2 B K )| ‘ ^(x’ -(2 A -L C -2 B K ) x+ A*-ALC-2ABK + 2BKLC)LoJ ( j . .B K ) J 'B (24) Substitute equations (12).[i •■[I] I” C = [C 0] .b k J [i -iJ _J_r AI -IB K ¥l 0] “ i4 a i. (3) '■[: . the transfer function of the system is..---------------------------.91 in the textbook and write the state equations of the system. (15) and (17).ALC .. (c) Find the transfer function of the system from r to y. Step 5 of 5 (c) Write the formula for the transfer function of the system. * .i J [l c a ....i Refer figure 7. x = 2/)x 1. A . r(s) = C [ l I .b *k J Substitute equations (9). Figure Block diagram for Problem 7..C®> Step 4 of 5 Substitute equations (4) and (10) in equation (13).... K = 1 X n. det|C| = 0 ...'A T .I 'B K I"B K I ” i * [ a i ’ . Write the general formula for q . BC(x-A+LC) r(x)= (x*-(2A ... C = 1 X n. L = /?x 1.a i + l ic + b ik J [i .’b ] ..(6) ^. (18) * =[ 0 A-LcJ* ...= [C (2) A -B K Hence. C 'S- C = [b AB .{ A -B K )J 'B [ ( j.. A = /?*/).L C -2 B K ) j + A '-A L C -2 A B K + 2BKLC) r f A ________________ BC(x-A+LC)_______________ ^ ' (x‘ -(2A -L C -2B K )x+ A ‘ -ALC-2ABK + 2BKLC) Hence. (7) Write state description matrix from equations (1).I ^ B K I’BK "I 0 A l’ .l i “c . (b) Let X = Tz...... (2)... X is transformation matrix..B K ) x .L C x + ( A . the system is uncontrollable and the system is decomposed in to controllable and uncontrollable parts in equations (18) and (19)... ' ‘ l ..2B K )i + A '. y = C x. i = A x + B r ..... >. B = /?X 1. + Br -[.!■ ' ' ‘ "(j ’ -( 2 A .iJ J^r A I'.b ’k J Calculate detjClfrom equation (22).A + LC) 0 = [C 0 )7 :7 - a 0 ( i -A * 2 B K ) ^( x '-(2 A .LC ..91 in the textbook and write the output equation of the system..l c . (20) Calculate > ^ fro m equations (8) and (9).... fA -B K BK 1 r ... x = Tz .B K BK 1 ( 12) "[ 0 A-LcJ Write the general formula for g .. [A .. [ a b .= C x .ib k .a i * + l i ' c A I“ . A = T .2ABK + 2BKLC) .L I* c J J^F a I ’ . (22.2ABK + 2BKLC) B (x-A +LC )l [--------.. L J « ^ C + l+ s C + J ^ C + C ^ l. (c) Interpret the conditions found In part (b) physically in terms of the time constants of the system. ^ _ jj[C (i^ -l)-£ Bi B^LC B ^B ^LC -B jC + B f. fl 1 1 „ 1 0) step 5 of 10 From Equations (1) & (5). 1 .B ^ L . B }LC Step 7 of 10 For the system to be uncontrollable. a 1' R^C R^L A .1 I ^ ■ Jt. system. C + jj^ £ C + l.F : d e t[fi/-F ] DENQMINATQR: ___l_ ~R^C ^ 1 -1 R.C + + s j^ £ C + j! i^ Z .39PP This problem is intended to give you more insight into controllability and observability. Figure Electric circuit Step-by-step solution step 1 of 10 y«) Step 2 of 10 From the above circuit.. c [ r.[W A cJ ' -1 B. (d) Find the transfer function of the system.iii+ £ s + /^ Z .A UJ 1 (9) L R^L ^ ■[-i ']H 5} Step 6 of 10 1 0 1 R^C R^C RG= 1_ 1 1 -A L R^L . fe )’ 3^.+ L For the sjrstem to be unobservable.1 R^L 1 A • k \ 1 ] +A [ lA + i± ^ l = R . ^A C . det s . R^-\-¥R^R^LC 1 --------- B ^ -\+ B iB ^ L C = L |fi^ ( i+ ^ Z c ) 7 l+ Il Step 8 of 10 1 ^ 1 -1 ' r c) F ' H ’ ' = ~R^C R^L 0 -R^ 1_ L a . and L that render the system uncontrollable.] r. B lL C . . Xi ' u (6) ^ R^C ^ Step 4 of 10 From Equations (4) & (6).X i + LX^ + =0 (■4) The second loop equation is: . s. with an Input voltage source u(t) and an output current y(t).B . y ( ‘ ) = ic + h (1) where ic = C ^ ” dt = C J fi (2) i n l y i n g KVL to the circuit. (3) Writing the above equation in terms o f state variables.£ B^C ^LC ^ si!i‘Z.F -O det H J T . (5) Step 3 of 10 Equating Equations (2) & (5).y = ^ + — +x^ (8) From equations (6). Consider the circuit In Fig. B iB ^ L C -B ^ C -^ B iC + L C = 0 B f H ^ B i.^ R^C ^ R^C R^C R^C Step 10 of 10 NU1£ERATQR: 1 S + ----- R^C ° B fl 1 -R . l j { ie . R2. (a) Using the capacitor voltage and inductor current as state-variables. A.^ i+ £ s + j! .i) = LC Step 9 of 10 s l. C. -C X ^ .L S + -L 0 R^C Iz A S+R 2 RiL = s^ + —^ s + J ^ s + .(7) & (8) ■ -1 o ' 1 R^C R^C a. write state and output equations for the system. ■ = ----. when the system is uncontrollable or unobservable).iC (fii-l)-£ 1 -B . Show that there is a pole-zero cancellation for the conditions derived in part (b) (that is. (b) Find the conditions relating R^.Xi + ------ . Problem 7. Find s similar set of conditions that result In an unobservable system.L + B f i{ B i..t t + Xj + io ^ i = 0 = ~Xj + tt ia = — + — s.\) B^LC -A 7* = H ’ ’ F ’’ / f ] __1_ j. .c s + i'l 1~ £ ± 5 l± 1 ] + "1 AC ) A J B. (35) -6®* 0 0 -t®’ Therefore..A = [0 0 0 1 ] .A ’ = [-6 o * 0 0 -4 o ’ ] . (27) Calculate C . 0 1 0 0 0 0 2d> A B . (12) and (13) in equation (10).... Step 5 of 9 (c) Write the general formula for observability matrix O- Write the observable condition from description matrices.[ 0 1 0 0 ] . and the outputs y1 and y2 are the radial and angular measurements.= .A*from equations (1) and (33).A ’ from equations (1) and (32).... (20) 0 0 . from equations (29) and (35).A from equations (1) and (30).A ’ = [0 0 0 1] 0 0 0 1 0 -2 o 0 0 C .A ’ from equations (1) and (27).. A '= [ 3 o ’ 0 0 2 < » ].value. (26) Calculate C.s 0 0 0 1 0 n =2& Q n —la ) 0 1 A * B . So this system is observable using both measurements.. C CA (21) Step 6 of 9 Calculate CA^rom equations (1) and (3). (33) Step 9 of 9 Calculate C .. . respectively. from equation (23).. c. (18) and (19) in equation (16). B= 0 0 0 1 0 0 [ o 0 1 Q J* 0 -2o» 0 0 0 1 . So this system is controllable only in |B...A -[1 0 0 0] 0 0 0 1 0 -2d) 0 0 C . Where ■0 1 0 0 ' ■0 0 ■ W 0 0 2a> 1 0 « _ r I 0 0 0 1 A= .= 0 0 0 1 1 0 -2fl) 0 0 0 2fl) 0 A ‘ B . C.. (13) 0 2a>‘ Substitute equations (9).. C . (11).= (19) ^ f l) * 0 Substitute equations (15). value from equation (3).... the rank of matrix Is [ ^ . 0 B ... 1 0 0 O' 0 0 10 Os 0 1 0 0 (23) 0 0 0 1 Therefore. y= c*.’ b ] ...[ 0 0 I 0] 0 0 0 1 0 -2 o 0 0 C .A ’ =[0 -2 o 0 0 ] .from equations (1) and (11)..la / .= (18) 0 -4fl)* Calculate A ^ jfrom equations (1) and (18).= [b .A 0 . (b) Show that the system is controllable using only a single input.A* Calculate C. the state-variables x1 and x3 are the radial and angular deviations from the reference (circular) orbit..* 0 0 1 -2<» 0 0 A B .s (12) -le a 0 Step 3 of 9 Calculate A ’ B..= 0 0 0 1 0 0 -2fl) 0 0 -4fl)* 0 -2fl)* A > B . ■0 1 0 0 ' 3®’ 0 0 2d) C . (29) 0 -d)* 0 0 Step 8 of 9 Write the C.. 0 1 0 0 3(»* 0 0 AB« 0 0 0 1 0 0 0 1 0 ABs (7) 0 -2 ® 0 Substitute equations (2) and (7) in equation (6).A ’ . = [0 0 I 0 ] . C. 0 1 0 0 ~ r~ 3<»’ 0 0 lO) 0 A * B ..... 1 0 0 0 0 2a) A B ... ■0 1 0 0' 3d)^ 0 0 2d) C . = C ... 0 0 0 3<»’ 0 0 As (1) 0 0 0 1 0 -le a 0 0 0 0 1 0 Bs (2) 0 0 0 1 :::] (3) '-K1 (5) The general formula for controllability matrix.. ■0 1 0 o ' fl 0 0 0 3d)* 0 0 2d) CA '[ o 0 1 0 0 0 0 1 0 -2d) 0 0 fo 1 0 0] CA (22) '[ o 0 0 ij Substitute equations (3) and (22) in equation (21).... A . 1 0 -4fl)^ 0 . (24) Modify equation (21).. From equations (14) and (20). (a) Show that the system is controllable using both control inputs..= 0 -2d) 0 0 . matrix is [^a n d rank of matrix in Q matrix Is..from equations (1) and (17).. (33) and (34) in equation (31).Affom equations (1) and (24).A . Which one is it? Step-by-step solution step 1 of 9 (a) Write the state description matrix. (26)..A .value. C 's. 0 1 0 1 0 -o ’ 0 ...= 0 0 0 1 -2fl) 0 -2 fl) 0 0 0 r -fl) 0 A > B . A -% ] (16) Calculate A B . ■ 0 0 0 0 I 0 0 o> = 3d)* 0 0 2d) . 0 0 2d) 0 0 2fl) 0 .* ■ (11) 0 -2a) Calculate A^B. ■0 0 1 0 0 0 0 1 o .* 0 0 0 1 0 -2 ^ 0 0 1 0 A B .. C=[b AB . (15) Modify equation (6)..A' (25) C.|value at zero u..to * .. O' B.A’ = [0 I 0 0] 0 0 0 1 0 -2d) 0 0 C .. The inputs u^ and u2 are the radial and tangential thrusts.. r 0 1 0 0 ]r 01 3<»’ 0 0 2fl) -fl)* A ’ B .matt1xis. from equations (1) and (15). respectively. and the outputs y1 and y2 are the radial and angular measurements. (9) Modify equation (6).. ..] (10) Calculate A B jfrom equations (1) and (9).. 0 1 0 0 0 3fl)* 0 0 2fl) 2fl) A ’ B . 0 1 0 0 ‘ 3o’ 0 0 2o C . AB .s .. C i= [B .A*from equations (1) and (26)...A o .(14) 0 0 -2a> 0 0 -2 « 0 2®’ Step 4 of 9 Writethe B^ value at zero u.’ B . ^ C.... C .. (27) and (28) in equation (25). the rank of matrix is [ ^ ... (17). (d) Show that the system is observable using only one measurement.. the rank of matrix in C. So this system is observable only in j^ jv a lu e . So this system Is controllable using both control inputs.. (32) Calculate C .(2 8 ) Substitute equations (24).. (32)...= [ l 0 0 0 ] .. (17) Calculate A^B. ■0 1 0 0' 3®’ 0 0 2d) C . . (31) C . (30) Modify equation (21).. the rank of matrix in matrix is [^a n d rank of matrix in 0.. (34) Substitute equations (30). A . ABj ... C . 0 1 0 0 2fl) 3®’ 0 0 2fl) 0 A > B . Problem 7. = C... 0 1 0 o' 3d)* 0 0 2d) C. 0 1 0 0 ' 3o’ 0 0 2o C ..A ’ =[3ffl^ 0 0 2«i] 0 0 0 1 0 -2d) 0 0 C.... Step 2 of 9 (b) Write the BiValueatzero upvalue.A ’ Calculate C . the state-variables x^ and x3 are the radial and angular deviations from the reference (circular) orbit.40PP The linearized dynamic equations of motion for a satellite are i = Ax + Ba.A’ = [o -o ’ 0 O ] ...A ’ =[0 -2 o 0 0] 0 0 0 1 0 -2d) 0 0 C .. (6) Calculate > ^ fro m equations (1) and (2). 0 0 1 0 1 0 0 lea ■(8) 0 0 0 1 0 1 -le a 0 From equation (8).[ S] - The inputs u^ and u2 are the radial and tangential thrusts..from equations (1) and (12). value from equation (3). Which one is it? (c) Show that the system is observable using both measurements. Step 7 of 9 (d) Writethe C. S .fhnl (a) Write the state-variable equations for the system.i? ) = ( s ’ + «?) + o’ + 2 ^) Step 4 of 4 D) There is no coi^ling between the spring mode (6^—@2) and the pendulum mode (S i+ a .41 PP Consider the system in Fig. using [ej $2 vector and F as the single input.02) . S te p -b y -s te p s o lu tio n step 1 of 4 A) The State vector J T = [e .fhnl K=kd 0.-e . 9 . Do so by first writing a new set of system equations involving the state-variables .) . . Hint If A and D are invertible matrices. = .02) .jr ( 0 .02>+ /7ml ^ + IT(0| .« ^ + jT(0| .+ ^ ^+6. Figure Coupled pendulums K = bt A. e. 0 0 1 O' 0 0 0 1 X -F 0 0 K 0 0 Step 2 of 4 B) We have r= [l 0 0 0]X Observability H 1 0 0 0 HF 0 0 1 0 e= ~ . (b) Show that all the state-variables are observable using measurements of 01 alone.( ® “ -i-j?:) i : HF^ 0 0 HF^ 0 0 -(® ^+ j^ r) K The state is observable with Step 3 of 4 Q Let X = y if 0 . . (c) Show that the characteristic polynomial for the system is the product of the polynomials for two oscillators. 6.0 2 ) + ^ ^ = . then [^ •o r= [r‘ ^]■ (d) Deduce the fact that the spring mode Is controllable with F but the pendulum mode is not. . S i. <r(0. ] 0 1 0 0 -<s“ 0 0 0 x + 0 0 0 1 0 0 -(o>“ + 2 ir ) The characteristic equation o f the system is det [ if l . Problem 7. ■yi ■ ■01+^ ■ yi ^ O l-0 2 . and -2.-1 ±y')and it is observable at (0 . It Indicates that the system is going to unstable mode.2 . Step 13 of 13 Therefore the poles of the system are at [origin|. Step 12 of 13 ^ Here all the points He on left half plane and there is no real pole.-l± 7)|. When the poles and zeros move towards the boundary. Step 11 of 13 X For a stable system. . the location of the poles must be both controllable and observable which occurs only in stable system.A)B for this system? (b) What are the poles of the reduced-order transfer function that includes only controllable and observable modes? Step-by-step solution Step-by-step solution step 1 of 13 (a) Step 2 of 13 Consider the equation of the system. the location of poles and zeros must be in left half plane.-1 . . it indicates that the system is going to unstable mode. A decomposition into controllable and uncontrollable parts discloses that the controllable part has a characteristic equation with roots 0 and -1 ± 1). When the poles and zeros move towards the boundary. i(>)=Cadj(sI-A)B Step 3 of 13 The equation is controllable at points (0 » -l± y)and it is observable at Step 4 of 13 To obtain the zeros of the system. the location of poles and zeros must be in left half plane. -1.-2 ) Step 10 of 13 To obtain the poles of the system. (a) Where are the zeros of b(s) = Cadj(sl . So the poles of the system are at origin. Step 8 of 13 (b) Step 9 of 13 Consider the equation is controllable at points (0. Step 5 of 13 For a stable system. -2. and -1 ± iy.42PP A certain fifth-order system is found to have a characteristic equation with roots at 0. the location of the zeros must be both controllable and observable which occurs only in stable system. A decomposition into observable and nonobservable parts discloses that the observable modes are at 0. .l .l ± y ) step 7 of 13 Therefore the zeros of the system are at the points |( -l^ -2 . Step 6 of 13 Here all the points lie on left half plane so the zeros of the system Is at ( . -1 . Problem 7. ... + B . -[o . i . x . i = land2. (1) .. ^ (s ) ' A W y(s) M ) ^ jv . ■ A .. ^..' Hence. y = c . C .... | AT. C . 0 B .*..(s) in equations (20). X/ = Ax/ + B/u/. where / = 1.. A. A.. A W A W AW AW M l AW a w a w -^a w a w AWAW AW AWAW AWAWAW+AWAW =___ AWAW___ . there are no cancellations in poles and zeros part of the transfer functions in feedback combination. X. C . the state equations of the parallel connection is irs '“ ' r B.+ B . X | . = A . (6) Write the output equation of series connection in matrix form from equation (6).(5)andZ)2(5) should be coprimel Step 7 of 7 ^ Refer 7. For controllability the parts | (j) a n d P 2( j ) should be ooprime o th e r w i s e ( j ) ~ | is covered from the output..*1 J~r7r~w5n j.23) *w a w a w +a w a w Therefore. (b) parallel....94(c) in the textbook and write the transfer function of the feedback connection. Give a brief reason for your answer in terms of pole-zero cancellations. 0 y=[0 c.] Step 3 of 7 Feedback connection: Refer figure 7.. Step 6 of 7 Refer 7.C jX j+ B .. y = [C... ( ^ ) and D j (y ) should be coppme|and controllability [Af2(j)andZ ^ W should be coprime| . * . ■ ( 12 ) '■ [ 4 ' ■ f t ] .( s ) Therefore. = A . (c) feedback Step-by-step solution step 1 of 7 (a) Series connection: Consider the controllable and observable realizations for any system.. X | .. .( s ) U (s ) D .+ B . = A ... For observability. Hence..94 (a) in the textbook and write state equations for series connection from equations (1) and (2). from equation (19).. .94(b) in the textbook and write the transfer function of the parallel connection.=A2AC2 + B... (11) Write the output equation of parallel connection in matrix form from equation {11). = (18) ^ f o r a (s) and ^ f o r CM ) in equations (18). Give a set of state equations for the combined systems in Fig. . X. > = C | X .. » .' Hence. the state equations of the parallel connection is 0 A.. ' A.MAT. A (*) ' A W ' ' y(s) Af. (c) feedback . A. employing series.( . there are no cancellations in poles and zeros part of the transfer functions in series combination.... parallel. the state equations of the series connection is i r s JC+ u B .94(a) in the textbook and write the transfer function of the series connection. 0 y = [C. (20) Substitute f ' . Figure Block diagrams. and feedback configurations. Problem 7.. = A .. (9) Write the state equation of parallel connection in matrix fonn from equations (8) and (9). Write the output equation of feedback connection. from equation (21). (a) series.. h (8) X.94 (c) in the textbook and write state equations for feedback connection from equations (1) and (2).. there are no cancellations in poles and zeros part of the transfer functions in parallel combination.. (22 ) R(s) 1+ G ... + C jX j.(5) (19) U (s) D .|[^-]| -=|C. (b) For each case... (5) Write the output equation of series connection...C... Ki-i*. 0] 0 step 5 of 7 (b) Refer 7.. oil (17) ■ A.. (a) Suppose we have controllable-observable realizations for each subsystem. (14) Write the state equation of feedback connection in matrix form from equations (13) and (14). For observability the parts [AT^(j)and£)..+ B . . X.' :][.(s ) A W y(s) A W A W i-JV jW P.C .. M for <7... x . W A W Substitute for GiW®'''^ G. Figure Block diagrams.. For observability and controllability the parts |D . .94 (b) in the textbook and write state equations for parallel connection from equations (1) and (2).. (13) x .. W 'n (22). 0 B.. y/ = Cixi.. from equation (23).]•& ( 10 ) Write the output equation of parallel connection. x .. x .2. (16) Step 4 of 7 ^ Write the output equation of feedback connection in matrix form from equation (16).[c... r ...A . (2) Where.C .+ B .. h (3) x . (b) parallel.. -A .x i . determine what condition(s) on the roots of the polynomials Ni and Di is necessary for each system to be controllable and observable.] step 2 of 7 Parallel connection: Refer figure 7. W should be coprime o th e rw ise Q (j) | is covered from the output.. (a) series. (15) y = C . (4) Write the state equation of series connection in matrix form from equations (3) and (4). A UJ A M l r(s) QW ^ W 'l ...) D . x ..U . = A .. Refer figure 7.W ■ ( 21 ) uW A W a W Therefore..43PP Consider the systems shown in Fig. ( 5) and M l G.. and draw vectors that correspond to the completely observable (xO) and the completely unobservable (xO) state-variables. G et help from a Chegg subject expert. Step-by-step solution There is no solution to this problem yet. (a) Find the state matrices Ac. Problem 7. Be. (d) Give the state matrices in observer canonical form and repeat parts (b) and (c) in terms of n n n tm llflh ilih / in<%tpaH n f nh. (c) Express xO and xCin terms of the observability matrix O.<?prvahilifv (d) Give the state matrices in observer canonical form and repeat parts (b) and (c) in terms of controllability instead of observability. x2) plane.44PP Consider the system y + 3 y + 2 y = u + u . (b) Sketch the eigenvectors of A c in the {x1. and Cc in control canonical form that correspond to the given differential equation. ASK AN EXPERT . Problem 7.= 5 3 o ’ . and design a full-order observer with poles placed at s = . u = engine thrust in the y-direction. and -3oj ± 3ojj. 1^=-51. If the orbit is synchronous with the earth’s rotation..2 cj. Figure Diagram of a station-keeping satellite in orbit Step-by-step solution step 1 of 2 A) There is not enough information to answer this question. Step 2 of 2 0 1 0 0 k 0 3ar‘ 0 0 2 ib 0 B) + 0 0 0 1 ^3 0 0 -2 o 0 0 1^4 1 z = [o 0 1 o]Jsr 1^=-44. (a) Is the state x = [ x x y y ] T observable? (a) Is the state x = [ x x y y ] T observable? (b) Choose x = [ x x y y ]7 a s the state vector and y as the measurement. where X = radiai perturbation.5W- 4 = llo . .45PP The dynamic equations of motion for a station-keeping satellite (such as a weather sateiiite) are x — 2ajy — 3af^x = 0t y + 2ew: = tt.5W . as depicted in Fig. y = longitudinal position perturbation. then w = 2 tt/(3600 * 24) rad/sec.3 cj. 7 (S '-20/7) ~ S ’ + 20S + 200 Step 4 of 4 D) d e t(S l-J i'+ O £ r) = 0 s ’ +£:jSf+oci‘ + i : . determine the state feedback gain K) so that the roots of the closed-loop characteristic equation are at s = . Problem 7.‘ £ . and pick the estimator roots to be at s = -10 ± 10j. Figure Pendulum diagram 1 Step-by-step solution Step 1 of 4 A) We have :■ m ' y= 0 \]x y = 0 T]X step 2 of 4 B) d e t(£ fl-i? + i« ) = 0 £ r^ + /jS '+ ii)’ ( . (b) Design an estimator (observer) that reconstructs the state of the pendulum given measurements of &. Assume w = 5 rad/sec.K ^ = S .46PP The linearized equations of motion of the simple pendulum in Fig. are 6 + w2d = u.4 ± 4/.l ) = 0 Step 3 of 4 C) We use the estimator equation i= F ^ + a u + L l^ y . (d) Design a controller (that is.L H ) X + O u + Iiy M = [l 0 ] ( S ^ i i' + £ « ) . (c) Write the transfer function of the estimator between the measured value of & and the estimated value of d. 6Stiffiaiea vaiue dr d.m ) = ( F . i .= l. = o K . (a) Write the equations of motion in state-space form. 1. 2 a n d ij = .1 and -0. and place the observer-error poles at -0. and place the observer-error poles at -0.47PP An error analysis of an inertial navigator leads to the set of normalized state equations liH l ‘i I][s]^l]' where x1 = east—velocity emor.1. Problem 7. Design a reduced-order estimator with y = x1 as the measurement. Design a reduced-order estimator witn y = x l as tne measurement. Be sure to provide all the relevant estimator equations.0 . Be sure to provide all the relevant estimator equations. x2 = platform tilt about the north axis.0 . 0 1 . u = gyro drift rate of change.1 and -0. Step-by-step solution step 1 of 1 Partitioning the system matrices yields 0 + 0 1 0 -1 0 ^1 fo = 1 0 1 + 0 0 0 0 ^3 [l y= 1 0 o]Jsr d e t(^ - Thus A = . x3 = north—gyro drift. . = 3 * .(j) “ 7 ^ .2 ± 2j.A + B K + L O .. using Eq. (3) Compare equations (2) and (3). 2 > .48PP A certain process has the transfer function G(i) = . j .i 4 : :i=" 4 ::i.'I)- * t + / | j + / . 0 .'L . using Eq.-K x The closed-loop characteristic equation is given by a .A + B K 4 L C ) ' L ‘■ • o .. B..{s )-0 a . H=tf([-264-133924]. 1 0 0 ...]*U a i - ^ [ [L-4+4A. Problem 7..2 ± 2 j a .“ <** .p(5 ) = j*+ 4 s+8 (2) The closed-loop characteristic equation is also determined by det ( 5 I ... and C for this system in observer canonicai fonn.l 0 j ) a .1 *+i .4 + 4 * | . |K = [3 1]| Step 3 of 6 ^ (c) Consider the equation to determine the estimator roots with estimator-emor poles at s = . 0 ... (4) Step 4 of 6 The estimator roots are also determined by a .. *. b 204 Thus. xJ+4A + 4 *.<^ --4 Consider the general observable canonical form in tenns of the transfer function coefficients is • 0 0 0 -a.A + L C ) det 1 "([. (c) Compute L so that the estimator error poles are located at s = -10±10y. (d) Give the transfer function of the resulting controller [for example.2 y ) o .4 .( » ) = d e t ( j I . the phase margin is |-137®| and gain margin is 1-53. " ( [ . ^ sl.< v A = 0 0 1 -fl.i ro I] fo i UJ-[4 oJ-U" y = 1 0 ]* 0 1 0 ■ A= . .( x ) — K ( x l . Step-by-step solution step 1 of 6 (a) Consider the transfer function -(1) V -V The order of the transfer function is n » 2 Consider the transfer function is represented as £7( 5) + By inspection. . ( i) = 5^ + 24^ + 292 Step 6 of 6 (e) Consider the MATLAB code to determine the phase and gain margin of the controller using nyquist plot.