Solution:

Given $\quad V(s)=\frac{3 s}{(s+2)\left(s^2+2 s+2\right)}...(1)$

zeros $\Rightarrow S=0$

poles $\Rightarrow s=-2, S=-1+j 1, S=-1-j 1$

By pole zero diagram, & eq $-^n$ (1) partial fraction expansion is:

$V(s)=\frac{A}{s+2}+\frac{B}{(s+1-1 j)}+\frac{C}{(s+1+1 j)}...(2) \\ $

$\begin{aligned} \\ A &=\frac{3\left(\overline{O P_1}\right)}{\left(P_2 \bar{P}_1\right)\left(\overline{\left.P_3 P_1\right)}\right.} \\\\ &\left.=\frac{3(21180)}{(\sqrt{2}(-135)(\sqrt{2} \lfloor 135}\right) \\\\ &=3 \lfloor180 \\\\ &=-3 \\ \end{aligned} \\ $

$\begin{aligned} \\ B &=\frac{3\left(\overline{P P}_2\right)}{\left(\bar{P}_1 P_2\right)\left(\bar{P}_3 P_2\right)} \\\\ &=\frac{3(\sqrt{2} \lfloor 135)}{(\sqrt{2} \lfloor 45)(2 \lfloor 90)} \\\\ B &=\frac{3}{2} \\ \end{aligned} \\ $

$\begin{aligned} \\ C &=\frac{3\left(\overline{O P_3}\right)}{\left(P_1 P_3\right)\left(\overline{P_2 P_3}\right)} \\\\ &=\frac{3\left(\sqrt{2} \mid-135^{\prime}\right)}{\left(\sqrt { 2 } \lfloor - 4 5 ^ { \circ } ) \left(2\left\lfloor-90^{\circ}\right)\right.\right.} \\\\ C &=\frac{3}{2} \\ \end{aligned} \\ $

Hence from (2)

$\begin{aligned} \\ V(s) &=\frac{-3}{s+2}+\frac{3 / 2}{(s+1-j 1)}+\frac{3 / 2}{(s+1+j 1)} \\\\ & \text { Taking I.L.T. } \\\\ V(s) &=-3 \cdot e^{-2 t}+\frac{3}{2} \cdot e^{(-1+j) t}+\frac{3}{2} e^{(-1-j) t} \\\\ &=-3 \cdot e^{-2 t}+\frac{3}{2} \cdot e^{-t}\left[e^{j t}+e^{-j t}\right] \\\\ &=-3 \cdot e^{-2 t}+\frac{3}{3} e^{-t}\left[\frac{e^{j t}+e^{-j t}}{2}\right] \\\\ V(s) &=-3 \cdot e^{-2 t}+3 \cdot e^{-t} \operatorname{cost} \\ \end{aligned} \\ $