Question

The feedback configuration and the pole-zero location of $G\left(s\right)=\frac{s^2-2s+2}{s^2+2s+2}$ are shown below. The root locus for Negative values of $K$, i.e for $-\infty <K<0$ , has breakaway/break-in points and angle of departure at pole $P$ (with respect to the positive real axis) equal to

• A. $\pm \sqrt{2}$ and $0^{\circ }$
• B. $\pm \sqrt{2}$ and ${45}^{\circ }$
• C. $\pm \sqrt{3}$ and $0^{\circ }$
• D. $\pm \sqrt{3}$ and ${45}^{\circ }$

Answer 1

$1+G\left(s\right)H\left(s\right)=0$
$1+\frac{K(s^2-2s+2)}{s^2+2s+2}=0$
$K=-\frac{s^2+2s+2}{s^2-2s+2}$
put $\frac{\delta K}{\delta s}=0$ we have
$(s^2-2s+2)(s+2)-(s^2+2s+2)$
$(s-1)=0$
$2s^2-4s^2+4=0$
$2s^2=+4$
$s=\pm \sqrt{2}$
Angle of departure is ${\varphi }_D=\varphi$
where $\varphi =\sum {\varphi }_Z-\sum {\varphi }_P$
$\varphi ={225}^{\circ }$
${\varphi }_D=\pm {225}^{\circ }$