( » ) = » ’ + 2 0 s+ 2 0 0 .. (5) Compare equations (4) and (5) /.]. (e) What are the gain and phase margins of the controller and the given open-loop system? Eq. V(j) D . ( j ) = (^ + 2 + 2 y ) ( 5 + 2 .. . + 4] . Step 5 of 6 ^ (d) Consider the equation to determine the transfer function of the resulting controller.]. NyiliiistDiagnB From Figure 1. compute K so that the closed-loop control poles are located at s = .. -2 6 4 J-6 9 2 j* + 2 4 s + 2 9 2 -2 6 4 J -6 9 2 Thus. -4 + 4 * . (a) Find A.-1 Thus. (d) Give the transfer function of the resulting controller [for example.0 . the phase margin is —137® and gain margin is —5 3 ^ dB Thus.] ] ® * ’ + 4 * .= 2 0 / ..2 d s l • . . the transfer function of the compensator is D .c g -•w -i’ g :])■'&] •^[212 .[1 24 292]) nyquist{H) The output after executing the MATLAB code is shown in Figure 1.. -.A + BK) s Q detl (■[. _ ^=[ 0 0 0 1 ] Consider the general observable canonical form of {1) in terms of the transfer function coefficients is r i. { s ) = {s + l 0 + l 0 j ) { s + l 0 .. (b) If u = -K x. c = [l 0] 4 0 -4 Step 2 of 6 (b) Consider u . 4706 1. Draw the corresponding Bode plot and compare the pole placement and SRL designs with respect to bandwidth.49PP The linearized longitudinal motion of a helicopter near hover (see Fig) can be modeled by the normalized third-order system ' -0 . and plot Its frequency response using Matlab. and place the estimator poles at -8 and -4 ± 4 V J y .3CT8 Step 2 of 5 (B) For the helicopter r a n k { q = ra iik [0 PO P '0 \ = 2 Step 3 of 5 (C) i : = [0. Draw a Bode plot for the closed-loop design.8130 15. What are the advantages and disadvantages of the reduced-order estimator compared with the full-order case? (f) Compute the compensator transfer function using the control gain and the full-order estimator designed in part (d). Problem 7. S te p -b y -s te p s o lu tio n step 1 of 5 (A) The open loop poles are the eigen values o f F d e t ( ^ n .9542f .2510 0.02 Figure Helicopter Suppose our sensor measures the horizontal velocity u as the output. stability margins. y = u. (b) Is the system controllable? (c) Find the feedback gain that places the poles of the system a ts = -1±1) and s = -2.0627] d e t ( 6 n . (g) Repeat part (f) with the reduced-order estimator. (h) Draw the SRL and select roots for a control law that will give a control bandwidth matching the design of part (c).7097 18. (e) Design a reduced-order estimator with both poles at -4. and select roots for a full-order estimator that will result in an estimator error bandwidth comparable to the design of part (d).6 5 6 5 and 0. step response.1183±^ 0.if „ + £ 7 r . and indicate the corresponding gain and phase margins.5800f Step 5 of 5 d e t ( 6 n . and control effort for a unit-step rotor-angle input. (d) Design a full-order estimator for the system. ) = 0 £= [14. Use Matlab for the computations.F ’) = 0 S = -0. that is.f + G « :) = 0 Step 4 of 5 (D) A = [44.0 0.4 0 -OJM I 0 0 -14 9J -0. (a) Find the open-loop pole locations. V = velocity of the cart. where 6 = angle of the pendulum. Find the feedback gains so that the resulting closed-loop poles are located at —1»—1 u = .K 3 v . indicating the corresponding gain and phase margins.K .K 3 v . The linearized and normalized equations of motion corresponding to this system can be put in the form e =e + v + u. plot the response of the system to an initial condition on d.50PP Suppose a DC drive motor with motor current u is connected to the wheels of a cart in order to control the movement of an inverted pendulum mounted on the cart.K . and draw the Bode plot of the closed-loop system.\e -K 2 d . v = 6 . (c) Give the transfer function of the controller. (d) Using Matlab. Construct an estimator for d and & of the form t = A l-(-L (y -C t). and -2. G et help from a Chegg subject expert. (a) We wish to control 6 by feedback to u of the fonn u = . —1 (b) Assume that 6 and v are measured. Select L so that the estimator poles are it -2. where x = [0 d\T and y = d.V . Step-by-step solution There is no solution to this problem yet. ASK AN EXPERT .\e -K 2 d . Treat both v and u as known. and give a physical explanation for the initial motion of the cart. Problem 7.u. Find the feedback gains so that the resulting closed-loop poles are located at —1. F + a K + L H y ''L -(54. and write state equations for the system.85+202.{ S j .5. (d) What is the transfer function of the controller obtained by combining parts (a) through (c)? (e) Sketch the root locus of the resulting closed-loop system as plant gain (nominally 10) is varied.5.9 0. Problem 7.2] Step 3 of 4 (Q £ = [14 21 i f Step 4 of 4 (D) The transfer function for the controller is D c { S )= . (b) Find /C1 and K2 so that u = -K ^ x ‘\ -K2x2 yields closed-loop poles with a natural frequency ojn = 3 and a damping ratio ^ = 0.n1 = 15 and ^1 = 0. Step-by-step solution step 1 of 4 The state equation are :a=[: .5) .= [1 o]x step 2 of 4 (B) jr = [0.51 PP Consider the control of Y(s) 10 C(*) = i '£/(«) 4 ( « + l) ‘ (a) Let y = x1 and x1 = x2.. (c) Design a state estimator that yields estimator emor poles with o. and compare the resulting bandwidth and stability margins with those obtained using the design of part (b). and indicate where x and appear. Design state feedback so that the closed-loop poles are at the same locations as those of the design in part (b). (a) Let u = -K x (position feedback alone). (e) Design an estimator for x and x using the measurement of x = y. arise in situations where the motion of an upside-down pendulum (such as a rocket) must be controlled. Step-by-step solution There is no solution to this problem yet. Draw a Bode plot for the closed-loop system.52PP Unstable dynamic equations of motion of the form x = x + u. (c) Sketch the Bode plot (both magnitude and phase) of the uncompensated plant. Sketch the root locus with respect to K. Problem 7. and select the observer gain L so that the equation for x has characteristic roots with a damping ratio ^ = 0. and estimate the phase margin. and sketch the root locus with respect to the scalar gain K. (d) Sketch the Bode plot of the compensated design. ASK AN EXPERT . (b) Consider a lead compensator of the form Select a and K so that the system will display a rise time of about 2 sec and no more than 25% Select a and K so that the system will display a rise time of about 2 sec and no more than 25% overshoot.5 and a natural frequency o)n = 8. (e) Draw a block diagram of your combined estimator and control law. G et help from a Chegg subject expert. M M.200y) = ( i + 2 0 0 )'+ (2 0 0 )' . kKy b 40 M ^ i. y= 1 0]jt Step 2 of 12 (b) It is clear that roots are —100±100y- Use the roots and calculate the equation. Calculate the characteristic equation for the robotic arm using the matrices shown In equation (1). Calculate the characteristic equation for the robotic arm using the matrices shown In equation (1). £ . the mass position.= 900 . ( i + 1 0 0 + 1 0 0 y)(j + 1 0 0 -1 0 0 y )= (i+ 1 0 0 )^ + (1 0 0 )' = « ’ + 10‘ +200s + 10‘ = « ‘ + 200«+(20000) It is clear that *'+ 2 0 0 « + (2 0 0 0 0 ) represents the characteristic equation. u = input. - kK.G K ) ] = 0 ■0 1 o' det .:] - It is clear from the value of the observability matrix that the system is observable.98 from the text book.G K ) ] = 0 It is clear from the equation (1) that.= X i k k X. ss— X + — tt Thus. where k/M = 900 rad/sec2. (c) Could both state-variables of the system be estimated if only a measurement of y was available? (c) Could both state-variables of the system be estimated if only a measurement of y was available? (d) Design a full-state feedback controller with roots at s = -20 ± 20/. andATj that large control levels are required so it is not desirable to design a system with the roots -200 ±200^- Thus. The formula to calculate the characteristic equation using the state space fonn to design the feedback controller is as follows. ( j + 200 + 2 0 0 y ) ( i+200 ..|i .044]: » syms s » l=[1 0. . the state matrix representation is as follows. = 0 ^ (» + A ) + [. 0 1].( K . ± ( x .K x Write the following code in MATLAB and obtain the transfer function of the compensator. 900 *87.111 Thus. (2) • ^ + £ . » G=[0.9 0 0 j = 0. (a) Write the equations of motion in state-space form.. the equations are. » L=[200. the position of the end of the spring. including a command input for y. (f) Write equations for the compensator.ij 0 \ .= 2 0 0 . It is clear that the value of — is 900 ra ^sfc * - (a) The equations of motion in state space form are as follows. the feedback controller is designed [[-0.= 40 M A : j= 4 0 o f . l..( K . bode plot is obtained for the compensator and the gain margin and phase margins are 1-10.= 2 0 0 900+1. the estimator is designed 19100 Step 5 of 12 (c) system. 0 1 O' = --i 0 fc * (1) .19100]: » K=[-0. det s I. i s " * * ') det E :]■ =0 -1 det = 0 ± ( i . Figure Simple robotic ann Step-by-step solution step 1 of 12 Refer to the Figure 7.) = 80000 M ^ _ 80000 . 1 det = 0 K :} s+L. State your reasons. Bode D ig ra m Step 12 of 12 Thus.L H ) ] = 0 0 1" det =0 det E :i. » F=[ 0 1:-900 0]. (i+20+20y)(i+20. x = (F-GK-LH)x+L> w= . + 40000 + 400s+40000 = s ' + 400s+(80000) Compare the equation s ' + 400s+(80000) with equation (4) and obtain and JC.444 -^ (l+ iC .^ l ' I.. 0 r F* A 0 L w J 0 <?= t »ndH = [l 0] IM I Step 7 of 12 Determine the characteristic equation of the system.0444 ^ ( l + i f . k [::} ± 0 M M .900]: » H=[1 0]. d e t [ r i. Draw a Bode plot for the closed-loop system and give the gain and phase margins for the design. ) = 800 800 /:.= 2 0 0 0 0 (3) Substitute the value of ^k :In equation (2) and (3) and calculate and - £ . ■0 1' 0 det E :]■ ± 0■ —( det E :i.( F . -1 det * . Consider that x ^ m y and Xj ■ y Thus.i-K :i M 1 e :i det — . 0 1' F* 0 L A/ J <7* 0] Step 3 of 12 Determine the characteristic equation of the system.) s.7 dB and respectiveiy. (e) Would it be reasonable to design a control law for the system with roots at s = -200 ± 200/. = 20000 l2 » 1 9 1 0 0 200 Thus.. » D=-K*inv{(s*l-F+G*K+L*H))*L D= (222*(5*s + 198))/{5*(10*s''2 + 2396*s + 278201)) + 8862049/(5*{10‘ s'^2 + 2396*s + 278201)) - (8404*(s + 200))/{10*s'‘2 + 2396*s + 278201) » simplify(D) ans = -(8182*s -100401 )/(10*s'^2 + 2396*s + 278201) Step 11 of 12 Draw the bode plot for the obtained transfer function of the compensator.M ^ .) (4) Step 8 of 12 Compare the obtained characteristic equation with + 4 0 5+ (800)- It is clear that.88 It is clear from the values of a:.( F . a :.. not desirable to design a control law for the system with roots -200± 200y • Step 10 of 12 The state space form generalized form to calculate the transfer function of the compensator is as follows. Problem 7.111 0. X . (b) Design an estimator with roots at s = -100 ± 100/. The formula to calculate the characteristic equation using the state space fonn is as follows.L H ) ] = 0 It is clear from the equation (1) that.i " 'I Substitute the tt value of H p In observability matrix. si [ .^ + A !]= o Step 4 of 12 Compare the obtained characteristic equation with * '+ 2 0 0 « + (2 0 0 0 0 ) It is clear that. 0. both the state variables of the system can be estimated using a measurement of * Step 6 of 12 (d) It is clear that roots are —2 0 ± 20y- Use the roots and calculate the equation.9 0 0 J = 0. The observability matrix is Calculate h F - I' " i [ J » i ] .111 0.20y)=(i+20)“+(20)' = 5*+400+40j +400 = 5 *+ 4 0 5 + (8 0 0 ) It is clear that 5^+4Q $+(800) represents the characteristic equation. -L .53PP A simplified model for the control of a flexible robotic arm is shown in Fig.044]| - Step 9 of 12 (e) It is clear that roots are -200 ± 200 J ■ Use the roots and calculate the equation. Thus. d e t [ jI. d e t [ jI. y = output. where 5/j1 = deviation of depth in tank 1 from the nominal level. y = 5h2). 6h2 = deviation of depth in tank 2 from the nominal level. = 146 . Using this measurement. 6h2 + a6h2 = a6h'\. Figure Coupled tanks (a) Level Controller for Two Cascaded Tanks: Using state feedback of the form 6u=-K■\6h^ -K25h2. + i^ ) = 0 = S=+46S+86==0 Gives /Ti = 26 and = 56 Step 2 of 2 = + ( 2 6 + /. are 5/j1 + aC/j1 = 6u. choose values of K^ and K2 that will place the closed-loop eigenvalues at s = -2a(1 ± j). Problem 7.)+ 6 ” = 0 = (S + S 6 + S 6 j)(S + S 6 -S e j) = J ’ +16&S'+1286’ = 0 Gives ly =1136 and i. compute and plot the response at y to an initial offset in Sh^. (d) Using Matlab. (b) Level Estimator for Two Cascaded Tanks: Suppose that only the deviation in the level of tank 2 is measured (that is. 6u = deviation in fluid in flow rate to tank 1 (control). design an estimator that will give continuous.54PP The linearized difTerential equations governing the fluid-flow dynamics for the two cascaded tanks in Fig. 1 (A) ldet| t [ -6 S+6j = S’ + ( 2 6 + i : i ) i'+ 6 “ + 6 ( ^ : . (c) Estimator/Controller for Two Cascaded Tanks: Sketch a block diagram (showing individual integrators) of the closed-loop system obtained by combining the estimator of part (b) with the controller of part (a). Assume a = 1 for the plot. with estimator error poles at -8a(1 ± j).)£ T + 6 ( /i+ i. k . Step-by-step solution step 1 of 2 r ^ r + e + i: . smooth estimates of the deviation in levels of tank 1 and tank 2. 0895) » +0.0439 -0.0895)(i + 0. ^1 I= I r -0. .0439J -0.A ] 'B > + 0.0895 0 0 *’ + 0. where ijjd is the desired heading.2.286 O' -0. are described by [ /. B Y = [0 0 1] V V f '] The matrix C is.0895)-0. Figure 1 View of ship from above r = yaw rate (see Fig.36I5* + 0.3615* + 0 . it is not possible to determine the roots of the system.272) .\ p .K 2 r-K 3 {iiJ .0439 -0.0145* (*+ 0. Figure 2 Ship control block diagram for Problem S te p -b y -s te p s o lu tio n step 1 of 3 The state space representation of the lateral motions of a ship is.3615*+0.0439 x 0.01l8)^® ■*(* + 0.272 0 r + -0.0145 =[0 0 1] 0.0439 -0.286 O ir^ ■0.0122 0 Step 2 of 3 (a) As it is required to find the transfer function from g to the output equation is. Hence.0145 r = -0.0145 Bs .36l5* + 0 . 0.0439)(0. G ( j) = C [ i I .0118) ■ j(i+0. determine vaiues of K^. moving at a constant velocity of 10 m/sec.286 O' 0.0439 -0.0439 -0. [ j j —A ]"' ■ _ 1 « [ ( « + 0.0 ll8 Step 3 of 3 Determine the matrix.272) -0 . 1).0895) * + 0.(0.272 0 0 I 0. and plot it frequency response. 1). (c) Design a state estimator based on the measurement of ijj (obtained from a gyrocompass. C » [0 0 1] The transfer function of the system is. Figure 1 View of ship from above (a) Determine the transfer function from 5to tp and the characteristic roots of the uncontrolled ship.0439i -0.286) i ( i ’ + 0 .286J ■ j(j+0.0439 ■0. K2. 2.272)+0. or to design a state estimator.286* *(* +0. and plot the step response of the system to a change in heading of 5°. 0.2). r = yaw rate (see Fig. or determine the state equations and to draw the Bode plot and step response of the system.000536 [0 0 1] -0.0895 0.272 0 .01l8) 0 As the transfer function is 0. -0.0145 "I ®I22 14. it is not possible to determine the state feedback. 'fi -0.286* *(* + 0.0895 0 0 ( i + 0. G (*) 0 1] “ *(*’ + 0.0122 0 -1 s 0 Determine the inverse matrix.55PP The lateral motions of a ship that is 100 m long.286 0 "I T ^ 1 -0.0122 0 0 *’ + 0.0122*(*+0. and compute the corresponding gain and phase margins.272) -0. (d) Give the state equations and transfer function for the compensator Dc(s) in Fig. Place the roots of the estimator error equation at s = -0.8 ± 0. (b) Using complete state feedback of the form 6 = .0 ll8 0 0.ijjd).8 and -0.K . Problem 7.3 6 l5 s + 0.0145' -0. (d).2 ± 0. for example).0895 -0.272) -0 . subparts (b).272 0 II I+1-0 [" 0.8).272) -0.0895 -0. Also.0439 J + 0.286J s ( f + 0.0439* -0.0895)(j + 0.-0. (e) and (f) cannot be solved.0122 S 0 1 oJ[»r 0 The matrix A and B ^re. 6 = rudder angle{deg). (f) Compute the feed-fonvard gains for a reference input.0895 -0. and K3 that wiii place the closed- loop roots at s = -0.0895 -0.0895) * + 0. (c).004147* *(*’ + 0. (fj = heading angle(deg). (e) Draw the Bode plot for the closed-loop system. where jS = side slip angle(deg). Footnote 11 A reasonable alternative is to select N such that. y(o) Mathematically. Now the DC gain is unity for the type-1 system. + Mr.M )J 1 N=- C (A . the system will give unity DC gain in the controller. « .+ 0 . Show that if the plant is Type 1. Hence the condition is proved..2.L C ) x . both the equations will give unity feedback controller design in the type-1 system. write the equation for general closed loop system block diagram.B K . « (0 ) Therefore..LC )->{B . Similarly in equation (6).B K . Eq.(6) Step 10 of 19 For this controller consider the DC gain from y to u is equal to the DC gain from rto u.56PP Asmentioned in footnote 11 in Section 7. The consequences o f this choice are that our controller can be structured as a combination o f error control and generalized derivative control... x . .B K .L C ) x „ + L y .B K .( A . 1 N = ‘ C (A .> ( B .K (A . G ( x ) = ie ( x ) Consider the formula to calculate the DC gain of the closed loop system.B K .L C ) i + L ^ + M r . Problem 7.LC) ‘ (5) Step 9 of 19 Substitute the values in equation (2) and rearrange equation..185a and 7.L .(1) u— Kx + Pr (2) Step 4 of 19 Assign the values and calculate the DC gain from y to u.K (A .9.L C ) '' + ^ ] r .b y . Step 11 of 19 Equate equation (4) and equation (6). y . value of closed loop system — will be unity. is to choose N such that when r and y are both unchanging.K [ ( A . Step 16 of 19 For Type-1 system.L C ) X. [ k ( a ..L C ) ‘ (3) Step 6 of 19 Substitute the values in equation (2) and rearrange equation.LC)"* L yo] = [ k A /( A . + A / r .[ k A /(A -B K -L C )" ' + w ]r. 0 = ( A .B K ). the DC gain from rto u is the negative o f the DC gain from y to u.48 and equations 7.185b in the text book. .B K . — ( A .K x . [ v G ( 0 ) s I (unity feedback)] ^ .M )J S t e p . (4) step 7 of 19 Assign the values and calculate the DC gain from rto u.(0 ) (6) « (0 ) D . when r and y are both unchanging. = K ( A . = K A /r .L C ) . a reasonable approach for selecting the feed- fon/vard gain in Eq. X*Xo y=0 Step 8 of 19 Substitute the values in equation (1) and rearrange the equation.B K .( 0 ) Refer equation 7. This equation helps to achieve unity DC gain in the controller design. if the gain and the feedback are same..L C ) " '+ . g (°) « (0 ) ' l-G (O )i)..L C )“' + Step 12 of 19 Rearrange and find .( A . ( A .. X*X p r =0 step 5 of 19 Substitute the values in equation (1) and rearrange the equation. and if the system is capable o f Type 1 behavior. « j= . if ND .s t e p s o lu t io n step 1 of 19 (a) Step 2 of 19 Refer figure 7.. > .B K .B K )-»B (1 .L C ) ''L . 0 = ( A . Step 17 of 19 Step 18 of 19 Step 19 of 19 Apply limit to the equation.— L y ( A . « (i) ' l-G (j)J > .B K . + 0 -(A -B K -L C )x „ -A /ji • (A -B K -L C ) X. = ... = [ l C A / ( A .B K . this choice is the same as that given by Eq.B K . « .. Step 3 of 19 Consider the equation of the system for the feed forward gains of the controller.B K .202 in the textbook.. M l * (A -B K -L C ) X . ^ = . that capability will be realized.L C )" ' ( L + A f) ] Step 13 of 19 ^ Consider the general closed loop system block diagram. Derive a formula for W based on this selection rule.lB [ l . Where. Step 14 of 19 Figure 1 step 15 of 19 ^ Take the value of p as the foot note. is the negative feedback path £) is the loop gain..K X j » . the DC gain from r to i/is the negative of the DC gain from yto u. Problem 7. Use Matlab (Simulink) software to simulate the system responses.x = w + w. Step-by-step solution Step-by-step solution step 1 of 2 The equations o f motion are given by x -x = u+W 0 = 0 A realization of these equation is »i[:] =y-r z = Xi X X '0 1 O' O' K= 0 0 1 0 _0 1 0 1_ Step 2 of 2 The design ofthe state feedback vector. Introduce integral error control. and select three control gains K = [K^ K2 K3]so that the closed-loop poles are at -1 and -1 ± j and the steady-state error to w and to a (step) position command will be zero.^ ) = 0 When K. K For closed loop poles o f 5 = -1 . Draw a block diagram of your design showing the locations of the feedback gains Ki.]= [2 5 3] The closed loop system is given by ■-1" Z = {F ^ -a ^ K )Z + 0 ^ (S > + I y ^ M .l ± j t d e t(5 1 -^'^ + G S . . Here w is a constant bias due to the power amplifier.57PP Assume that the linearized and time-scaled equation of motion for the ballbearing levitation device is x . Assume that both x and x can be measured. Verify that the system is Type 1. Plot the response of the closedloop system to a step command input and the response to a step change in the bias input. /:. Let y = X and the reference input ^ ^ be a constant.z For the closed loop system we have 0 1 0 0 0 1 1 -^ j -^ 3 This immediately gives Z z—0 AndZz—y —0 . as well as the sensitivity function (S) and the complementary sensitivity function (7).y. the system has zero steady-state error. the system is not robust under changes to the system parameters in A.tpm n n r a m p t p r s in A (c) Show that if A changes to A+ 5A. (d) The system steady-state error performance can be made robust by augmenting the system with an integrator and using unity feedback—that is. (f) For part (d). the system is not robust under changes to the Ru<. Therefore. (b) Choose N so that if r is a constant. by setting x l = r . where x l is the state of the integrator. where A/ is a nonzero scalar. Draw Bode plots of the controller. use Matlab (Simulink) software to plot the time response of the system to a constant input. G et help from a Chegg subject expert. where 5A is an arbitrary 2*2 matrix.K^xl so that the poles of the augmented system are at -3. —2 (e) Show that the resulting system will yield y (^) = r no matter how the matrices A and B are changed. ASK AN EXPERT . S te p -b y -s te p s o lu tio n There is no solution to this problem yet. then your choice of N in part{b) will no longer make yC“ ) = r. To see this.1 ] ’ »=[!]