Solution:

$ \begin{aligned} \\ &N(s)=s^3+8 s^2+15 s\\\\ &\begin{array}{l|ll} s^3 & 1 & 15 \\\\ \hline s^2 & 8 & - \\\\ \hline s^1 & 15 & - \\\\ \hline s^0 & - & \end{array} \\ \end{aligned} \\ $

N(s) is Hurwitz's polynomials.

$$ D(s)=s^2+5 s+4 $$

$$ \begin{array}{c|cc} s^2 & 1 & 4 \\\\ \hline s & 5 & - \\\\ \hline s^0 & 4 & \end{array} \\ $$

$D(S)$ is Hurwitz's polynomial

$ \begin{aligned} &D(S)=0 \\\\ &\therefore \quad S^2+5 S+4=0 \\\\ &\therefore S_1=-1 \quad \& \quad S_2=-4 \end{aligned} \\ $

roots are real \& -ve hence residues are not required.

$ A(\omega)=m_1 m_2-n_1 n_2 \\ $

where

$ \begin{aligned} m_1 &=8 s^2 ; & n_1=s^3+15 s \\ m_2 &=s^2+4 ; & n_2=5 s \end{aligned} \\ $

$ \begin{aligned}\\ \therefore A(s) &=\left(8 s^2\right)\left(s^2+4\right)-\left(s^3+15 s\right)(5 s) \\\\ &=\left(8 s^4+32 s^2\right)-\left(5 s^4+75 s^2\right) \\\\ &=3 s^4-43 s^2 \\\\ & \text { put } s=j w \\\\ \therefore \quad A(w) &=3(j w)^4-43(j w)^2 \\\\ &=3 w^4+43 w^2 \\ \end{aligned} \\ $

It is clear that $A(\omega)\gt0$ for all values of $w^{-}$ All conditions are satisfied, hence ' $F(S)$ is P.R. function.