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Convert the following filters with system functions. $ H(s)=\frac{(s+0.1)}{(s+0.1)^2+9} $
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Solution:

$ \begin{aligned}\\ &S=\frac{2}{T}\left(\frac{1-2^{-1}}{1+2^{-1}}\right)\\\\ &=2\left(\frac{1-z^{-1}}{1+z^{-1}}\right) \text {. }\\\\ &\left[2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+0.1\right]^2+9\\\\ &=\frac{2\left(1-z^{-1}\right)+0.1\left(1+z^{-1}\right)\left(1+z^{-1}\right)}{\frac{\left(2\left(1-z^{-1}\right)+0.1\left(1+z^{-1}\right)\right)^2}{\left(1+z^{-1}\right)^2}+9}\\ \end{aligned}\\ $

$ H[z]=\frac{\left[2\left(1-z^{-1}\right)+0.1\left(1+z^{-1}\right)\right]\left(1+z^{-1}\right)}{\left[2\left(1-z^{-1}\right)+0.1\left(1+z^{-1}\right)\right]^2+9\left(1+z^{-1}\right)^2}\\ $

By Impulse Invariant method:

$ H(s)=\frac{s+0.1}{(s+0.1)^2+9}\\ $

The inverse L.T. of the given function is

$ \begin{aligned}\\ h(t) &=e^{-0.1 t} \cos (3 t) & & \text { for } t \geqslant 0 \\\\ &=0 & & \text { otherwise. }\\ \end{aligned}\\ $

Sampling the function gives,

$ \begin{aligned}\\ h[n T] &=e^{-0 \text { nT }} \cos (3 n T) & & \text { for } n \geqslant 0 \\\\ &=0 & & \text { otherwise. }\\ \end{aligned}\\ $

$ \begin{aligned}\\ H[2] &=\sum_{n=0}^{\infty} e^{-a \text { e.nT }} \cos (3 n T) z^{-n} \\\\ &=\sum_{n=0}^{\infty} e^{-0.1 n T}\left[\frac{e^{j 3 n T}+e^{-j 3 n T}}{2}\right] z^{-n} \\\\ &=\frac{1}{2}\left[\sum_{n=0}^{\infty} e^{-(0.1-j3) n T} z^{-n}+\sum_{n=0}^{\infty} e^{-(0.1+j 3) n T} z^{-n}\right]\\ \end{aligned}\\ $

$ \begin{aligned}\\ &=\frac{1}{2}\left[\frac{1}{1-e^{-(0.1-j 3) T} z^{-1}}+\frac{1}{1-e^{-(0.1+j 3) \tau_z^{-1}}}\right] \\\\ &=\frac{1}{2}\left[\frac{1-e^{-(0.1+j 3) T} z^{-1}+1-e^{-(0.1-j 3) T} z^{-1}}{\left(1-e^{-(0.1-j 3) T} z^{-1}\right)\left(1-e^{-(0.1+j 3) T} z^{-1}\right)}\right]\\ \end{aligned}\\ $

$ =\frac{1}{2}\left[\frac{2-e^{-0 . I T} \cos \left(e^{j 3 T}+e^{-j 3 T}\right) z^{-1}}{1-2 e^{-0.1 T} \cos (3 T)\\ z^{-1}+e^{-0.2 T} z^{-2}}\right]\\ $

$ H[z]=\frac{1-e^{-0.1 T} \cos (3 T) z^{-1}}{1-2 e^{-0.1 T} \cos (3 T) z^{-1}+e^{-0.2 T} z^{-2}}\\ $

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