• (a) Use feedback of the form u(t) = -K x{t) + Nr(t). to move the pole to -3 ± 3). where 5A is an arbitrary 2*2 matrix. first use state feedback of the form u = -K x . then your choice of N in part{b) will no longer make yC“ ) = r.58PP Consider a system with state matrices ^=[ 0 . as long as the closed-loop system remains stable. Problem 7. that is. (c) Show that if A changes to A+ 5A. Therefore. y (^) = r. 59PP Consider a servomechanism for following the data track on a computer-disk memory system. Problem 7. (a) Let (uO = 1. Because of various unavoidable mechanical imperfections. G et help from a Chegg subject expert. and place the poles of the error system for an internal model design at ac(s) = (s + 2 ± J2){s + 1 ± y i) and the pole of the reduced-order estimator at ae(s) = (s + 6). . and the pole of the reduced-order estimator at ae(s) = (s + 6). Also verify the presence of the blocking zeros at ±j(jjO. ] ■ >'■ The sinusoidal reference input satisfies y — —o j^r. the data track is not exactly a centered circle.[ . (c) Use Matlab (SImulink) software to plot the time response of the system to a sinusoidal input a frequency ojO = 1. ASK AN EXPERT . and thus the radial servo must follow a sinusoidal input of radian frequency ojO (the spin rate of the disk). (d) Draw a Bode plot to show how this system will respond to sinusoidal inputs at frequencies different from but near wO. and clearly show the presence of the oscillator with frequency ojO (the internal model) in the controller.[ . Step-by-step solution There is no solution to this problem yet. A ]. (b) Draw a block diagram of the system. The state matrices for a linearized model of such a system are * . 72 Motor Speed system with extended esQmaton (a) block diagram. The sys­ tem augmented with equivalent external input p. 1 = -3S-I-« + p + fe(e . To besin.5 sec) are shewn in Rg. Problem 7. is ^ v e n 1^ p=0.x ) . (7.256c) f = 0.iT . K = 2. The estimator error gain is found to be L ^ [ 225 27 from the characteristic equadon * •[1 *+ + A block diagram of the system is given in Fig.i). 7.7 ^ ^ .256a) y = x + w.3 . i = . w e design die ctmtiol law by ignoring the equivalrat disturbance. To begin.x ) x = -3 x + p + 4 + i3 u = -K x -p p l^ r o -I. (7.3 i + h.2 7 9 ( £ f + 4 .72(bX Step-by-step solution step 1 of 1 The related equation p = /i ( c .256b) w = 0» (7.256d) Place the control pole at 5 = —5 and the tw o extended estimator poles at 5 = -1 5 .8 2 Time (fee) W> Figure 7. Therefore. and the step responses to input at the command r (applied at t = 0 sec) and at the disturbance w (applied at / w 0. 7 .4 ? X The controller transfer function is r{S) S{S+3+K+l. Solution.2 x + u + pt -4 r -5 0 02 04 a s 04 1. Rather. (b) command step response and (fistufbMce step response The extended estimator equatims are ^ = / i( e .03 23 ) 5 (5 + 3 2 ) . (7. What is the prominent feature of the controller that allows tracking and disturbance rejection? Example Steady-State Tracking and Disturbance Rejection o f Motor Speed by Extended Estimator Construct an estimattN' to ccxitFol the state and cancel a constant bias at the output and track a constant reference in the rootof speed system described x = .J « = [-! .60PP Compute the controller transfer function [from Y(s) to tyfsj] in Example. Solutioii.4 14 1.0 U 1.i.) . w e destitn die control law by iiniorins the equivalent Place the control pole at s = —5 and the tw o extended estimattMr poles s = -1 5 . we notice by inspectitMi that a gain c f —2 will move the single pole from —3 to the desired —5. which replaces the actual disturbance w and the le fn ra c e r . ][p jJ [o . 61 PP Consider the pendulum problem with control torque Tc and disturbance torque T&. B. and write the estimator equations for Pick estimator gains [ I I 12 /^Tto place all the roots of the estimator error characteristic equation a t -10. plant. and C. that is. but with a constant unknown bias b. derive a control law t place the closed-loop poles at -2 ± j2. b changes from 0 to some constant value. G et help from a Chegg subject expert. S te p -b y -s te p s o lu tio n There is no solution to this problem yet. (f) Introduce the estimated bias into the control so as to yield zero steady-state error to the outpu bias b. (b) Using state-variable methods. and controller) using integrator blocks. Demonstrate the performance of your design by plotting the response of the system to a step change in b. B. and C. (Here = 4. &-+ 40 = Tc + Td. ASK AN EXPERT .) Assume that there is a potentiometer at the pin that measures the output angle 0. (a) Take the “augmented” state vector to be where w is the input-equivalent bias. Problem 7. show that the characteristic equation of the model is s(s2 4) = 0. (e) Draw a block diagram of the complete closed-loop system (estimator. (d) Using full-state feedback of the estimated (controllable) state-variables. for the matrices A. (c) Show that w is observable if we assume that y = 0. Write the system equations in state-space form. Thus the measurement equation is y = 0 + b. Give values for the matrices A. 0000 3.. (7). y = [l 0]* (6) Write the state description matrix from equations (5) and (6). title(Tracking performance demonstration').. •*.t. Bm=[0..2)C].x) AAm=[Am zeros(2. (1) Consider the state space equation of the system. B=[0. the model control law is designed and its tracking performance is shown in [Figure 1 . Problem 7..N. K= 8.eye(N)) kron(eye(x).* . (4).0 0]. M=reshape(xx(1 :N*x). ylabel('r..A)-kron(Am'. plot(t. (17) Step 2 of 5 ^ Write the MATLAB program for design the control law and plot the tracking performance of the system from equations (3).. *-=[o 3 “-= [i] Consider the output equation of the system...t): plot(t..B. kron(eye(x). nicegrid.CCm. Design a model-following control law and demonstrate its tracking performance.zeros(2.pc) [N. Step-by-step solution step 1 of 5 Write the general state space equation of the system.1]: C=[1 0]: D=[0]: pc=[-2+sqrt(-1 )*2. Place the closed-loop poles at s = .1]: Cm=[1 0].. C ......C . "-=[o Write the state description matrix from equations (9) and (10).1)]... (15). Dm=[0]: A=[0 1..B).. y= C x . aa=[kron(eye(x). K=place(A. [x..2 ± J 2 ...62PP Consider the servomechanism problem where we wish to track a ramp reference signal..'-’) xlabel('Time (sec)'). DDm=[0]: sysm^ss(AAm. (4) Write the general output equation of the system. *■[: -J. + B ^ .t]=impulse(sysmf. s = ...m]=size(B): [p. x . The plant and the desired model equations are y= [ 1 o ]i *“ = [ 0 0 ]* ■ • y« = [ 1 0 ] i« .01:5..m).. cic Am=[0 1.( 9 ) Consider the value of matrix jfc^in the system.Cm(:)]: xx=aa\bb. (12). = [l 0] (15) D .ymf) hold on. . C = [l 0] (7) D -[0 ] (8) Consider the matrix equation in the system.= [l 0 ]* . bb=[zeros(x*N.DDm): t=0:. J=zeros(p.D)]. hold on.-2-sqrt(-1 )*2].N]=size(C). y .2). y ..x]=size(Am)..0000 M= 10 01 Step 4 of 5 ^ The tracking performance from MATLAB output is given below. = [ 0 ] . BBm=[Bm. Step 3 of 5 The output of the MATLAB program is given below. *■[! Write the state description matrix from equations (1) and (2).y'). (11). ..BBm. (13) Consider the output equation of the system.. CCm=[zeros(1.C) kron(eye(x). (5) Consider the output equation of the system.2 ± j2. (14) Write the state description matrix from equations (13) and (14).B*(N+K*M) A-B*K]. x = A x + B u . [ymf.. (16) and (17). (8)...1). Figure 1 Step 5 of 5 ^ Hence.. (16) Consider the value of closed loop poles. = A .0-1]. + D . . Rewrite the equation. equivalent to the standard one with the addition of a cross­ Show that this performance Index Is > weighting term between the control sand the state of the form <x> where " J 1 0 = (CA . (CA-A. both the equations are in scalar quantities and are equal..A«C)’'Q i (CA .A .C ) x ]+ (7) [ii’' ( C B ^ -Q .. J = J ( ( y .A . y = J(x ’^Qx+2a^Sx+i/^Ru)<*. called the implicit model Z = A/77Z.A .A«C).C y Q. We may minimize a modified LQR performance index j f “ [ (J . .A .A ..CB. ( C A . Problem 7.. J = J ^ (C x .(1) y = C r .y ) + a.y ) + (5) Substitute equations (2). C ) x ] + y = [ [2 « 4 (C B y Q . C B (C A ..A . [ x ^ ( C A . in term 2. -C B b + ii’'R ii] Here.. f i = R + B’'C’’Q.A ^ / q . (3) In equation (5). (2) > = C i .C ). C ) x ] + y = J [x ^ ( C A -A ..C ) ’ -Q .c y Q.weighting term In the equation.C B Rewrite equation (7)..A ^ C x ) + ( 6) Substitute equation (1) in equation (6).C)] S . it is proved that LQR performance index equation is equivalent to the standard equation after adding the cross. S= (CA . (y .63PP Suppose we wish the closed-loop system to behave like a desired model. Rearrange the equation. Q . C B ( C A .. C B (C A -A ..C ) ’ -Q . x = A i + B i / . Step-by-step solution step 1 of 1 Consider the general state variable form of vector equations.A .C ) R . .A . (8) Equation (8) is similar to standard form of LQR equation after adding the cross-weighting term between the control and state form... » .A ..R + B ^ C ’ Q . •C B u + ii ’'R b ] Now substitute the following vectors in equation (7)... (3) Consider the general form of LQR design.C )x ]+ dt [ii’' ( C B y -Q .( C A .. [ x ^ ( C A . Therefore.. CBH+i(’ (C B y Q.. y f Q . Q -[(C A -A . J = J (* ’'Qx + u’'R « ) * (4) 0 Consider the modified form of LQR performance index.A ^ C x ) Q j (C x .A .. A + B K i)-*K 2 (*I-A *)-*B p . Ooxd-loop ityHamics Ftet^ntwti dynamics This indicates that the feedfonward dynamics may be used to improve the transient response of the system. Suppose we now drive the model as follows i = A **+ where up may be the pilot input in an aircraft system.C ( j I . that is. ASK AN EXPERT .K .Q iC Qi J- (b) Which state variables of the system are uncontrollable? Is this result surprising? (c) The optimal control is of the form H = . Problem 7. G et help from a Chegg subject expert. (a) Show that this performance index can be converted to the standard one by augmenting the states of the plant and the model and again choose the augmented state vector. ] ' ' " L .64PP Explicit Model-Following: Suppose in the LQR problem. we wish the closedloop system to behave as close as possible to a system of the form which represents the model of desirable dynamics. with no explicit model? Step-by-step solution There is no solution to this problem yet. which means that the model’s equations must be implemented as part of the control law. f = [xT zJ]T and write down the system equations to show that write down the system equations to show that J= j + 0 where r C ^Q iC -C ^ Q .i)’ 'Q |C y -l) + ll’ ’K«j<it.X -K 2 Z . (d) What are the transmission zeros of the overall system? (e) What is a possible disadvantage of this scheme compared to the standard LQR. We may choose a performance index of the form 00 7= ^ j(J r . Show that = . where C(s) = ^ s ( i + l ) ( i + 2) The Smith compensator for this system is given by Piot the frequency response of the compensator for 7 = 5 and Dc(s) = 1.24 Step-by-step solution There is no solution to this problem yet. ASK AN EXPERT . and draw a Bode piot that shows the gain and phase margins of the system. G et help from a C hegg subject expert.65PP Consider the system with the transfer function e-TsG(s). Problem 7. l ) Step 2 of 5 Sampling period T=lsec to convert H (z) to H (s ) putting 2+sT z = ------- 2-sT _ 2+s 2-s Step 3 of 5 3(6+s)(2+s) “ (2+3s)(S+2s) _3(s^+8s+12) ~ 6s^+28s+16 3(6+s)(2+s) (2+3s) (8+2s) _ 3 ( s ’+8s+12) 6s^+28s+16 3 (s“+8s+12) 1 (s’+8s+12) Step 4 of 5 Thus the natural frequency L«1.6329rad/sec 14 A nd the Damping ratio is given by 2^ cd^= .l4 k .01 PP The z-transform of a discrete-time filter h(k) at a 1 Hertz sample rate is l + (l/2 ). we get 3 . l y ( l C . 2 ) = u ( k ) 4 . Problem 8. Find a difference equation reiating u(k) andy(k). ( k ) = .-' H& = [1-(1/2)J->U1 + (1/3)Z-']' (a) Let u(k) and y(k) be the discrete input and output of this filter.‘Y ( z ) 4 a '" Y ( a ) = U ( z ) + ^ z ‘U (z ) 0 0 Z Taking inverse z-transform of both tiie sides.lz . tile transfer function Y (a )_ '■'I 1+ '+1 TJ(z) U(z) Y ( z ) .j 3x2 1.6329 |t= 14289| Step 5 of 5 Poles of the filter are 2 8=— 3 and : 2 As botii the poles are in left half plane. tiie filter is stable! . ly ( k . As u (k) and y (k) are the di screte input and output of the filter. (c) Is the filter stable? Step-by-step solution step 1 of 5 (a). (b) Find the natural frequency and damping coefficient of the filter’s poles. l ) . ----.l .2) [ ( .+ 24 2 (^ -0 z '.l ) ' 2z ( z . .. ib <0.3 z .\ ) '* B ( z .k ) | . » ( Apply z-transfonn to this condition. 2 for x ^nd _2 for C in the equation (3).. 4 ( 2 .’l / ) ( r ) y (z ) 2 ( z .l) ' n o “ ( z ..2 ) (3) Step 4 of 5 Put 2 = 2 in equation (3).l) + C (2 -2 ) ( 2 .z .2 ) ( 2 . y ( i) = 0 .> (* ) -3 .2 ) ( l.2 ) it ^ O fc< 0 > (k ) = 0 k<0 Step 2 of 5 Apply the z-transform to the difference equation. Problem 8.2 ) 2^A A=2 Put z — 1 in equation (3). .2 ) ( 0 .1 ) . (2) ( z .2) 2 .2 ) ( z .1 ( 2 .2 ) 2+2B +4=2 -I B ^ -2 Step 5 of 5 Now substitute 4 for ^ .(t..< (l-l) ' + a ( l ..l ) ' + B ( 0 .l) ' > y (^ ) 2 Z z-2 ( z . K(t) = 0..>(z) = 2(z-' .02PP Use the z-transform to solve the difference equation y (t) .2 ) ( 0 .\ ) + C { z ..+ .2 ) ( 2 ..2»(* .2) = 2«(* . (1) ‘ z-2 Step 3 of 5 We know that.l ) + C ( l .2 ) 2— C C = -2 Now substitute z . y (z )-3 z-'y {z)+ 2 z-^ y { i) = 2 !-'U {z)-2 z-^U (i) ( l..' + 2 r-’ ).l ) ( z .2 ) = 2 i(( * -l) -2 » ( * .2 ) ( 2 .'.< ( 2 . .3 z + 2 _ 2 (£ -1 ) ( z . ( z . the solved difference equation using z-transform is |y ( k ) = 2 ( 2* . Step-by-step solution step 1 of 5 Step 1 of 5 Consider the following data.l) '+ g ( 2 . where *> o . *< 0 .l ) ' ” z .3y(* .l) Substitute -------T t / f z ) in the equation (1)..r ^ ) U (z) l-3 z -'+ 2 z -* A 1.l ) + ( . Difference equation.l) + 2 y (* . to obtain the constant j i ..1)* 2 .24 . V 2z 2z 2z Now apply the inverse z-transfonn above equation.T . to obtain the constant C ■ 2 = . 4 for ^ and « 2 for C 'n the equation (2).l) * + 5 ( 2 .l ) ' Apply partial fractions to above equation.. A B C Z + .2). 2 = ( 2) ( 0 . 2 = .2 ) ( z .2 2 . y(*) =2(2‘ -l-* ) Therefore. 1 ..1) + 2y(* .1 ) '“ (z -2 )(2 -1 )' 2 ^ A { z . ( z .l ) + C ( 2 .2) ( z . . / ) z ^ " ‘. ^ ( 0 )| Step 2 of 5 0 >) We know that. if u(k) represents the kth Fibonacci number...q z '^ l------------- .6 1 8 | Step 4 of 5 We know that U (z) = . where i + l = J Z [ / { k + \ ) } = z Z / O ) z .c ii E7(z)= ^--------- -----I=r T ----.^ " ) a ^ -O i . z ^ U (z ) .618.618 Thus.O jZ ‘ ^ Finding the invers e z transform of ^ ( z ) : u [k ) = . (c) Compute the pole locations of the transform of the Fibonacci numbers..j^ Thus.z -1 (c) Find the poles ofthe system U (z) = 2 .za (1) . the poles are = 1.l)C /(^ ) = z)a(O ) + z a (l) Step 3 of 5 Since.U { z ) = 0 { z ^ .j ---------^ We can rewrite this as Z/ (z) = ----.. O j.. Hence. Step-by-step solution step 1 of 5 w Find the z-transform o f / (^+ 1 ) z { /{ k + i) ] = ± / { k + i) z .z .o f+ —S — of O i.z^u (0) . a ( * ) = U . Let u(0) = u(^) = [Hint: You will need to find a general expression for the transform of f(k + 2) in terms of the transform of f[k).0 0 .^ Z { / ( * + l ) } = 2 / ( .C ^ Thus.. I < 1. Step 5 of 5 Apply the principle o f partial fraction to U (z) = ----------. a ( i+ 2 ) .: ^ f ( 0) 0 Thus. (b) Use the one-sided transform to solve for the transforms of the Fibonacci numbers generated by the difference equation u(k+2) = u(k+^)+ u(k).. and ot^ = -0 ...3. for large jbthe second term is s 0 .z f ^ ). we get U (2 ) = z^. we get the quotient as l-fz “*-Fz"® +z~^ +.. z^ —z —1 U (z) = ( 1 .5.618. 2.2 V (0) .a ( jt ) = 0 ■We have Z [ / ( i + 2)] = I V (z) .z .03PP The one-sided z-transform is defined as 00 F ® = !]/( * » ■ * ■ 0 (a) Show that the one-sided transform of f(k+ ^) is Z{ f(k + 1)} = zF(z) ..^ ^ . we get («) S i n c e . W e g e ty ( 2 ) = .] (c) Compute the pole locations of the transform of the Fibonacci numbers. and the ratio o f a (^+ 1 ) to u {k )is .1 = 0 1 ± V5 s 1.z . (d) Compute the inverse transform of the Fibonacci numbers.JeT (1) Obtain the z transform and simplify further. then the ratio u(k + 1)/u(k) will approach • This is the golden ratio valued so highly by the Greeks. a (0 ) = a ( l ) a n d a (l) = l .' . and -0 .O i O i... we g et |z(/(A + l)) =zF(s) .[ z U (z) . z z By the long division rule.1 z^ . Problem 8. (e) Show that.a ( i+ l) .za (0 )] .T ^ ^ 1 . . go] The m q ^ in g in z-plane stability gives rise to.04PP Prove the seven properties of the s-plane-to-z-plane mapping listed in Section 8. I f is decreased. Step 9 of 10 (7) Considering N yquist f r e q u e n t and decreasing the value o f O ' from 0 to OD.d ie p o k w illb e sh ifie d a lo n g d ie p o s itiv e r e a I a x is in d ie z - fd a iie to w a r d sth e p o in tz = l. d ie m ^iping o f z-plane is. z=eT^ = ^ ( c x 3 B d r + ysmtsET) Since.3. = e . ml Step 7 of 10 A t z s ± y — . z-^lane p ole at Z^Zj at the wwnplwg o f equivaleiit s i l a n e pole The equivalent characteristics in tfie z-plane are related to s i l a n e b y the Where T is the sanq)le period The equivalent characteristics in d ie z-plane are related to s-plane b y the Where T is d ie sanqile period The follow ing is the m^>ping betwe en the s-plane and z-plane b y z —^ Imagioafy part o f s ila n e Imaginary part o f Z-plaoe step 2 of 10 ^ (1) The stability boundary in s^^ane is given by.f lane w ould b ^ z=0 I f G)g is in c ie a s e d .f = 0 . Hence. Problem 8. w e can infer diat the negative real z-axis always represents s fiequeiKty cd' ^ in s-plane. Step 6 of 10 (5 ) N o w considering different vertical points in d ie s-plane w e can prove diatd iese vertical poiiits map w idiin d ie unit d i c k o f d w z-plane. Step 3 of 10 ^ (2) h i d ie sm all victnity around z = 4-1 is essentially identical to d ie v id n ity around s = 0 in d ie s-plane.a „ . z= e * (coBJT—7 s m x ’) 0*0 ->G0 . . M = (c o s n '+ y s in . Step 4 of 10 (3 ) The stability boundary in s^^ane is given by. = ( coB^ + j s in x ’) If -> o o .ff')+ sin ^(ffiir) =Ji =1 H ence d ie poles located o n the unit circle in d ie z-plane are equivalent to the pole location on d ie imaginaty axis in the s il a n e Thus the unit d i c k s in z-plane rqnesent d ie stability boundary.2.r ) =-i+yo From d ie above expression.oo.f lM ii« " = l The magnitude o f z gives [z| = ^oo8^(fi.- = {C 0 8 X ± J ^ X ) =-l Thus. each period w ill be sanipled less disn tw ice die average.^ l= e ^ This resuh is true udien s = 0 or the sampling tim e T = 0 . s = 0 ‘+ j& F<xall <9 between [ . radier dian to tim e as in s-plane Step 5 of 10 ^ (4 ) Consider a value o f in d ie s-plane and substitute inth erelatian o f sa n d z-plane w e have . A t z = 0 . z . I f d ie sampling rate is small.J ^ The m ^iping in z-plane cane^m iiding to this point is. as d ie sanqiling tune cannot be considered as zero w e crmsider die v id n ity around d ie s-plane a s zero. the pole w ill b e shifted aknig the z-plane towards die p oin tz = 0 . d ie mapping in z-t>lane w ould z= 0 Step 10 of 10 Atpoiiil s = . d ie mapping in z . vertical lines in the le d h a lf o f the s^dane m ^ into circles within d ie unit ctccle o f z-plane. the sanqiling rate is or greater than Nyquist frequency diere w in be over lap. Hence. 5 = 0 *+jeo F<»all a> between [-<x>.oo] The m q ^ in g in z-plane stability gives rise to.d ie m ^ ip in g o fz -iila iie is . = e '( c o 8 « ( r + j ^ a i r ) I f c r = 0 . die z-plane locations g ive reqxm se information normalized to d ie sample rale. A t point The m q ^ in g in z-plane ccHie^Kinding to this point is. w e can infer diat. Step 8 of 10 ^ (6 ) Tn tK w iy n n ta l ling-g t i a w I w m d r a w n d i a t T n y rg a w it a r o n g t a n t t f n a g tn a f y part s = ja> z=e^^ s ( gos<0 ^ + y s tii< 9 r) ( 2js = cosm— + j8tn<9T L ) These lines are mapped onto to the radial lines in z-planes as the phase in the m qiping o f z-plane is. e ' i s a crmstant value. form die above points w e can infer diat. form d ie above pdn ts. d ie vicinity around z = + l is identical to vidnftyaround. d ie horizontal points in the a- plane radial lines in the z-^lane. Step-by-step solution step 1 of 10 ^ A.. Problem 8.25 + 1 D c (s ) = 50s+ l Using the matched pole-zero approximation.05PP A unity feedback system has an open-loop transfer function given by 250 The following lag compensator added in series with the plant yields a phase margin of 50°. 1-e® |z . 1-e® =K.e “ ” | .25) HI 1. z-c’ ' At low frequency ency 1 (-1." 50 . The MPZ iq>proximation is 1.25x50 or K. s/1. Step-by-step solution Step-by-step solution step 1 of 1 The compensator transfer function is given by ” r-l D (z>. determine an equivalent digital realization of this compensator. Thus the digital compensator transfer function is I 7 M 7 ^ . I. = i2 5 _ 50s-l s-1.25 i. ) (b) Using a log-log scale for the frequency ranges = 0.082z‘^) (z-0.082) Step 7 of 7 lm(z) .111 and zero is: 9-7z’^=0 2^=1=0.=-0. : 4. and compare with H(s).06PP The following transfer function is a lead network designed to add about 60° of phase at o. and compute and plot in the z-plane the poie and zero iocations of the digital implementations o1 H(s) obtained using (1) Justin’s method and (2) pole-zero mapping.ls+l and T=0. {Hint Magnitude Bode plots are given by \H(z)\ = \H(eja)T)\. l^ vAere. o ^ ’** and P=e** -O T7Q••1^ H(z)=4. compute the amount of phase lead provided by the network at z 1 = ej(jj1 T. = 0. Step 6 of 7 ^ Let the digital i^proximation of (s) be Hp (z ) . (b) Using a log-log scale for the frequency rangeo..1 tocu =100 rad/sec.(1) '■ ' 1. H (z)^ # )* ' .) Step-by-step solution Step 1 of 7 Given that s+l O.2 5 sc c d is given by 0+1 ^ o+ioj 1-e-'" =lx 1 K.25 sec. and compare with H(s).s) = O .ls + l (a) Assume a sampling period of T= 0...25sec Step 2 of 7 (1) Tustin's Method 2 ” 0.