Determine the order and the poles of a type I lowpass Chebyshev filter that has a $1 \mathrm{~dB}$ ripple in the passband and passband frequency $\Omega_p=1000 \pi$, a stopband frequency of $2000 \pi$ and an attenuation of $40 \mathrm{~dB}$ or more.

Solution:

Given data $\alpha_p=1 \mathrm{~dB} ; \Omega_p=1000 \pi \mathrm{rad} / \mathrm{sec} ; \alpha_s=40 \mathrm{~dB}$ $\Omega_s=2000 \pi \mathrm{rad} / \mathrm{sec}$

$ N \geq \frac{\cos h^{-1} \sqrt{\frac{10^{0.1 \alpha_s}-1}{10^{0.1 \alpha_p}-1}}}{\cos h^{-1} \frac{\Omega_s}{\Omega_p}} \geq \frac{\cos h^{-1} \sqrt{\frac{10^4-1}{10^{0.1}-1}}}{\cos h^{-1} \frac{2000 \pi}{1000 \pi}}=4.536 $

$ \begin{aligned} & then N=5 \\ & \varepsilon=\sqrt{10^{0.1 \alpha_p}-1}=0.508 ; \quad \mu=\varepsilon^{-1}+\sqrt{1+\varepsilon^{-2}}=4.17 \\\\ & a=\Omega_p\left[\frac{\mu^{1 / N}-\mu^{-1 / N}}{2}\right]=289.5 \pi ; \quad b=\Omega_p\left[\frac{\mu^{1 / N}+\mu^{-1 / N}}{2}\right]=1041 \pi \\\\ & \phi_k=\frac{\pi}{2}+\frac{(2 k-1) \pi}{2 N} \quad k=1,2, \ldots 5 \\\\ & \phi_1=180^{\circ} ; \phi_2=144^{\circ} ; \phi_3=180^{\circ} ; \phi_4=216^{\circ} ; \phi_5=252^{\circ} \\\\ & s_k=a \cos \phi_k+j b \sin \phi_k \quad k=1,2, \ldots 5 \\\\ & s_1=-89.5 \pi+j 989 \pi ; \quad s_2=-234.2 \pi+j 612 \pi ; \quad s_3=-289.5 \pi \\\\ & s_4=-234.2 \pi-j 612 \pi ; \quad s_5=-89.5 \pi-j 989 \pi \\\\ & \end{aligned} $