Convert the analog filter with system function,

$$ H_a(s)=\frac{s+0.5}{(s+0.5)^2+16} $$

into a digital IIR filter using the bilinear transformation. The digital filter should have a resonant frequency of $\omega_r=\pi / 2$.

Solution:

From the system function, we observe that $\boldsymbol{\Omega}_c=4$. The sampling period T can be determined using the equation,

$ \Omega_c=\frac{2}{T} \tan \frac{\omega_r}{2}\\ $

$ T=\frac{L}{\Omega_c} \tan \frac{\omega_r}{2}=\frac{L}{4} \tan \frac{\pi}{4}=0.5 \mathrm{~s}\\ $

$ H(z)=\left.H(s)\right|_{s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}=\left.H(s)\right|_{s=4}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\ $

$ H(z)=\left.\frac{s+0.5}{(s+0.5)^2+16}\right|_{s=4}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\ $

$ \begin{aligned} & =\frac{4\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+0.5}{\left[4\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+0.5\right]^2+16} \\ & =\frac{\left[4\left(1-z^{-1}\right)+0.5\left(1+z^{-1}\right)\right]\left[1+z^{-1}\right]}{\left[4\left(1-z^{-1}\right)+0.5\left(1+z^{-1}\right)\right]^2+16\left[1+z^{-1}\right]^2} \\\\ & =\frac{4.5+z^{-1}-3.5 z^{-2}}{36.25+0.5 z^{-1}+28.25 z^{-2}}\\ \end{aligned}\\ $