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 & …