Solution:

$H(s)=\frac{1}{(s+1)(s+3)} \cdots(1)$

In Bilinear Transformation, put $s=\frac{2(z-1)}{T(z+1)}$ in $\mathrm{H}(\mathrm{s})$ to get $\mathrm{H}(z)$

$\therefore s=\frac{2(z-1)}{2(z+1)}(\text { Given }, \mathrm{T}=2)$

$\therefore$ From $(1),$ Transfer function of the digital filter

$H(z)=\frac{1}{\left(\frac{z-1}{z+1}+1\right)\left(\frac{z-1}{z+1}+3\right)}$

$=\frac{1}{\left(\frac{z-1+z+1}{z+1}\right)\left(\frac{z-1+3 z+3}{z+1}\right)}$

$=\frac{(z+1)^{2}}{2 z(4 z+2)}$

Transfer Function of the digital filter $=\frac{z^{2}+2 z+1}{4 z(2 z+1)}$