# ECE382/ME482 Spring 2005 Homework 8 Solution December 11,

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2 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 1: Open Loop for P1.21 Gm = 7.96 db (at 3.46 rad/sec), Pm = 2.7 deg (at 5.29 rad/sec) alfa = (1+sin(33*pi/18))/(1-sin(33*pi/18)) to obtain α = , so we ll let α = 3.4. Note that we must not call our variable alpha because Matlab has a function called alpha. Next, we find the frequency at which the magnitude for the uncompensated network is 1log 1 α or 5.3 db. This gives us ω m = 6.8 rad/s, as shown in Figure 2. We must also multiply by α to compensate for the attenuation, therefore each of our twin compensators is α(s + ωm / α) 3.4(s + 6.8/ 3.4) G c (s) = = s + ω m α s (5) 3.4 Our combined open loop transfer function is then G 2 c(s)g(s) = 2(3.4)(s + 6.8) 2 s(s ) 2 (s/2 + 1)(s/6 + 1) for which the phase margin is degrees, as shown in Figure 3. (6)

3 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 2: Finding ω m in the Open Loop for P1.21 Gm = 7.96 db (at 3.46 rad/sec), Pm = 2.7 deg (at 5.29 rad/sec) System: untitled1 : 6.81 : Now we have to make a decision based on our application. Is degrees close enough to 45 degrees? If not, we would adjust the estimated phase margin needed, recompute α, and repeat the procedure above. Not having a specific application in mind, we ll assume that degrees will do. Finally, we must check to see that we have achieved the required closed loop bandwidth. As shown in Figure 4, we have achieved a bandwidth of ω BW = 7.75 rad/s which exceeds 4 rad/s so we have met our design criteria. Here s the Matlab script that I used, so that you can see how I automated the steps of checking the Bode plot to obtain the needed phase margin and the new crossover frequency. P1_21.m solves part of problem P1_21 from Dorf and Bishop 1th ed. 3 April 5 --sk clear all

4 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 4 Figure 3: Open Loop for Compensated System of P1.21 Gm = 14.5 db (at 12.9 rad/sec), Pm = deg (at 4.26 rad/sec) G = tf([2],[1/12 1/2+1/6 1 ]); figure(1) margin(tf([2],[1/12 1/2+1/6 1 ])) print -deps p1_21a having determined that the phase margin of the open loop system is -2.7 degrees, we solve for alpha to obtain half the needed additional phase margin of 66 degrees. (To automate things, I ve reiterated the margin command, this time with arguments to obtain the phase margin in the variable Pm.) [Gm,Pm,Wcg,Wcp] = margin(g); ph_needed = (45-Pm)/2; alfa = (1+sin(ph_needed*pi/18))/(1-sin(ph_needed*pi/18)) mmag = -1*log1(alfa)

5 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 5 Figure 4: Closed Loop for Compensated System of P1.21 Gm = 12.7 db (at 12.9 rad/sec), Pm = 76.3 deg (at 6.9 rad/sec) System: untitled1 : 7.75 : having obtained alpha I automate the process of finding the frequency omega_m at which the magnitude is mmag and then determine Gc wrange = logspace(-2,3,1); [mag,ph] = bode(g,wrange); wm = interp1(2*log1(squeeze(mag)),wrange,mmag) Gc = tf(sqrt(alfa)*[1 wm/sqrt(alfa)],[1 wm*sqrt(alfa)]) GcGcG = series(series(gc,gc),g) figure(2) margin(gcgcg) print -deps p1_21ol figure(3) margin(feedback(gcgcg,1)) print -deps p1_21cl

