Solution:

$ \begin{gathered} \text { i) } F(s)=\frac{s^3+6 s^2+7 s+3}{s^2+2 s+1}=\frac{N(s)}{D(s)}....(1) \\\\ N(s)=s^3+6 s^2+7 s+3 \end{gathered} $

$ \begin{aligned} \\ &\begin{array}{c|c} s^3 & 1 \\\\ \hline s^2 & 6 \\\\ \hline s^1 & \frac{13}{2} \\\\ \hline s^0 & 3 \\\\ \end{array}\\\\ &N(S) \text { is Hurwitz's polynomial } \end{aligned} $

$ \begin{aligned} &D(s)=s^2+2 s+1\\\\ &\begin{array}{c|cc} s^2 & 1 & 1 \\\\ \hline s^1 & 2 & - \\\\ \hline s^0 & 1 & \end{array} \\ &D(S) \text { is Hurwitz's polynomial } \end{aligned} \\ $

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

roots are real hence residuals are not required.

$ A(w)=m_1 m_2-n_1 m_2-2....(2) \\ $

$ where \\ $

$ \begin{aligned} m_1 &=6 s^2+3 & & n_1=s^3+7 s \\ m_2 &=s^2+1 & n_2 &=2 s \end{aligned} \\ $

$ \therefore A(s)=\left(6 s^2+3\right)\left(s^2+1\right)-\left(s^3+7 s\right)(2 s) \\ $

$ =\left(6 s^4+6 s^2+3 s^2+3\right)-\left(2 s^4+14 s^2\right) \\ $

$ =4 S^4-5 S^2+3 \\ $

put $ S=j \omega \\ $

$ \therefore A(\omega)=4(j \omega)^4-5(j \omega)^2+3 \\ $

$ =4 w^4+5 w^2+3 \\ $

it is clear that for all values of $w$ $ A(\omega)>0 \ $ $\therefore$ All conditions are satisfied, hence F(S) is P . R. function.