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note4 - x6. Frequency-Response Design Methods 1. Frequency...

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Unformatted text preview: x6. Frequency-Response Design Methods 1. Frequency Response: We recall inverse Laplace transform: 2 3 K5 K 4 L,1 6 s , j! + s + j! 7 = 2jK j cos!ot + 6 K : o o Consider sinusoidal input signal ut = Au cos!ot applied to a plant Gs. Then the output in s-domain is s Y s = GsU s = Gs s2Aus 2 = Gs s + j!Aus , j!  + !o o  o  K + K + terms generated by Gs: = s , j!o s + j!o The K -value can be computed from sG 1 K = s!j!os , j!oY s = s!j!o Au+ j!s = 2 AuGj!o lim lim s o = jK j = AujGj!oj=2; 6 K = 6 Gj!o: 1 By inverse Laplace transform, yt = 2jK j cos!ot + 6 K  + terms generated by Gs = AujGj!oj cos!ot + 6 Gj!o + terms of Gs: If Gs is stable, then the steady-state response is ysst = AujGj!oj cos!ot + 6 Gj!o: Observation: The steady-state response is also sinusoidal. But its amplitude is ampli ed by a factor of jGj!oj, and its phase is shifted by 6 Gj!o. Example: The gure below shows ut = cos!ot dashed line, and ysss = 2 cos!ot + =4 solid line with !o = 1. Thus Gj!o = 26 45o. 2 1.5 1 0.5 0 −0.5 −1 −1.5 −2 0 2 4 6 8 2 10 12 14 16 De nition: Gj! is termed frequency response with jGj!j magnitude response, and 6 Gj!o phase response. Because every signal is a composition of sinusoidal functions, jGj!j and 6 Gj!o characterizes spectral information of the system, and they uniquely determine the behavior of the system. Consider feedback system: r(t) e(t) D(s) G(s) y(t) η(t) Then there hold DGs 1 E s = 1 + DGs Rs; Y s = , 1 + DGs N s: Some primitive goals are: tracking reference signal in operating frequency range: 0; !r , and noise rejection in frequency range !n; 1. It thus requires ideally 1 j1 + DGj!j = 0; if 0  !  !r ; jDGj!j = 0; if !  ! : n j1 + DGj!j 3 Performance in Frequency Domain Since perfect tracking and noise rejection is not possible, and 1 DGs = 1; 8! 2 R; + 1 + DGs 1 + DGs performance requirement are often given by 8 1 jS j!j = j1 + DGj!j :  ;  1; 8 DGj!j jT j!j = j1 j+ DGj!j :  1; ; for ! 2 0; !r for ! 2 !n 1; for ! 2 0; !r for ! 2 !n 1: The performance requirement on S s is shown in the following gure by assunming !n !r : | S(j ω)| 1+ δ 1- δ δ ε ωr ωn 4 Sensitivity Consideration: Recall sensitivity:  change in T 1 S s =  change in G = 1 + DGs ; that is the same as transfer function from reference input to tracking error output. Hence the tracking error performance is consistent with sensitivity requirement. Translation to Loopshaping: We have that jS j!j  is guaranteed by jDGj!j  1 + 1 ; and jT j!j  is guaranteed by jDGj!j  1 + : We thus translated performance requirement on sensitivity to loop transfer function: 8  1 + ,1; ! 2 0; !r ; jLj!j = jDGj!j :  1 + ,1; ! 2 !n; 1: 5 Loopshaping Design A typical loopshape is shown next: |DG(jw)| The lower-left shaded area represents the performance requirement on sensitivity, and tracking, and the upper-right shaded area represents the performance requirement on noise rejection. The loop transfer function jLj!j = jDGj!j need avoid the forbidden region shaded areas in order to satisfy the performance requirement. This is called loopshaping design. Compensator Ds will be designed to satisfy the design speci cations. 