Tài liệu bài giảng 4 bode diagram

The slides used material the source PowerPoint files of Anthony Rossiter the transfer function between a loop input and a signal somewhere in the loop GC  forward _ path M (s)   return _ path  M ( s )G ( s ) s 1 G( s)  ; M ( s)  s2 s For the loop above, find the transfer function relationship between the signal r and the signal u Where are the closed-loop poles? u M r 1 G ( x) M ( x) GC  ( s  1)( s  2) ( s  1)( s  2)  s ( s  2)  3( s  1) s  5s  For the loop below, find the transfer function relationship between the signal r and the signal y What is the closed-loop pole polynomial? Are the poles all in the LHP? 0.5 s  0.4 ( s  10 ) G( s)  ; M ( s)  ; H ( s )  0.1 s 1 s ( s  1) forward _ path M (s) GC    return _ path  M ( s )G ( s ) H ( s )  0.5( s  0.4) ( s  0.4)( s  1) s ( s  1) GC    0.1 0.5( s  10)( s  0.4) s ( s  1)  0.1( s  10)( s  0.4) 1 s ( s  1)  PC  s  s   0.4 s  0.4 For the loop below, find the transfer function relationship between the signal r and the signal z What is the closed-loop pole polynomial? Are the poles all in the LHP? Where are the poles? 0.5 s  0.5 G( s)  ; M ( s)  ; K ( s)  0.5; s2 s ( s  8) H ( s)  0.1 ; T (s)  ( s  0.8) s5 0.5 s  0.5 ; M ( s)  ; K ( s)  0.5; s2 s ( s  8) H ( s)  0.1 ; T (s)  ( s  0.8) s5 G( s)  GC  forward _ path K (s)M (s)   return _ path  M ( s ) K ( s )G ( s ) H ( s )T ( s ) ( s  0.5) s GC  ( s  0.5) 0.5 ( s  8) 1    s s2 ( s  0.8) ( s  5) The syntax for simple loops is as follows: Gc = feedback(FP,RP) Where • FP is blocks between loop input and signal • RP is remaining blocks in the loop >> %% Question >> G=tf(3,[1 2]) G= s+2 Continuous-time transfer function >> M=tf([1 1], [1 0]) M= s+1 s Continuous-time transfer function >> Gc=feedback (M,G) Gc = s^2 + s + s^2 + s + Continuous-time transfer function >> [p,z]=pzmap(Gc) p= -4.3028 -0.6972 z= -2 -1 A more efficient and simple method for determining A(w) and (w) is from the system transfer function G(s) It can be show that: A  G( jw) ;   arg(G( jw)) That is, substitute s=jw and compute the modulus and argument of the resulting complex number G s  3s  4 G ( jw)   ( jw)  jw  jw   w2 3w 1 G ( jw)  ; G ( jw)   tan 2 2  w 9w  (2  w ) Use frequencies of rad/s and rad/s G ( jw)  9w  (2  w ) 2 3w G ( jw)   tan  w2 1  4 G ( j1)    1.26 10  (2  1)  G ( j1)   tan  1.24 1 4 G ( j 2)    0.63 40 36  (2  4) G ( j 2)   tan  1.89rad 2 1 G s  3s  Overlay the input and output for a system G(s) and a single frequency (here w = 1rad/s) u  sin t y  1.2 sin(t  1) 0.5 Analytic solution u  sin t -0.5 y  1.26 sin(t  1.24 ) -1 45 50 55 Time (secs) 60 65 G s  3s  Overlay the input and output for a system G(s) and a single frequency (here w = 2rad/s) u  sin 2t y  0.62 sin(2t  1.8) 0.8 0.6 0.4 Analytic solution 0.2 u  sin 2t -0.2 -0.4 y  0.63 sin(2t  1.89 ) -0.6 -0.8 -1 50 51 52 53 54 55 56 Time (secs) 57 58 59 60 2s  G s  s  3s  Find the frequency response (gain and phase definitions) for G(s) jw  jw  G ( jw)   3 ( jw)  ( jw)  jw  j (3w  w )   w G ( jw)  4w2  (3w  w3 )  (2  w2 ) w ( w  w ) 1 1 G ( jw)  tan  tan 2w Be careful to take solution in the correct quadrant A( )  G ( j )  Re G ( j )2  ImG ( j )2 ImG ( j )  ( )  G ( j )  tan Re G ( j ) 1 We will not tend to use this means of finding the frequency response because it is not efficient and does not use the easy rules that come from multiplying/dividing complex numbers 31 An efficient method for determining the gain and phase A(w) and (w) is from the system transfer function G(s) However, the following expressions tend to be cumbersome to use A( )  G ( j )  Re G ( j )2  ImG ( j )2 ImG ( j )  ( )  G ( j )  tan Re G ( j ) 1 Consider simple factors with LHP roots (numerator or denominator) such as G(s)=(s+a) or H(s)=1/(s+b) G ( jw)  jw  a w G ( jw)  w  a ; G ( jw)  tan a H ( jw)  jw  b 1 w H ( jw)  ; G ( jw)   tan b w2  b 2 1 You may find it useful to convince yourself that the phases given here are indeed in the correct quadrant given a,b>0 Students are reminded of the following rules for