Control_Basics

March 26, 2018 | Author: Vijay Indukuri | Category: Telecommunications Engineering, Signal Processing, Applied Mathematics, Cybernetics, Systems Science


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Fall 2010Control Systems (EEE 226) Unit-3 Frequency Response Analysis Faculty: Ramesh Babu. N A.P(Sr)/SELECT/VIT Frequency Domain Analysis of Control Systems Frequency Response Specifications 1. Resonant or peak value Mr : The maximum value of T(jw )  2. Resonant or peak frequency w r : The frequency at which the peak  response Mr occurs.  3. Bandwidth Bw or w B : The frequency at which drops to 70.0%  of , or -3db, its zero frequency or DC value.  4. Slope Roll-off or Cut off rate: The slope of at bandwidth w B  Frequency Response Specifications Polar Plots Magnitude vs Phase in polar coordinates as w varies from zero to ¥ .Draw all vectors then connect the tip off all of them as shown in red .At each frequency w determine the corresponding vectors (mag & phase) .FREQUENCY RESPONSE PLOTS. polar plot for G(s) = 1/(s +2) . Phase-angle curves can easily be drawn if a template for phase-angle curve of (1 + j!) is available. Multiplication of magnitudes get converted into addition. and a plot of phase angle = \G(j!) versus frequency !. . Advantages of Bode plot: An approximate bode plot can always be drawn with hand.Bode Plot         A Bode diagram consists of two graphs: a plot of 20 log jG(j!)j (in dB) versus frequency !. w 2/w n 2)± r Bode plots are straight line plots connected at corner or break frequencies Bode plots (in dB and degrees) simplifies the plots by adding and subtracting complex terms.Constant k 2.Construction of Bode plot          The factors are 1. .Complex Poles and Zeros of order r: (1 + j2z w /w n .Poles or Zeros at the origin of order p: (jw )± p 3.Poles or Zeros at s = -1/T of order q: (1 + jw T)± p 4. BODE PLOT FACTORS . BODE PLOT FACTORS . BODE PLOT FACTORS . 01 jw  1)  ( jw) ( jw  10) ( jw) (0.1 jw  1) 2 2 .Example H ( jw)  640( jw  1)(0.01 jw  1) 64( jw  1)(0. the output remains bounded with increasing time (all system poles must lie in the left half of the s-plane).  Marginal Stability: A system is marginally stable if some of the poles lie on the imaginary axis. while all others are in the LHS of the s-plane. Some inputs may result in the output becoming . for every bounded input.Stability Definitions  Bounded Input Bounded Output Stability: A system is BIBO stable if.  Methods are available for testing for roots with positive real parts. the roots of the characteristic equation. i.  Also.  .e.Stability Analysis To test the stability of a LTI system we need only examine the poles of the system. methods are available for testing the stability of a closed-loop system based only on the loop transfer function characteristics. which do not require the actual solution of the characteristic equation. Relative Stability Analysis The Nyquist Criterion The Routh-Hurwitz Criterion provides a check of absolute stability based on the closed-loop characteristic equation.  .  The Nyquist Criterion may be used to analyse the relative stability of the closedloop system based on the loop characteristics.  Relative stability refers to how close the system the system is to the absolute stability boundary. Relative Stability Analysis The Nyquist Criterion  Consider the general feedback system C(s) R(s) G(s) + – H(s)  The closed-loop transfer function is C( s)  G( s) R( s) 1  GH ( s) The characteristic equation is 1  GH ( s)  0  Define F(s) as F ( s)  1  GH ( s) . The Nyquist Criterion  F(s) is a rational polynomial in s and can be written generally as  z1)( s  z2 )( )L ( s F ( s)  ( s  p1)( s  p2 )( )L where -zi are the zeroes of F(s) and -pi are the poles of F(s) . . Note: the zeroes of F(s)  the roots of the characteristic equation  poles of the closed-loop system.  