Question

If $G\left(s\right)H\left(s\right)=\frac{10}{(s+1)(s+20)}$ then the gain margin and phase margin of the system are respectively.

• A. $\frac{1}{2},\infty$
• B. 0, 180
• C. $\infty ,\infty$
• D. $\infty ,180{}^{\circ }$

Answer 1

The correct option is C $\infty ,\infty$

$\angle G\left(j\omega \right)H\left(j\omega \right)=\angle \frac{10}{(j\omega +1)(j\omega +20)}=-180{}^{\circ }$

$=-{\tan }^{-1}\left(\omega \right)-{\tan }^{-1}\left(\frac{\omega }{20}\right)=-180{}^{\circ }$

$-{\tan }^{-1}\left(\frac{\omega +\frac{\omega }{20}}{1-\frac{{\omega }^2}{20}}\right)=-180{}^{\circ }$

$\frac{\omega +\frac{\omega }{20}}{1-\frac{{\omega }^2}{20}}=0$

The above expression to become zero $1-\frac{{\omega }^2}{20}=\infty$

$\therefore {\omega }_{pc}=\infty$

$|G\left(j\omega \right)H\left(j\omega \right)|=\frac{10}{\sqrt{1+{\omega }^2}\sqrt{{\left(20\right)}^2+{\omega }^2}}$

$|G\left(j\omega \right)H\left(j\omega \right){|}_{{\omega }_{pc=\infty }}=0$

Gain margin $=\frac{1}{|G\left(j\omega \right)H\left(j\omega \right)|{\omega }_{pc}}=\infty$

$\therefore$ Gain crossover frequency does not exist.

$\therefore$ Phase margin is = $\infty$