complex numbers Modulus of the product is the product of the moduli Phase (or argument) of the product is the sum of the phases wzy  w z y ; wz wz  ; ykm y k m 2 w z w z  3 yk y k w2 z wzy  w  z  y;   2w  z  3y  k yk Use rules of complex numbers to find the frequency response for the following sa G (s)  sb jw  a jw  a   a2 G   jw  b jw  b   b2 jw  a 1   1   ( )  ( jw  a)  ( jw  b)  tan    tan   jw  b a b Given G(s)=2(0.5+s)/(2+s), find the gain and phase for a frequency of G ( j )    22     16   1    1    G ( j )  tan    tan    tan   2 3 4 1 Which is the amplitude and phase of the output? G ( j )  G(s)  ; u  sin 2t ( s  6)( s  4) (  ) (  )   1    1   1   G ( j )  tan    tan    tan    tan   6 4 3 2 1 Write the gain and phase expressions ( s  2) G(s)  ( s  3)( s  4) G ( j )      22   32     1    1     G ( j )  tan    tan    tan    tan   2 3 4 1 1 Write the gain and phase expressions G ( s)  s ( s  4) G ( j )     42  G ( j )  900 [...]... number 4 G 2 s  3s  2 4 4 G ( jw)   2 ( jw)  3 jw  2 3 jw  2  w2 4 3w 1 G ( jw)  ; G ( jw)   tan 2 2 2 2 2  w 9w  (2  w ) Use frequencies of 1 rad/s and 2 rad/s G ( jw)  4 9w  (2  w ) 2 2 2 3w G ( jw)   tan 2  w2 1  4 4 G ( j1)    1.26 2 10 9  (2  1) 3  G ( j1)   tan  1. 24 1 1 4 4 G ( j 2)    0.63 40 36  (2  4) 2 6 G ( j 2)   tan  1.89rad 2 1 4 G 2... function >> M=2*tf([1 0 .4] , [1 0]) M= 2 s + 0.8 s Continuous-time transfer function >> H=0.1*tf([1 10], [1 1]) H= 0.1 s + 1 s+1 Continuous-time transfer function >> Gc=feedback (M*G,H) Gc = s^2 + 1 .4 s + 0 .4 -s^3 + 2.1 s^2 + 2. 04 s + 0 .4 Continuous-time transfer function >> [p,z]=pzmap(Gc) p= -0.9226 + 0. 847 7i -0.9226 - 0. 847 7i -0.2 548 + 0.0000i z= -1.0000 -0 .40 00 >> %%Question 3... transfer function >> T=tf(2, [1 5]) T= 2 s+5 Continuous-time transfer function >> Gc=feedback (M*K,G*H*T) Gc = s ^4 + 8.3 s^3 + 19.5 s^2 + 15.8 s + 4 s ^4 + 7.8 s^3 + 15.7 s^2 + 8.85 s + 0 .4 Continuous-time transfer function >> [p,z]=pzmap(Gc) p= -5.0210 -1.8656 -0.8 640 -0. 049 4 z= -5.0000 -2.0000 -0.8000 -0.5000 • What is frequency response? • Why is it useful? • How do I represent frequency... 0.5 Analytic solution 0 u  sin t -0.5 y  1.26 sin(t  1. 24 ) -1 45 50 55 Time (secs) 60 65 4 G 2 s  3s  2 Overlay the input and output for a system G(s) and a single frequency (here w = 2rad/s) u  sin 2t 1 y  0.62 sin(2t  1.8) 0.8 0.6 0 .4 Analytic solution 0.2 0 u  sin 2t -0.2 -0 .4 y  0.63 sin(2t  1.89 ) -0.6 -0.8 -1 50 51 52 53 54 55 56 Time (secs) 57 58 59 60 2s  1 G 3 2 s  s  3s ... frequency (here w = 1rad/s) 4 G 2 s  3s  2 Output amplitude about 1.2 1 0.5 0 u  sin t -0.5 Phase shift about 1 second which -1 corresponds to 1 45 radian 50 y  1.2 sin(t  1) 55 Time (secs) 60 65 1 Output amplitude about 0.62 0.8 0.6 0 .4 0.2 0 -0.2 -0 .4 Phase shift about -0.6 0.9 second which -0.8 corresponds to -1 50 51 52 53 1.8 radian u  sin 2t y  0.62 sin(2t  1.8) 54 55 56 Time (secs) 57 58...  1   ( )  ( jw  a)  ( jw  b)  tan    tan   jw  b a b Given G(s)=2(0.5+s)/(2+s), find the gain and phase for a frequency of 4 G ( j )   2  22  2  9  2  16   1    1    G ( j )  tan    tan    tan   2 3 4 1 ... 57 58 59 60 2s  1 G 3 2 s  s  3s  2 Find the frequency response (gain and phase definitions) for G(s) 2 jw  1 2 jw  1 G ( jw)   3 2 3 2 ( jw)  ( jw)  3 jw  2 j (3w  w )  2  w G ( jw)  4w2  1 (3w  w3 ) 2  (2  w2 ) 2 3 2 w ( 3 w  w ) 1 1 G ( jw)  tan  tan 2 1 2w Be careful to take solution in the correct quadrant A( )  G ( j )  Re G ( j )2  ImG ( j )2 ImG ( j ... s^2 + 1 .4 s + 0 .4 -s^3 + 2.1 s^2 + 2. 04 s + 0 .4 Continuous-time transfer function >> [p,z]=pzmap(Gc) p= -0.9226 + 0. 847 7i -0.9226 - 0. 847 7i -0.2 548 + 0.0000i z= -1.0000 -0 .40 00 >>... Gc = s ^4 + 8.3 s^3 + 19.5 s^2 + 15.8 s + s ^4 + 7.8 s^3 + 15.7 s^2 + 8.85 s + 0 .4 Continuous-time transfer function >> [p,z]=pzmap(Gc) p= -5.0210 -1.8656 -0.8 640 -0. 049 4 z= -5.0000... 9w  (2  w ) 2 3w G ( jw)   tan  w2 1  4 G ( j1)    1.26 10  (2  1)  G ( j1)   tan  1. 24 1 4 G ( j 2)    0.63 40 36  (2  4) G ( j 2)   tan  1.89rad 2 1 G s  3s