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I am studying control systems, and I am studying the lead and lag compensator. I have seen than if I use a lead compensator for the closed loop transfer function,

$T(s)=\frac{s+2}{(s+1)(s+8)},$

with the designed lead compensator,

$lead(s) = \frac{3s+1}{0.1s+1}$

and I use it as a pre-compensator, so that the lead compensator is outside the loop,

$lead(s)\cdot T(s),$

and plot the frequency response, I see that I have a frequency response in which there is a resonance peak, which increases as I decrease the frequency at which the zero is present.

enter image description here

In this case, the red line is a lead compensator defined as,

$\frac{3s+1}{0.1s+1},$

and the green line is the system in which I have used the following compensator,

$\frac{6s+1}{0.1s+1}.$

Can somebody explain to me why?

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  • $\begingroup$ Hint : Plot the frequency of both compensators. Which one has the highest gain ? $\endgroup$
    – Ben
    Feb 24 '20 at 20:24
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The first compensator $\frac{3s+1}{0.1s + 1}$ has a high-frequency gain of 30.

The second compensator $\frac{6s+1}{0.1s + 1}$ has a high-frequency gain of 60.

The difference is 6.02 dB, it looks pretty close to the difference between the green and red line.

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  • $\begingroup$ Thanks for answering. I would like to ask if there is a reason of why this happen, or it is just a computational issue? $\endgroup$
    – J.D.
    Feb 24 '20 at 22:06
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    $\begingroup$ Do you know how to compute a transfer function gain at infinite frequency? It's basically L'Hospital rule $\endgroup$
    – Ben
    Feb 24 '20 at 22:26

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