(i) $H(s)=\frac{1}{(s+2)(s+0.6)} \cdots(1)$

Let $\frac{1}{(s+2)(s+0.6)}=\frac{A}{s+2}+\frac{B}{s+0.6}$

$\therefore A=(s+2) \times\left.\frac{1}{(s+2)(s+0.6)}\right|_{s=-2}=\frac{1}{-2+0.6}=\frac{-5}{7}$

$B=(s+0 .6) \times\left.\frac{1}{(s+2)(s+0.6)}\right|_{s=-0.6}=\frac{1}{-0.6+2}=\frac{5}{7}$

$\therefore H(s)=\frac{-5 / 7}{s+2}+\frac{5 / 7}{s+0.6}$

Here, $s=-0.6$ and $s=-2$ are the poles in s-plane.

Case 1: By Impulse Invariance Transformation

$\frac{1}{s-p_{i}} =\frac{z}{z-e^{p_{i} T}}$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{-5 z / 7}{z-e^{-2 T}}+\frac{5 z / 7}{z-e^{-0.6 …