ch11 - E L E V E N Design via Frequency Response SOLUTIONS...

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E L E V E N Design via Frequency Response SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Gain Design a. The required phase margin for 25% overshoot ( ζ = 0.404), found from Eq. (10.73), is 43.49 o . From the solution to the Case Study Challenge problem of Chapter 10, G(s) = 50.88K s(s+1.32)(s+100) . Using the Bode plots for K = 1 from the solution to the Case Study Challenge problem of Chapter 10, we find the required phase margin at ω = 1.35 rad/s, where the magnitude response is -14 dB. Hence, K = 5.01 (14 dB). 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) 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)
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Solutions to Case Studies Challenges 463 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
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464 Chapter 11: Design via Frequency Response Antenna Control: Cascade Compensation Design a. From the solution to the previous Case Study Challenge in this chapter, G(s) = 50.88K s(s+1.32)(s+100) . For K v = 20, K = 51.89. Hence, the gain compensated system is G(s) = 2640.16 s(s+1.32)(s+100) Using Eq. (10.73), 15% overshoot (i.e. ζ = 0.517) requires a phase margin of 53.18 o . Using the Bode plots for K = 1 from the solution to the Case Study Challenge problem of Chapter 10, we find the required phase margin at ω = 0.97 rad/s where the phase is -126.82 o . To speed up the system, we choose the compensated phase margin frequency to be 4.6 * 0.97 = 4.46 rad/s. Choose the lag compensator break a decade below this frequency, or ω = 0.446 rad/s. At the phase margin frequency, the phase angle is -166.067 o , or a phase margin of 13.93 o . Using 5 o leeway, we need to add 53.18 o - 13.93 o + 5 o = 44.25 o . From Figure 11.8, β = 0.15, or γ = 1 β = 6.667. Using Eq. (11.15), the lag portion of the compensator is G Lag (s) = (s+0.446) (s+ 0.446 6.667 ) = s+0.446 s+0.0669 . Using Eqs. (11.9) and (11.15), T 2 = 1 ω max β = 0.579. From Eq. (11.15), the lead portion of the compensator is G Lead (s) = s+1.727 s+11.51 The final forward path transfer function is G(s)G Lag (s)G Lead (s) = 2640.16(s+0.446)(s+1.727) s(s+1.32)(s+100)(s+0.0669)(s+11.51) b. Program: %Input system ***************************** K=51.89; numg=50.88*K; deng=poly([0 -1.32 -100]); G=tf(numg,deng); Po=15; z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); %Determine required phase margin************** Pmreq=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi) phreq=Pmreq-(180)%required phase w=0.1:0.01:10; [M,P]=bode(G,w); for i=1:1:length(P);%search for phase angle if P(i)-phreq<=0; ph=P(i) w(i) break end end wpm=4.6*w(i)
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Solutions to Case Studies Challenges 465 [M,P]=bode(G,wpm);%Find phase at wpm Pmreqc=Pmreq-(180+P)+5%Find contribution required from compensator+5 beta=(1-sin(Pmreqc*pi/180))/(1+sin(Pmreqc*pi/180)) %Design lag compensator*************** zclag=wpm/10; pclag=zclag*beta; Kclag=beta; %Design lead compensator********** zclead=wpm*sqrt(beta); pclead=zclead/beta; Kclead=1/beta; %Create compensated forward path********* numgclag=Kclag*[1 zclag]; dengclag=[1 pclag];
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ch11 - E L E V E N Design via Frequency Response SOLUTIONS...

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