Question

A PD controller is tuned to get desired phase margin for second order control system. Phase crossover frequency for compensated system is

• A. 2.52 rad/sec
• B. 3.94 rad/sec
• C. 1.45 rad/sec
• D. does not exist

Answer 1

The correct option is D does not exist


At phase crossover frequency,

$\angle G\left(j\omega \right)H\left(j\omega \right)=-180{}^{\circ }$

Given, $G\left(s\right)H\left(s\right)=\frac{(9+0.5s)(s+2)}{s(s+3)}$

$G\left(j\omega \right)H\left(j\omega \right)=\frac{(9+j0.5\omega )(j\omega +2)}{j\omega (3+j\omega )}$

$\angle G\left(j\omega \right)H\left(j\omega \right)={\tan }^{-1}\left(\frac{0.5\omega }{9}\right)+{\tan }^{-1}\left(\frac{\omega }{2}\right)-90{}^{\circ }-{\tan }^{-1}\left(\frac{\omega }{3}\right)$

$At\omega ={\omega }_{pc}\angle G\left(j\omega \right)H\left(j\omega \right)=-180{}^{\circ }.$

$-180{}^{\circ }+90{}^{\circ }={\tan }^{-1}\left(\frac{0.5\omega }{9}\right)+{\tan }^{-1}\left(\frac{\omega }{2}\right)-{\tan }^{-1}\left(\frac{\omega }{3}\right)$

$-90{}^{\circ }={\tan }^{-1}\left(\frac{0.5\omega }{9}\right)+{\tan }^{-1}\left(\frac{\frac{\omega }{2}-\frac{\omega }{3}}{1+\frac{{\omega }^2}{6}}\right)$

$-90{}^{\circ }={\tan }^{-1}\left(\frac{0.5\omega }{9}\right)+{\tan }^{-1}\left(\frac{\omega }{(6+{\omega }^2)}\right)$

$-90{}^{\circ }={\tan }^{-1}\left(\frac{\frac{0.5\omega }{9}+\frac{\omega }{6+{\omega }^2}}{1+\frac{0.5{\omega }^2}{9(6+{\omega }^2)}}\right)$

$\Rightarrow \left(\frac{\frac{0.5\omega }{9}+\frac{\omega }{6+{\omega }^2}}{1+\frac{0.5{\omega }^2}{9(6+{\omega }^2)}}\right)=-\infty$

$1=\frac{0.5{\omega }^2}{54+9{\omega }^2}$

$9{\omega }^2-9.5{\omega }^2+54=0$

$\omega =\sqrt{-6.35}$
Frequency can not be imaginary. Thus, phase corssover frequency does not exist. Alternatively, for second order system, Bode phase plot never crosses -180° axis. Thus phase crossover frequency is infinite.