The normalized transfer function of an analog filter is given by $$ H_a\left(s_n\right)=\frac{1}{s_n^2+1.414 s_n+1} $$ Convert the analog filter to a digital one with a cut-off frequency of $0.4 \pi$ using the bilinear transformation.

Solution:

$ \begin{aligned} \Omega_c & =\frac{2}{T} \tan \frac{\omega_c}{2}=2 \tan (0.2 \pi)=1.45 \\\\ H_a(s) & =\frac{1}{\left(\frac{s}{1.45}\right)^2+1.414\left(\frac{s}{1.45}\right)+1}=\frac{(1.45)^2}{s^2+2.055 s+(1.45)^2} \\\\ H(z) & =H_a(s) s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\ \end{aligned} $

$ \begin{aligned} H(z) & =\frac{2.1}{\frac{4\left(1-z^{-1}\right)^2}{\left(1+z^{-1}\right)^2}+(2.055) 2 \frac{\left(1-z^{-1}\right)}{1+z^{-1}}+2.1} \\\\ & =\frac{2.1\left(1+z^{-1}\right)^2}{4\left(1-z^{-1}\right)^2+4.11\left(1-z^{-2}\right)+2.1\left(1+z^{-1}\right)^2} \\\\ & =\frac{2.1\left(1+z^{-1}\right)^2}{4\left(1-2 z^{-1}+z^{-2}\right)+4.11\left(1-z^{-2}\right)+2.1\left(1+2 z^{-1}+z^{-2}\right)} \\\\ & =\frac{2.1\left(1+z^{-1}\right)^2}{10.21-3.8 z^{-1}+1.99 z^{-2}} \\\\ & =\frac{0.2\left(1+z^{-1}\right)^2}{1-0.37 z^{-1}+0.195 z^{-2}}\\ \end{aligned} $