Answer to Question #209003 in Electrical Engineering for gopal

Question #209003

 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


1
Expert's answer
2021-06-22T05:10:09-0400

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 ωup = 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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