6 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 6 P1.22 We are asked to design a phase-lag network for the same system, having open loop transfer function K G(s) = (7) s(s/2 + 1)(s/6 + 1) and unity negative feedback, such that the closed-loop system has velocity error constant K v = 2, phase margin near 45 degrees, and closed loop bandwidth at least ω = 2 rad/s. What s changed from P1.21 is that the bandwidth requirement has been relaxed. Solution: As in the previous problem, we first note that the definition of the velocity error constant is K v = lim sg(s), (8) s (see Section 5.7, page 262, of the textbook). Thus, to meet the design specification K v = 2 we must choose K = 2. As determined in the previous problem from the Bode plot shown in Figure 1, the phase margin of the uncompensated system is 2.7 degrees at 5.29 rad/s. From the Bode plot in Figure 1, we find the frequency ω c at which the phase margin (plus 5 degrees for the phase lag network) would be satisfied if the magnitude curve crossed db there. As shown in Figure 2, we choose ω c to be that frequency at which the phase is = 13 degrees, namely ω c = 1.12 rad/s. We see that we will need to adjust the attenuation by a gain of 23.7 db in order to have the magnitude curve cross db at ω c. We place the zero of the compensator one decade below ω c, i.e. at ω z =.11. We determine the value of α from 2log 1 α = 24 (9) to be α = The pole of the lag network is then placed at ω z /α.74, so lag compensator has transfer function G c = (s +.11) 15.22(s +.74) (1) and the closed-loop transfer function will be T(s) = G c(s)g(s) 1 + G c (s)g(s) = 2(s +.11) 15.21(s +.74)(s/2 + 1)(s/6 + 1) + 2(s +.11). (11) As shown in Figure 5, the phase margin requirement is met. Figure 6 shows that the bandwidth requirement is met to within one percent, which we will declare to be adequate. Otherwise, we d have to fiddle with gains and pole locations to improve it. The Matlab script used to generate these results is shown below.

7 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 5: Open Loop for Compensated System of P1.22 Gm = 15.1 db (at 3.34 rad/sec), Pm = 44.6 deg (at 1.13 rad/sec) P1_22.m solves part of problem P1_22 from Dorf and Bishop 1th ed. 3 April 5 --sk clear all G = tf([2],[1/12 1/2+1/6 1 ]) figure(1) margin(tf([2],[1/12 1/2+1/6 1 ])) print -deps p1_22a wrange = logspace(-2,3,1); [mag,ph] = bode(g,wrange); wc = interp1(squeeze(ph),wrange,-13) mmag = 2*log1(interp1(wrange,squeeze(mag),wc)) alfa = 1^(-mmag/-2)

8 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 8 Figure 6: Closed Loop for Compensated System of P1.22 Gm = 13.4 db (at 3.34 rad/sec), Pm = 62 deg (at 1.67 rad/sec) System: untitled1 : 1.99 : Gc = tf([1 wc/1],alfa*[1 (wc/1)/alfa]) figure(2) margin(series(gc,g)) print -deps p1_22ol figure(3) margin(feedback(series(gc,g),1)) print -deps p1_22cl P1.23 We are again given the system with unity negative feedback and forward path transfer function open loop transfer function G(s) = K s(s/2 + 1)(s/6 + 1). (12) We are asked to design a lead-lag compensator such that the closed-loop system has

10 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 7: Open Loop for System of P1.23 Gm = 7.96 db (at 3.46 rad/sec), Pm = 2.7 deg (at 5.29 rad/sec) System: G : 1.54 : z = 2.35/ 1 and obtain phase margin of 47 degrees, as shown in Figure 9, and closed-loop bandwidth of around 6.35 rad/s as show in Figure 1. The lead-lag compensator obtained has transfer function G c = The Matlab script used is given below. (s +.154)(s +.154). (17) s +.743)(s ) P1_23.m solves part of problem P1_23 from Dorf and Bishop 1th ed. 3 April 5 --sk clear all G = tf([2],[1/12 1/2+1/6 1 ]) figure(1)

11 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 8: Open Loop for Lag Compensated System of P1.23 Gm = 11.2 db (at 3.3 rad/sec), Pm = 32.8 deg (at 1.54 rad/sec) System: untitled1 : 2.35 : margin(g) print -deps p1_23a alfa=1; mmag = -2*log1(alfa) wrange = logspace(-2,3,1); [mag,ph] = bode(g,wrange); wcp = interp1(2*log1(squeeze(mag)),wrange,-mmag) Gc_lag = tf(1/1*[1 wcp/1],[1 (wcp/1)/alfa]) figure(2) margin(series(gc_lag,g)) print -deps p1_23ol1 wm=2.35 where phase is =-165 Gc_lead = tf(alfa*[1 wm/sqrt(alfa)],[1 wm*sqrt(alfa)])

12 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 9: Open Loop for Compensated System of P1.23 Gm = 1.7 db (at 7.69 rad/sec), Pm = 47 deg (at 3.69 rad/sec) figure(3) margin(series(gc_lead,series(gc_lag,g))) print -deps p1_23ol2 figure(4) margin(feedback(series(gc_lead,series(gc_lag,g)),1)) print -deps p1_23cl

13 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 1: Closed Loop for Compensated System of P1.23 Gm = 7.74 db (at 7.69 rad/sec), Pm = 41.5 deg (at 5.59 rad/sec) System: untitled1 : 6.37 :

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