Determine the lowest order of Chebyshev filter that meets the following specifications:

(i) $1 \mathrm{~dB}$ ripple in the passband $0 \leq|\omega| \leq 0.3 \pi$

(ii) Atleast $60 \mathrm{~dB}$ attenuation in the stopband $0.35 \pi \leq|\omega| \leq \pi$ Use the bilinear transformation.

Solution:

Step 1 Bilinear transformation is to be used.

Step 2 Attenuation constant

$$ \varepsilon=\left[\frac{1}{A_1^2}-1\right]^{-\frac{1}{2}}=\left[\frac{1}{(0.891)^2}-1\right]^{-\frac{1}{2}}=0.509 $$

Step 3 Ratio of analog edge frequencies

$$ \frac{\Omega_2}{\Omega_1}=\frac{\frac{2}{T} \tan \frac{\omega_2}{2}}{\frac{2}{T} \tan \frac{\omega_1}{2}}=\frac{\tan \frac{0.35 \pi}{2}}{\tan \frac{0.3 \pi}{2}}=1.2 $$

Step 4 Order of the filter

$$ \begin{aligned} & N \geq \frac{\cosh ^{-1}\left[\frac{1}{\varepsilon}\left(\frac{1}{A_2^2}-1\right)^{\frac{1}{2}}\right]}{\cosh ^{-1}\left(\frac{\Omega_2}{\Omega_1}\right)} \geq \frac{\cosh …