# Why if I place the zero of the lead compensator at lower frequencies, do I obtain a resonance peak?

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,

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

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

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.

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?

• Hint : Plot the frequency of both compensators. Which one has the highest gain ? – Ben Feb 24 '20 at 20:24

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.