6 Logrithm plots: In engineering pratice, linear scale plots are not used. Bode invented logrithm plots where both frequency and magnitude jGj!j in logrithm scale. In horizontal frequency axis, the length is measured with ratio: log!2 , log!1 = log!2=!1: In vertical axis, the length is also measured with ratio that has unit dB: 10 dB of jGj!j = 10  20 log10jGj!j: See the picture below: dB 80 (10 4 ) 60 (10 3 ) 40 (10 2 ) 20 (10) ω 2ω 4ω 7 8ω 16 ω Real factor s + z , z 0: 8 20 log10z ; 20 log10jj! + z j = : 20 log10!; ! ;! z; z: Straight line approximation gives with GL = 20 log10z : 20 dB GL z Real factor s + p,1, p 10z 0: 8 20 log10jj! + pj,1 = : ,20 log10p; ! ,20 log10!; ; ! GL 20 dB p 10p 8 p; p: largest approximation error takes place at corner frequency !o = z or !o = p that has gain p 20 log10j!o + j!oj = 20 log10 2!o = 3 + 20 log10!o = 3 + GL: Thus the approximation error is at most 3dB. The recti ed curve is depicted with dashed line as in previous page. Summary: the magnitude bode plot has a gain GL for small frequency with zero slope, and it increases slope by 20dB dec after corner frequency if corresponding zero; and it decreases slope by 20dB dec after corner frequency if corresponding pole. Example: Find magnitude Bode plot for 10 + 5 + 100 Gs = s + 0s1s +s1s + 50 : Solution: We sketch magnitude Bode plot as follows: Step 1: We mark corner frequency on horizontal axis with  corresponding to pole, and to zero. Step 2: We compute low frequency gain: GL = 20 log10 jG0j = 20 log105000=5 = 60dB: 9 Step 3: We begin at frequency ! = 0:01 with zero slope, and decrease its slope by ,20dB dec after it comes across 0.1 corresponding to pole, and decrease its slope by ,20dB dec again after it comes across 1 corresponding to pole. Increase its slope by 20dB dec after it comes across 5 corresponding to zero    until high frequency ! = 1000. Rectify the plot with 3dB correction. dB 60 40 20 log( ω) -2 -1 0 1 -20 -40 10 2 3 Magnitude Bode Plots for Type ` Systems Type ` system has a transfer function: 1s Gs = s` K + + z1  s + zm ; s p1 s + pn where zi 6= 0 and pk 6= 0 for 1  i  m and 1  k  n. We have the following modi cations: Low frequency gain is replaced by 0 jKz1z2    zmj 1 : @ A GL = 20 log10 B !` jp p    p j C n L 12 Initial slope is now ,`  20dB=dec. Example: Find magnitude Bode plot for s Gs = 10s + 5s + 100 s + 1s + 50 Solution: We sketch magnitude Bode plot as follows: Step 1: We mark corner frequency on horizontal axis with  corresponding to pole, and to zero. Step 2: We compute low frequency and its gain: !L = 0:1 minfjz1j;    jzmj; jp1j;    ; jpnjg = 0:1; 0 10  5  100 1 GL = 20 log10 @ 0:1  1  50 A = 60dB: 11 Step 3: We compute its initial slope: ,20dB=dec. Step 4: We begin at frequency !L; GL = 0:1; 60 with slope of ,20dB=dec, and draw straight line by increasing !. Step 5: We decrease its slope by ,20dB dec after it comes across 1 corresponding to pole, and increase its slope by 20dB dec after it comes across 5 corresponding to zero    until high frequency ! = 1000. Rectify the plot with 3dB correction. dB 60 40 20 log( ω) -1 0 1 -20 -40 12 2 3 Magnitude Bode Plots for Complex Poles Consider proto-type system: 2 !n Gs = s2 + 2 ! s + !2 : n n Straight line approximation: 2 !n ; Gs  s + ! 2 n if  1. Thus we have zero slope straight line for ! and has a ,40dB=dec slope after ! crossing !n. dB 0 -20 ωn /10 ωn -40 13 10ωn !n, Correction at !n: To rectify the plot, 6= 1 needs take into consideration. Since Gs =  s 2 + 1 s + 1 : 2 !n !n The maximum error at !n is: 1 20 log10 jGj!nj = 20 log10 : 2 dB 0 -20 ωn /10 ωn -40 14 10ωn Example: Sketch maginitude Bode plot for s Gs = ss2002s+ 1 : 2+ + 100 Corner frequencies are 1; 10; 10, and 2 s2 + 2s + 100 = s2 + 2 !ns + !n with !n = 10, and = 2=2!n = 2=20 = 0:1: The low and high freqeuncies are thus !L = 0:1; !H = 100: Compute low frequency gain: GL = 20 log10200=10 = 26dB: Since it is type 1 system, the initial slope is ,20dB=dec. We begin at !L; GL = 0:1; 26 and draw stright line with initial slope ,20dB=dec. After crossing corner frequency 1 corresponding to zero, we increase slope by 20dB=dec, and after crossing frequency 10 corresponding double poles, we decrease slope by 40dB=dec. 15 After completion of straight line approximation, we make correction at corner frequencies. At 1, the correction is 3dB, as it corresponds to real zero. At 10, the correction is 20 log102 ,1 = 20 log105 = 14dB: The complete plot is shown next with dashed line for corrected Bode plot. 20 0 -20 0.1 1 10 -40 16 100 Phase Bode Plot: Straightline approximation for real pole: Gs = s K p ; K 0; p = !o 0: + Then its argument at s = j! is given by 8 ,5:7o; ! = 0:1!o; 6 Gj!  = , tan,1!=!o  = ,45o; ! = !o ; : ,84:3o; ! = 10!o: Since 6 Gj! = 0 for ! !o, and 6 Gj! = ,90o for ! !o, a stright line approximation is given as follows: 0.1ω o -45 -90 ωo o o 17 10ω o Correction: 8 6 ,5:7o; ! = 0:1!o; Gj! = : ,84:3o; ! = 10!o; which are at the two end points: one decade above, and one decade below ! = !o. We also note that the following three points are coincide with the staight line approximation: 8 6 ,45o; ! = !o; Gj! = ,9o; ! = 0:16!o; : ,81o; ! = 6:4!o : Thus we have the following correction, together with the straight line approximation. 0 −10 −20 −30 −40 −50 −60 −70 −80 −90 −1 10 0 10 18 1 10 Phase plot for real zero: Gs = s + z , z = !o 6 0: Gj! = tan,1!=!o that is positive. Thus we easily get a phase plot as follows with the correction: 90 80 70 60 50 40 30 20 10 0 −1 10 0 10 1 10 For zero corner frequency: Gs = K=s, or Gs = s, then 8 90o; 6 Gj!  = : ,90o; if corresponding to zero; if corresponding to pole: 19 Discussion of Phase Plot: We consider only the transfer function having all poles and zeros on negative real axis, including the origin. Each corner frequency corresponding to zero contributes a slope of 45o=dec, and the slope is e ective for only two decades centered at the corner freqeuncy. Each corner frequency corresponding to pole contributes a slope of ,45o=dec, and the slope is e ective for only two decades centered at the corner freqeuncy. Initial phase at the low freqeuncy is 8 0o ; ,180o; PhL = l  90o; : ,l  90o; if G0 0; if G0 0; if Gs has l zero at the origin; if Gs has m pole at the origin: The initial slope is zero if !L is smaller than 10 of the smallest of the corner freqeuncy. 20 Example: Sketch phase Bode plot for Gs == 10s + 5s + 100 ss + 1s + 50 Solution: We sketch phase Bode plot as follows: Step 1: We mark corner frequency on horizontal axis with  corresponding to pole, and to zero. Step 2: Since we have one pole at the origin, the low frequency phase is given by: PhL = ,90o: Step 3: We begin at frequency ! 0:1 with a phase of ,90o, and draw straight line with zero slope. If ! crosses ! = 0:1 that is within a decade of the corner frequency 1 corresponding to a pole, decrease its slope by ,45o=dec. As frequency increases to ! = 0:5, it is within a decade of corner frequency of 5 corresponding to zero, increase its slope by 45o=dec. 21 As frequency arrives at ! = 5, the corner frequency at 50 has the e ect, that corresponds to a pole. In this case, decrease the slope by ,45o=dec. If ! crosses ! = 10 that is beyond a decade of the corner frequency 1, the slope generated by this pole has no e ect that eliminates the ,45o=dec generated by the corner frequency at 1. Since ! = 10 is now within a decade of corner freqeuncy at 100 corresponding to a zero, we need increase its slope by 45o=dec. The total e ect is that we increase the slope by 90o=dec. As frequency increases to ! = 50 that is beyond a decade of the corner frequency at 5 corresponding to a zero, its slope has no e ect. Thus we decrease its slope by 45o=dec. As frequency arrives at ! = 500, the corner frequency at 50 has no e ect, that corresponds to a pole. We thus increase its slope by 45o=dec. Finally we reach ! = 1000. The corner frequency at 100 now has no e ect. Hence we increase its slope by 45o=dec. 22 Note that after ! = 1000, no corner frequency has no contribution to the slope of the straight line approximation. We should come back to zero slope. Rectify the plot with corrections. Another way to make correction is to compute actual phase at ! = 0:1; 0:5; 1; 5; 10; 50; 100; and connect them smoothly with straight line approximation as the guidance. 