a unity feedback system g(s)=k/s(s+1)(0.2s+1).design a suitable phase lag compensator to give velocity error constantof 8sec'and phase margin 40 degrees
e^{ramp}_{ ss} =\frac{ 1}{ lim_{s \to 0}0 sG(s)}= \frac{ (20)(85)}{ 22010} = 0.0772
We \space want \space e^{ramp} _{ss} = 0.00772. But \space e^{ramp} _{ss} = \frac{(20)(85)} K , so \space K = 220100.
We increase the phase margin to
account for the phase of the lag compensator.
The desired phase margin depends only on the desired overshoot, so the answer as the same as in. From the Bode plot, we find that the frequency at phase φ = -180°+53°+10° = 117° is ω ≈ 7:85 rad/s.
The magnitude at ! ≈ 7:85 rad/s is 20 log| jG(j7.85)|= 23.69 dB . Then to shift the magnitude
to 0 dB at this frequency we design the lag compensator to have high-frequency asymptote at -23.69 dB, and low-frequency asymptote at 0 dB. We set the upper break frequency to 1 decade below ω = 7:85, that is at ω_{up} = 0:785. The difference between the upper and lower break points is -23.69 dB × \frac{1 dec}{ -20 dB} = 1.18 dec , so the lower break frequency is at ω = \frac{0.785}{ 10^{1.18}} = 0.0517 . So the compensator has the form Gc(s)= \frac{K_c( s + 0.785)}{s+0.0517} . Since we want this to be 0 dB for low frequency, |Gc(s = 0)| =|\frac{K_c(0.785)}{0.0517} | = 1 . So the lag compensator is Gc(s) = 0.0066* \frac{(s + 0.785)}{ s + 0.0517}
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