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For the constraints $0.8 \leq\left|\mathrm{H}\left(\mathrm{e}^{\mathrm{jw}}\right)\right| \leq 1 \quad, 0 \leq \mathrm{w} \leq 0.2 \pi$

For the constraints $0.8 \leq\left|\mathrm{H}\left(\mathrm{e}^{\mathrm{jw}}\right)\right| \leq 1 \quad, 0 \leq \mathrm{w} \leq 0.2 \pi$

$\left|\mathrm{H}\left(\mathrm{e}^{\mathrm{jw}}\right)\right| \leq 0.2 \quad, 0.6 \pi \leq \mathrm{w} \leq \pi\\$

With $\mathrm{T}=1$ sec.Determine system function $\mathrm{H}(\mathrm{z})$ for a Butterworth filter using the impulse invariant method.

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Solution:

Given $\frac{1}{\sqrt{1+\varepsilon^2}}=0.8$ from which $\varepsilon=0.75$

$\frac{1}{\sqrt{1+\lambda^2}}=0.2$ from which $\lambda=4.899$.

$\omega_s=0.6 \pi \mathrm{rad}, \omega_p=0.2 \pi \mathrm{rad}$

\begin{aligned}\\ &\frac{\omega_s}{\omega_p}=\frac{\Omega_s T}{\Omega_p T}=\frac{\Omega_s}{\Omega_p}=\frac{0.6 \pi}{0.2 \pi}=3 \\\\ &\begin{aligned}\\ N=\frac{\log d_{\varepsilon}}{\log 1 / k} &=\frac{\log 4.899}{\log 3} \\\\ &=1.71\\ \end{aligned}\\ \end{aligned}\\

\begin{aligned}\\ &\therefore N=2\\\\ &H(s)=\frac{1}{s^2+\sqrt{2} s+1}\\\\ &\Omega_c=\frac{\Omega_p}{\varepsilon^{1 / N}}=\frac{0.2 \pi}{(0.75)^{1 / 2}}\\\\ &=0.231 \pi\\ \end{aligned}\\

\begin{aligned}\\ H_a(s) &=H(s) \mid \\\\ s \rightarrow s / 0.231 \pi \\\\ &=\frac{0.5266}{s^2+1.03 s+0.5266}\\ \end{aligned}\\

\begin{aligned}\\ &=\frac{0.516 j}{s+0.51+j 0.51}+\frac{0.516 j}{s+0.51-j 0.51} \\\\ &=\frac{0.516 j}{s-(-0.51-j 0.51)}-\frac{0.516 j}{s-(-0.51+j 0.51)}\\ \end{aligned}\\

\begin{aligned}\\ &H(z)=\frac{0.516 j}{1-e^{-0.51 T} e^{-j 0.51 T} z^{-1}}-\frac{0.516 j}{1-e^{-0.51 T j 0.51 T} e^{-1}} \\\\ &T=15 e \\\\ &H(z)=\frac{0.3019 z^{-1}}{1-9.048 z^{-1}+0.36 z^{-2}} .\\ \end{aligned}\\