It can be shown that the frequency where the phase is maximum is given by

$\displaystyle \omega_{\max } = \frac{1}{{\tau \sqrt \alpha }}$

The maximum phase corresponds to

$\begin{displaymath} \begin{array}{l} \sin \phi _{\max } = {\dfrac{{1 - \alpha... ...} )}}{{1 + \sin (\phi _{\max } )}}} \right) \\ \end{array} \end{displaymath}$

The magnitude of

Example 1: Consider the following system

$\displaystyle G(s) = \frac{1}{{s(s + 1)}},\,\,\,\,\,\,H(s) = 1$

Design a cascade lead compensator so that the phase margin (PM) is at least 45° and steady state error for a unit ramp input is ≤ 0.1 .

The lead compensator is

$\displaystyle C(s) = K \frac{\tau s+1}{\alpha \tau s+1},$ where, $\displaystyle \alpha < 1$

Steady state error for unit ramp input is

$\displaystyle \frac{1}{\lim_{s\to 0} s C(s) G(s)} = \frac{1}{C(0)} = \frac{1}{K}$

PM of the closed loop system should be 45°. Let the gain crossover frequency of the uncompensated system with K be ω_g.