Question

The transfer function of a compensator is given as
$G_c\left(s\right)=\frac{s+a}{s+b}$

The phase of the above lead compensator is maximum at ____ when a = 1 and b = 2

• A. $\sqrt{2}rad/s$
• B. $\sqrt{3}rad/s$
• C. $\sqrt{6}rad/s$
• D. $1/\sqrt{3}rad/s$

Answer 1

The correct option is A $\sqrt{2}rad/s$

For phase to be maximum
$\frac{d\varphi }{dt}=0$
$\Rightarrow (1+\frac{{\omega }^2}{2})(\frac{1}{a}-\frac{1}{b})-(\frac{\omega }{a}-\frac{\omega }{b})\left(\frac{2\omega }{ab}\right)=2$
$(1+\frac{{\omega }^2}{2})(\frac{1}{1}-\frac{1}{2})=(\frac{\omega }{1}-\frac{\omega }{2})\left(\frac{2\omega }{2}\right)$

$\frac{1}{2}+\frac{{\omega }^2}{4}=\frac{{\omega }^2}{2}\Rightarrow \frac{{\omega }^2}{4}=\frac{1}{2}$
$\Rightarrow \omega =\sqrt{2}$ rad/sec.