Question

The transfer function of a phase lead compensator is given by
$D\left(s\right)=\frac{3(s+\frac{1}{3T})}{(s+\frac{1}{T})}$
The frequency (in rad/sec), at which $\angle D\left(j\omega \right)$ is maximum is

• A. $\sqrt{\frac{3}{T^2}}$
• B. $\sqrt{3T}$
• C. $\sqrt{\frac{1}{3T^2}}$
• D. $\sqrt{3T^2}$

Answer 1

The correct option is C $\sqrt{\frac{1}{3T^2}}$

$T\left(s\right)=\left(\frac{1+3Ts}{1+Ts}\right)$

Frequency at which $\angle T\left(j\omega \right)$ is maximum
$\left({\omega }_m\right)=\frac{1}{T\sqrt{\alpha }}$

$\alpha =\frac{1}{1/3}=3$

${\omega }_m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}$