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 45o and the velocity constant is 50. The bode plot shows that the PM for the open loop system is 64.2o at ω = 1.931rad/s, at the same frequency, we see that the P.M. of the closed loop system is 129o. The velocity constant indicates the steady state error to a ramp input must be ess =1/Kv = 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 Kv 50, the compensator must have a gain of at least 50/2.5 = 20.
If our compensator Gc(s) = Kc = 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 and the open and closed loop phase mar…
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 45o and the velocity constant is 50. The bode plot shows that the PM for the open loop system is 64.2o at ω = 1.931rad/s, at the same frequency, we see that the P.M. of the closed loop system is 129o. The velocity constant indicates the steady state error to a ramp input must be ess =1/Kv = 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 Kv 50, the compensator must have a gain of at least 50/2.5 = 20.
If our compensator Gc(s) = Kc = 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 and the open and closed loop phase mar…