Solution homework7 - Indiana...
Transcript of Solution homework7 - Indiana...
P 8.27 %P8.27 %Bode plots K=0.75 p=K*[1 50]; q=[1 10 25]; sys=tf(p,q); bode(sys) grid on %Bandwith for K wn=sqrt(50*K+25); damping=(10+K)/(2*wn); wb=wn*(-1.1961*damping+1.8508) Script run K = 0.7500 wb = 8.2028
Figure P8.27 K=0.75
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Figure P8.27 K=1
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Figure P8.27 K=10
K 0.75 1 10 L Gain at w=0 db 3.52 6.02 26 wb rad/s 8.20 9.4 30.44 wc rad/s 3.42 5.48 22.5
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AP8.5 %AP8.5 %Log-magnitude-phase curve phase=[101.42 250.17 267.53 268.93 269.77]; magnitude=[40 4.85 -13.33 -20.61 -33.94]; plot(phase,magnitude) xlabel('phase, degree') ylabel('Magnitude GcG, db') grid on
Figure AP8.5
By checking the poles, we can conclude that the open-loop system is unstable and that the closed-loop system is stable.
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DP8.6 %DP8.6 syms s K p A=[0 1; -1 -p] B=[K ;0] C=[0 1] % closed-loop transfer function C*inv(s*eye(2)-A)*B %Bode plot p=-1; q=[1 1.38 1]; sys=tf(p,q); bode(sys) grid on Script run A = [ 0, 1] [ -1, -p] B = K 0 C = 0 1 ans = -K/(s^2 + p*s + 1) Using the transfer function from the Matlab script, we determined that the steady error is zero for K=-1. For a percent overshoot of 5%, the damping ratio is 0.69 and so p is determined to be 1.38. The natural frequency is wn=1rad/s. Using the approximation wb=(-1.19*ζ+1.85)*wn=1.028rad/s. The Bode plot is shown on Figure DP86. The bandwidth is wb=1.02rad/s.
Figure DP86
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System: sysFrequency (rad/sec): 1.02Magnitude (dB): 3.01
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CP8.3 %CP8.3 %Bode plot a) p=2000; q=[1 110 1000]; sys=tf(p,q); figure (1) clf; bode(sys) grid on %Bode plot b) p=100; q=[1 11 12 2]; sys=tf(p,q); figure (2) clf; bode(sys) grid on %Bode plot c) p=[50 5000]; q=[1 51 50]; sys=tf(p,q); figure (3) clf; bode(sys) grid on %Bode plot b) p=100*[1 14 50]; q=[1 503 1502 1000]; sys=tf(p,q); figure (4) clf; bode(sys) grid on
Figure CP83 a) the crossover frequency is 17.1rad/sec
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Figure CP83 b) the crossover frequency is 3rad/sec
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Figure CP83 c) the crossover frequency is 70.7rad/sec
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Figure CP83 d) the crossover frequency is 3.2rad/sec
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