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

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?
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.