Design Butterworth filter using the impulse invariant method for the following specification,

$$ \begin{array}{r} 0.8 \leq\left|H\left(e^{j \omega}\right)\right| \leq 1, \quad 0 \leq \omega \leq 0.2 \pi \\ \left|H\left(e^{j \omega}\right)\right| \leq 0.2, \quad 0.6 \pi \leq \omega \leq \pi \end{array} $$

Solution:

$ \begin{aligned} & \frac{1}{\sqrt{1+\varepsilon^2}}=0.8 \Rightarrow \varepsilon=0.75 \\\\ & \frac{1}{\sqrt{1+\lambda^2}}=0.2 \Rightarrow \lambda=4.899\\ \end{aligned} $

Ws $=0.6$ pi rad: $w_p=0.2$ pi rad . then assume $\mathrm{T}=1 \mathrm{sec}$

$ \frac{\omega_{\mathrm{s}}}{\omega_{\mathrm{p}}}=\frac{\Omega_{\mathrm{s}} \mathrm{T}}\\{\Omega_{\mathrm{p}} \mathrm{T}}=\frac{\Omega_{\mathrm{s}}}{\Omega_{\mathrm{p}}}=\frac{0.6 \pi}{0.2 \pi}=3\\ $

$ \mathrm{N} \geq \frac{\log \left(\frac{\lambda}{\varepsilon}\right)}{\log \left(\frac{\Omega_{\mathrm{s}}}{\Omega_{\mathrm{p}}}\right)} \geq \frac{\log \left(\frac{4.899}{0.75}\right)}{\log 3} \geq 1.71\\ $

For $\mathrm{N}=2$ the transfer function of normalized Butterworth filter is

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

Cut off frequency,

$ \begin{aligned} & \Omega_{\mathrm{c}}=\frac{\Omega_{\mathrm{p}}}{(\varepsilon)^{1 / \mathrm{N}}}=\frac{0.2 \pi}{(0.75)^{1 / \mathrm{N}}}=0.231 \pi \\ & \mathrm{H}_{\mathrm{a}}(\mathrm{s})=\left.\mathrm{H}(\mathrm{s})\right|_{\mathrm{s} \rightarrow \frac{\mathrm{s}}{}} ^{\Omega_{\mathrm{c}}} \\ & =\frac{0.5266}{(\mathrm{~s}+0.51+\mathrm{j} 0.51)(\mathrm{s}+0.51-\mathrm{j} 0.51)} \end{aligned} $

Using the impulse invariant method, if

$ \begin{aligned} & \mathrm{H}_{\mathrm{a}}(\mathrm{s})=\sum_{\mathrm{k}=1}^{\mathrm{N}} \frac{\mathrm{c}_{\mathrm{k}}}{\mathrm{s}-\mathrm{p}_{\mathrm{k}}} \\\\ & \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}} \\\\ & \mathrm{H}(\mathrm{z})=\frac{\mathrm{j} 0.516}{1-\mathrm{e}^{-0.51-\mathrm{j} 0.51} \mathrm{z}^{-1}}-\frac{\mathrm{j} 0.516}{1-\mathrm{e}^{-0.51+\mathrm{j} 0.51} \mathrm{z}^{-1}}\\ \end{aligned} $

By solving,

$ \mathrm{H}(\mathrm{z})=\frac{0.3022}{1-1.047 \mathrm{z}^{-1}+0.36 \mathrm{z}^{-2}}\\ $