Question

The open-loop transfer function of a unity feedback configuration is given as $G\left(s\right)=\frac{K(s+4)}{(s+8)(s^2-9)}$
The value of a gain $K(>0)$ for which $-1+j2$ lies on the root locus is

• A. 25.54

Answer 1

The correct option is A 25.54

$G\left(s\right)=\frac{K(s+4)}{(s+8)(s^2-9)}$

$=\frac{K(s+4)}{(s+8)(s+3)(s-3)}$

For the point (-1+2j) to lie on the root locus the angle condition must be satisfy

$i.e\angle G\left(s\right)H\left(s\right)=\pm (2Q+1){180}^{\circ }$

Taking LHS:
$\angle G\left(s\right)H\left(s\right){|}_{s=-1+2j}$

$=\frac{\angle K+\angle (s+4)}{\angle (s+8)+\angle (s+3)+\angle (s-3)}$

$=\frac{\angle K+\angle (-1+2j+4)}{\angle (-1+2j+8)+\angle (-1+2j+3)+\angle (-1+2j-3)}$

$=\frac{0+ta{n}^{-1}\left(\frac{2}{3}\right)}{ta{n}^{-1}\left(\frac{2}{7}\right)+ta{n}^{-1}\left(1\right)+{180}^{\circ }-ta{n}^{-1}\left(\frac{1}{2}\right)}$

$=ta{n}^{-1}\frac{2}{3}-ta{n}^{-1}\frac{2}{7}-ta{n}^{-1}-{180}^{\circ }+ta{n}^{-1}\frac{1}{2}$
${33.69}^{\circ }-{15.945}^{\circ }-{45}^{\circ }-{180}^{\circ }+{26.565}^{\circ }$

$\simeq {180}^{\circ }$

As angle condition is satisfy the value of system gain $K$ can be obtained by using magnitude condition

$i.e.|G\left(s\right)H\left(s\right){|}_{s=-1+2j}=1$

$=\frac{K\sqrt{(-1+4{)}^2}+2^2}{\sqrt{(-1+8{)}^2+2^2}.\sqrt{(-1+3{)}^2+2^2}.\sqrt{(-1-3{)}^2}+2^2}$

$=1$

$\Rightarrow \frac{K\sqrt{9+4}}{\sqrt{49+4}\sqrt{4+4}\sqrt{16+4}}=1$

$\Rightarrow \frac{K\sqrt{13}}{\sqrt{53}\sqrt{8}\sqrt{20}}=1$

$K=\frac{\sqrt{53\times 8\times 20}}{\sqrt{13}}=25.54$