Question

The open loop transfer function of a unity gain negative feedback control system is given by

$G\left(s\right)=\frac{s^2+4s+8}{s(s+2)(s+8)}$

The angle $\theta$, at which the root locus approaches the zeros of the system, satisfies

• A. $|\theta |=\pi -ta{n}^{-1}\left(\frac{1}{4}\right)$
• B. $|\theta |=\frac{3\pi }{4}-ta{n}^{-1}\left(\frac{1}{3}\right)$
• C. $|\theta |=\frac{\pi }{2}-ta{n}^{-1}\left(\frac{1}{4}\right)$
• D. $|\theta |=\frac{\pi }{4}-ta{n}^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is D $|\theta |=\frac{\pi }{4}-ta{n}^{-1}\left(\frac{1}{3}\right)$

$G\left(s\right)=\frac{s^2+4s+8}{s(s+2)(s+8)}$

$Zero:s^2+4s+8$

$s=\frac{-4\pm \sqrt{4^2-4\times 8}}{2}$

$=\frac{-4\pm \sqrt{16-32}}{2}$

$=\frac{-4\pm 4j}{2}=-2\pm 2j$

poles : s = 0, -2, -8

Pole zero plot:


${\theta }_{arrival}=\pm \pi -sum\left({\theta }_2\right)+sum\left({\theta }_p\right)$

$-\pi -[{\theta }_1+{\theta }_3+{\theta }_2]-{\theta }_4$

${\theta }_1=\pi ={\theta }_a$

$=\pi -ta{n}^{-1}\left(\frac{2}{2}\right)=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

${\theta }_2=ta{n}^{-1}\frac{AC}{BC}$

$={\tan }^{-1}\left(\frac{2}{6}\right)=ta{n}^{-1}\left(\frac{1}{3}\right)$

${\theta }_3=\frac{\pi }{2}$

${\theta }_4=\frac{\pi }{2}$

$\therefore {\theta }_{arrival}=\pm \pi -[\frac{3\pi }{4}+\frac{\pi }{2}+ta{n}^{-1}\left(\frac{1}{3}\right)]-\frac{\pi }{2}$

$=\pi +\frac{3\pi }{4}+ta{n}^{-1}\left(\frac{1}{3}\right)$

$or-\pi +\frac{3\pi }{4}+ta{n}^{-1}\left(\frac{1}{3}\right)$

$=\frac{7\pi }{4}+ta{n}^{-1}\left(\frac{1}{3}\right)$

$or-\frac{\pi }{4}+ta{n}^{-1}\left(\frac{1}{3}\right)$

$\therefore$ for ${\theta }_{arrival}=-\frac{\pi }{4}+ta{n}^{-1}\left(\frac{1}{3}\right)$

$Hence,|{\theta }_{arrival}|=\frac{\pi }{4}-ta{n}^{-1}\left(\frac{1}{3}\right)$