( s ) = .1497 / (l-0.1497 (z-0..25 [ l + z ‘ 3 Step 3 of 7 Therefore.1497 T h « e fo « H ( z ) = K .2z‘ Thus the poles of H (z) are 1. here T ^ .1 tocu =100 rad/sec. . then H. {Hint Magnitude Bode plots are given by \H(z)\ = \H(eja)T)\.778) 4.1 = 3 rad/sec: s+l H(.8+0. Problem 8.8-+O.778 Step 4 of 7 Step 5 of 7 (2) Matched pole zero method s+l H . plot the magnitude Bode plots for each of the equivalent digital systems you found in part (a).2 2.2z'^=0 z ‘= H 0. plot the magnitude Bode plots for each of the equivalent digital systems you found in part (a). For each case. Problem 8.07PP The following transfer function is a lag network designed to introduce a gain attenuation of 10(-20 db) at w = 3 rad/sec: lOs+l H{s) = 100s+ 1 (a) Assume a sampling period of T= 0.97531 / / ( z ) = 0.01 to 10 rad/sec. (b) For each of the equivalent digital systems in part (a). (b) For each of the equivalent digital systems in part (a).25 sec. Step-by-step solution step 1 of 2 w Let lOff+l H {s y - lOOs+1 |/ / p ® ) | = ® = 3 = 0 . plot the Bode magnitude curves over th frequency range w = 0. compute the amount of gain attenuation provided by the network at z1 = eyojl T .99750 Step 2 of 2 (b) All there are es sentially the same and indistinguishable on the plot because the range o f intere st is below the half s an^ le frequency .01 to 10 rad/sec. plot the Bode magnitude curves over th frequency range w = 0.- + 0. and compute and plot in the z-plane the poie and zero iocations of the digital implementations o1 H(s) obtained using (1) Justin’s method and (2) pole-zero mapping. For each case.10112.1 0 0 1 (-2 i^ ) Tustin's method -0. 1.8 ± y ^1.01)sysd=feedback(sys.i. Figure 2 Step 10 of 10 Write a MATLAB code for step response at 7 = 0.28.'+ z .2.888 .den=[1 0 -1].1 .S ±J.2886 .sys=tf(num.6(1 + CTj ) + 8C)y = 0 Equate the real and imaginary parts to zero. (1) o . r = o .^ j 8 : .422z-' + 12.71.97-284." ) + 0 .422z-' +12.2.284.. 1 7 .0 .+ f .z " ' 1 .404]. 3 6 r „ + 0 . for the matched pole zero approximation add a pole at 2 s -1 . ^ + 0 .1 1 .042 .r ).28 *0. 1 :^ 6 . 8 1 7 j r T C (r) = Step 5 of 10 Calculate the expression D ( z ) at r = i .>i— >0. Problem 8.7 ')J {c o 8 (2 a > ^ )-l} Step 7 of 10 Equate the DC gains." ) + l : ^ ( l .042 -28.den=[1 0 -1]. i .den=[1 0 -1].7’)+ e '^ * * ^ j{c o s (2 ffl|. 0 5 r ( l + 2 z .8 8 :+ — = 0 2| 1. Map poles and zeros according to the relation z — (z + l)(z -l) There is no DC gain for this transfer function D ( z ) . Substitute —[-i— for every occurrence of s in Z)(5).3 4 5 4 z ‘' l. 7} ■ 10.jz .0.z .367„+ 0 .1)sysd=feedback(sys.2886 . Step 9 of 10 Write a MATLAB code for step response at 7 = 0. 7 = 0. 1 <?w =- 5 ( 5 + 1) Continuous PID controller: Evaluate the characteristic equation. r = 0. Calculate the values of ^ . l+ z " 'J l.den.7432Z-' .97 . match the gain of D ( z ) at 5 = y<v^(closed loop natural frequency).1)step{sysd) Step response for 7 = 0.345].7432 -0. 8 1 7 ( l.r)-(e~^+e"^)co s(iD .3454].0. (z + l)(z -l) N -1— 4< = |cos(2ffl. a: = 1.3912 Therefore.1..sys=tf(num.1 sec • num=[16.422 12.den.) . 3.528-(0. 8 1 7 + i^ ^ + 0 . Therefore.97 . the required specifications are met with s ~ -O .Therefore.r) B = {c o s ( 2 a i.1)step(sysd) Step response for 7 = 0.6 ( i+ 8 :r „ ) + 8 := o 0. 3.5 and wn > 1 rad/sec. ii. ^ -------------^ --------------.0 . D ( z ) = K . = 0 . r = o .345Z-' 1 -z ^ Step 8 of 10 Sketch the step response for each of the digital implementations at 7 = 1. = 0.0.404z-“ 143. = 1 .01 sec using MATLAB.0. 7 = 0.2886 -2.284. S t ^ Response Figures .36(1+ATo ) .32z-' + 140. and 0. 0 5 r ..5 sec Select the dominant closed loop poles to exceed the specifications. TD.r ) + e '^ * * ^ } s in ( 2 ffl. 0 5 7 . 7 = 0.1 >s shown in Figure 2.'+ ^ i.r ) + {s m (2 < » ._L_=o j(j+ l) * ’ + ( l+ A T o ) s ’ + « j + — . Discretize the PID controller using (a) Tustin’s method (b) The matched pole-zero method Use Matlab to simulate the step response of each of these digital implementations for sample times of 7 = 1. Write a MATLAB code for step response at 7 = 1sec • num=[3.36(1 + ATj.01 is shown in Figure 3.9 2 .^ ® ^ ^ ^ + 0.817. and 7/ so that the closed-loop poles satisfy ^ > 0..0.345Z-* 1 -Z -" Step 6 of 10 (b) Observe the PID controller transfer function G ( z ) = 8 :( i+ v + ^ ) There is one more zero than pole. the transfer function of continuous PID controller is.0 . num=[143.1) sysd=feedback(sys.2 z .8 ± y with (ta .74325-' -0.8 ± J .042 ." 16.625 Therefore.32 140.34542-* l-z -' 16.1.32z-' +140.92 + 8: 1+8T„ (2) 1.3 9 1 2 s + ^ j Step 4 of 10 (a) Discretize the PID controller using Tustin’s method.01 sec..01 sec.888 .>- Calculati Calculate the expression Z )(z) at r = l.r)-l}4 jsin (2fii'.. Solve equations (1) and (2) by substituting 7} s 10.6 Arbitrarily choose the value.den.0 . l + D ( j) G ( i) = 0 l + Xf--------- . D (5 ) = 1 .r) + « "^ )s in sin (2fl).404z'' l-z 143. Consider s = . step 3 of 10 Evaluate the characteristic equation at » = . Therefore. 8 .r)-(c “^ + < “^ )s m ((» . 9 2 -1. i ^ 1.2 $ 2 !t sec Compare » = . Figure Control system Step-by-step solution step 1 of 10 Write the plant transfer function. 8 ) 8 : + ^ = 0 ‘t 1.(e “^ + e “^ ) c o s ( fi> .1) step(sysd) Step response for j* = 1 is shown in Figure 1.0 . ®.1.z . 3 9 1 2 [|( ) ) h 10 T ( .0. r ( ^ i+ 2 " 'J ''' = 1.08PP For the system shown in Fig.8 1 7 ^ 1 + 0 . find values for K.sys=tf(num.01 sec.817 1+ 0'■ .528 = ^ 0 .28.' + z-*) l-z -' . 0 Step 2 of 10 Consider the required specifications. 1 .8A:+ + ( 0 .o i sec. Figure Control system Step-by-step solution There is no solution to this problem yet.4 ) ‘ (a) Find the transfer functionGfzJ for r = 1 assuming the system is preceded by a ZOH. G et help from a Chegg subject expert. ASK AN EXPERT . (b) Use Matlab to draw the root locus of the system with respect to K. where the sampling switch is always closed). where. Problem 8.5 for the continuous case. Which system has a larger allowable value of K7 the case in which an analog controller is used (that is. where the sampling switch is always closed).1 . (c) What is the range of K for which the closed-loop system is stable? Compare your results with the case in which an analog controller is used (that is. Which system has a larger allowable value of K7 (d) Use Matlab to compute the step response of both the continuous and discrete systems with K chosen to yield a damping factor of ^ = 0. 40(j + 2) G(i) = - ( i+ 1 0 ) ( P .09PP Consider the system configuration shown in Fig. ^ -------. Figure 1 Satellite control schematic 6 = angle of the satellite axis with respect to an inertial reference with no angular acceleration. The equations of motion of the system are given by I6 = Mc + M d . Figure 1 Satellite control schematic MD = disturbance torques.383 for r= ls e c ={ 0.63) (c) D ( z ) = . To gain Insight into the three-axis problem.1 OPP Single-Axis Satellite Attitude Control: Satellites often require attitude control for proper orientation of antennas and sensors with respect to Earth. we often consider one axis at a time.692 (d) k= 1. and obtain e = u+Wd- Taking the Laplace transfonn yields 1 e {s ) = ^ [H ( s ) + W d ( s ) l.^ (z+0. We normalize the equations of motion by defining M e M d u = — . without disturbance. (c) Add discrete velocity feedback to your controller so that the dominant poles correspond to ^ 0. where motion is allowed only about an axis perpendicular to the page.\ T^(^+l) step 3 of 4 \ k(z~ 0. MC = control torque applied by the thrusters. we can use the methods described In this chapter to obtain the discrete transfer function r z+l 1 «(z) 2 L f e . (d) What is the feedback gain if T= 1 sec? If 7 = 2 sec? (e) Plot the closed-loop step response and the associated control time history for 7 = 1 sec.D ^ J ' (a) Sketch the root locus of this system by hand. Figure 2 Communications satellite Source: Courtesy Space Systems/Loral (SSL) Step-by-step solution step ^ of A ^ (a) The loci braches depart vertically &om z s 1 Step 2 of 4 y-» I.5 and wn = ZnAOT.-0. becomes In the discrete case in which u is applied through a ZOH.113 Step 4 of 4 0.3458 for 7 . Wd = — .44y. Problem 8. (b) Draw the root locus using Matlab to verify the hand sketch. which.44) z*0. Figure 1 depicts this case. MD = disturbance torques. assuming proportional control.2 sec . Figure 2 shows a communication satellite with a three-axis attitude-control system. where / = moment of inertia of the satellite about its mass center.44±0. and neglect higher-order terms. Problem 8.4 N/A. IC\ = 20 N/m. and let the sampling period be 0.1 sec. expand f about x = 0 and / = 10. and K2 = 0. (8. and show both X and the control current /. and discuss the possibility of using your closed-loop system to balance balls of various masses. neglecting the nonlinear terms in f(x. 2. and K2 = 0.02 sec. Compute the transfer function of the equivalent discrete-time plant.54) are m = 0.11 PP It is possible to suspend a mass of magnetic material by means of an electromagnet whose current is controlled by the position of the mass (Woodson and Melcher. ASK AN EXPERT . we obtain the linearized equation Reasonable values for the constants in Eq. The equations of motion are where the force on the ball due to the electromagnet is given by f(x. (d) Plot a root locus with respect to /(I for your design. G et help from a Chegg subject expert. and a photo of a working system at Stanford University is shown in Fig. lc\ = 20 N/m. (c) Design a digital control for the magnetic levitation device so that the closed-loop system meets the following specifications: fr< 0. Suppose we let 10 represent the current at equilibrium.02 kg.4 N/A.02 kg. If the sensor can measure x only over a range of ±1/4 cm and the amplifier can provide a current of only 1 A. The schematic of a possible setup is shown in Fig. (a) Compute the transfer function from / to x. 1)7 Figure 1 Schematic of magnetic levitation device for Problem 8. (e) Plot the step response of your design to an initial disturbance displacement on the ball. te< 0. 1968). (8. If we write 1 = 10 + /. At equilibrium the magnet force balances the gravity force. Reasonable values for the constants in Eq.4 sec. 10. (b) Assume that the input is passed through a ZOH. and draw the (continuous) root locus for the simple feedback / = -Kx. where / represents a deviation from the nominal current.1.11 o Figure 2 Magnetic ball levitator used in the laboratory Source: Photo courtesy o f Gene Franklin Step-by-step solution There is no solution to this problem yet.54) are m = 0. what is the maximum displacement possible for control. and overshoots 20%. I). Step-by-step solution There is no solution to this problem yet. (e) Give the Matlab response of your final design to a reference step. Problem 8 . Assume that the output y is sampled. • Theresponse to a reference stepinput is to have a rise time of no more than 0.4 sec. G et help from a Chegg subject expert. (a) Design a lead compensation that will cause the system to meet the dynamic response specifications.1 2PP Repeat Problem in Chapter 5 by constructing discrete root loci and performing the designs directly in the z-plane. and the sample rate is 15 Hz. • The steady-state emor to a unit ramp at the reference input must be less than 0.05. • Theresponse to a reference stepinput is to have a rise time of no more than 0. • Theresponse to a reference stepinput is to have no more than 16% overshoot.4 sec. ASK AN EXPERT . (d) Give the Matlab plot of the root locus of your final design. the input u is passed through a ZOH as it enters the plant. Problem A servomechanism position control has the plant transfer function 10 G(i) = - f( j+ l) ( f+ l( ^ You are to design a series compensation transfer function Dc(s) in the unity feedback configuration to meet the following closed-loop specifications: • Theresponse to a reference stepinput is to have no more than 16% overshoot. ignoring the error requirement. (b) What is the velocity constant Kv for your design? Does it meet the emor specification? (c) Design a lag compensation to be used in series with the lead you have designed to cause the system to meet the steady-state error specification. these arise to some extent from bearing and aerodynamic friction. Assume that J = 600.0265.1 and 2 and described in Problem. and 2000.m.6299 . The equations of motion are J0 + B$=Te. but mostly from the back emf of the DC drive motor.400. Problem 8.^'= 0. where Tc is the torque from the drive motor.« * — / 30 B We get u (s ) s(30ff-l-l) From the specifications ^ > 0 . (c) What is the maximum value of K that can be used if you wish to have an overshoot Mp < 10%? (d) What values of K will provide a rise time of less than 80 sec? (Ignore the Mp constraint.000 kg m2 B = 20.5 ( f i) ) s = -0. Do the plots to confirm your calculations in parts (c) and (d)? Figure 1 Satellite-tracking antenna Source: Courtesy Space Systems/Loral S te p -b y -s te p s o lu tio n Step 1 of 2 (a) The equation o f m otion is "Where J = 600000 kgm^ B = 20000N. (a) What should the sample rate be? (b) Use discrete equivalent design with the matched pole-zero method. where K is the feedback gain.0225 Step 2 of 2 (b) u ( s ) = * ( 5 ( f i) .0205^ The corresponding natural frequency and damping » .36 You wish to control the elevation of the satellite-tracking antenna shown in Fig. Figure 1 Satellite-tracking antenna Source: Courtesy Space Systems/Loral ve: uounesy apace aysiems/Lorai Problem 3.sec I f we define B 1 7! a * —* — . (a) Find the transfer function between the applied torque Tc and the antenna angle 6 . Find the overshoot and rise time of the four step responses by examining your plots.13PP Design a digital controller for the antenna servo system shown in Figs.2. The design should provide a step response with an overshoot of less than 10% and a rise time of less than 80 sec. (c) Use discrete design and the z-plane root locus.0167±0.) (e) Use Matlab to plot the step response of the antenna system tor K = 200. (b) Suppose the applied torque is computed so that 9 tracks a reference command 9r according to the feedback law Tc = K ( d r -d ).000 N m sec. The antenna and drive parts have a moment of inertia J and a damping B.1 and Fig. = 0. Find the transfer function between 0rand 6 .1000.5 4 (21^ >0. 8564±0.70 D (z ) = i _ 5 _ ^ ^ z -1 .1278j = 2. ___ __ z+0.07rad/sec.7 4 0 8 ) ( z -0 . Problem 8.9 9 ) ^ Is e c ^ 1 .1)(s+3) ’ . Using a z- plane root locus.9019 ■ ^ ( z .8 rad/sec z = 0.1 )(* + 3 ) is to be controlled with a digital controller having a sampling period of 7 = 0.^=0.14PP The system 1 G(*) = ■ (i+ 0 . design compensation that will respond to a step with a rise time tr< 1 sec and an overshoot Mp < 5%.0 . What can be done to reduce the steady-state error? Step-by-step solution step 1 of 1 Step 1 of 1 Continuous plant 0 W = (s+0.1 sec. « (jt) .15PP The discrete transfer function for pure derivative control is D d (z )= K T D ^ -j^.\ a (z ) Z)(z) = * . where the pole at z = 0 adds some destabilizing phase lag. Problem 8. Can this phase lag be removed by using derivative control of the form D4(0 = KTd ^ ^ ^ 1 Support your answer with the difference equation that wouid be required and discuss the requirements to implement it. Step-by-step solution Step-by-step solution step 1 of 1 ' (a) No we can't use derivatiTe control of the form To remove the phase lag The difference equation corresponding to 2.7 i T E{z) e ( A + l ) . For small values o i9 . (/ + R9) s / s in ^ = d ^= 0 We can rewrite this as + g d = 0. and . Let3c = ^5 Find Xi. we get (/ + R0) = I Simplify further.( / ? ^ + g s in ^ J ' 1+R9 gsin . Thus. 9 -\-(g /i) = 0. Step 3 of 4 w This is a second order non linear equation in 9.01 PP Figure shows a simple pendulum system in which a cord is wrapped around a fixed cylinder. Problem 9. and show that for small values of 6 . is the radius o f the cylinder. Figure Motion of cord wrapped around a fixed cylinder (a) Write the state-variable equations for this system. it is proved that ^ + • ^ ^ = 0 I . the state variable equations for this system are -(/Z x j+ g s in x a ) / + Rx2 Step 4 of 4 0 >) Linearize the equation around the point ^ = 0. the system equation reduces to an equation for a simple pendulum—that is. + Aca x^=9 = ^1 Thus. Step 2 of 4 The motion of the system that results is des cribed by the differential equation { l+ R 0 ) e + g s m 0 + R ^ = O Where / is the length in the vertical position. (b) Linearize the equation around the point 9 = 0 . R = radius of the cylinder. where /= length of the cord in the vertical (down) position. The motion of the system that results is described by the differential equation ( i+ R 9 ) e + g ^ 9 + t i0 ^ = 0. Step-by-step solution step 1 of 4 Sketch the given simple pendulum S3rstem in which a cord is w n ^ e d around a fixed cylinder. ^ v ) .4 ) for ig in equation (1). the linearized model about the other two equilibrium points is. (5) = -S i + g (l + Su . (b) Find the linearized model of the system about the equilibrium point u = 1. i and v are the state variables If is the input Step 2 of 5 (a) At equilibrium state.7 in the text book.v ) . /1 = v1 = 0.v ) = « ( l.v . ^ = -/ + g (a -v ) Where.1){vG - 4). 'o .( l.Sv)[Su .^ U J" -5 ± 2 ^3 ^If . Si = .02PP The circuit shown in Fig has a nonlinear conductance G such that iG = g(vG) = vG(vG . Svy a nd Si ■ Tom the differential equations.. + v.S i + .s C v o ) The differential equations are. (a) One equilibrium state occurs when i/ = 1.v ..tfv ) .. (c) Find the linearized models about the other two equilibrium points.i + g (tt-v ) = 0 (2) Substitute ( v « ..5 v ) .... = -g i + .(l-v ) ( -v ) (. . ^^3(-i±S)\A{-X±S)-2 .2 ) = 0 . yielding /1 = v1 = 0..(6) Simplify the expression s -S i + ^ ( l + ^ if .4 ) + v = 0 .-l± > /3 From equation (1). ( " ... ^ =5w 2S dv And. v = 0 .. Therefore.^ v ) further. 'hus. (4) Solve the equation (4).l) ( l.v ) ( l.i + v= 0 ( 1) ^ = 0 dt .A + 3 ^ if + 3 ^ v . the pairs of y and i ^^at will produce equilibrium is. du dv m -5 ± 2 S The linearized model about the other two equilibrium points is.v) ( 8) -g '{ u .dv.\ ± S Therefore.= . ^ = v’ + 2 v . and / by 1 ^if.l) ( if. where / and v are the state variables and u is the input. From Figure 9. .lf ..v ) ( if.i + g ( ii.l± S i = 0 .v for vg in equation (3). V c ..v .*') g '{u -v ) (9) Similarly. « ( " .3 .2 + v(2v + 2) dv = 3 v*+ 4 v -2 Substitute -1 ± for v . d[Sn ■-1 1 r^ /1 0 -1 5 T 2 .. (7) Vrite the equations (5) and (7) in matrix form..5 ± 2> 5J Thus. -1 I ■ O' [s n Su -1 ^ dv.4 ) + v = 0 Substitute 1 for If ....v .57 in the textbook. .Sv){Su . Su -1 Step 4 of 5 (c) Write the general linearized model equation..v ) = v ^ + 2 v -2 Apply partial differentiation.i+ v = 0 i= v 1= o . v = 0 . Figure Nonlinear circuit R=l dv = .v ) + v = 0 v(v* + 2v . The current flowing through the resistor is. g g (« .^ v ) [ ( l + Su .( if. 4 -0 .3) ^ v s .5 t 2 -^ Step 5 of 5 It is know that..4 ] = -S i + (1 + 5if .V Substitute i f ... When u = 1.l ) ( v o .v) And. Step-by-step solution step 1 of 5 Refer to Figure 9. W * " . v. Find the other two pairs of v and /that will produce equilibrium. Problem 9. The state difTerential equations are di . the linearized model of the system at equilibrium point is 'F '. 4 1 ■' i M i .l ] [ ( l + Su ..l± y /3 Step 3 of 5 leplace u. * = Su -1 m u 5 T 2 a^ J L ^ vJ ’ ' [ . . 6 x2 + 2 1 +S u 2 = .. x .= x ... i.l ^ . the initial state (capacitance.(15) 0 = x . Draw the linear circuit using the values of the elements.x ^ ..3 + S u ^-S3 c^)-S3C2 + 2 7 + S u2 . G = ((-3 + * ..j | ) ..* . * .0 + 2 7 = 0 ( l ... m <5 ) Where. ..) Step 3 of 9 Thus. C is the capacitance.(16) Consider the following equation for the G.n t . | .. x .r ( x .-x . G = ( l. i.A.. = L — x... Figure 2 Nonlinear resistance (a) Show that the circuit equations can be written as i | =C<*i + ^ « j t i ...) .. the circuit equation is |i^ = G ( l l | .+ u ^ = C ^ x . The linear circuit model is shown in Figure 1....* ” )’ + 27 = 0 (18) The roots of the Equation (18) are 4 and ^ ( . (c) Draw the circuit diagram that corresponds to the linearized model.= x . Find the equilibrium state = [ •*!• ■*2* ^ for the circuit.x . the value of is 4. + 6 U21. Consider the following equation for the voltage v. Sx^ -{. and inductor current) is slightly different from the equilibrium and so are the independent sources.2 7 ^ x j ..x ^ .58 in the textbook. ^ ... equation (12) and equation (13).) .. = 3^ | and | i . ^ = 27).. the linear circuit model is drawn and it is shown in Figure 1.. .. ^and ^ is represented as loop currents.^ -2 4 =2 Consider the foiiowing formuia for the Equiiibrium state fomi.-< 5 x .59 in the text book and find value.r ( jl^ ) | Step 4 of 9 Consider the following equation to form the Equilibrium state equation.. L is the inductance... C S |= G ( ii | . For a particular input uo.59 in the textbook. (20) S i k j^ ji i+ S x ^ . G { a . ..... / j.))' (26) Substitute « 0 and equation (26) in equation (23).S x 2 .... || ( 0 = I ^ + <U(().... Problem 9. ) .{ 2 + S x2) = SX f-S X 2 -S X 2 Consider the following equation for the G. u^ and U2 are voltage and current sources.r ( j^ + t f a ^ ) (22) Substitute . (b) Due to disturbances. Substitute s 4 and r ( 0 ) = 2 in equation (16). ti|and represented as voltage and current source. (2) (3) 1. s 2 and s 0 in equation (19) x‘ =[4 2 of Thus. | ^ ^ = ^ ] a n d Step 8 of 9 (c) Determine the values of the following elements.. respectively. and R^ and R2 are nonlinear resistors with the following characteristics.. i( W = A < b ) + ii( ib ) ...g . 0 = G ( 1 . an equilibrium state of the system is defined to be any constant state vector whose elements satisfy the relation i| = i2 = i3 = 0 . Resistor 2: v2 = r(i2). ^ " + 2 7 .jcs. any system started in one of its equilibrium states will remain there indefinitely until a different input is applied.j ^ + i i. iif = i. u^ = l and * 2 7 in equation (11). = G ( » .(14) 0 = . voltages.x . equation (8) and equation (10). Do a small-signal analysis of the network about the equilibrium found in (a). the value of r( 0 ) is 2.x * ). ) .. 0 = 4 -..03PP Consider the circuit shown in Fig. i [ i ) .2 . Step-by-step solution step 1 of 9 (a) Refer Figure 9. ( 11) . . Refer Figure 9. x . • V... displaying the equations in the form iii +fai!ct +fasi3 +giSui +gisii2..I± 3 iV 3 ^ - Therefore. (8) Substitute equation (6) in equation (5).59 in the text book... (21) & ^ = ( x | * + 5 x .) (10) Substitute and s i in equation (7). (7) Substitute equation (4) in equation (3). = ^ ...r ( x .I* f (17) Substitute « 0 and equation (17) in equation (14)..* .2 and s Q in equation (20). = ^ "2 “ 27 Thus.