Determine the system function $H(z)$ of the lowest order Chebyshev IIR digital filter with the following specifications: $ \begin{aligned} & 3 \mathrm{~dB} \text { ripple in passband } 0 \leq \omega \leq 0.2 \pi \\ & 25 \mathrm{~dB} \text { attenuation in stopband } 0.45 \pi \leq \omega \leq \pi \end{aligned} $

Solution:

Let T = 1 and bilinear transformation is used

$ \varepsilon=\left[\frac{1}{A_1^2}-1\right]^{\frac{1}{2}}=\left[\frac{1}{0.707^2}-1\right]=1\\ $

The ratio of analog 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{0.45 \pi}{2}}{\tan \frac{0.2 \pi}{2}}=2.628\\ $

$ 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\}}\\ $

$ \begin{aligned} & \geq \frac{\cosh ^{-1}\left\{\frac{1}{1}\left[\frac{1}{0.0562^2}-1\right]^{\frac{1}{2}}\right\}}{\cosh ^{-1}(2.628]} \\ & \geq \frac{3.569}{1.621} \geq 2.20 \approx …