# ch11 - ELEVEN Design Using Sinusoidal Tools SOLUTIONS TO...

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E L E V E N Design Using Sinusoidal Tools SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Gain Design a. The required phase margin for 25% overshoot   0.404 , found from Eq. (10.73), is 43.49 . From the solution to the Case Study Challenge problem of Chapter 10,      50.88 1.32 100 K G s s s s . Using the Bode plots for 1 K from the solution to the Case Study Challenge problem of Chapter 10, we find the required phase margin at 1.35rad/s , where the magnitude response is 14dB . Hence,   5.01 14dB K . b. Program: % Input system numg=50.88; deng=poly([0 –1.32 –100]); G=tf(numg,deng); %Percent Overshoot to Damping Ratio to Phase Margin Po=input('Type %OS '); z=(–log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); Pm=atan(2*z/(sqrt(–2*z^2+sqrt(1+4*z^4))))*(180/pi); fprintf('\nPercent Overshoot = %g',Po) fprintf(', Damping Ratio = %g',z) fprintf(', Phase Margin = %g',Pm) %Get Bode data bode(G)

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11-2 Chapter 11: Design Using Sinusoidal Tools pause w=0.01:0.05:1000;%Step size can be increased if memory low. [M,P]=bode(G,w); M=M(:,:); P=P(:,:); Ph=–180+Pm; for i=1:1:length(P); if P(i)–Ph<=0; M=M(i); K=1/M; fprintf(', Frequency = %g',w(i)) fprintf(', Phase = %g',P(i)) fprintf(', Magnitude = %g',M) fprintf(', Magnitude (dB) = %g',20*log10(M)) fprintf(', K = %g',K) break end end T=feedback(K*G,1); step(T) Computer response: Type %OS 25 Percent Overshoot = 25, Damping Ratio = 0.403713, Phase Margin = 43.463, Frequency = 1.36, Phase = –136.634, Magnitude = 0.197379, Magnitude (dB) = –14.094, K = 5.06641
Solutions to Problems 11-3 Antenna Control: Cascade Compensation Design a. From the solution to the previous Case Study Challenge in this chapter,      50.88 1.32 100 K G s s s s . For 20, 51.89 v K K . Hence, the gain compensated system is      2640.16 1.32 100 G s s s s Using Eq. (10.73), 15% overshoot (i.e. 0.517 ) requires a phase margin of 53.18 . Using the Bode plots for 1 K from the solution to the Case Study Challenge problem of Chapter 10, we find the required phase margin at 0.97rad/s where the phase is 126.82 . To speed up the system, we choose the compensated phase margin frequency to be 4.6*0.97 4.46rad/s . Choose the lag compensator break a decade below this frequency, or 0.446rad/s .

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11-4 Chapter 11: Design Using Sinusoidal Tools At the phase margin frequency, the phase angle is 166.067 , or a phase margin of 13.93 . Using 5 leeway, we need to add 53.18 13.93 5 44.25      . From Figure 11.8, 0.15 , or 1 6.667 . Using Eq. (11.15), the lag portion of the compensator is     Lag 0.446 0.446 0.446 0.0669 6.667 s s G s s s . Using Eqs. (11.9) and (11.15), 2 max 1 0.579 T w . From Eq. (11.15), the lead portion of the compensator is   Lead 1.727 11.51 s G s s The final forward path transfer function is               Lag Lead 2640.16 0.446 1.727 1.32 100 0.0669 11.51 s s G s G s G s s s s s s b.
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## This note was uploaded on 11/07/2011 for the course MEM 355 taught by Professor Ani during the Fall '08 term at UPenn.

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ch11 - ELEVEN Design Using Sinusoidal Tools SOLUTIONS TO...

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