Solution:

$z(s)=\frac{(s+1)(s+3)}{(s)(s+2)}=\frac{s^2+4 s+3}{s^2+2 s}--------(1) \\ $

$\text { Cauer-1 form - Using C.F.D. }$

$\begin{aligned} \\ \therefore z_1(S) &=1 \\\\ y_2(s) &=\frac{1}{2} S \\\\ z_3(s) &=4 \\\\ y_4(S) &=\frac{1}{6} S \\ \end{aligned} \\ $

$\text { Hence circuit in cauer-1 form is } \\ $

$\text { Cauer }-2 \text { form - Using C.F.D. } \\ $

$\begin{aligned} \\ \therefore z_1(s) &=\frac{3}{2 s}=\frac{1}{\frac{2}{3} s} \\\\ y_2(s) &=\frac{4}{5} v \\\\ z_3(s) &=\frac{25}{2 s}=\frac{1}{\frac{2}{25} \cdot s} \\\\ y_4(s) &=1 / 5 v \\ \end{aligned} \\ $

$\text { Required circuit in Cauer }-2 \text { form is } \\ $