Question

The transfer function of a phase-lead compensator is given by
$G_e\left(s\right)=\frac{1+3Ts}{1+Ts}whereT>0.$
The maximum phase-shift provided by such a compensator is

• A. $\pi /2$
• B. $\pi /3$
• C. $\pi /4$
• D. $\pi /6$

Answer 1

The correct option is D $\pi /6$

Maximum phase shift
${\varphi }_m=\angle G_e\left(s\right)$
$\varphi =ta{n}^{-1}3\omega T-ta{n}^{-1}\omega T$
For maximum phase shift
$\frac{d\varphi }{d\omega }=0$
$\Rightarrow \frac{3T}{1+(3T\omega {)}^2}=\frac{T}{1+(T\omega {)}^2}$
$3[1+(T\omega {)}^2]=1+(3T\omega {)}^2$
$3+3T^2{\omega }^2=1+9T^2{\omega }^2$
$2=6(\omega T{)}^2$
$(\omega T{)}^2=\frac{1}{3}$
$\omega T=\frac{1}{\sqrt{3}}$
${\varphi }_{max}=ta{n}^{-1}3\times \frac{1}{\sqrt{3}}-ta{n}^{-1}\frac{1}{\sqrt{3}}$
$=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$