ps9_soln - -50 50 100 Magnitude (dB) 10-3 10-2 10-1 10 10...

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Unformatted text preview: -50 50 100 Magnitude (dB) 10-3 10-2 10-1 10 10 1-180-135-90 Phase (deg) Bode Magnitude and Phase Plots of K o G p (s) Frequency (rad/sec) EE 428 Problem Set 9 Solutions -5- Problem 40: The Bode plot of K o G p (s) reveals: gc = 1.09 rad/sec PM = 5.23 PM = 5.23 gc = 1.09 rad/sec 10-3 10-2 10-1 10 10 1-180-160-140-120-100-80 Phase [deg] Frequency [rad/sec] 10-3 10-2 10-1 10 10 1-50 50 100 Bode Plot of K o G p (s) Magnitude [dB] EE 428 Problem Set 9 Solutions -9- Problem 40: Bode plot of K o G p (s). The horizontal line is draw at ( ) ( ) 20log 9.97 o c gc dB K G j K dB = = . The frequency gc at which the equality holds is 1.96 rad/sec, the estimated gain crossover frequency of the compensated loop transfer function. gc = 1.96 rad/sec-100-50 50 100 Magnitude (dB) 10-3 10-2 10-1 10 10 1 10 2-180-135-90 Phase (deg) Bode Magnitude and Phase Plots of G c (s)G p (s) Frequency (rad/sec) EE 428 Problem Set 9 Solutions -11- Problem 40: The Bode plot of G c (s)G p (s) reveals: gc = 1.93 rad/sec PM = 57.7 PM = 57.7 gc = 1.93 rad/sec 1...
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ps9_soln - -50 50 100 Magnitude (dB) 10-3 10-2 10-1 10 10...

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