For the analog transfer function

Solution:

$\mathrm{H}(\mathrm{s})=\frac{2}{(\mathrm{~s}+1)(\mathrm{s}+2)}$

Applying partial fractions,

$\begin{aligned} & \mathrm{H}(\mathrm{s})=\frac{2}{(\mathrm{~s}+1)}-\frac{2}{(\mathrm{~s}+2)} \\\\ & =\frac{2}{(\mathrm{~s}-(-1))}-\frac{2}{(\mathrm{~s}-(-2))} \\\\ & \mathrm{H}(\mathrm{z})=\sum_{\mathrm{k}=1}^{\mathrm{N}} \frac{\mathrm{c}_{\mathrm{k}}}{1-\mathrm{e}^{\mathrm{p}_{\mathrm{k}} \mathrm{T}} \mathrm{z}^{-1}} \\\\ & \left(\mathrm{~s}-\mathrm{p}_{\mathrm{k}}\right) \rightarrow\left(1-\mathrm{e}^{\mathrm{p}_{\mathrm{k}} \mathrm{T}} \mathrm{z}^{-1}\right)\\ \end{aligned}$

Here , $\mathrm{p}_{\mathrm{k}} \rightarrow \mathrm{p}_1, \mathrm{p}_2$

$\begin{aligned} & \mathrm{p}_1=-1 ; \quad \mathrm{p}_2=-2 ; \quad \mathrm{T}=1 \mathrm{sec} \\\\ & \mathrm{H}(\mathrm{z})=\frac{2}{1-\mathrm{e}^{-1} \mathrm{z}^{-1}}-\frac{2}{1-\mathrm{e}^{-2} \mathrm{z}^{-1}} \\\\ & =\frac{2}{1-0.3678 \mathrm{z}^{-1}}-\frac{2}{1-0.1353 \mathrm{z}^{-1}} \\\\ & \mathrm{H}(\mathrm{z})=\frac{0.465 \mathrm{z}^{-1}}{1-0.503 \mathrm{z}^{-1}+0.0497 \mathrm{z}^{-2}}\\ \end{aligned}$