Question

A unity feedback control loop with an open loop transfer funciton of the from $\frac{K}{s(s+\sigma )}$ has a gain cross-over frequency of 1 rad/sec and a phase margin is ${60}^o$. If an element having a transfer function $\frac{s-\sqrt{3}}{s+\sqrt{3}}$ is inserted into the loop, the phase margin will become

• A. $0^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

The correct option is A $0^o$

$T\left(s\right)=\frac{k}{s(s+\sigma )}$
$P_m={60}^o={180}^o+\angle T\left(j\omega \right){|}_{\omega =1}$
${60}^o={180}^o(-{90}^o-ta{n}^{-1}\frac{\omega }{\sigma })$
$\Rightarrow ta{n}^{-1}\frac{\omega }{\sigma }={30}^o$
$\Rightarrow \frac{\omega }{\sigma }=tan{30}^o=\frac{1}{\sqrt{3}}$
$\Rightarrow \sigma =\sqrt{3}\omega$

$\therefore T^{\prime }\left(s\right)=\frac{s-\sqrt{3}}{s+\sqrt{3}}.\frac{k}{s(s+\sigma )}$

$\angle T^{\prime }\left(j\omega \right)=-2ta{n}^{-1}\frac{\omega }{\sqrt{3}}-{90}^o-ta{n}^{-1}\left(\frac{\omega }{\sigma }\right)$

$\therefore P_m={180}^o+\angle T^{\prime }\left(j\omega \right){|}_{\omega =1}$

$={180}^o-2ta{n}^{-1}\frac{1}{\sqrt{3}}-{90}^o-ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)=0^o$