Question please give explanation complex Zero, 9-8. Find the angles of departure and arrival for all complex poles and zeros of the open-loop trans- fer function of K(s+2s+2) K>0 G(s)H(s)= s(s?+4) the real

Given, G(s) H(s)=\frac{K\left(s^{2}+2 s+2\right)}{S\left(s^{2}+4\right)}= Open loop transfer function,Openloop poles are:\begin{array}{l}S\left(s^{2}+4\right)=0 \Rightarrow s=0 \text { and } s^{2}+4=0 \\\Rightarrow s^{2}=-4 \Rightarrow s= \pm \sqrt{-4} \\\therefore \text { Poles are } s=0,2 j,-2 j\end{array}Open loop zeroes:\begin{array}{ll}s^{2}+2 s+2=0 \Rightarrow s^{2}+2 s+1+1=0 \Rightarrow & (s+1)^{2}+1=0 \\& (s+1)^{2}=-1 \Rightarrow s+1= \pm \sqrt{-1} \\& s+1= \pm j \\\therefore \text { zeroes are } s=-1+j,-1-j & s=-1 \pm j\end{array}\therefore zeroes are s=-1+j,-1-jPlot of poles and zeroes:Angle of Departure: \left(\phi_{d}\right)A root locus branch starts at complex pole with an angle \phi_{d}.\left.\phi_{d}=180+\phi \quad \text { where } \phi=-\sum_{1} \text { (angle made by poles }\right)+\sum_{1} \text { angle made by } \text { zeroes }\phi=-\sum_{1} \phi_{p}+\sum_{1} \phi_{z}There are two complex poles:For s=+2 j\begin{aligned}\sum_{1} \phi_{p}=\theta_{1}+\theta_{2} & =90+90=180^{\circ} \\\sum_{1} \phi_{z}=\theta_{3}+\theta_{4} & =\tan ^{-1}\left(\frac{3}{1}\right)+\tan ^{-1}\left(\frac{1}{1}\right) \\\sum_{1} \phi_{z} & =71.565+45^{\circ} \\\sum_{1} \phi_{z} & =116.565^{\circ}\end{aligned}\begin{aligned}\phi=-\sum \phi_{p}+\sum_{1} \phi_{z} & =-180+116.525 \\\phi & =-63.435\end{aligned}\begin{array}{c}\phi_{d}=180+\phi=180-63.435 \\\phi_{d}=116.565^{\circ}\end{array}For s=-2 j\phi_{d}=-116.565^{\circ}[\because Angle of departure of a complex conjugate pole is \left.-\left(\phi_{d}\right)\right)Angle of Arrival \left(\phi_{a}\right) :A root locus ends at complex zero with an angle \phi_{a}.\phi_{a}=180-\phiThere are two complex zeroesFor s=-1+j :\begin{array}{c}\sum_{1} \phi_{p}=\theta_{1}+\theta_{2}+\theta_{3} \\\sum_{p}=\left(90+\tan ^{-1}\left(\frac{1}{3}\right)\right)+\left(90+\tan ^{-1}\left(y_{1}\right)\right. \\-\left[90+\tan ^{-1}\left(\phi_{1}\right)\right] \\\phi=-\sum_{1} \phi_{p}+\sum_{1} \phi_{z}=-g 6-\tan ^{-1}\left(\frac{1}{3}\right)+g 6 \\\phi_{a}=180-\phi_{a}=\theta_{4}=90^{\circ} \\=180-\left(-\tan ^{-1}(1 / 3)\right)=180+\tan ^{-1}(1 / 3)=180+18.435 \\\phi_{a}=198.435^{\circ}\end{array}Qa for its complex conjugate s=-1-j is -198.435^{\circ} ...