Obtain digital filter transfer function by applying impulse invariance transfer function,,

$$ H(s)=\frac{s}{(s+5)(s+2)} \quad \text { if } \mathrm{Ts}=0.1 s $$

Solution:

$$ \text { Given } H(s)=\frac{s}{(s+5)(s+2)} $$

step 1- Obtain $h(t)$ by I.L.T By PFE,

$ \begin{aligned}\\ & H(s)=\frac{A}{S+5}+\frac{B}{S+2} \\\\ & A=\left.(s+5) H(s)\right|_{s=-5} \quad B=\left.(s+2) H(s)\right|_{s=-2}\\ \end{aligned}\\ $

$ \begin{aligned}\\ & A=5 / 3 \quad B=-2 / 3 \\\\ & H(s)=\frac{5 / 3}{5+5}-\frac{2 / 3}{5+2} \\\\ & n(t)=\frac{5}{3} e^{-5 t} u(t)-\frac{2}{3} e^{-2 t} u(t)\\ \end{aligned}\\ $

Step 2:- Replace $t=n T$

$$ h(n)=\frac{5}{3} e^{-5 n T} u(n)-\frac{2}{3} e^{-2 n T} u(n)\\ $$

Step 3:- Apply Z.T.

$ \begin{aligned} H(z)=\sum_{n=0}^{\infty} h(n) z^{-n} & =\sum_{n=0}^{\infty} \frac{5}{3} e^{-5 n z^{-n}} -\sum_{n=0}^{\infty} \frac{2}{z} e^{-2 n \pi} \cdot z^{-n} \cdot \\\\ & =\frac{5}{3}\left(\frac{1}{1-e^{-5 T} z^{-1}}\right)-\frac{2}{3}\left(\frac{1}{1-e^{-2} z^1}\right)\\ \end{aligned} $

$ \begin{aligned}\\ H(z) & =\frac{5}{3}\left(\frac{1}{1-e^{-5 T} z^{-1}}\right)-\frac{2}{3}\left(\frac{1}{1-e^{-2 T} z^{-1}}\right) \\\\ T & =0.1 \mathrm{sec}\\ \end{aligned} $

$ \begin{aligned}\\ H(z) & =\frac{5 / 3}{1-0.6065 z^{-1}}-\frac{2 / 3}{1-0.8187 z^{-1}} \\\\ & =\frac{1.667-1.3645 z^{-1}-0.667+0.4043 z^{-1}}{\left(1-0.8187 z^{-1}-0.6065 z^{-1}+0.4965 z^{-2}\right)}\\ \end{aligned}\\ $

$ =\frac{1-0.9602 z^{-1}}{1-1.4252 z^{-1}+0.4968 z^{-2}}\\ $