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Obtain digital filter transfer function by applying impulse in variance transfer function.
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ANS

$H(s)=\frac{5}{(S+5)(S+2)}$ if $T=0.1 s$

Solution: Given:

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

Step 1: Obtain h(t) By I.L.T.

By PFE.

$H(s)=\frac{A}{s+5}+\frac{B}{s+2}$

$A=\left.(s+5) H(s)\right|_{s=-5}$ $A=5 / 3$

$B=(S+2) H(S) | S=-2$ $B=-2 / 3$

$H(s)=\frac{513}{s+5}-\frac{213}{s+2}$

$\therefore\left[h(t)=\frac{5}{3} e^{-5 t} u(t)-\frac{2}{3} e^{-2 t} u(t)\right]$

Step 2: Replace t = nT

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

Step 3: Apply Z.T.

$H(z)=\sum_{n=0}^{\infty} h(n) \tilde{z}^{n}=\sum_{n=0}^{1} \frac{5}{3} e^{-5 n T} z^{-n}-\sum_{n=0}^{\infty} \frac{2}{3} e^{-2 n T} z^{-n}$

$=\frac{5}{3} \frac{1}{1-{e}^{-5 T} z^{-1}}-\frac{2}{3} \frac{1}{1-e^{-2 T} {z}^{-1}}$

$H(z)=\frac{5}{3} \frac{1}{1-e^{-5 T} {z}^{-1}}-\frac{2}{3} \frac{1}{1-e^{-2 T} {z}^{-1}}$

$T=0.1 \mathrm{sec}$

$H(z)=\frac{5/3}{1-0.6065 {z^{-1}}}-\frac{2 / 3}{1-0.8187 {z}^{-1}}$

$=\frac{1 \cdot 667-1.3645 {z}^{-1}-0.667+0.4043 \zeta^{-1}}{\left(1-0.8187 z^{-1}-0.6065 E^1 +0.4965{\zeta}^{-2}\right)}$

$H(2)=\frac{1-0.9602 {z}^{-1}}{1-1 \cdot 4252 {z}^{-1}+04965 {\zeta}^{-2}}$

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