Block diagrams, SFG, Time response Analysis, RH criterion, Root Locus technique 1. For the electric network shown below determine R L such that V0 ( s ) Vi ( s ) = 1 1 2s + 3R + s + 2 R 2 (a) 1Ω 2. Determine (b) Determine (a) (c) 4. (c) R (d) None of these C C1 and 2 for the block diagram shown? R R (a) 0, 1 3. R 2 (b) 1, 1 (c) 0, 0 (d) 1, 0 C for the SFG shown below R G1G2G3 ..............Gn (1 + H2 )(1 + H3 )(1 + H4 ) ........ (1 + Hn+1 ) G1G2G3 ..............Gn (1 + H1 )(1 + H2 )(1 + H3 ) ........ (1 + Hn+1 ) (b) (d) G1G2G3 ..............Gn (1 + H2 )(1 + H3 )(1 + H4 ) ........ (1 + Hn ) G1G2G3 ..............Gn (1 + H1 ) (1 + H2 ) (1 + H3 ) ........ (1 + Hn ) For the SFG shown below determine: No. of forward paths = _______________ No. of Loops =________________ No. of two non touching loops = ________________ 1|Page http://g-krishna.blogspot.in (a) Z1 Z1 . x8 = x3 + x5 + x6 . x8 ? x1 11 13 Determine x5 = x 4 (b) 13 12 (c) 12 13 (d) None of these (c) 1 2 (d) 2 C ? R (b) 1 The open loop transfer function of a unity negative feedback system is G ( s ) = k 10 (s + 2) . Z1 + Z 2 + Z 4 Z1 + Z 2 (c) Z1 Z1 . Z1 + Z 3 + Z 4 Z1 + Z 3 (b) Z1 Z1 .5. Z1 + Z 2 + Z 3 Z1 + Z 3 A system is represented by the following equations: x 2 = x1 − x 3 . x7 = x6 .blogspot. x 6 = x1 − 3x 7 . Z1 + Z 3 + Z 4 Z1 + Z 2 (d) Z1 Z1 . x3 = x2 .in . The range of K for closed loop system stability is 2|Page http://g-krishna. x 4 = x1 − 2x 5 . (a) 0 8. Determine (a) 7. A SFG representation of an electrical network is shown below: Determine G2 and H ? 6. The range of parameter ‘a’ for which the closed loop system time response oscillates for two positive values of gain K is (a) 0 < a < 2. (b) 0 < K < 1500 (c) 0 < K < 1291.36 (d) 0 < K < 1691.828 10. The root locus diagram of the negative feedback control system as K varied from 0 to ∞ is: 3|Page http://g-krishna.in .2 and ω = 1r /sec 2) K = 0.(a) 0 < K < 1000 9. What are the values of K and frequency of oscillation ω ? 14. What are the number of roots of the equation which lie to the left of the S + 1 = 0? (a) 2 13.59 r/sec (c) 0.36 Consider a negative feedback control system shown below. then the closed loop system time response oscillates with a frequency of (a) 0. (b) a > 5 (c) a > 5. 1) K = 0. The closed loop system is marginally stable for a positive value of ‘k’ if the range of parameter T is (a) T > 0 11. (b) 3 (c) 4 (d) 5 The system having characteristic equation S + (1 + K ) S + 5KS + 6K = 0 is to be used as an 3 2 oscillator.29 r/sec 12.blogspot.5 and ω = 2r /sec 4) K = 0. (b) 0.828 (d) 0 < a < 1 The open loop transfer function of a unity feedback system is G ( s ) = K ( S + 1) S ( S + 2) ( S + T ) 2 . [c] (b) T > 1 (c) T > 2 (d) T > 3 For the feedback control system shown below the value of gain k is adjusted such that the closed loop system is marginally stable.2 and ω = 2r /sec The open loop transfer function of a negative feedback control system is G ( s ) H (s ) = K ( s + 2) (s + 3 ) s 2 ( s + 1) .99 r/sec 3 The characteristic equation of a control system is S + 15S + 85S + 225S2 + 274S + 120 = 0 .5 and ω = 1r /sec 3) K = 0.79 r/sec 5 4 (d) 0. The root locus diagram as K varied from 0 to ∞ is (a) (b) (c) (d) The open loop transfer function of a unity negative feedback system is G ( S) H ( S) = K (1 − S2 ) (S 2 − 4) . The root locus diagram as K varied from 0 to ∞ is (a) (b) (c) (d) 4|Page http://g-krishna.in .blogspot. (a) (b) (c) (d) The open loop transfer function of a negative feedback control system is G ( S) H ( S ) = 16. K (1 − S ) (1 + S ) ( 2 − S ) (2 + S) .15. But these acceleration measurements need to be supplemented by visual signals. with your hands resting on your hips and your elbows bowed outward.F. (d) Stable with your eyes open unstable with yours eyes closed.S = −3 A unity feedback has a loop frams for function G ( S ) = 4 ( S2 + 1) S (S + a ) (d) K = 17. This orientation system is primarily run by the information received in the inner ear. Did you experience a low frequency oscillation that grew until you lost balance? Is this orientation position stable with and out use of your eyes? (a) Stable with your eyes open and closed. (b) Close your eyes.in . (b) K = 27. The root locus diagram of an open loop transfer function G ( S ) H ( S ) = (a) (b) (c) (d) K ( S + 1) ( S + 2 ) S3 We all use our eyes and ears to achieve balance our orientation system allows us to sit (or) stand in a desired position even while in motion.17. of a unity feedback system is G ( S) = K ( S + 1) S2 ( S + 9 ) Find the gain when all three roots are real and equal also find the roots when all the are equal? (a) K = 17. (b) Unstable with your eyes open and closed. S = −3 Sketch the root locus for 0 ≤a ≤ ∞? (a) 5|Page (b) (c) (d) http://g-krishna.blogspot.S = −2 (c) K = 27. Try the following experiment: (a) Stand with one foot infront of another. The open loop T. 18. (c) Unstable with yours eyes open and stable with your eyes closed. S = −2 20. where the semicircular canals sense angular acceleration and the otoliths measure linear acceleration. 19. (b) 2.32 (c) 10. determine the gain K and time constant T of the controller such that the closed loop poles are located at S = −2 ± j2 6|Page http://g-krishna.25 (c) 5. The root locus diagram of G ( S) H ( S ) = (a) 22. Determine the range of ‘K’ for stability? (a) K > 0 (b) K > 1 (c) K > 2 (d) No value of K that allows systems stability. 23. (b) 0.5 ) Determine the value of K that gives the system characteristic equation a dampingratio of 0.in .blogspot.15 24. (b) is S ( S2 + 10 ) (b) The root locus diagram of G ( S) H ( S) = (a) K ( S2 + 2) (c) K ( S + 1) (d) 2 S3 ( S + 10 ) 2 is (c) (d) The open loop transfer function of a unity feedback control system is G ( S) = K ( S2 − 2S + 5 ) ( S + 2) ( S − 0. 26.0625 4 3 (d) None of these 2 Consider the following characteristic equation: S + KS + S + S + 1 = 0 .5? (a) 0. For the control system shown below.32 (d) 20.32 The root locus plot of a negative feedback control system is shown below The value of K for which the negative deedback system time response is non–oscillatory is (a) 1.21.5 25. (c) 3. 7.33 min (d) 0.4 31.16.β > 0 . 0. ab + k ab + k k (s+ ∝ ) (s + β ) 2 .25 min (c) 0.76.16. how much error does the thermometer show? (a) 2 degrees 29. –1 7|Page (b) 5. –1 http://g-krishna. T = 1 2 (c) K = 8. 28) A thermometer require 1 minute to indicate 98% of the response to a step input. 0. 1 (d) 1. −k k .5 min If the thermometer is placed in a bath.2 (d) 2.16. 1 Find the sensitivity of the steady state error with unit step input to changes in parameter K and parameter a for the system in Fig.(a) K = 4. ab + k ab + k (c) k k . Assuming the outputs steady state can be approximated by a ramp.8 (c) 2. T = 1 4 (d) K = 4. ab + k ab + k The open loop transfer function of a unity feedback system is G ( s ) = (d) k −k . No: 27.3 degrees (d) 5 degrees Consider a unity feedback control system with the closed loop transfer function C (s ) R (s ) (a) 30. (a) 32.in . 7. α and β such that.1.5 degrees = ks + b .2 min 28. 1 (c) 1. (b) 2. the ess for a unit step input is 0.16. the temperature of which is changing linearly at rate of 100/min. The steady state error in the unit ramp response is given by s 2 + as + b k−a b (b) a −k b (c) k b (d) a b Fig shows the ramp input r(t) and the output c(t) of a system. T = 1 2 Statement for Linked Answer (Q. T = 1 4 (b) K = 8. 1. (b) 0. ab + k ab + k (b) −k −k . (b) 2.5. 1. 27. Determine k. Assuming the thermometer to be a first order system.74.74.blogspot. Find time constant? (a) 0. 0.76. to meet the following specifications. and natural frequency is 10 r /sec? (a) 5. damping ratio is 0. Find the steady state error and the steady state error if the input becomes r(t) = t u(t)? (a) 2. π 2 (b) K = π 3 (c) K = π (d) K = π 4 A particular control system yielded a steady state error of 0. A unit integrated is cascaded to this system and unit ramp input is applied to this modified system. What is the value of steady state error for this modified system? (a) 0.33.in . the steady state 2 error is 0.2 (d) 0.25 A unity negative feedback control system has a loop transfer function G ( s ) = k ( s s + 2k ) Determine the range of K so that the settling time (2% criterion) is less than 1 second? (a) k > 8 8|Page (b) k > 64 (c) k > 32 (d) k > 4 http://g-krishna.blogspot.2 will be (a) K = 34. A position control system having ufb has G ( s ) = 10k s (1 + 0.1s ) The value of amplifier gain ‘k’ so that when the input shaft rotates at 1 rps .15 (c) 0.2 for unit step input.1 35. (b) 0.