0.1 1 10 100 1000 -90 -180 We note the nal phase, or phase at the high frequency is: PhH = ,n , m  90o; where n is the number of zeros, and n the number of poles. For our example, n = 3 and m = 2. Thus the nal phase is ,90o. 23 Phase Bode Plot for Complex Roots: The complex roots are treated as double roots at natural frequency !n. The correction is more involved for the phase plot as the damping ratio determined the steepness at the corner frequency. An easy way to make correction is to compute the phase angles at several frequencies near the natural frequency, and connect these points with a smooth curve. Example: Sketch phase plot for s Gs = ss2002s+ 1 : 2+ + 100 The corner frequencies are: 1; 10. Thus we set !L = 0:1; !H = 100; and mark corner frequencies, as well as !L and !H . Since we have one pole at the origin, PhL = ,90o. We begin at !L; PhL. The left of this point is a straight line with zero slope. The straight line to the right has a slope of 45o=dec, generated by the zero corresponding to corner freqeuncy at 1. 24 As ! increases to ! = 1, the double corner frequency at 10 corresponding to poles generates ,90o=dec which yields ,45o=dec for the slope. As ! passess 10, the slope generated by the corner frequency at 1 is not e ective. We thus have ,90o=dec for the slope which disappears after !  100. Since n = 3, and m = 1, the nal phase is ,180o. Correction are easily made at ! = 0:1 that should be 5:7o higher than ,90o. It is also easy to make correction at 1 that should be 11:4o lower, and at ! = 100 that should be 11:4o higher due to double corner frequency. 0.1 1 10 -45 -90 -135 -180 25 100 Example: Sketch both magnitude and phase plots for 1000ss2 + s + 10 Gs = s + 12s + 10s + 100 : The numerator has a pair of complex roots with p pp !n = 10; = 1=2 10: Thus the corner frequencies are 1; 1; 10; 10; 10; 100. The low and high frequencies are chosen as !L = 0:1, and !H = 100. For magbitude plot, the low frequency gain is GL = 20 log101000  0:1  10=10  100 = 0dB: Since there one zero at the origin, we begin with slope of 20dB=dec. We change the slope at 1 by decreasing slope by 40dB=dec as it corresponds to double pole. Hence the slope is ,20dB=dec after ! = 1. p The slope comes back to 20dB=dec at ! = 10 due to double zero. After ! = 10, the slope decreases to zero due to pole. For !  100, it rolls o at ,20dB=dec due to the pole at 10. 26 For phase plot, we have that PhL = 90o; !  !L; due to one zero at the origin. p For ! 2 0:1; 0:1 10 , the slope is ,45o=dec due to double corner freqeuncy at 1. p For ! 2 0:1 10; 1 , the slope is zero due to the cancellation p e ect of the double corner frequency at 1 and 10. For ! 2 1 10 , the corner frequency at 10 is also in e ect that gives ,45o=dec slope. p For ! 2 10; 10 10 , the double corner frequency at 1 is not e ective, but all others are that give zero slope. p For ! 2 10 10; 100 , the slope is generated by only pole at 10. The nal phase is ,90o as n = 4 and m = 3. 27 The magnitude and phase plots are given as follows: dB 20 1000 0.1 1 10 100 -20 degree 90 -90 28 Read Example 6.5 of page 355. 29 ...
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This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.

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