Here's a problem I had to do for homework. Some of the equation formatting didn't copy over, sorry... Given an system G(s) = 10/(s(s+4)) and it’s open loop bode plots, design an appropriate compensator so that the closed loop phase margin is at least 45 o and the velocity constant is 50. The bode plot shows that the PM for the open loop system is 64.2 o at ω = 1.931rad/s, at the same frequency, we see that the P.M. of the closed loop system is 129 o . The velocity constant indicates the steady state error to a ramp input must be e ss =1/K v = 1/50 = 0.02. First, the velocity constant of the uncompensated system is: Kv=limit as s->0 of sG(s) =limit as s->0 of s10/s(s+4)=10/4=2.5 since we want K v 50, the compensator must have a gain of at least 50/2.5 = 20. If our compensator G c (s) = K c = 20, we get the following new responses by simply shifting the open loop bode up by about 26dB. Thus the new open loop 0dB crossing is at ω = 13.83 rad/s

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