Design a Chebyshev filter with a maximum passband attenuation of $2.5 \mathrm{~dB}$; at $\Omega_p=20 \mathrm{rad} / \mathrm{sec}$ and the stopband attenuation of $30 \mathrm{~dB}$ at $\Omega_s=50 \mathrm{rad} / \mathrm{sec}$.

Solution:

Given,

$ \begin{array}{lll} \Omega_p=20 & \mathrm{rad} / \mathrm{sec} ; & \alpha_p=2.5 \mathrm{~dB} \\ \Omega_s=50 & \mathrm{rad} / \mathrm{sec} ; & \alpha_s=30 \mathrm{~dB} \end{array} $

We know about,

$ \begin{aligned} N & =\frac{\cosh ^{-1} \lambda / \varepsilon}{\cosh ^{-1} 1 / k} \\\\ \lambda & =\sqrt{10^{0.1 \alpha_s}-1}=31.607 \\\\ \varepsilon & =\sqrt{10^{0.1 \alpha_p}-1}=0.882 \\\\ k & =\frac{\Omega_p}{\Omega_s}=0.4\\ \end{aligned}\\ $

After,

$ N \geq \frac{\cosh ^{-1} \frac{31.607}{0.882}}{\cosh ^{-1} \frac{1}{0.4}}=2.726 $

Then N = 3,

$ \begin{aligned} \mu & =\varepsilon^{-1}+\sqrt{1+\varepsilon^{-2}}=2.65 \\\\ a & =\Omega_p \frac{\left[\mu^{1 / N}-\mu^{-1 / N}\right]}{2}=6.6 \\\\ b & =\Omega_p\left[\frac{\mu^{1 / N}+\mu^{-1 / N}}{2}\right]=21.06 \\\\ s_k & =a \cos \phi_k+j b \sin \phi_k ; \quad k=1,2,3 \\\\ \phi_k & =\frac{\pi}{2}+\left(\frac{2 k-1}{2 N}\right) \pi ; \quad k=1,2,3 \\\\ \phi_1 & =120^{\circ}, \phi_2=180^{\circ}, \phi_3=240^{\circ} \\\\ s_1 & =-3.3+j 18.23 \\\\ s_2 & =-6.6 \\\\ s_3 & =-3.3-j 18.23 \\ \end{aligned} $

Denominator of ,

$H(s)=(s+6.6)\left(s^2+6.6 s+343.2\right)$

Numerator of.

$H(s)=(6.6)(343.2)=2265.27$

Transfer function .

$H(s)=\frac{2265.27}{(s+6.6)\left(s^2+6.6 s+343.2\right)}$