+ u . the Equilibrium state form is 2 O f I- Step 7 of 9 (b) Consider the following equation from Equilibrium state equation. = G { . . | j . U2 -2 7 < 3^ . ( 12) x .i |') ’ ( l. that is. equation (21)and equation (22). (23) 6 X2 » SXy . = 1 ^ .. < .= x ...H q)- Suppose we have a constant voltage source of 1 volt at u^ and a constant current source of 27 amps (that is. -> L = \H Refer Figure 9.= 4 -4 -r(:^ ) (13) Substitute iij s s s 0..63C2 + 2 7 S u ^ + 6 U 2 Thus. Figure A nonlinear circuit Here the function ris defined in Fig.= C { u .x .. C x j .6 x j+ 2 7 6 u ..... 2.. Step 9 of 9 Thus.. Step 5 of 9 From Figure 9..( t ^ + 5 x ...r i^ = L ^ I ^ (9) Substitute equation (4) in equation (9). = -276x^ . x *= [x * X* x j ] '' (19) step 6 of 9 ' Substitute s 4.3 + S U f-S 3 c ^ f . the resultant equation is | ^ i. . ( l .') ..(6) Step 2 of 9 Substitute equation (1) and equation (4) in equation (2).r ( 0 ) ... Consequently. Resistor 1: i*i = G (V |) = V j. = . the values of the elements are calculated.-n c .x . Apply KVL in the given circuit and the corresponding equation is given below..x . Give the values of the elements. L x .x . 3^ .) (1) = C -^ x .. 4 .. (b) Find the linearized model about the nominal solution In part (a). Step-by-step solution There is no solution to this problem yet. ASK AN EXPERT .04PP Consider the nonlinear system jfc= — x( 0) =l . (b) Find the linearized model about the nominal solution In part (a). (a) Assume uo = 0 and solve forxo(fj. G et help from a Chegg subject expert.04PP Problem 9.9. where a > 0 is the feedback gain. Find an expression for y(t) as a function of r(t) for the closed- loop system. a ^ 0 is the feedback gain. Find X and y .+ —/ 3 3 4 4 Step 6 Of 11 ^ Sketch the open-loop v/s closed loop gri^h for a s 0. So.05PP Linearizing effect of feedback: We have seen that feedback can reduce the sensitivity of the input-output transfer function with respect to changes in the plant transfer function. In this problem we explore another beneficial property of feedback: It can make the input-output response more linear than the open-loop response of the plant alone.y ( T ) ) d r .y for u ^ 1.y j) d t oo. The e^»ression infers that y^ = r . u £ 1 ^ (0 = u +1 -. .y y -^ y -r Find.\ ir .u > \ w The given proportional feedback is « (i) = r ( i ) + ” . since. on further simplification we obtain the e^sression —^ ^— . y y^- Hence. Thus. y =r+ a { r . w hena < 1. (2+ a)^ = l + (l+ a )r l + ( l+ a )r y = -------i --------.1.) Sketch the nonlinear transfer characteristic for a = 0 (which is really open loop). and a = 2 . Hence. {This function is called the nonlinear characteristic at the system. Show that if r(t) is a constant. in steady-state condition.y when and 2. (b) Suppose we use integral control: t h( 0 = r(t) + J ( r { r ) . Problem 9. the integral control makes the steady-state transfer characteristic of /-►oo' the closed-loop system exactly linear.i— 2+a Step 4 of 11 Sketch the response of the open-loop sjrstem v/s the closed-loop system 2h 3 Find the eg ressio n for. y «) = a f i. II > 1 . r is a constant . Open loop closed loop lor winous a a=^ z y y' X' z z (b) Sketch the non-linear system with saturation inte^'al control output.r . we get Y= R.y ) Step 11 of 11 From the equation we get 2y y ^ r. where r is a constant Findj'. Thus. y = r^ \[{r-y ) R -Y (l + ff)r = ( l + s ) 5 Thus. Thus we get I 1 2 y\ . Step 10 of 11 > In stable condition. where a > 0 is the feedback gain.a ) y y= (l-ii)r Thus. Find. y . we infer that lim .u Obtain y for u > 1 u + \ Simplify the eg ression further.y w h e n a > 1. The closed-loop system is therefore nonlinear and dynamic.^ = r. and 2. R (s) 2s + 1 Therefore. Find y -^ y y =r. a = 1.y stays bounded. and reduce the effects of a disturbance acting on the plant.y(t)). we get lim = lim r ( t ) . Step 2 of 11 ^ The nonlinear system with saturation proportional control is sketched. then lim y ( /) = r . Thus. let us ignore all the dynamics of the plant and assume that the plant is described by the static nonlinearity « < 1. 2y s \ + r + a r . we get y . Find an expression for y(t) as a function of r(t) for the closed- loop system.. ) I V J in p u t step 8 of 11 ^ Given that The closed loop system is non-linear and dynamic. (a)Suppose we use proportional feedback u(t) = r(t) + a(r(t)-y(t)). Can the closed-loop system be described by a transfer function from rto y? Step-by-step solution >1 of11 ^ Given that the plant is described by the static non linearity as u.y (t) = r . Hence.y ) y = { l. = . We know that =j^a).) Sketch the u(tf= r(t) + a(r(t) .y (0 ] • Here. For simplicity.£^. we get ^ = r + (r . say r. (This function is called the nonlinear characteristic ot the system. if y ^ * r . we getu = r -I. (a) Find the equilibrium point and solve toxx(t). Jt(0) 0. Problem 9. Consider the system X = w ?. (b) Assume that a = 1. Is the linearized model a valid representation of the system? (c) Assume that a = -1 . G et help from a Chegg subject expert. ASK AN EXPERT .06PP This problem shows that linearization does not always work. Is the linearized model a valid representation of the system? Step-by-step solution Step-by-step solution There is no solution to this problem yet. Sketch the figure given. Q r= H 6x H [ax dv ds\ Step 3 of 3 Thus. The system equations are where z = constant. we get « ■ [ ? • . The output o f this system can be written as = r. r = V P T ? . X = constant = Vq.] ■ We know that ± = Fx.07PP Consider the object moving in a straight line with constant velocity shown in Fig. Step 2 of 3 0 1 0 The system equations are 0 0 0 0 0 0 where z = constant. Derive a linear model for this system. Figure Diagram of the moving object f2 Object Step-by-step solution step 1 of 3 Consider the object moving in a straight line with constant velocity as shown in the given figure. Therefore. 0 - . we get Sr = . Figure Diagram of the moving object Derive a linear model for this system. r = ^ . X = constant = vO. The only available measurement is the range to the object. vAere r = A(x) Find R 6 r = — (5x dx. Problem 9. The root locus plot is shown in Figure 1.\ J ' (y<»)*-i-2(ya»)tli I 0«>)' ] (yV»)’ * K 2 ( j o )+AT = 0 -J a ?-K a^+ K 2Jo)+ K = 0 (7) Equate the real and imaginary terms. the breakaway point is [g^ = . the two zeros are -1 and -1. To find poles put denominator D(s) = 0 • *•= 0 Thus. . the system is conditionally stable with saturation which can be expected stable for the small input signals.. Figure 1 Hence. Step 2: Consider the formula for the number of paths N um ber o f path s N um ber o f poles »3 Step 3: Consider the formula for the angle of asymptotes.S which corresponds to ^ = 0.S Thus.» ’ ( 2 j+ 2 ) ( j ’ + 2s + l) ’ iifC Consider — s Q. i+ jc ii^ = o (2) Consider the following formula for the roots of the general form of an equation by the root locus method. ^ __ d f ^ ds < is(i* + 2 s + l j ( j ' + 2 i+ l) 3 s ’ . (3) D is ) Where.. locate carefully the point at which the locus crosses the imaginary axis. due to some unknown mechanism.l) ■ (3 )-(2 ) = 2 Thus. • Mark the poles on the real axis.000867 on the root locus. as the reference input size gets larger. ds . 1 8 0 °+ 3 6 0 » (1 -1 ) 3 -2 = 180* Thus.. departure angles. the equivalent gain can get smaller due to the saturation and the system is expected to become less well damped.. s’ j ’ + 8 :(j+ iy = o j ’ + J C ( i '+ 2 i + l ) = 0 (5) «' + 2 *+ l step 5 of 9 Differentiate Equation (5) with respect to s . the root locus is plotted for the given system and it is shown in Figurel.75 Thus. However. the centre of asymptotes is 2. (sum o f finite p o Ie s )-(s u m o f finite zeros) (n u m b e ro f finite poles )-(nunU >er o f finite zeros) ( O ) . Consider the following formula for the general fonn of characteristics equation. (-3)’ ( _ 3 )’ + 2 ( . -l+ 2 ^ = 0 K = 0. -1 . Step 7 of 9 (b) Consider the following formula for the locus crosses the imaginary axis. Figure Control system -O Y Step-by-step solution step 1 of 9 (a) Refer Figure 9. Rearrange Equation (2). (b) Using graphical techniques. the system is expected to be unstable at some point for the large inputs. the amplifier output is given by the following saturation non-linearity (instead of by a proportional gain K): i«i < 1.l.61 in the textbook. the three poles are 0 . I«l < 1..l) ■(4) n —m Where. Step 8 of 9 The graphical technique for the locus crosses the imaginary axis is shown in Figure 2.. and so on. Qualitatively describe how you would expect the system to respond to a unit-step input. Substitute the value -3 for s in the Equation (5). K G ( jo ) H ( ja ) = .s|. 3 for n and 2 for m in equation (4).. Step 2 of 9 Step 1: Consider the number of poles and zeros from the characteristics equation.s ’ (2 j + 2 ) ] = 0 -s* -4 s’ -3 s’ .0 . Problem 9. » = { 1. To find zeros put numerator JV(j) = 0 ( j+ l) ’ =0 Thus...08PP Consider the third-order system shown in Fig. (1) Substitute ( f ^ l) _ f o r G (j)and1 for f f ( j) in Equation (1). The roots of D{s) = 0 are the poles. Step 4 of 9 Step 5: Consider the following formula for the breakaway points. the locus crosses the imaginary axis which is shown in Figure 2.l. the locus crosses the imaginary axis at |ttis lra d /s e c |fo r |jj^ aQ . . (6) The roots of the Equation (6) are 0. Number of poles is n Number of zeros is m Substitute 1 for /. Therefore. and draw the asymptotes from centroid at an angle of | 3q* • Locate the breakaway point on the real axis.0 and 0.[ ( i ’ + 2 i + l)3 s ’ . • Draw the root locus. « > 1. K =. showing your calculations for the asymptote angles.( .. What is the value of K at that point? (c) Assume that. -land -3. Thus. • Locate the centroid on the real axis. the system to be expected conditionally stable or unstable depending upon the input signals.s ’ (s’ +4s+3) = 0 . the ahgle of asymptotes is .. Step 9 of 9 (c) If s 0. ^ I8 0 * + 3 6 (r ( /. 0. -K a ^ + K = 0 a=\ .a ^ + K 2a = 0 (») Substitute the value Ifo r uiin the Equation (8).. e < . Step 3 of 9 Step 4: Consider the formula for the centre of asymptotes. Figure 2 !.3 |- Step 6 of 9 Step 6: Procedure to draw root locus plot for the given system: • Take real and imaginary lines on X axis and Y axis respectively. l + A:G(5)ff(j) = 0 .. The roots of iV( 5) = 0are called the zeros of the problem. (a) Sketch the root locus for this system with respect to K. Compare Equation (2) and Equation (3).3 ) + 1 = 6. Plot both the step response and the control effort using Simulink. ASK AN EXPERT . G et help from a Chegg subject expert. to control this system. Problem 9. Compare the system response for a step input of size 10 with and without antiwindup circuit. Qualitatively describe the effect of the antiwindup circuit. Step-by-step solution There is no solution to this problem yet.09PP Consider the system with the plant transfer function 1 GW = ?T T - We would like to use PID control of the fonri D tW = I0 ^ 1 + ^ + 2 i^ . Qualitatively describe the effect of the antiwindup circuit. It is known that the system’s actuator is a saturation nonlinearity with a slope of unity and |r/| < 10. 1 OPP Compute the describing function for the relay with dead-zone nonlinearity shown in Fig. Yi = . the described function is DF -s fl . relay with dead zone. Fig. D F =^ 4V cos(<ai) TTa Hence.V 'y { t ) s m ^ a t ) d [ e x ) 7T^ (« *) 7T^ sin {a t) d (a*) Thus.ntg + sin QiX + Step 3 of 5 Find Yi. we get The described function is then given by D F = — Step 5 of 5 Find the DF. we get Yi = ^ < t o t { a t ^ ) Step 4 of 5 We know that k — a Thtis. Step 2 of 5 To find the describing fimction for the relay with dead-zone non-linearity The standard equation is u ( 0 ~ osin(dV) and . Step-by-step solution Step-by-step solution Step 1 of 5 Sketch the given figure. Problem 9. (ZUj s sm ' —' Step 3 of 4 Simplify the e^>ression further ?r y The described function is then given by DF = — Step 4 of 4 Find the DF. Problem 9.l]sin(atf) ( izk) Since. DF: 2K.asin(<zv) and sin QiX + This is an odd nonlinearity so that all the cosine terms are zeroes and the DF is real yrJO ( “ ) ■='('“ ) = ^^J^^[j4sin((Zif) . the describing function is 2K^ DF = . DF = Thus. gain with dead zone. fj IkpeXb -k 7 Step-by-step solution Step-by-step solution Step 1 of 4 Sketch the given figure. h = asin(4:u^) We have. Fig. Step 2 of 4 ^ The standard equation is u (t) .11 PP Compute the describing function for gain with dead-zone nonlinearity shown in Fig. m f W h. • . or coulomb plus viscous friction. AT step 3 of 3 Here. + ^ yra Hence. 2 ka . 4N DF=K.1 2PP Compute the describing function for the preloaded spring or Coulomb plus viscous friction nonlinearity shown in Fig. there is a combination o f a gain. Y The described function is then given by D F = — . preloaded spring. Step 2 of 3 We know that. a Obtain the describing function. Problem 9 . the describing function is 4}7 D F = K f. plus relay nonlinearity. ______ ^ . Fig. y 'Slope JCo Steo-bv-steo solution Step-by-step solution Step 1 of 3 Sketch the given figure slope fC.+ — . 0 <a < — 2 K m («3) = ' J-----. 0 <a < - 2 1 1* 2« . c le a r a l l . Problem 9. p l o t ( l i n s p a c e (1 / 2 . K e q ).1 2» + 1 M l w "\ [ 2a V 2 2 ^ Step 5 of 7 ^ Obtain the MATLAB code for generating the describing function.2 ^ ■ V '1i.v — i \ / P 's / . Step 2 of 7 The abscissa brealq>oints are denoted by ^ We know that b^=— J [a sin(ZU)sin d[cx) Step 3 of 7 Simplify the e^>ression further. c lo s e a l l . that resembles a staircase. (2 * n + l)/ 2 . n n = 6.» Step 4 of 7 ^ Obtain the des cribing function (a ) = — . f o r n i= l:n a f o r k = l..( ( 2 * k . 2 .5L 0. the staircase can be ^>proximated by a straight line. + — x siacatd(cot) jrJO — ( C 0 S ^ + C08<l^ + .. . a a 0. y l a b e l ( 'K _ { e q } ' ) . the describing function is 0. end.---------. Figure Quantizer nonlinearity Step-by-step solution step 1 of 7 Given figure for the quantizer nonlinearity is... ^ = sin d n ( ^ ) f o r i = 1 . .27 corresponding to —« 0. Desofeing funcOonfor quantizer nonineatny i 1...n K e q ( ( n . ( 2 * n n + l)/ 2 . the slope is one. end.O L (^2i-l V 2 »-l 2» + l Thus.. ^ a Since. .7. f o r n « l:n n a i = li n s p a c e ( ( 2 * n . it is seen that the DF will in the limit approach the slope o f the linear ^proximation. n a = 99.l) * n a + n i) * K e q ( (n -1 )* n a + n i) + (4 / ( p i * a i ( n i ) ) ) * s q r t ( 1 .l ) / ( 2 * a i ( n i ) ) ) ^ 2) . + C O S ^ ) Here. Hence.m function to generate it.' a/q • Step 7 of 7 The descnbing function is plotted in the figure as a function o f —.l) / 2 . The maximum o f the DF occurs at « 1. n a ) . Step 6 of 7 Sketch the obtained gnq>h for nonlinearity after execution o f the code.n a * n n ). g r id on. K e q = z e r o s (1^ n n * tia ) . x l a b e l ( 'a / q ').4 — 02 -■ : ! ' i■ 0 3 4 • . Find the describing function for this nonlinearity and write a Matlab . l \ = — ( " o X sinatf d lo jt ) + — \ ’^ q x ana > td (at) + .13PP Consider the quantizer function shown in Fig. t i t l e ( 'D e s c r i b i n g fu n c tio n f o r q u a n tiz e r n o n l i n e a r i t y ') . end. Step 2 of 7 ^ The standard equation is u (t) . Problem 9.1 and T = 1. Step 5 of 7 Find the D F when d < a . DF = ^ 4T Thus. Figure Contactor for Problem Step-by-step solution Step 1 of 7 Sketch the given figure. a <d The descnbing function is not frequency dependent Step 6 of 7 Frequency dependence will be introduc ed with the introduction o f a time delay. the describing function is D F = 0. Is it frequency dependent? Would it be frequency dependent if it had a time delay or hysteresis? Graphically sketch the time histories of the output for several amplitudes of the input and determine the describing function values for those inputs.0.asin(dv) and sin Q at + Step 3 of 7 This is an odd nonlinearity so that all the cosine terms are zeroes and the DF is real — J|j*^[j4sin(atf)-l]sin(aif) d(ait) Step 4 of 7 AT On further simplification we get 2^ = — cos(<zK^) Since then cos(4Xi) ■ft The described function is then given by D F = — . Sketch the describing function for<i = 0. . D e s c r ib in g f u n e f e n : D F Step 7 of 7 ^ Sketch the time histories for the several amplitudes ofthe input frequencies.14PP Derive the describing function for the ideal contactor controller shown in Fig. The fi^quency ofthe limit cycle is fixed by the phase. 4 Find a . we get — — « ---------. Be Kh V + i . the result is verified successfully.Ols. O.0 l5 + l) _ Step 5 of 17 Sketch the bode plot o f the obtained transfer function BocteUngran I % Step 6 of 17 Sketch the root locus ofthe transf^ function. ____ 1 +1 B c O . M o t o r and c o n tro lle r step 2 of 17 Given that / = 0. a = K<p.m . = -----------. e From the figure it is obtained that 7i = 0. XX„ = 100n m .( .= O.1 2 5 x 1 0 ^ x 1 0 "’ a s 0.= 8 x 1 0 "* Step 17 of 17 For the analysis obtain a suitable value for a. d = 10~^ m l.5 7 x 1 0 -“ rad/s Thus..125 rad Here.lff+ 1 On further simplification we get f V l :-l. T=IN-m. which is same as before. • = 0 .01 10 X 10“^ ^ ^ ^ Simplify the equation further. Obtain the eg ressio n for <p.1 sec. . B * = 1. Step 3 of 17 >*■ The limit cycles depend on the natural behavior of a closed-loop part The describing function of the switch is . the ea>ression for X „is obtained as K „ = — L fl . the maximum is at iZ ! trd Sketch the response of the non-linear system.15PP A contactor controller of an inertial platform is shown in Fig. Step 9 of 17 Step 10 of 17 ^ The output is a constant for levels of m > 3 d . - m i \ a ) ‘\ \a ) • xIO' ttx lO '^ U J Find —. rrx • • •.rfa t at =-95 ra d /s l. we get X* = 10.m^. Problem 9.. r = iN . U -O lJ s(10s + l)(0 . Hence. 3.= 10s. d a=- 8x10" ^ 0 = 0 . Step 11 of 17 > Sketch the time histories representing the anq^litudes. r «• constant „ This IS consistent with = constant for K. * 2^ A t= — Aa . So.1 kg. and frequency of possible limit cycles as a function of the gain K and the DF of the controller.01 sec. Step-by-step solution S te p i of 17 Sketch the given figure. for large enough K to insure the K<p » d . Find — . = 100 (rad/s)(nm sec/rad) X X .8 x l0 -® x l0 .iiL The required stabilization resolution is approximately 10-6 rad.14 A t * ------ 200 At = 1 . So (P is a constant The result is independent o f K. . 21 <p= (5 rad/s^)(Af)^ Step 13 of 17 We know that At = —.then » dt is satisfied.= 0. ^ = 0. Step 4 of 17 The characteristic equation is iTG+1 sO . a n d d = 10^ ra d . c . where / = 0. Hoot Locus step 7 of 17 >*■ The conditions for limit cycle is phase equal to -180° v4iich occurs at and when the magnitude is one. . Figure Block diagram of the system Discuss the existence.s l0 0 n m Step 8 of 17 Thus. W eknow that K x 10“* = d. KK 1 Thus. Step 16 of 17 We know that K . amplitude. and r = 1 N . Hence. 0 = 1 2 . The required stabilization resolution is approximately 10-6 rad. w eg etX „ s lO n m . Since.* ^ ^ ^ m i \a Here. we must consider the mid frequency model because the limit cycle is at m = 100 rad /s.5 X 10"* X 100 rad Thus.1 kg • m^. Repeat the problem for a deadband with hysteresis.01 sec. but goes through the lead circuit for a gain o f 10 andX = 10.1 s. iTT o » = ---- p (P= 100 rad/s Step 14 of 17 Find At using the obtained values. K<pm > d for (pm > 10-6 rad. c = 0. we get <P= 12 x 10"*rad Step 15 of 17 ^ If the resolution o f platform pickoff should be ~ 10~^rad and short term s ensor noise «10~ ^rad. d = 10^rad. 0. at gets inte^ated in the gyro below its break fi^quency. we get a = 0. ^=1. = ^ Jl .125 ra d .Ols.1 Let us assume that 4x1 1. x. if e(t) ^0. It can be implemented with op-amps and diodes. G et help from a Chegg subject expert. crosses zero {that is. ASK AN EXPERT . It can be implemented with op-amps and diodes. The Clegg integrator can be described by x(t) = e(t). it resets its output to zero. X. Step-by-step solution There is no solution to this problem yet. Problem 9. x . to zero whenever the input to the integrator. changes sign). Clegg suggested that we modify the linear integrator to reset its state. A potential disadvantage of the Clegg integrator is that it may induce oscillations. x(t+) = 0. In 1958. is l e s e i lu z e i u irn m e u ia ie iy after e changes sign. if e{f) = 0. A potential disadvantage of the Clegg integrator is that it may induce oscillations. The Clegg integrator has the property that it acts like a linear integrator whenever its input and output have the same sign. J. C.16PP Nonlinear Clegg Integrator: There have been some attempts over the years to improve upon the linear integratorA linear integrator has the disadvantage of having a phase lag of 90° at all frequencies. (b) Prove that the DF for the Clegg integrator is = — . w n e t e m e la u e i e q u a u u i i im p lie s ir ia i m e s i a i e u i m e iiiie y ia iu i. where the latter equation implies that the state of the integrator. is reset to zero immediately after e changes sign. (a) Sketch the output of the Clegg integrator if the input is e = a sin(cuf). e.A and this amounts to a phase lag of only 38°. Othenwise. l) = z l n ( l . 0 0 Xi = . time reversal means that we let r = . '" In (^ “ 1) = T + Since (O) = 0 .1 . .1 Eliminate T to get Xj = .I n ( . x^ = In ^1 + jx^p . X2 = . Eliminate z to get. = ln (l) Find z.a n d that changes the time of the system ai dx the input matrices — = .* . given that . l n ( .l . integrate to findxi. Xj= . * .^ + 1) = T + Cl Step 5 of 7 Since (O) = 0 .1. we get Xi = .7 ^ + tt For u = + 1 .. I 'i T S n i . when Xj (0) = 0. .^ = . and > 0.Xj.x^. Problem 9 . —2.^ + 1.1) .1 .i + I .Ibc . So. we get . for all Xj = sgn (.1 7PP Compute and sketch the optimal reversal curve and optimal control for the minimal time control of the plant l«l < 1- Use the reverse-time method and eliminate the time. we get . find C\.Xj. Step-by-step solution step 1 of 7 Step 1 of 7 Given that X2.In (1 .t^)ln^l + |x^p . = dz. dz . ^ . We have C i = l n ( .x^ ■ Therefore.s g n [j. ^ .i^) = r T = ln (l .e* + T + Ca Step 7 of 7 ^ Thus.G « . * . .