Solution:

Foster −1 from:

$ \begin{aligned} \\ &z(s)=\frac{3(s+2)(s+4)}{(s)(s+3)}....(1)\\\\ &=\frac{3 s^2+18 s+24}{s^2+3 s}\\\\ &\therefore z(s)=3+\frac{9 S+24}{s^2+3 S}\\\\ &=Z_1(s)+Z_2(s) ...(2)\\ \end{aligned} \\ $

Where

$ \quad z_1(s)=3 \\ $

And

$ \left.\begin{array}{rl} \\ z_2(s) & =\frac{9 s+24}{(s)(s+3)} \\\\ & =\frac{A}{s}+\frac{B}{s+3} \\ \end{array}\right\}....(3) \\ $

$ \begin{aligned} \\ &A=\left.\frac{9 S+24}{(s)(S+3)}\right|_{S=0}=\frac{0+24}{0+3}=8 \\\\ &B=\left.\frac{9 S+24}{S}\right|_{S=-3}=\frac{-27+24}{-3}=1 \\ \end{aligned} \\ $

from (3)

$ \begin{aligned} \\ z_2(s) &=\frac{8}{s}+\frac{1}{s+3} \\\\ &=Z_2^{\prime}(s)+z_2^{\prime \prime}(s)...(4) \\ \end{aligned} \\ $

Where

$ Z_2^{\prime}(s)=\frac{8}{s}=\frac{1}{\frac{1}{8} s} \\ $

And

$ \begin{aligned} \\ z_2^{\prime \prime}(s) &=\frac{1}{s+3}=\frac{1}{y_2^{\prime \prime}(s)} \\\\ \therefore y_2^{\prime \prime}(s) &=s+3 \\ \end{aligned} \\ $

Foster $-1$ form is

Foster −2 from:

$ \begin{aligned} \\ V(s) &=\frac{1}{z(s)}=\frac{(s)(s+3)}{3(s+2)(s+4)} ....(1)\\\\ &=\frac{s^2+3 s}{3 s^2+18 s+24} \\ \end{aligned} \\ $

$\because$ polarity of remainder is -ve, divide both side of equation by S.

$ \left.\begin{array}{rl}\therefore \frac{y(s)}{S} \\& =\frac{S+3}{3(S+2)(S+4)} \\ & =\frac{A}{S+2}+\frac{B}{S+4}\end{array}\right\}...(2) \\ $

$ A=\left.\frac{s+3}{3(s+4)}\right|_{s=-2}=\frac{-2+3}{3(-2+4)}=\frac{1}{6} \\ $

$ B=\left.\frac{s+3}{3(s+2)}\right|_{S=-4}=\frac{-4+3}{3(-4+2)}=\frac{1}{6} \\ $

from $ \mathrm{eq}^{-1} \\ $

$ \frac{y(s)}{s}=\frac{1 / 6}{s+2}+\frac{1 / 6}{s+4} \\ $

$ \frac{y(s)}{s}=\frac{1}{6 s+12}+\frac{1}{6 S+24} \\ $

$ Y(S)=\frac{S}{6 S+12}+\frac{S}{6 S+24} \\ $

$ y(s)=y_1(s)+y_2(s)...(3) \\ $

Where $ y_1(s)=\frac{s}{6 s+12}=\frac{1}{\left[\frac{6 s+12}{s}\right]} \\ $

$ =\frac{1}{6+\frac{12}{\operatorname 5}}=\frac{1}{Z_1(s)} \\ $

$ \therefore Z_1(s)=6+\frac{12}{9}=6+\frac{1}{\frac{1}{12} s} \\ $

And

$ \begin{aligned} \\ y_2(s) &=\frac{s}{6 S+24}=\frac{1}{\left[\frac{6 s+24}{s}\right]} \\\\ =& \frac{1}{\left[6+\frac{24}{s}\right]}=\frac{1}{Z_2(S)} \\\\ \therefore Z_2(S) &=6+\frac{24}{s}=6+\frac{1}{\frac{1}{24} s} \\ \end{aligned} \\ $