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A digital filter with a 3 dB bandwidth of 0.4 is to be designed from the analog filter whose system response is, $H(s)=\frac{\Omega_c}{s+2 Q_c}$...

A digital filter with a $3 \mathrm{~dB}$ bandwidth of $0.4$ is to be designed from the analog filter whose system response is:

$ H(s)=\frac{\Omega_c}{s+2 \Omega_c} $

Use the bilinear transformation and obtain $H(z)$.

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Solution:

We know that,

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

$ \Omega_c=\frac{2}{T} \tan \frac{0.4 \pi}{2}=\frac{1.453}{T}\\ $

This system response of the digital filter is given by,

$ H(z)=H_a(s) \mid=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\ $

$ =\frac{\Omega_c}{\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+2 \Omega_c}=\frac{\frac{1.453}{T}}{\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+2\left(\frac{1.453}{T}\right)}\\ $

$ \begin{aligned}\\ & =\frac{1.453\left(1+z^{-1}\right)}{2\left(1-z^{-1}\right)+2\left(1+z^{-1}\right) 1.453} \\\\ & =\frac{1+z^{-1}}{3.376-0.624 z^{-1}}\\ \end{aligned}\\ $

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