^ < 0. dz So. tJT dx^ Here. This is the reversal curve forn = . the control law is u = .1 -I- dx^ =f-1+«*) dz Integrate both sides to getxi. In (ji^ + 1) = T -I.sgn(jrj)ln(l + |j^D] .l ) . Use the reverse time method and eliminate the time.T . In (:(^ + 1) = T ji^ + 1 =e* T = ln ( . and Cj = 1.Cl C. . = dz Step 2 of 7 Integrate both sides. = 1 . find z. X2 = -X2 + U.:i^) Step 3 of 7 Simplify further. We can then write in general.1. d{x^ + \) Integrate both sides ln (.^ + l) Step 6 of 7 Now.e* Simplify further. J^Xj =J(“l +«*) dz 0 0 Xj = « * .t. Step 4 of 7 This is the reversal curve for u s 1. For w = .«* + t +1. Step-by-step solution step 1 of 3 Given that X i= X2. . x p ( : .1] . and that changes the sign on the system and the input matrices. 1 ) . g rid . p l o t (x p ( : . the required reverse optimal curve is sketched. H = [ l 0] . We simulate the system using the MATLAB Isim function with u = +1 and store and X2 .3X2 “ • We reverse time that means T =. i i = -2*1 .1 and store and X2 and plot the result to obtain the optimal reversal curve. and repeat with a = . Step 2 of 3 Write the M AILA B program to plot the reverse optimal curve for the give system. %u= +l.F . t=Q: .^ . t ) . p l o t (xm ( : . . % U sin g t h e r e v e r s e t i n e m ethod s y s = s s ( . Given that * i= * a . x la b el(» x _ l» )? y l a b e l (’x_2’ ).2 X i . 2 ) ) } h o ld on. t . G =[0.18PP Sketch the optimal reversal curve for the minimal time control with |u| < 1 of the linear plant 1| =J(2. dz dr Here. . X2 = . F=[0 l . J ) . t i t l e ( ' O ptim al r e v e r s a l c u r v e * ) / Step 3 of 3 ^ Sketch the optimal reverse curve for the given system. t / x p ] .l * o n e s (101.3x2+ M. % u=-l [ y m . Problem 9. = .3 ] .I s i m ( s y s ^ o n e s ( l O l r 1) r t ) . 1) .2 . 2 ) ) .a . and X2= 2^1 + 3xj .1) . xm ( : . Thus. J=[0] . x m ] = l s i m ( s y s . 0 1 : 1 . [ y p . s in ( i) . Xj (<) = cos (i) . Similarly for a = . > 0. we get x^ = 1 .cos ( i). Find the Li^lace transform of the given functions. the required figure is sketched.1.{s )= - '+ 1 Step 2 of 5 Find the inverse Ls^lace transform for the obtained functions. and show a trajectory for x^ (0) = 3 and x2(0) = 0.1 we get. = . Step 4 of 5 Sketch the reverse curve for the given system. X .19PP Sketch the time-optimal control law for XI =X2.-1 Step 3 of 5 ^ Therefore.l)^ + = 1. This is called the Bushaw problem in optimal control literature. JC 2=-X l+U .Xj + a. Step 5 of 5 The trajectories for this system are circle s centered at (±1. (t) ~ sin ( t) . which is a circle with center at (l. = . Thus.. 0). Problem 9. Step-by-step solution step 1 of 5 Step 1 of 5 Given that . IkI < i . .c o s ( /) = X . and We know thata = 1. and (xj + 1)^ + =1 Which is a circle with center at (-1 . 0). x^ < 0.0 ) . We see that (xj . it is piecewise linear We have T + oT = BNsgti (e) with hysteresis with % = 800°C. (a) What is the limit-cycle period? (b )if Tr is commanded as a slowly increasing function.058 + 0. T. aP { boo) l. % = 0. Time (sec) . we get P = 23. Problem 9. limit cycle period is sq>proximately equal to ^ -(T r~ giw s as a) ^ j and gives as follows. Show the solution for T r“large.3 0 0 j P = 100(0. Step 2 of 4 ^ w This is a first order system so use {T.20PP Consider the thermal control system shown in Fig.176) P = 23. A ~ 100®C. etc. The physical plant can be a room.4 s . For an oven. an oven.4 8 Thus. sketch the output of the system. Step 4 of 4 Sketch the ten^erature plot for the given system. BN -----w 1000®C above 7* = 0 (say room temperature) a Thus. we get limit cycle period ^ • Step 3 of 4 (b) Given that a = 0.01 s“^plot vs at. plot.” Figure Thermal system Step-by-step solution Step 1 of 4 Sketch the giveti figure. we get A 5 -f TO = 0. Find the change in slope L&. Step 8 of 8 ^ Sketch the phase portrait o f the sjrstem with dead band and a constant disturbance. 9 s a. and $ = 500®*. (b) deadband plus hysteresis. a spring-mass system with resonant frequency well below the frequency of switching. (c) deadband plus time delay T.<^ + A + TO * 0) + 7*.21 PP Several systems. (d) deadband plus a constant disturbance. 2 Thus. (c) deadband plus time delay T. The switching function is e = 0 + roj. Thus. step 5 of 8 (<? The system has a dead band as well as a constant We know that T a = L $ .r a j + ( d . and the control signal s 10~^rad/s^. step 3 of 8 0 >) Obtain the phase portrait o f the system where there is deadband as well as hysterisis. Step-by-step solution step 1 of 8 The given switching function is e ^ 0 + 1 It is assumed that r = 10 s. and a large motor-driven load with very small friction can be modeled as just an inertia. Step 4 of 8 (c) The phase portrait o f the system with dead band and the time delay T is sketched. when t. sketch the phase portraits of the system.k). a = u + D . Find the e^>ression for a and a . da Find de da _ u dd a d a _ lOr^rad/sec* d& a We know that — = u&. L9 = -z a + Ta L B = -(r-T )a i Step 6 of 8 Find 9 when u tends to zero for (^ . Now sketch the results with (a) deadband. . For an Ideal switching curve. We know that 9^ = . #3 9 s + D'^ — + a ^ + 9j Find 9 when t = T. Step 7 of 8 Find 9. we get ^ « 0.05 when a = 10^ Step 2 of 8 (a) Sketch the phase portrait o f the system with dead band. ^ 9^ a = {u + D )t +Of. such as a spacecraft. Assume that r = 10 sec and the control signal = 10-3 rad/ sec2. Problem 9. ^ ^ tin the expressions u sin g s and 6^. the delay is A seconds.j . a = — ( S .r ] T0J^=O After solving for 0j^ we get From the equation 0j^ + . using ii = -sgQ(r^+0). LN \ ^ A = j[2 e A ^ . Problem 9.* we get the result . Step 9 of 9 Sketch the time history o f 0 6 From the figure we infer that it is non-linear oscillatioa . Delay o f the sateUite altitude is given as AAT Since.A ) Step 2 of 9 Eliminate t. Step 4 of 9 From the geometry o f the limit cycle we obtain the e :^ e s s io n ^ j + . and ^ s 0 Eliminate 0^ from the equations ^ ( 217) Thus.A). Step 7 of 9 • L ( .j —units during the delay.A ) 2 f-A f I4l= 8/ t -A J 1^1=14.22PP Compute the amplitude of the limit cycle in the case of satellite attitude control with delay 19 = N u { t. Sketch the phase-plane trajectory of the limit cycle and time history of 0 giving the maximum value of 0.a„ = — Thus. Step-by-step solution Step 1 of 9 The equation for the satellite is li = —sgn {z& + . we get step 6 of 9 Now eliminate 0^ from the equation A 1 t „ AATl Eliminate 0^ from the equation 0^-¥ 0.A ) ] 2 /(t . AAT[A + 2 ( r . we g/st0j^+ T0j^ = 0 Similarly at point S we get 0^ Step 5 of 9 Solve for the expressions w h e n c e = 0^. ATA A^ATf 1 1 Simplify the equation ^ 2/ a J / _ Step 8 of 9 F in d |^ |.£ |. ) + — Simplify further to obtain T 0+T0=O 1 r=— step 3 of 9 Sketch the phase plane trajectory o f limit cycle.^ = 0^ The geometry of the limit cycle at the point A is a ^ . Obtain the e}q>ressions for 0 and 0 from / ^ = A 6 i( i. 0 must travel . Figure Pendulum Step-by-step solution step 1 of 5 Sketch the given figure o f the pendulum. O btain^ and a . ± 2?r. step 4 of 5 The centre is obtained at d » ± ^ .23PP Consider the point mass pendulum with zero friction as shown in Fig.. Using the method of isoclines as a guide. using the isoclines. .4a. W = mg/ sin 6 m /^d = ^ / s i n d d = ^ s in d / Step 2 of 5 When 0 ~ 0 ... sketch the phase-plane portrait of the motion. Problem 9. We know that when 5 ~ zr. Indicate a trajectory corresponding to spinning of the bob around and around rather than oscillating back and forth. e = ^0 i i S± step 3 of 5 The saddle point is at d s 0. Obtain the equations governing the motion o f the pendulum. e^. we get sin 5 — 6 Obtain the eg ressio n corre spending to sin ^ ~ ^. l 0 )*nth«conpiiaiHn«MM. we have sin ^ ~ 5 Simplify the obtained eiq^ressionby substitution.. d& d = —sm d I d$ a t= — d$ ac= —sin5 / The isoclines are sinusoidal curves. 27T. The i ^ e r and the lower portraits correspond to the whirling motion with the circular motion o f the pendulum. Obtain the phase portrait using MATIAB Step 5 of 5 1htMCk«vd»itlli«n(-l. Pay particular attention to the vicinity of 0 = rr.. ^-e i s = ± j. 028 * 0. V is the average v in a given interval of ^ Divide x into five intervals going from left to the right.*x):plot(x. by graphical integration. » syms x{t)D2x=diff(x. write the expression for change in time using above equation.2 “ 0. B A/.'Dx(0)=0'.02 * 0.04 * 0. Draw the phase portrait plot for the given system using MATLAB.4x10'’ 0. + Afj + Afj -I-A/4 + A/5 _ 0.2 0.2 0.8x10-’ IxlO-* V 0.2). {v Time can also be found by calculating the s v(x)"' plot.028 0.6x10-’ 0.035 0.1):» dsolve('D2x=1E-6'. Find the transition time ffb y graphical means from the parabolic curve by comparing your solution with two different interval sizes and the exact soiution. Othenvise. Ax To calculate the transition time.045 = 10+7.04 0. Figure 1 Step 3 of 4 Write the expression for velocity.035 0.29 sec Therefore.24PP Draw the phase trajectory for a system between i(0) = 0 . is [32.and x(t) = 1 mm. we can obtain the exact transition time using the following formula.71+5+4.000001:1 E-3. .39 see].'x(0)=0')ans =t''2/2000000» x=0:0. the transition time.t=sqrt{200000.44 = 32.2 0.02 0. Write the MATLAB script for drawing the phase portrait plot.045 Step 4 of 4 Now calculate the transition time.Dx=diff(x. Step-by-step solution step 1 of 4 Step 1 of 4 Consider the following system.14+5. X 0. Problem 9.2x10'’ 0.x (0 ) = 0.2 0. Better approximation can be obtained by finer division for x .t) Step 2 of 4 Draw the phase portrait plot. Here. G et help from a C hegg subject expert. (a) What physical system does this correspond to? (b) Draw the phase portraits for this system. (c) Show a specific trajectory for 0(0) = 0. Problem 9. Step-by-step solution Step-by-step solution There is no solution to this problem yet. ASK AN EXPERT .5 rad and e = 0.25PP Consider the system with equations of motion 9 + 9+sine = 0. When a lead network is used we get We know that 6 = sin ^ + a . Step-by-step solution step 1 of 2 The giTen sjrstem is a non-linear p endiilum with a motor at its base as an actuator. Design c feedback controller to stabilize this system.4xj -3x3 “ 2xi Hence.26PP Consider the nonlinear upright pendulum with a motor at its base as an actuator. We know that 9 = sin ^ + a . die equation for the lineaiized system is 0 1 0 -3 0 -4 -2 0 -3 The poles of the system are at det(s’I . X' = sin Xj .4xj . = step 2 of 2 Obtain die matrix form of the sjrstem equations. When a lead network is used we get We have x^=9 x^ = 9 X.F ) * (s +1)* If die system starts near the iq u i ^ position dien die system is asynqitotically stable near the oiigia . Problem 9. Step-by-step solution step 1 of 2 The given system is X = .x^(3x.27PP Consider the system i s —siox. The Ly^mnov function for P = 1 is step 2 of 2 Find V ( x ) . it is evident that the origin is an asymptotically stable equilibrium point . w e choose Q = \ Therefore. ^ (x) = 2xx P^(x)= -2 x s in x We know that P^(x) ^ X.^ x . f o r 0 ^ x ^ l . for 0 i X ^ 1 Since.sin X To prove that r ( x ) 5 . Prove that the origin is an asymptotically stable equilibrium point. s in x ^ . Problem 9.sin X The given system is X = . / ( x ) satisfies the conditions /(0 ) = 0 / ( x ) > 0 for X <0 f (x) < 0 for X > 0 The given L3ra5>nov function is V (x) = — To prove that the system is stable near the origin. we conclude that the system is stable. m = o. T < 0 Thus. we getP^(x) = 0 Here. for X > 0. Step 2 of 2 ^ Find V [x ) r ( x ) = XX "Whenx > 0 and / ( x ) > 0. we getf^(x) > 0 Whenx < 0 and / ( x ) < 0 . = -f{x). by sq>plying Lys^novs stability criterion. Step-by-step solution Step-by-step solution step 1 of 2 The descrytion o f the first order non line ar system is x = 2 / (x) Where f (x) is the continuous and non linear function Here. where ^x) is a continuous and differentiable nonlinear function that satisfies the following. for x<0. f(x)<0.28PP A first-order nonlinear system is described by the equation jp. we getP^(x) > 0 Whenx = 0 and / ( x ) = 0. Problem 9. f(x) > 0. . for all x ^ 0. Use the Lyapunov function V(x) = x2J2 to show that the system is stable near the origin (x = 0). • All the terms in the first column of the Routh’s array must be positive sign. Substitute n s 2 in equation (4). the value of K is K > 3 and IC > 4 - X ’ -10JC’ +42Ar-72 >0 36(K-4) (iC-4)(jr»-6iC-M8) ^ 36(JS:-4)" p r .* ' ’- . Thus. V H .i... r(» ) G M w (» ) (2 ) R(s) 1 + G ( i) « ( i) Substitute G ( * ) = » ( » ) = • in equation (2). will be stable. *■[?-'r>] * .6 p = .l (7) 2 q -6 p = -\ (8) p {4 -K )+ r-3 q = 0 (9) Rearrange the equation (7).a:) Step 4 of 6 Consider the two stability conditions for /> > 0 is given below.4 ) = 0 . 2{4-K) ( z p ) ] . Thus. step 5 of 6 ^ From equation (4). ( 10 ) From equation (10).l (4 -J C ) -6 p = ..6 ^ + 1 8 ) > 0 ...] - Rearrange the equation (6) and simplify the equation is given below. the value of K \s K > 4 ^nd it is satisfied the stability condition. the stability condition is satisfied when K > 4 . Find the closed loop transfer function from the given Figure. the characteristics equation is given below.2 1 6 ( 4 .a:) 6 (4 -A :)^ > U (4 -A T )J .29PP Use the Lyapunov equation A7P + PA = -Q = -I.. *-[T ?] Substitute —4and oj s 3 in equation (5). the stable condition is > 4.since IS^is always positive. (11) Apply Routh-Hurwitz criteria to equation (11). i r ‘r ].. .H.\ (4 -A T ) -1 -6 /? = . *-[T T ] From equation (3). in equation (8). [? :]•[: t . Therefore.r :-. 2? (4-A T ) = ..1 Substitute matrix A = P= in equation (1). to find the range of K for which the system in Fig. Compare your answer with the stable values for K obtained using Routh’s stability criterion.A T ) „ ^ = 6 (4 -A )^® Thus. :]' . 4 0 5 0 3 ( K .4 ) ( A '* .g * = ( ^ : . A ''P + P A = .if ) 6 ( 4 .[ ? ‘ -. Figure J«f)=0 -o m Step-by-step solution Step-by-step solution step 1 of 6 Consider the following formula for the Lyapunov equation.. l( 3 . »’ + 3 s + ( A : .67 in the textbook. (1) Consider the following matrix for the positive definite matrix ■[” ] Refer Figure 9. the highest exponent of s is 2. 5 2 1 K .. From Figure 1.a: ’ h-7 a : .I . .l) ' I(£ )_ j£ ± i)M L 1+ 7 ( j+ 4 ){ j-l) y (» ) K (3) R{s) s‘ +3s+(K-A) Step 2 of 6 Consider the formula for the closed loop system matrix A in controller canonical form. Problem 9.l + {4 -K ) ^ 6 (4 -A) Substitute q ^ i M i 2 (4 . Step 3 of 6 ^ Substitute q ^ . ( i+ 4 )( s .. the value of K is I a: > 4 | same for the Lyapunov equation and Routh-Array criteria . the value of is > 4 and it is satisfied the stability condition.4 ) Figure 1 Step 6 of 6 The system is stable if the equation satisfies the following condition. ( a + 2 ) j+ ( 2 a + l) = 0 There are no values of a which produce stable roots. Output equation is. The input of the system is. ) ’2 .y ...l ) ( .a y .. So it is concluding that. . Step 5 of 6 Case 2: Consider the equilibrium point.+ )i (15) The characteristic equation of the linearized system is. (12) 0 + P = l .y. s 1 and X2..1 Step 3 of 6 ^ Case 1: Consider the equilibrium point.+ a y .>2+ 1 ^ = . ( l) .. .. + a + ^ ) = 0 ] 4 + x . -1 y .( a ( l) + /9 ) I + j( ^ ( a + ^ ) = 0 I (5) From equation (4).( a + f i) = 0 Substitute _ ] for x .+ fi) (3) i j = x 2 { x j+ a x .'i ..-t-x. . closed loop system equations are. + y ^ { a + P ) .= y .^2 =(.y j (9) Substitute for ^ + l f o r aT| and y 2 -\-\^ox in equation (4)...a ) * y i * a y ^ i ( I'l) Substitute i |2 for ^ * i> lfo r JC| and I f o r X2 in equation (4).( l+ a ) .*'i+ “ + /? ) = y .• y .+ y i+ a y .l ) ( “ U + 0 + /®) = ... s ^ + a s + { 2 a ..a y ...a > .>'2. '2 + O U 2 +0+^) = { y i+ ^ ) ( y . = y i ( ^ . ^ + . Problem 9.J ’l + l + ( j ’2 .1 y 2 = a y . (7) 0 + P = .(-l)+aj>.> ’2 .. .. s 1 and s i Is an unstable equilibrium point.+ a y .. >■2= ( . (6) y i = x 2 .+ ^ + a y i+ a * P ) ." 2 + « J ’i>’2 + ..M .+ P Substitute oX| for in equation (1).{ax..>... Step-by-step solution step 1 of 6 Step 1 of 6 Consider the following system. + f i) (4) Step 2 of 6 To find the equilibrium points of for the desired output of ^ substitute X | s l ..—I Let. the values of or and are.. > )(. Step 6 of 6 Substitute jc^ for ^ + 1 for x^ and y 2 —\ for X2 in equation (3). a (/) = c u .> (f)+ P ( 2) Substitute X| for in equation (2).. ( a + / J ) j^ .0+^ 2 + i+ a j> . > ’i+ > )+ t» ) =(^2 .i ) ( “ J’i + ‘* + ^ ) =>>.y t + 0 W 2+>’2(«+/9)->’2 ~{a+P) = A -y 2 +W 2 + ^ '2 (> )-J ’2 + > -o '.1 + a>’i + «+> 9 ) = A . * a + p My. ’2( . i*|s O and s O in equations (3) and (4). (13) Differentiate the equation (11). + l+ ay. >1=^1 Differentiate the equation (12). 0 = J i( jC j+ a ( l) + j9 ) j^ ( * .. a (/)= a .. j'. ' ! + i+ ( > ’2 + i ) ( “ ..+ l+ « y .i .( a + p ) My. + ^ 2 + ‘*>’1^2+^2 ( « + ^ ) + .i) = 0 The system is stable for small signals near the equilibrium point if. + l + a y . . s 1 and X2 —1• Let. a > —and a+ fi= \ Therefore. . o r > — a n d a + > 9 = l . 0 = l+ * . y j + y . >■1 . y i+ } i °) Step 4 of 6 ^ The characteristic equation of the linearized system is. Differentiate the equation (7).l .y .(8) Differentiate the equation (6). -1 y. y . i .-y . = x. 'i+ i + ( > ’2 . (11) y i = x i + l . + \ + a y j/ t + y t( a + p ) * a y . From equation (3).. The nonlinear.'+ a ( ..l= 0 jt J .>>1+ i + ( > ! + > )(“ (:> 'i+ > )+ ^ ) = ..30PP Consider the system Find all values of a and for which the input u{t) = ay(t) + fSwill achieve the goal of maintaining the output y(t) near 1. +. Substitute for ^ + l f o r aT| and y 2 -\-\^ox In equation (3). 1 for a:|...(2 1 i-2 C .y iy i+ y i Recall equation (2). y > -^ step 5 of 7 Recall equation (1). Recall equation (3). — y . | 5 l. 0 = :tf-l JCj = ± 1 Substitute 0 for jc^ in equation (2).a n d as 0. 0.l.= 0 The equilibrium points are [±1. (2) = . for x. the linearized system about the equilibrium.0 ] + 0 = 0 3 * ( s + 1 ) + 2 ( s + 1) = 0 ( 3 * + 2 ) ( j + 1) = 0 From the characteristic equation observe that the system has two poles on the ja> axis.1. y .{ x . o f Assume.’h Differentiate the equations. <3) Now. Substitute ^ ■and +1 for x^.. the Lyapunov theory does not tell whether the system is stable or not. the equilibrium points are |[^ ■ o ■ o rl and »■ o n step 2 of 7 (b) (i) Consider the following equilibrium point. x . ) Substitute for ^ .+ \ y i .( x . for jtj.Vi = 3 ’2[3’3 . the nonlinear terms affect the stability at the equilibrium point [1. iy^-JC^X.j f .0 ] .2 ) ( j + 1 ) .+ y . * l' X . Calculate the characteristic equation.x . set the values i j . 0..l) .( j ' . The non. 0.. 0 = 3^( a^ . x = [. o f Therefore.0 ] .X . the system at equilibrium point . for x^. y^Fy 0 I O' = -2 0 0 y 0 0 1 Therefore.. . Problem 9. 2 0 0 =0 0 0 s 0 0 -1 3 I 0 -2 » 0 = 0 0 0 3+ 1 3 [ s (s + 1 ) . ) J^=0 Substitute 0 for in equation (2). + l for The linearized system is. . (b) Find the linearized system about each equilibrium point.a n d .x fy x. Substitute y^ for jfcj.l) [ ( 2 ) ( j. o f Assume. These poles make the system neutrally or marginally stable. (c) For each case in part (b).( .!.1 [ ( . |d . the system has two poles on the Ja> axis and one on the real Therefore. i^ . x = [l. o f . = X j( x . = y t-2 y. y^^y 0 -1 0 2 0 0 y 0 0-1 0 -1 0 Therefore. Substitute 0 for in equation (1)..F | = 0 s 0 0 0 -1 o ' 0 1 0 .= x . and solve the non-linear equations...0 ] = y 2 y 2 . Hence.a n d y^ for Xy.1 ) + 2 ( j . point is. Substitute for i^ .) •«2 = -X . what does Lyapunov theory tell us about the stability of the nonlinear system near the equilibrium point? nonlinear system near the equilibrium point? Step-by-step solution step 1 of 7 (a) The non-linear autonomous system is.l) .. step 3 of 7 Recall equation (1). ..x . ic.= x. 0 10 y= -2 0 0 y 0 0 1 Step 6 of 7 (c) (i) The linearization is developed to determine the stability of the system near the equilibrium conditions. The linearized system is.1 ) = 0 (j “+ 2 )( 3 . for and y .* . J>|-1 for x^. the linearized system is 2 0 0 y 0 0 -1 Step 4 of 7 (ii) Consider the following equilibrium point. =-xfy Substitute for i .I f o r a:|.F | = 0 s 0 0 0 I 0 0 4 0 -2 0 0 0 0 4 0 0 1 s -1 = 0 0 0 s- j[s ( s . 0. and ^ .and ji . y i= ^ Differentiate the equations. Recall equation (2).+ 1 for jr .y .l .1 ) = 0 From the characteristic equation. Therefore.0 ] + 0 = 0 3*( j . _yj for jtj.= x .) Substitute for i | . 0 = ..31 PP Consider the nonlinear autonomous system j r 1 r 1 (a) Find the equilibrium point(s). .linear autonomous system can be written as follows: x . Recall equation (3). ji ..) (1) i i . X . o f Step 7 of 7 (ii) Calculate the characteristic equation. 0. y f( v ) (9) Where.j^ 2 dt 2 '■ dt wv ^ dv di. Consider the following equation for the inductance and capacitance equation. Apply KCL in the given circuit and the corresponding equation is given below. V = Cv— + L i . (5) Rearrange equation (5).. M c c dt = Cv >L / ( > ’) ' c c = . the system is stable for any positive DC diode characteristics. /< . C ^ * k + f { y ) = 0 .32PP Consider the circuit shown in Fig.W i. -(1) -(2 ) 4 - . 2 y.. y = .. c f = .4-/(v) dv" / ( ”) (6 ) d e c Step 2 of 2 Consider the following equation for the energy equation. dii _ v dt ^ L Thus. C is the capacitance. V = Cv ± . .. is the capacitance current. Problem 9. L is the inductance.68 in the textbook.^ (8 ) dt '■ dt Substitute equation (6) in Equation (8)... . is the diode current. . dv 2 ..ty ( v ) + v ij r = . di. F = ic v “ + iiiJ (7) Take differentiation on both sides.H + « o -0 (4) Substitute equation (1) and equation (3) in Equation (4). (3) Where.L i. from the equation (9).C v — + . For what diode characteristics will this system be stable? Figure Circuit diagram Step-by-step solution Step-by-step solution step 1 of 2 Refer Figure 9. y a y u \lu v lu i lu u u i i y = ^ o ivciu i i u i c i c y n j i i u i s ia u i ii iy a o p i c u i o i c u u y this Vin the Lienard plane.y) plane] i =y+s ( H (b) Use the Lyapunov function y = and sketch the region of stability as predicted by this Vin the Lienard plane. (c) Plot the trajectories of part (b) and show the initiai conditions that tend to the origin. G et help from a Chegg subject expert. ASK AN EXPERT . Step-by-step solution There is no solution to this problem yet. ^u/ w o cj u 1C L . Simulate the system in Simulink using various initial conditions on x{0) and A ( 0 ) Consider two cases.33PP Van der Pol’s equation: Consider the system described by the nonlinear differential equation 1 + «(1 + J i^ ) i+ x = 0 with the constant £ > 0. with £ = 0.5 and £=1. (a) Show that the equations can be put in the form [Lienard or (x. Problem 9.0. Show that the system response can be very sensitive to slight perturbations on the initial conditions x ^ ). (e) Repeat part (c) for /c = 0. Plot the Poincare sections for parts (c)-(e). and A = 7. 