poles of F(s) D (poles of GH(s) (loop transfer s)  N ( s) = 0 L L function)  zeroes of F(s)  roots of the characteristic . L L L L     F ( s) GH ( s )  If D ( s) L .The Nyquist Criterion: Notes on Zeroes and Poles  The zeroes of F(s) are the values of s that make F(s) = 0 and the poles are the values of s that make F(s) N ( s) D ( s)  N ( s) N =( s ) . then 1 D ( s) L D ( s) L  The denominator and numerator order of F(s) are equal to the order of the loop transfer function GH(s) .  This is achieved through the application of the principle of the argument.The Nyquist Criterion The Principle of the Argument The stability analysis of the closed-loop system now becomes the task of determining if there are any zeroes of F(s) in the right hand side (RHS) of the s-plane.  .  This involves the mapping of a function from the complex s-plane to the complex F(s) or GH(s) plane. which is a result from general complex number theory. Im   Path  is a Re clockwise encirclement of point A A Im  A counterRe clockwise encirclement Enclosements in the complex plane.The Principle of the Argument: Some Definitions  Encirclements in the complex plane. Im  Im  Re Re The area to the right of the path  is the area enclosed by  . . then F(s) will trace a closed path in the F plane starting at F(s1) and terminating at F(s1) .The Principle of the Argument Complex Function Mapping  s-plane  F(s) complex plane mapping. . j s2  s3  Im{F} unique s1 F(s1 )  Re{F} F(s2 non unique ) F(s3 )  path If F(s) is analytic along the (no poles of F(s) on ) and s starts at s = s1 and traces a closed path terminating at s1 .  .The Principle of the Argument The principle of the argument states that for an arbitrary closed path  in the s-plane.  The direction of the encirclements of origin is the same as the path  if the number of zeroes of F(s) is greater than the poles. the corresponding closed path in the F plane will encircle the origin as many times as the difference between the number of zeroes of F(s) and poles of F(s) located in the area enclosed by the path  . but does not go through any poles of F(s). j j    -j   .The Nyquist Path  Define the Nyquist path  such that it encloses the right Poles of F(s) hand side of the s-plane.  Z. corresponds to the number of poles of the closed.The Nyquist Criterion Map F(s) along the Nyquist path enclosing the RHS. where Z is the number of zeroes of F(s) in the RHS. P.loop system in the RHS. corresponds to the number poles of the loop transfer function in the RHS. .  Then the number of clockwise encirclements of the origin of the F(s) plane is N = Z – P . P is the number of poles of F(s) in the RHS.  If P is non zero. If the loop transfer function is stable. then P = 0 and N must be zero for a stable closed-loop system. then there must be P counterclockwise encirclements of the origin. P is known.The Nyquist Criterion Normally. This works since F(s) = 1 + GH(s) .  A slight modification of the process is to map the GH(s) function rather than F(s) and then check for encirclements of the –1 point in the complex GH(s) plane.  . The Nyquist Criterion: A Simple Example  Consider a system with the loop TF s-plane j K GH ( s)  s( s  a) j GH-plane Im{GH} j mapping  Pole of GH   Re{GH} -1 -j -j .  The number poles of the loop transfer function GH(s) in the RHS.The Nyquist Criterion: A Simple Example The number of encirclements of the –1 point in the GH-plane is zero. N = 0.  Therefore. Note: the Nyquist path excludes the pole at the origin. is zero (P = 0). is Z = N + P = 0 + 0 =0. the number of poles of the closed-loop system = the zeroes of 1 + GH(s).  . in the RHS. j Section I : s =  e j . R  = +90  –90 Section IV : s = – j  j0– .The Nyquist Criterion: Sketching the GH(s) Function In most cases only an approximate sketch of the GH(s) mapping is required.   0  j  = –90  +90 II III   I IV -j Section II : s = j0+  j Section III : s = R e j .  Consider various segments of the Nyquist path.   0 .k k j k  e p1p2 L As  varies from – 90º +90º (CCW) /GH varies from k 90º  – k 90º (CW) .  = – 90º  + 90º ) j   z1)( L ) K ( e j GH (  e )  k j k  e (  ej   p1)( L ) as   0 GH  Kz1z2 L   / .The Nyquist Criterion: Sketching the GH(s) Function  Consider a general loop transfer function K( s  z1)( s  z2 )( )L GH( s)  k s ( s  p1)( s  p2 )( )L  Section I: ( s =  e j .  = 0+    0       GH ( j ) / k90    GH ( j ) 0 / (n m)90 where.The Nyquist Criterion: Sketching the GH(s) Function  Section II: (s = j . . n = the order of the denominator of GH m = the order of the numerator of GH This represents the normal frequency response of the loop system GH(j).  Section III: ( s = Re j . R   . .  = +90º  –90º ) j  z1 )( L ) K ( R K e j  ( n.m) This corresponds to (n-m)180 º counter-clockwise rotations about the origin.The Nyquist Criterion: Sketching the GH(s) Function  Section IV: This section is the mirror image of Section II.m) j ( n.(n .m) GH ( Re )  k j k j R e ( Re  p1 )( L ) R e  0 /. . II: s = j 0+ . Sec. III: no effect. GH(j)  0 /-270º Sec. GH(j)   /-180º s = j . I: (k = 2) Then GH(j0) rotates 360º clockwise from GH(j0–) to GH(j0+) with a magnitude of  . Sec. IV: mirror image of II.The Nyquist Criterion: Example 1 Nyquist Path j j II III   I IV -j GH ( s)  K s2 ( s  a) Sec. Z = N+P = 2 which means there are two poles of the closed-loop system in the RHS.e.The Nyquist Criterion: Example 1 j GH(j0+) GH(j0–) -1 -j Im{GH}  GH ( s)  K s2 ( s  a) Stability Analysis: ◦ There are 2 clockwise rotations of GH(s) about the “–1” point. the system is unstable. Re{GH} ◦ No poles of GH(s) in the RHS.  P=0 . i. ◦ Then. .  N = 2. I: (k = 1) Then GH(j0) rotates 180º clockwise from GH(j0–) to GH(j0+) with a magnitude of  . IV: mirror image of II. GH(j)  0 /-270º Sec. . Sec. GH(j)   /-90º s = j . Sec. III: no effect.The Nyquist Criterion: Example 2 GH ( s)  K s( s  a)( s  b) j j II III   I IV -j Sec. II: s = j 0+ . The Nyquist Criterion: K Example 2  GH ( s) s( s  a)( s  b) j Im{GH} Where is the “–1” GH(jc) point?  If the “–1” point is Re{GH} inside the GH path. -1 then N=2 .  If the “–1” point is to Find |GH(jc)| where c is the left of the GH path. defined by Im{GH(jc)} =then 0  . P= 0 and Z=2  the system is -j unstable. j  c3  ab j  c  0   c  ab Check if this is less than –1 .( a  b) c .j  3  ab j  . GH ( j  c )  2 . K K  Then.The Nyquist Criterion: Example 2 GH ( s)  GH ( j  )   K s( s  a)( s  b) K j  ( j   a)( j   b) K .( a  b) 2 For.( a  b)ab . Im{ GH ( j  c )}  0 . Im{GH} |GH(jc)| -1 fm Re{GH}  20log10 1 GH ( j  c ) Phase Margin: the additional phase lag in degrees. that will make the system marginally stable. that will make the system marginally stable.Relative Stability Gain and Phase Margins  Gain and phase margins are a measure of how close the system is to instability. fm  180 . . Gain Margin: the additional gain in db. II: s = j 0+ . GH(j)  0 /-180º Sec.2  j 2 2e j   2e j 2  2 GH ( s)  . GH(j)  K/2 s = j .   -90o  90o j II  IV  -j then III I  K j 2   (j 2 e ) 2 K  .The Nyquist Criterion: K Special Example GH ( s)  2 s 2 j 2  e Sec. I: Let s  j 2  e j  j j where   0 . Z = 0 and the closed loop system is stable for all values of K.(  90) GH ( s)  Re{GH} -1 K/2 -j Section III: no effect  Section IV: mirror image of II. N = 0. therefore. Since P = 0 ( no roots of GH in the RHS). .The Nyquist Criterion: Special Example GH (s)  jIm{GH} K 2 s 2 For   0 K j (  90)  2 2 e   /.  There are no net rotations about the “–1” point.
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