Question

The phase margin of a system with the open loop transfer function$G\left(s\right)H\left(s\right)=\frac{(1-s)}{(1+s)(3+s)}$, is

• A. $\infty$
• B. ${90}^o$
• C. ${68.3}^o$
• D. 0

Answer 1

The correct option is A $\infty$

$G\left(s\right)H\left(s\right)=\frac{(1-s)}{(1+s)(3+s)}$
So, $|G\left(j\omega \right)H\left(j\omega \right)|=\frac{\sqrt{1+{\omega }^2}}{\sqrt{1+{\omega }^2}\sqrt{9+{\omega }^2}}=\frac{1}{\sqrt{9+{\omega }^2}}$

Phase angle is,

$\varphi =-2ta{n}^{-1}\omega -ta{n}^{-1}\frac{\omega }{3}$

At $\omega =0;$ $|G\left(j\omega \right)H\left(j\omega \right)|=\frac{1}{3}$ $\varphi =0^o$

At $\omega =4;$ $|G\left(j\omega \right)H\left(j\omega \right)|=\frac{1}{5}$ $\varphi =-{205}^o$

At $\omega =\infty ;$ $|G\left(j\omega \right)H\left(j\omega \right)|=0$ $\varphi =-{270}^o$


Polar plot does not cross unit circle, for all values of $\omega \left(0to\infty \right)$, magnitude will be less than unity. In this case, gain crossover frequency does not exist and phase margin becomes $\infty$.