### Control Systems in Python - Phase Lag Compensator Design

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 Ï‰ = 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  Ï‰ = 13.83 rad/s and the open and closed loop phase margins are 16.12o and  23.2o respectively.

However since the specification states that the closed loop phase margin be at least 45o, this simple controller will not work.

Since we are trying to adjust the steady state error of the system instead of transient response, a phase-lag controller of the form Gc(s) = K(s+z)/(s+p) will be used. Following the phase lag design procedure for bode plots, the controller gain must be 20 or more, thus we have the condition that:
(1) Kz/p = 20

The desired crossover point will give a 45o phase margin, and the controller will add about  5o and decrease the gain by 10log(20)   13dB. So i’ll pick the frequency where the open loop phase is about 130o, at around Ï‰c = 3 rad/s.

Then z =  Ï‰c/10 = 0.3 and p becomes

p=0.3*20/|G(3j)|= 0.0225.
then using (1) K becomes,  20*0.0225/0.3=1.5.  The controller is then:
Gc(s) =

the response then becomes:
which shows that the P.M. of the open loop system is 47.8o and the P.M. of the closed loop system is 82.6o which is well above the desired amount of 45o

Recalculating the velocity constant with the compensator we get:

Kv=limit as s->0 of sGc(s)G(s)=limit as s->0 of s*1.5(s+0.3)/(s+0.0225) 10/s(s+4) =0.45*10/(0.0225*0.4)=50

which is the desired value.

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