Question

The transfer function of a compensator is given as $G_c\left(s\right)=\frac{s+1}{s+2}$
The phase of the above lead compensator is maximum at

• 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,
$\frac{dP}{d\omega }=0$
$\frac{1/1}{1+\frac{{\omega }^2}{1^2}}-\frac{1/2}{1+\frac{{\omega }^2}{2^2}}=0$
$\frac{1}{1+{\omega }^2}-\frac{1/2}{1+\frac{{\omega }^2}{4}}=0$
$\frac{1}{1+{\omega }^2}-\frac{2}{{\omega }^2+4}=0$
${\omega }^2+4-2-2{\omega }^2=0{\omega }^2=2$
$\omega =\sqrt{2}rad/sec$