10))= 1 for f = 200 sec. Problem 9. G et help from a Chegg subject expert.25. Let k = 0. the spring gets stiffer as the displacement increases. e = 1. In other words rather than looking at the system continuously. Compare the two results.01 and 1(0)= 4.01. and A. Draw the phase-plane plot for the system.05. and ^ = 11. This equation represents the model of a hard spring where k is the spring constant and if £ > 0. Plot the time response of the system for x(0) = 1. Show that unlike the phase-plane plots in parts (c)-(e). Find the solution to the Duffing equation for x(0) = -1. the points fall on a well-structured plot referred to as a Poincare section (also called a strange attractor). i(0 ) (the system is said to be chaotic). we “strobe” the system and plot the behavior at strobe times only. S te p -b y -s te p s o lu tio n There is no solution to this problem yet. Simulate the response of the system with xfO) = 3 and 4 for f = 30 sec.5. 10))= 1 for f = 30 sec. e = 1.000 sec in order to plot the Poincare sections. (f) We can get more insight into the system by plotting H t j ) versus x(tj) at several hundred points at 2 n periodic observation times. Plot the time resoonse of the svstem for xfO) (b) Consider the unforced Duffing equation {u = 0). Show that the origin is an equilibrium point.1.34PP Duffing’s equation: Consider the system described by the nonlinear differential equation where u = A cos{t). Simulate the system using the initial conditions x(0) = -1 and A ( 0 ) = 1 for f = 10.5. ^ = 1. fb) Consider the unforced Duffino eouation (u = 0 ). e. Draw the phase-plane plot versus x(t)) for this case. ASK AN EXPERT . (c) Now consider the forced Duffing equation {u_= 0). (d) Repeat part (c) for k = 0. Repeat the simulation for slightly perturbed initial conditions xfO) = 3. and A = 8. (a) Build a simulation of the system in Simulink. (g) What can you conclude about the nature of the solution of the Duffing equation from the results of the previous parts? (h) Characterize the system behavior in terms of the ranges of the system parameters k. Using the above block diagram. = 1« . M = li SB(s) = l i p s f — !— .Y = R . Problem 10. Y ^O {W -\-B D c) E = R . or derivative). • ^ ^ ^ s-»o^ ll+ £ ) p O \+ D fP Step 3 of 3 Writing G'(-S') = and using a step input R { S ) = — and a step disturbance ' ' dO{s) ^ ' S 0)(^) = ^ we can show that integral control leads to zero steady state error. While proportional and derivative control in general.. do not Bitegral control. Step-by-step solution step 1 of 3 Integral control is he most effectiTe n reducing the error due to constant disturbances. D q (S) = ^ 0 Derivative control (s) = s . ■W \+ D c O \-D (D O . (^S) = — = U E[S)= i-* 0 S+0 dgS + n g ) If Proportional control. ^ 0 if i^ p ( 0 ) ^ 0 This analysis assumes that there are no pole zero cancellations between the plant G and the con^ensator. integral.01PP Of the three components of the PID controller (proportional. which one is the most effective in reducing the error resulting from a constant disturbance? Explain. proportional or derivative control will not have zero steady state error .0 { W + EDc) 1 « O E *. Step 2 of 3 Block diagram for showing integral control is the most effective means o f reducing steady errors. in general. D . Problem 10. In Figure 2.5.26. Step-by-step solution step 1 of 3 Step 2 of 3 PD control o f satellite: Non . the sensor and the actuator are collocated.collocated Yes for con^arison. resulting in a stable closed- loop system with PD control in fig 2. see the following two root loci which were taken from the discussion in the text on satellite attitude control. the sensor and the actuator are not collocated creating an unstable system with the same PD control Step 3 of 3 .02PP Is there a greater chance of instability when the sensor in a feedback control system for a mechanical structure is not collocated with the actuator? Explain. Step-by-step solution Step-by-step solution step 1 of 3 Gain. hence the closed loop system will be unstable with F M = -1 0 ° to have P i/« 7 0 ° we nee d . (a< b). . Contributes phase lead of a i^proximately 80°. together with any number of lead compensators.03PP Consider the plant Gfsj = 1/s3. S+ D (a) What is the maximum phase margin of the resuiting feedback system? (b) Can a system with this plant. The masimum phase lead from a compensator (i?) = k is 90° with — = oo S+ b a In practice a lead compensator with —= 100. Ifw e try positiTe feedback one pole departs at 0° so again one pole starts into the rig h t-h a lf-p la n e . for exan^le. a double lead (S + a y compensator Dc(S) vith = 100 Step 3 of 3 b) No diis plant cannot be made unconditionally stable **Because the Root Locus Departure angles from the three poles at the origin are ±60°. the poles are alwa3rs in the right half plane. Problem 10. for low enough. Step 2 of 3 Q‘(dS') = has phase angle o f -270'* for all frequencies. Determine whether it is possible to stabilize this plant by adding the iead compensator D As) = K ^ . For lo w -enough gain the system will be unstable. be made unconditionally stable? Explain why or why not. 2y«)) tan''(0.250-A" 115 5 1 . calculate the value of Phase margin. = |G (» | = 2 0 lo g ( l+ l) = 6. < O js]0.1778 Therefore.1y«>) (o. -> 0 173. But. F . 0. there is the possibility for transient behavior of the system.02SA :+866.x lo g ^ + A .1 -90 5. Second pole at break frequency.43 45 5. a: >160.000 115 5 0 Figure 2 Step 16 of 18 From figure 2..52 1 -90 45 11. 100c Therefore.28 -263. 1 <?(*) (7) i(i+5)(j-tl0) A ( f F l) KX j { s ) D { s ) = . Corner Frequency Slope Change in slope Term g) mm (S) -1 20 - ( y ® + i) -20 0 1 -20 -20 (o.a 2 .020db Step 6 of 18 i)a t .424. Consider figure 1 and find the magnitude corresponding to . ®caj 7. The magnitude corresponding to 70* is -7dB.000and^<0.92db Calculate gain at s . di^slOO Substitute Ja> for s in Equation (2). (8 ) t(f +S)(s+10)(«+100) Substitute Substitute equations (7) and (8) in equation (6) and simplify the equation.56 2.86 -85..42 84.iy v » + i)(o .to « } ^ x lo g ^ + A ^ ^ » 0 + 6.42 87.88. Figure Control system W(s) m • ns) Step-by-step solution step 1 of 18 (a) Step 2 of 18 Refer Figure 10.000+A" 52 173.72 Therefore. ® . the system moves to stable condition at Al > 0 ^nd -155.98 db 10J +A.. IF (* (9) « (« + S )(l + 1 0 )(s + 1 0 0 )+ A (s+ l) If q t(/) is consfant.020db s6.17781- Step 11 of 18 (d) Determine the value of K to yield the phase margin of q».88 in the textbook.) (g) Suppose the specifications require you to allow larger values of w than the value you obtained in part (f) but with the same error constraint [|y(°°)| 0. and determine what value of K causes (e) Sketch the root locus with respect to K for the system. Use the formula and find the value of K. a:( j + i ) X . Phase margin is I80’ . the value of gain margin is |7 <|B|- Step 10 of 18 (c) Determine the value of K to yield the phase margin of 7q* . . 000. Step 3 of 18 Derive the transfer function from figure 10.13 84..72 -119. 3. 2y<»+ i) 1 <» rt-1 0 -20 -40 (O .7 _7_ k>gA:» '2 0 a: -2 .71 1. g ( j+ i) G (*)J > W = .69 45 26.57 0.1? (Assume r= 0.424.250.o iy o + i) Step 5 of 18 Calculate gain A using table 1.020db 1 1> at a>= +A.x lo g ^ + A _ ®c4j “ 1-13. (5) Substitute equation (2) in equation (5). G (» =r (4) ' ' [ y<»(o.2 3 8 Therefore..04PP Consider the closed-loop system shown in Fig.250. i+ .057 -82.. and (3). 1 1. r-o s{s+S){s + 10)(i +100) Simplify the equation. \ + K J > {s ) G {s ) = 0 . Problem 10. c<10 Hence. ®c4j — l-1 3 .87 100 -90 89.0]) Step 13 of 18 Figure 2 Step 14 of 18 Consider the formula and calculate the range of K.^ A . Therefore.155. (a) What is the phase margin if /C= 70.71 0. .88 in the textbook. Third pole at break frequency.xlog— 1+A ^ +6.250.28 45 -216.58 5 -90 78. tan"' * -90 tan"' ( J a ) tad‘ '{0. what is the maximum allowable value for w if yC“ ) is to remain less than 0.4 to < » . Number of zeros present in the system is 1 One pole at break frequency. the range of K for the system to be stable is |0 < A <160. K {\+ s M G (* (3) ( i+ iM ) 0 + V < » i) ( i+ V < » > ) Compare Equations (2). ( 1 + 100) y (« )= .47 Draw the Bode diagram with phase and mac nitude values as shown in fic ure 1.000.000 is l i E Step 9 of 18 (b) From figure 1. ( 10 ) H 4 )= 7 Consider the formula for steady state output.250-A : Solve -AT* -155.. By adding the integral controller to the circuit reduces the steady state output error to zero. Consider figure 1 and find the magnitude corresponding to 7g».ly-ffl+l) I -1 0 0 -20 -60 ( o .000 in equation and rewrite equation (1).lta M * ) (11) Substitute equation (9) and (10) in equation (11) and apply limits.025 K +866.000? (b) What is the gain margin if /C = 70.550 53 115 5.G ( i ) Take the values from figure 10.99 1000 -90 89.. the gain K can be increased.721 Step 17 of 18 (f) Refer Figure 10. Consider J ^ * 0 write the transfer function. -201ogA: = .5 j -33. the maximum allowable limit a> occurs when lOOc step 18 of 18 (g) 100c Consider the equation y Assign AT = 10. the value of phase margin at AT= 70.1 and find the value of c. o step 15 of 18 Apply Routh-Array criteria to the equation.5000. Use the formula and find the value of K.57 -62.oiy<») de deg deg deg deg g ) deg g 0.025AT+866. -2 0 \o g K ^ \S I ^ *5 lo g A :. Assign lower frequency ^ and higher frequency ^ slO O O '^^- sec sec Calculate gain (A) at ■ A=\G{je>)\ -2 0 Io g M = 42. ( 1) < (« + S )(« + 1 0 )(s + l0 0 ) Substitute 70.30 5. Gain margin is 7 dB. G ( j) D ( j) = i ( j + 5 ) ( i + 1 0 ) ( f + 100) _________7 0 0 0 0 (» + l)_________ (2) Plot the straight Bode diagram for the transfer function <?(#) • Step 4 of 18 Consider the standard form of transfer function for a third order system. and determine what value of K causes the system to be on the verge of instability.o iy « » + i)] Find the magnitude of G { j 0 ) in dB using Equation (4) as listed in table (1). >'(4) G (4 (6) B '( i ) l + A J 3 ( s ) . calculate the value of gain margin.73 10 -90 84. (f) If the disturbance w is a constant and K = ^Q.1].23^- Step 12 of 18 (e) Write the matlab program and draw the root locus of the transfer function. rlocus([1 1].959db a - n f l» r tto < » .9 8 d b 10 J n a » . 10' 10' tor Frequency (ladtsec) Figure 1 Step 8 of 18 From figure 1.14 0. the value of K that yields 7q* phase margin is (AT =0.[1.1 6 3 '= 1 T Gain margin is 7 dB. by adjusting the values of ^^and c.. to«>.020db Calculate gain (A) at c t fs o . Discuss what steps you could take to alleviate the problem. The magnitude corresponding to 70* is 15 dB.1550.2y < » + i)(o . the value of K that yields 7q* phase margin is |Af = 2.71 89.94 89.28 63. 5^1154^+I550S*+(5000+ a:)4 + a: . fhen q t ( / ) s c - Converting q )(/) in S-domain yields. Therefore.98db Step 7 of 18 Find the phase plot for equation (4).88 in the textbook.000 >0and find the value of K. -|s lo p e fix ) m a )^ . 000? (c) What value of K will yield a phase margin of ~ 70*? (d) What value of K will yield a phase margin of ~ 0*? (e) Sketch the root locus with respect to K jo r the system.115. we can utilize the magnitude criterion at the desired dominant closed .7 so that our lead design is ^ s+n i To find the compensator gain. a{s) So we have .05PP Consider the system shown in Fig. D. As the plot shows ®. we get P s 11. = 3 and PAf = 67-3".15+ 4) Is shown on the next page using li^at lab’s bode command.7 ) ( ^ “ + 0. which represents the attitude rate control for a certain aircraft. we can write an e ^ e s s io n fort the angle. and lead network pole . k. Contribution from the lead network zero. at the closed loop pole location S '= .( S ) 1 7 .1j + 4 r Rateg Step-by-step solution step 1 of 3 b) The Bode plot the system loop transfer functions. Figure Block diagram for aircraft-attitude rate control when w/7 2 ^ 5 and ^>1^5^ Figure Block diagram for aircraft-attitude rate control Compensator Hydiuulic servq r Aircraft 2J + 0. and select the compensation so that the crossover frequency is at least 2 ^ rad/sec and PM > 50*.2 ± j i z = l = > t = 17.5.s) With section of z s 0.0 So die lead design is D .4^ =117® PHI=.pt po^yval{d. Let Using the angle criterion. =180®=> +134®-180®-135®-116® = -180® . '^0 angle(polyval (n. ^ ^ ' s+n i step 3 of 3 With a constant gain compensator ^ >die root locus of D .4 .loop pole location we find that | A W L = . (c) Sketch the root locus for your design.2 + 2 j . Therefore we need condensation of at least a lead network.( S ) 0 [S ) = = = + 0-15+A) dm Does not pass through-2 ± 2 J. 3 4 (^ r+ 0 -4 ) (^r + 0-05) D ^ [S ) 0 [S )^ S ( ^ '+ ll. Problem 10.1 s 5^ + 0. Step 2 of 3 The velocity constant is most easily found from either the bode plot or from ^ .. (b) Sketch the Bode plot for your design.{s).s) pi . (a) Design a compensator so that the dominant poles are at -2 ± 2j. and find the velocity constant when ojn 2 ^ 5 and 0. (b) The gain K must be negative for the system to be stable. thereby making the final value of e negative. Step 2 of 7 b) True Step 3 of 7 c) True Step 4 of 7 d)True Step 5 of 7 e) False Step 6 of 7 £) True Step 7 of 7 U False . not important The Actuator pole dramatically alters the root locus plot o f the system to be controlled.1 N*m) S te p -b y -s te p s o lu tio n Step 1 of 7 a) Tme. the error signal e must have a final value greater than zero. and the disturbance results in a speed emor whose steady-state magnitude is less than 5 rad/sec. (e) If K must be negative for stability. the control system cannot counteract a positive disturbance (f) A positive constant disturbance will speed up the load. (g) With only a positive constant command input r. Which of the following claims are true? (a) The actuator dynamics (the pole at 1000 rad/sec) must be included in an analysis to evaluate a usable maximum gain for which the control system is stable. (c) There exists a value of K for which the control system will oscillate at a frequency between 4 and 6 rad/sec. Problem 10. even though it is tempting to ^proxim ate the actuator as infinitely &st and hence. tfW_The_system is Iin s ta h ip if IAC1 i n (d) The system is unstable if |/C| 10. (h) For K = -1 the closed-loop system is stable.06PP Consider the block diagram for the servomechanism drawn in Fig. Figure Servomechanism Distmbance (0. 07PP A stick balancer and its corresponding control block diagram are shown in Fig. '■ 'U o + v Let Dcz[S) = U . we have the compensator ' S + 451 Step 2 of 2 We need a lag network in addition to a lead network to get the required . (a) Using root-locus techniques. we can utilize the magnitude criterion at the desired dominant closed . From the lead network zero. Figure Servomechanism 1 (j 2 -6 4 ) Step-by-step solution step 1 of 2 To have the cotnpens ated plant root locus go through the pole location S = . We find that \ D ^ { S ) a [ S ) \ ^ = .001 for a constant input torque Td= • Phase margin > 50°. and lead network pole 4i^.o i+ ijU o o o + ij This compensator will meet our design specifications.loop pole location S = .S ± ^ = 'i=t^ k = A7\ Therefore. The control is a torque applied about the pivot. We have. (b) Use Bode plotting techniques to design a compensator Dc(s) to meet the following specifications.707). ^ = 0. The bode plot of D X s )a (s ) . Figure Servomechanism • Closed-loop bandwidth~= 7 rad/sec. To fmd the compensator gain.loop pole locations.S + ^ we can write an expression for the angle contribution. Problem 10. design a compensator Dc(s) that will place the dominant roots e s = -5 ± 5) {corresponding to ojn = 7 rad/sec. and • Closed-loop bandwidth~= 7 rad/sec. • Steady-state 6 displacement of less than 0. we employ a led con^ensator Using the angle criterion = -1 8 0 At the closed . k.S ± ^ . Step 6 of 9 b) For ilf.^ 1 1. and plot the root locus of the compensated system. (f) Find the root locus of the compensated system in part (e). (b) Using the plant transfer function from part (a). Problem 10.7+lJ U 2 I+ 1 J U 035 + 1A 2 2 1 +U The system bode plot and step response 2 ^ ear. Therefore..o ^ ( ^ ) = -1 7 2 And conq^ute the pole to be at ^ = .5 + lJ U o + U Step 4 of 9 Now select the zero o f D q (£f) one decade below vtiiich implies CD = 0-25 rad/sec This Results in . With d of the lag equal to 20.4 6 ± jll0 5 we set the lead zero at = 1-5 . Let t = 0 .0072 2 34-7 The Lag netwoilc is thus S 0-0072+1 And the loop gain is o f . Figure Block diagram of a standard feedback control system D ^s) — * G(5) S te p -b y -s te p s o lu tio n step 1 of 9 ^ v * 1 0 0 ^ ^ s l0 0 We have Dlead{S) = 22 . 2 s)* and let the performance specifications now be /Cv = 100 and PM > 40°. The velocity const system.y ^ A 0 0 0 3 + 1J U + 1J U 0 + 1J Step 3 of 9 e) Again the design specification o f steady . and compare your findings with those from part (c).+ i CD p Such that o = = Z = 37 and — = /7»110. .0-25+ lj D . . Step 2 of 9 That the requires the plant gain to be equal to 100. we have S Dlog{S) = -2-2± L 0-035+1 The Lead .08PP Consider the standard feedback system drawn in Fig. = ^ sO(S) = 1 0 0 i = > j t = l The Bode plot of 2500 . the steady- state error due to a ramp should be less than or equal to 0. (a) Suppose 2500AT G(j) = - s (i + 2 5 )‘ Design a lead compensator so that the phase margin of the system is more than 45°.= = 0.state error provides information for tile design o f k. the con^ensator is s+m Dc(S) =74-65 ■ S+\2S4 Step 7 of 9 c) TheD esignspecificationof steady state eiror provides information for Step 8 of 9 a) The Design specification of steady state eiror provides information for the design ofk.Lag con^ensator is 1.F o r i. = 115 rad/sec And S = . (d) Using G(s) from part (c).1 sec. = 25%. since k^ = U SO{S) = jt= > it= 1 0 0 Therefore 100 G {S ): S r(l+ 0 1 5 )(l+ 0 -2 5 ) 5000 5r(2T+5)(5+10) Hence.0.1+1 Now we select the zero o f the lag at last one decade lower than W^. The Design o fk = Im s 0{S) = k = > k =100 The Bode plot of 100 a (^ : 1 . (e) Repeat part (c) using a lead-lag compensator. < 0-1 4 -6 Let Co. Step 9 of 9 The Lead con^ensator is ^+1 ^+1 = -I— — +1 .4 . design a lag compensator such that the peak overshoot is less than 20% and/Cv=100.01. 1 = 0 0 1 ^ J t= 1 0 0 it. (c) Suppose (c) Suppose G (s)- j (5 + 1 + 0 . S D a { S ) = 01+1 J ^ 0-003+1 And the loop gain is 100 f— 1 O c (S )= - / g Y g Y . we have cc S D c (S ) = 110+1 The Bode plot o f the conq^ensated.1 j ) ( H .(s y a {s ) = s{ ^ ^0 0072+ i A 5 + i A 1 0 + V Step 5 of 9 d) We can design a lag con^ensator using root locus methods.1284 using the angle criterioa The Bode plot and step response show the specifications are net with an additional gain of 20. design a lead compensator so that the overshoot is less than 25% and the 1% settling time is less than 0. Therefore.0 . « — = 46 • 01 Thus o . Is the lead compensation effective for this system? Find a lag compensator. • Zero steady-state error for step inputs.wj'lonifinafinnc (b) Design a suitable compensator Dc(s) to meet the specifications." ) . (a) What kind of compensation should be used and why? (b) rie. Figure Block diagram of a standard feedback control system D ^s) ds) Step-by-step solution ste p 1 of 2 a) Since we need 5 5 °-1 0 ° s 45° ofphase lead.( s ) a ( s ) = K ill.09PP Consider the system in Fig.225)(J+ 4)(f + 180)■ The compensator Dc(s) is to be designed so that the closed-loop system satisfies the following specifications. Step 2 of 2 b) From a phase lead requirement of 45° we have —m 10 a The co n ^ ensator and the loop gain are S 3-5+1 D . Problem 10.flw. • PM = 55% GM > 6 db..sipn.=uitabJfifCoroDeosatocJ)c''sTtoji>©<at tbe. • Gain crossover frequency is not smaller than that of the uncompensated plant. where 300 G (s) = - s(s + 0. a single lead network will do the job. Whereas the state variable technique requires a controller of complexity comparable to that of the plant right from the start. and the state-variable pole-assignment method.locus or state variable pole assignment are the most direct for control over dynamic response. percent overshoot. (d) This method provides the most direct control over the steady-state error constants Kp and Kv (e) This method is most likely to lead to the least complex controller capable of meeting the dynamic and static accuracy specifications. . Step 4 of 7 d) The Frequency response method of bode shows the error constant directly on the graph. Step 3 of 7 c) The state variable pole assignment is most easily programmed because once the specifications are given. G(I) = (j + 3)2' S te p -b y -s te p s o lu tio n step 1 of 7 a) Frequency response method is the most convenient for experimental data because the sinusoidal steady . Explain which of these methods is best described by the following statements (if you feel more than one method fits a given statement equally well. Step 5 of 7 e) The Root locus or bode method will give the least complex controller. and settling time.10PP We have discussed three design methods: the root-locus method of Evans. Step 6 of 7 f) Either the root locus. say so and explain why). (c) This method lends itself most easily to an automated (computer) implementation. the design is completely algorithmic. (g) This method can be used without modification for plants that include transportation lag terms —for example. Compensation only as necessary to meet the specifications. Step 2 of 7 b) Either the root . where by the locus is required to be entirely in the left half plane up to the operating gain. (b) This method provides the most direct control over dynamic response characteristics such as rise time. Problem 10. These techniques begin with the gain alone and then add network. (a) This method is the one most commonly used when the plant description must be obtained from experimental data. while the root locus requires a small modification and the state variable design method requires an approximation. Either the root locus or state variable design generally requires a separate system identification effort between the experimental data and the construction of a model suitable for the design method.state records can be obtained directly in the laboratory. Step 7 of 7 G) The Frequency response technique can be used immediately for transportation lag. the frequency- response method of Bode. (c) This method lends itself most easily to an automated (computer) implementation. (f) This method allows the designer to guarantee that the final design will be unconditionally stable. Assuming a Type 1 system.” or “a decrease. (d) Percent overshoot. In each case. Step-by-step solution Step 1 of 1 ^ Lead Lag k.” “substantially unchanged. indicate the effect as “an increase. Problem 10. (b) Phase margin. (c) Closed-loop bandwidth. (c) Closed-loop bandwidth. unchanged increased Phase Margin increased Unchanged Closed Loop Bandwidth Increased Unchanged Percent overshoot Decreased Unchanged Settling Time Decreased Unchanged .” Use the second-order plantG(s) = /C/[s(s+1)] to illustrate your conclusions. indicate the effect of these compensation networks on each of the listed performance specifications. {a)Kv. and (e) Settling time.11PP Lead and lag networks are typically employed in designs based on frequencyresponse (Bode) methods. e ^ q u 'a ^ 'w S g f iT . 