Question

Consider a control system having open-loop transfer function of
$G\left(s\right)H\left(s\right)=\frac{1}{s^2(s-a)(s-b)(s-c)}$
If a,b,c are positive, then the polar plot of the system will be

• A.
• B.
• C.
• D.

Answer 1

The correct option is A


$G\left(s\right)H\left(s\right)=\frac{1}{s^2(s-a)(s-b)(s-c)}$
Put s = j$\omega$, $G\left(j\omega \right)H\left(j\omega \right)=\frac{1}{\left(j\omega {)}^2\right(j\omega -a\left)\right(j\omega -b\left)\right(j\omega -c)}$


$G\left(j\omega \right)H\left(j\omega \right)=\frac{1}{{\omega }^2\sqrt{{\omega }^2+a^2}.\sqrt{{\omega }^2+b^2}.\sqrt{{\omega }^2+c^2}}$

when, $\omega =0$, Magnitude $=\infty$; phase $=0^o$

$\omega =\infty$, Magnitude $=0$; phase $={270}^o$


$\angle G\left(j\omega \right)H\left(j\omega \right)=-[{180}^o+({180}^o-ta{n}^{-1}\left(\frac{\omega }{a}\right))+({180}^o-ta{n}^{-1}\left(\frac{\omega }{b}\right))+({180}^0-ta{n}^{-1}\left(\frac{\omega }{c}\right))]$

$\varphi =ta{n}^{-1}\left(\frac{\omega }{a}\right)+ta{n}^{-1}\left(\frac{\omega }{b}\right)+ta{n}^{-1}\left(\frac{\omega }{c}\right)$


Note : Since a, b and c are positive values, for $\omega =1$, phase of $G\left(j\omega \right)H\left(j\omega \right)$ will be a positive quantity,
So, the polar plot will start into the first quadrant.