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(jc) point? If the “–1” point is Re{GH} inside the GH path. -1 then N=2 . If the “–1” point is to Find |GH(jc)| where c is the left of the GH path. defined by Im{GH(jc)} =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(jc)| -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 2e 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) jIm{GH} K 2 s 2 For 0 K j ( 90) 2 2 e /. There are no net rotations about the “–1” point.