Lead condition and maximum phase lead frequency

For $G_c(s) = \dfrac{s+\tau}{s+a\tau}$ the zero is at $s = -\tau$ and the pole is at $s = -a\tau$.

Condition for a lead compensator

The network gives phase lead when the zero lies closer to the origin than the pole:

$$a\tau > \tau \quad \Rightarrow \quad a > 1$$

Then the phase angle is

$$\phi(\omega) = \tan^{-1}\frac{\omega}{\tau} - \tan^{-1}\frac{\omega}{a\tau} > 0 \quad \text{for all } \omega > 0$$

(If $a < 1$ the pole is nearer the origin and the network becomes a lag compensator.)

Frequency of maximum phase lead

Differentiate $\phi$ with respect to $\omega$ and set it to zero:

$$\begin{aligned} \frac{d\phi}{d\omega} &= \frac{1/\tau}{1+\omega^2/\tau^2} - \frac{1/(a\tau)}{1+\omega^2/(a\tau)^2} = 0 \\ \frac{\tau}{\tau^2+\omega^2} &= \frac{a\tau}{a^2\tau^2+\omega^2} \\ a^2\tau^2 + \omega^2 &= a\tau^2 + a\omega^2 \\ \omega^2 (a-1) &= a\tau^2 (a-1) \\ \omega_m &= \tau\sqrt{a} \end{aligned}$$

So the maximum phase lead occurs at $\omega_m = \sqrt{\tau \times a\tau} = \tau\sqrt{a}$, the geometric mean of the zero and pole frequencies (the midpoint on the log-frequency scale of the Bode plot).

The maximum lead is

$$\phi_m = \sin^{-1}\frac{a-1}{a+1}$$

Example: $\tau = 2$, $a = 4$ gives $\omega_m = 2\sqrt{4} = 4$ rad/s and $\phi_m = \sin^{-1}(3/5) = 36.9^\circ$.