2). z = altitude of the balloon. linearized about vertical equilibrium. 2002. (e) Select a gain K for the lead-compensated system to give a crossover frequency of 0. Use Routh’s criterion (or let s = jo ) and find the roots of the characteristic polynomial) to determine the value of the gain and the associated frequency at which the system is marginally stable. he could afford a more complex controller implementation.. The equations of vertical motion for a hot-air balloon (Fig. will be used.) S te p -b y -s te p s o lu tio n There is no solution to this problem yet. ^ .z(s)]. ASK AN EXPERT .06 rad/sec. 1). becoming the first solo balloonist to circumnavigate the globe (see Fig.3 m/(sec .12PP Altitude Control o f a Hot-Air Balloon: American solo balloonist Steve Fossett landed in the Australian outback aboard Spirit o f Freedom on July 3rd.. Because Steve Fossett was a millionaire. (f) If the emor in part (e) is too large.72. Problem 10. (b) Our intuition and the results of part (a) indicate that a relatively large amount of lead compensation is required to produce a satisfactory autopilot. z = altitude of the balloon. w = vertical component of wind velocity.z). (d) Select a gain K for the lead-compensated system to give a crossover frequency of 0. ri T2? + Z = o8 T -\-W t 6T = deviation of the hot-air temperature from the equilibrium temperature where buoyant force equals weight.06 rad/sec. 5q = Dc(s)(zd . An altitude-hold autopilot Is to be designed for a balloon whose parameters are r1 = 250 sec t2 = 25 sec a = 0. where zd Is the desired (commanded) altitude.. (c) Sketch the magnitude portions of the Bode plots (straight-line asymptotes only) for the open- loop transfer functions of the proportional feedback and lead-compensated systems. r1. how would you modify the compensation to give higher low- frequency gain? (Give a qualitative answer only. Figure 1 Spirit of Freedom balloon Source: Steve Holland/AP Images Figure 2 Hot-air balloon Only altitude is sensed. Sketch a root locus of the closed-loop eigenvalues with respect to the gain K for a double-lead compensator. a = parameters of the equations. so a control law of the form 6q(s) = Dc(s)[zd(s) . G et help from a Chegg subject expert. are S t + —ST = Sq.I2.”C).. (a) Sketch a root locus of the closed-loop eigenvalues with respect to the gain K for a proportional feedback controller. where /«+araV U+0. 5q = -K(z-zd). 6q = deviation in the burner heating rate from the equilibrium rate (normalized by the thermal capacity of the hot air).. and Z. 'q>. G et help from a Chegg subject expert.Zd). where J = moment of inertia of the wheel. Select the gains of a state-feedback controller Tc = to locate the closed-loop poles at j = -0 . Where should the zero of the lead network be located? Draw the root locus of the compensated system. (a) Suppose Dc(s) = KO. Dc(s) = compensation. / = satellite inertia (1000 kg/m2).Q 2 :t0 . What is the type of this system with respect to rejection of Tex? (h) Write state equations for the open-loop system. (d) For what range of K^ is the system stable? (e) What is the steady-state error (the difference between Z and some reference input Zd to a constant disturbance torque Tex for the design of part (c)?) (f) What is the type of this system with respect to rejection of Tex7 (g) Draw the Bode plot asymptotes of the open-loop system.0 ^ ^ . a damping ratio ^ = 0. /■= wheel speed.04 rad/sec. a = sensor constant (1 rad/sec). Tex = disturbance torque. Draw the root locus with respect to KO for the resulting closed-loop system.5. Problem 10. Add the compensation of part (c). S te p -b y -s te p s o lu tio n There is no solution to this problem yet. W heel: = —Tc. and compensation given by . ASK AN EXPERT . The equations of motion for such a system are Satellite: l ^ = Tc + Ta. Tc = control torque. q> = angle to be controlled. with the gain adjusted for the value of K^ computed in part (c). and compute the phase margin of the closed-loop system. Z = measurement from the sensor. Zd = reference angle. using the state variables g>. s+z De{s) = K i ^ ^ ^ . and give the value of K^ that allows the specifications to be met. Tc = control torque.13PP Satellite-attitude control systems often use a reaction wheel to provide angular motion. a constant. Measurement: Z = ^ — Control: Tc = -Dc(s)(Z . (b) For what range of KO is the closed-loop system stable? (c) Add a lead network with a pole at s = -1 so that the closed-loop system has a bandwidth wBW = 0. L \ = 1 4* i 64 . d = l . ^ R S’ + ( l+ r f ) S ‘ + ( * „ + fl( + * i) S + 4 1 ^ r ^ < i( g + 4 ) R (S + 4 )^ A nd k j. a = 4 ^ A_ R S ^ + {U k ^ )S + k 2 = ^^1 ^= 16. s 4 fo r all the design . System Kv -1 i. .2 . K2. ^ ^+4s+U S^+A s •R.14PP Three alternative designs are sketched in Fig. d. Problem 10. . = U e (t) = U SB(s) = 1 k . K3. k j . = 3 { d . *. 1 1 . (b) Complete the following table. a pole is to be placed at s = -4. 16 m -. * j2 + 4 i + I6 ‘ Note that in system III. KT. a.r1l t ^ 1 II III (c)Rank the three designs according to the following characteristics (the best as “1 the poorest as “3”): Performance Tracking Plant-noise rejection Sensor-noise rejection Figure Alternative feedback structures Step-by-step solution step 1 of 2 R ^ + aS + k^ :> A i=16. ^ 3 m. 1 w I j. (a) Compute values for the parameters K \ . 4 = 1 = 1 W ^L . The signal w is the plant noise and may be analyzed as if it were a step.4 ) => A^ = 16. expressing the last entries as A/sk to show how fast noise from V is attenuated at high frequencies. = 9 Step 2 of 2 £ fs) = R .y = R . for the closed-loop control of a system with the plant transfer function Gfsj = l/sfs + 1). S" +4S+16 e. and KD so that in each case (assuming w = 0 and v = 0). = — ^ k . expressing the last entries as A/sk to show how fast noise from V is attenuated at high frequencies:__________________________________________________ (b) Complete the following table. k j. the signal v is the sensor noise and may be analyzed as if it contained power to very high frequencies. where the output is stick angle.9 -9 S ) Step 5 of 6 e) L = [19-85983f Step 6 of 6 2055(8’+5-58) (8’-3-69) 0 D.F + LH) Step 4 of 6 ^ The conq^ensator transfer function can be obtained from = . with wn = 4 rad/sec.216 8.521 ± 1.832 and -0. and the control input is voltage on the motor that drives the cart wheels.i( S 7 .832 and -0.068j. (a) Compute the transfer function from u to y.521 ± 1. Problem 10. (g) Compute the frequency response of the two compensators.33 0 -0. Step-by-step solution step 1 of 6 a) The Transfer Function is -0-649 (^+0-0028) a{s): [S-5-59) {S-5-606) [S+0-2) Step 2 of 6 b) With Oc=(-Sr+2-832) (S^+2 084 ±4-272y) The feedback gains are calculated using the ACRERMAIIN’S formula Step 3 of 6 ^ c) The Estimator gains with ot^ (5^) = [S +10)^ with det (57 . (b) Determine the feedback gain K necessary to move the poles of the system to the locations -2. with wn = 4 rad/sec.068j.649 | k. (d) Determine the transfer function of the estimated-state-feedback compensator defined by the gains computed in parts (b) and (c). (b) Determine the feedback gain K necessary to move the poles of the system to the locations -2.[S) (S +48-2)(S-21-4) .016 -0. and cart velocity are r ® 1 0 ±= 3133 0 0. What Is the required estimator gain? (f) Repeat part (d) using the reduced-order estimator. and determine the poles and zeros. stick angular velocity.649 y = [ 10 0 0 Jx. L -31.P + a K + L H Y ' l 0-239S{S+5-60){S-30e) ~ (S+23 A ± J 2 2 \) {S .15PP The equations of motion for a cart-stick balancer with state variables of stick angle. (c) Determine the estimator gain L needed to place the three estimator poles at -10. (e) Suppose we use a reduced-order estimator with poles at -10 and -10. 151(5-1.327 -0 .2 1220 1 1 2 -0 .IO 9)(s-hO . and choose the estimator poles to be five times faster at ae(s) = {s + 5.12)(s + 0.1 0 ) Step 4 of 4 d) Use Mat lab bode command .0 5 ) (j+ 0 .0 3 2 8 ± 0 . (e) Note the similarity of this design to the one developed for different flight conditions earlier in the chapter.198).I5 I(J -H l.1441 ' ■ P ' ' 0.9 7 5 0 P 0.1478 0.].0 3 2 8 ± 0 .0 6 4 6 ± 0 .40/). If we use the state given in the case study (Section 10. ♦ .812 ± 3.0148 ' t 0. (c) Compute the transfer function of the SRL compensator.85187) ( S + 0.0 5 )(J -H 0 . 0 0.6 0 3 1 9 ± j 3 .O 425)(a-h0i)646± O .731/)‘ The corresponding transfer function is f ( j) -O .s ) = 0 Using Mat lab’s root locus command * = [0-0308 .7 3 y )* (a) Draw the uncompensated root locus [for 1 + /C6(sJ] and the frequency response of the system.0 4 2 5 )(s -h 0 . then the lateral-direction perturbation equations are p ■ ' -0 .3) and assume a velocity of 221 ft/sec (Mach 0. Problem 10.1 0 9 )(J -h 0 .1 6 6 0 r -0 .825)(s + 0.16PP A 282-ton Boeing 747 is approaching land at sea level. The corresponding transfer function is r ( j) -0.149 1 0 0 ■P ' r ^ = [0 1 0 0] P .0636 * .4 l4 fl “ (j-H .2 1 7 -0 .3 3 0.58)(s + 0.168 -0 .681/).165)(s + 0.0 8 9 0 -0 .151 — + p -1 . What type of classical controller could be used for this system? (b) Try a state-variable design approach by drawing a SRL for the system.034] The estimator gains are i = [1546-75 39-53 9 7 3 -9 8 f Step 3 of 4 c) The con^ ensator transfer function is given by -38-247(5+0-94479) (5r+0-9851±yl-CT13) D c (s ) = (S + 6 -2 9 8 7 )(£ ’ + 0. Choose the closed- loop poles of the system on the SRL to be ac(s) = {s+ 1.4 1 4 0 G(5) = s m ~ (5-H . What does this suggest about providing a continuous (nonlinear) control throughout the operating envelope? Step-by-step solution step 1 of 4 a) A clas sical lag network could be used to lower the resonant gain Step 2 of 4 The Symmetric Root locus l+ j tG ( s ) a ( .162 ± 0.9 8 9 0. (d) Discuss the robustness properties of the system with respect to parameter variations and unmodeled dynamics. A key component of this system is an armature-controlled DC motor.) Next 0 1 0 So 0 e.0 65. [if iJ . Note that we have assumed that both potentiometers have the same proportionality constant.0 -24. and the load damping is BL. The error signal Ei .2 54. = -N % Step 3 of 6 First we will find the transfer function from ^ to Sg and then we will find the closed loop transfer function.0 -10. and a back emf constant Ke. which in turn produces an armature voltage that drives the motor. L.. (c) The open-loop frequency-response data shown in Table were taken using the armature voltage va of the motor as an input and the output potentiometer voltage EO as the output. a torque constant Kt. Assuming that the motor is linear and minimum-phase.0 60.0 80.0 19.0 -48.0 34. 0 K .0 4. 1.0 5.0 40. is proposed as a position control system.{S ) step 4 of 6 O (S’) . Similarly. Swindlehurst) The feedback control system shown in Fig. + J .1 60. Figure A servomechanism with gears on the motor shaft and potentiometer sensore Frequency-Response Data Frequency (rad/sec) Frequency (rad/sec) 0. the moment of inertia of the load is JL. Finaiiy. (a) Write the differential equations that describe the operation of this feedback system.0 1.0 3.0 30. an armature inductance La.0 -20. Design Dc(s) to meet these specifications.A. (aj Wnife tne ainefenfiai dquaiio'nS tnar'deschOS thfe operation bt inis reeaPaciCsystem. k S(£r+5)(S+70) step 5 of 6 1p ^ * — * 0-3 sec " 6 Step 6 of 6 ^ ' (S + 2 0 0 f . T is proportional to the angular speed. The moment of inertia of the motor shaft is Jm. (d) Determine a set of perfonnance specifications that are appropriate for a position control system and will yield good performance. Ei = Kpdi. We have r.5 200. Problem 10.3 50. and the rotational damping due to bearing friction is Bm. the gear ratio is A/.0 -21.0 -72.8 42. (b) Find the transfer function relating 00 and 6i(s) for a general compensator Dc(s).0 23. => e .0 500.0 27. 0m(s) G(s) = where dm is the angular position of the motor shaft.0 20.0 10.0 0.0 40. = 0 -a -P + 0 e. 4.0 300. we have JA = T -P r.0 -59.4 .0 2.0 7.0 0.17PP (Contributed by Prof. fee radius of the gear is connected to the motor shaft and is the radius of fee gear connected to the ou^ut shaft. The input potentiometer produces a voltage Ei that is proportional to the desired shaft position.0 0.EO drives a compensator.0 -30.0 14.5 46. The motor has an armature resistance Ra.0 2. J Where r^. 1 ^ A. (e) Verify your design through analysis and simulation using Matlab. the output potentiometer produces a voltage EO that is proportional to the actual shaft position: £0 = Kpd.5 Step-by-step solution step 1 of 6 Step 2 of 6 Using KIRCHHOFF’S voltage laws we can write The torque of the motor.0 0. .0 100.a {s )D . N k. And so the closed loop transfer function is M £1 _ k . = M-. make an estimate of the transfer function of the motor. hence dt Using NEWTON Law OF Motion.. Problem 10. It uses colls surrounding permanent magnets as the actuator to move the beam. Figure Ball-balancer design example Source: Photo courtesy o f David Powell Step-by-step solution step 1 of 1 STEP RESPONSE . and a hall-effect device to sense the beam position. Research other possible actuators and sensors as part of your design effort. Compare the quality of the control achievable for ball-position feedback only with that of multiple- loop feedback of both ball and beam position. solar cells to sense the ball position. An example of such a device is shown in Fig.18PP Design and construct a device to keep a ball centered on a freely swinging beam. Problem 10. ASK AN EXPERT . G et help from a Chegg subject expert.19PP Design and construct the magnetic levitation device shown in Fig. You may wish to use LEGO components in your design. Figure Magnetic bail levitator used in the laboratory Source: Photo courtesy o f Gene Franklin fl 3 Step-by-step solution There is no solution to this problem yet. ASK AN EXPERT . Problem 10. G et help from a Chegg subject expert.20PP Design and build a Sun tracker using an Arduino board and related software. Step-by-step solution There is no solution to this problem yet. where b is known. Problem 10. 4r Oxide diickiiess = J pe~'^dt. The system reference input R is a desired step in temperature {700° C). replaces R by (i? +b) to cancel the sensor bias. Step-by-step solution step 1 of 2 a) We just increase R by + b Le. Normally. The model of the system is first order. and an integral controller is used as shown in Fig. where ff is the process duration. and p and c are known constants. and control the wafer surface temperature accurately using rings of tungsten halogen lamps. we are trying to control the thickness of the oxide film grown (Ox) on the wafer and not the temperature. In reality. Figure RTP system (a) Suppose the system suddenly develops a sensor bias b ^O . The semiconductor process engineer must use off-line equipment (called metrology) to measure the thickness of the oxide film grown on the wafer. Step 2 of 2 ■y-T 5R T = s a . Ox = 5000 A) by employing the temperature controller and the output of the metrology. there is no sensor bias {b = 0). At present. What can be done to ensure zero steady-state tracking of temperature command R.21PP Run-to-Run Control: Consider the RTP system shown in Fig. despite the presence of the sensor bias? (b) Now assume b = 0. The output of the system is temperature Tas a function of time: y = T(t). and the control input is lamp power. We wish to heat up a semiconductor wafer. The relationship between the system output temperature and Ox is nonlinear and given by the integral of the Arrhenius equation. no sensor can measure Ox in real time. Suggest a scheme in which the center wafer oxide thickness Ox can be controlled to a desired value (say. A pyrometer is used to measure the wafer center temperature. ^ A = le d B = ■4p. Step-by-step solution step 1 of 1 Let us define the normalized temperature T x = — andRe-w rite Eq(12) as the non linear.opC£^„ i = 4 A 4 j: .B 1 ic d A A 4 16s o 7 ? The Kadioactive power 7 = . First order system % X ——A —x*^) ^ ~IT 4 e o T. Problem 10.22PP Develop a nonlinear model for a tungsten halogen lamp and simulate it in Simulink. = 3-7419 X 10r“ ffl/W.23PP Develop a nonlinear model for a pyrometer. frequencies or wave length. <j 7-‘ Step 2 of 2 The Black body Emissive flux is given by Where C. is obtained by integrating over all frequencies or wave length. Step-by-step solution step 1 of 2 The total black body radiation intensity . Problem 10. Cj = 14388 0 = a’ c r * The temperature may be determined as r = ^ ’ Where . Show how temperature can be deduced from the model. Demonstrate the performance of the linear design. Problem 10.24PP Repeat the RTP case study design by summing the three sensors to form a single signal to control the average temperature.S) ■ S . and validate the performance on the nonlinear Simulink simulation. Step-by-step solution step 1 of 2 A Linear model for the system was dehTed in the text as r = Y= H^T + J ^ A Linear model for the system was derived in the text as 7* = Y= H^T + J ^ Where ^ =L and ■-0-0682 0-0149 0-0000 F = 0-0458 -0-1181 0-0218 0-0000 0-4683 -0-1008 0-5122 G= 0-5226 0-4185 Step 2 of 2 a (^ r) = ■ 0 ■4344 (3-+0.0878) (S + 0-1485) ” (S + 0 1482) ( J + 0 0527) (S +0 0863) We may try a simple PI controller of the form (8' + 0 0527) Doi. fr< 20 sec. Mi. Laboratory experiments have shown that the transfer function from the heater power. 3IS me Toiiowing iime-aomain specincaiions i. y.25PP One of the steps in semiconductor wafer manufacturing during photolithography is performed by placement of the wafer on a heated plate for a certain period of time. . ASK AN EXPERT . to the wafer temperature. Steady-state error to a 1*C step input command < 0. Draw the 180° root locus for the compensated system. Dc(s). Mp < 5%. design a dynamic compensator.00018) (a) Sketch the 180* root locus for the uncompensated system. u. ts < 60 sec. u(s) (j + 0 .1 *C.78)( s + 0 . is given by ^ = C(s) = __________ 2:5?__________ .1 9 )(5 + 0. Mp < 5%. G et help from a Chegg subject expert. (b) Using the root-locus design techniques. iv. such that the system meets the following time-domain specifications i. S te p -b y -s te p s o lu tio n There is no solution to this problem yet. ii. Problem 10. 0 . then plot the disturbance rejection response of the system for a unit step disturbance input.2 + 0. Step 2 of 5 Sketch the Feedback loop representatioa Step 3 of 5 ^ Find the Lr^lace transform of the given equation ___________ 0 .8 Step 5 of 5 ^ The disturbance response is shown in the following figure. It rejects constant disturbances in a robust &shion FindG(s) fo ra = 0.) = s’ + 1.5 X 0.26PP Excitation-Inhibition Model from Systems Biology {Yang and Iglesias.2) 0. Draw the feedback block diagram of the system showing the locations of the disturbance Input and the output. and y is the output which Is the fraction of active response regulators. It is known that a ^ Ifor this version of the model. Step Response Time (sec) Thus. Show that there is an alternate representation of the system with the “plant” transfer function ____________ O jlf ! ) __________ j2 + ( l + o + y )« + (a + y + o K ) ’ and the “feedback regulator" j^ + ( l + o + K)5 + (« + y + a y ) and the “feedback regulator" It is known that a # 1 for this version of the model.5 and y = 0.2. Step-by-step solution step 1 of 5 The external disturbance to the ou^ut transfer function is given by wM ( s + a ) ( s + l) ( s + y ) Where w is the external disturbance signal proportional to chemo attractant concentration.5 and y = 0.“ )___________ ^ (^ ) _ + (1 + Of + y)s+{a+ y+ay) ay 1+ + (1 + Of + y)s + ( a + y + ay) W( s ) (s + a ) ( « + l) ( f f + y) Step 4 of 5 The significance of this particular representation is that it reveals the internal model. (1 .5 + 0. 2005): In Dictyostelium cells. The external disturbance to the output transfer function is: wd) ' ' ( I + o ) ( j + 1)(J + y ) ' where. the disturbance rejection response of the system is sketched.2.2)s + (0.5 + 0. . Problem 10. the activation of key signaling molecules involved in chemoattractant sensing can be modeled by the following third order linearized model. Hence the system is Type I with respect to disturbance rejection. and^ is the output which is the fraction of active response regulators.7s + 0. namely the pure integrator.5s 0 ( .5 ) b 0 (s) = s" + (1 + 0. What is the significance of this particular representation of the system? What hidden system property does it reveal? Is the disturbance rejection a robust property for this system? Assume the system parameter values are a = 0. w is the external disturbance signal proportional to chemoattractant concentration. 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