# why if I place the zero of the lead compensator at lower frequencies I have 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 compensatpr in the following way:

consider a closed loop transfer function:

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

and I design a lead compensator as:

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

and I use it as a precompensator, so the lead compensator is outside the loop:

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

and plot the frequency response, I see that I have a frequency respose I have that there is a reasonance 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 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.