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**Solution:**

The analog Chebyshev filter is designed by approximating the ideal frequency response using an error function.

The approximation function is selected such that the error is minimized over a prescribed band of frequencies.

There are two types of Chebyshev approximation. In type-1 approximation, the error function is selected such that, the magnitude response is equiripple in the passband and monotonic in the stopband.

In type-2 approximation the error function is selected such that, the magnitude response is monotonic in the passband and equiripple in the stopband.

The type-2 magnitude response is also called the inverse Chebyshev response. The type-1 design is presented in this book.

The magnitude response of the Type-1 lowpass filter is given by,

$ \left|H_a(\Omega)\right|^2=\frac{1}{1+\epsilon^2 C_N^2\left(\frac{\Omega}{\Omega_c}\right)}\\ $

where $\in$ is attenuation constant and $\mathrm{C}_{\mathrm{N}}\left(\Omega / \Omega_c\right)$ is the Chebyshev polynomial of the first kind of degree $\mathrm{N}$.

The attenuation constant, $\in=\left[\frac{1}{A_1^2}-1\right]^{\frac{1}{2}}$

where $A_1$ is the gain or magnitude at passband edge frequency $\Omega_1$ For small values of $\mathrm{N}$ the Chebyshev polynomial is given by

$ C_N(x)= \begin{cases}\cos \left(N \cos ^{-1} x\right) & ; \text { for }|x| \leq 1 \\ \cosh \left(N \cosh ^{-1} x\right) & ; \text { for }|x|\gt1\end{cases}\\ $

For large values of $\mathrm{N}$ the Chebyshev polynomial is given by the recurrence relation,

$ C_N(x)=2 C_{N-1}(x)-C_{N-2}(x)\\ $

with initial values $\mathrm{C}_0(\mathrm{x})=1$ and $\mathrm{C}_1(\mathrm{x})=\mathrm{x}$

The transfer function of the analog system can be obtained from equation (4.53) by substituting $\Omega$ by $s / j$.

$ \therefore \mathrm{H}_{\mathrm{a}}(\mathrm{s}) \mathrm{H}(-\mathrm{s})=\frac{1}{1+\epsilon^2 \mathrm{C}_{\mathrm{N}}^2\left(\frac{\mathrm{s} / \mathrm{j}}{\Omega_{\mathrm{c}}}\right)}\\ $

For the normalized transfer function, let us replace $s / \Omega_c$ with $s_n$.

$ \therefore H_a\left(s_n\right) H\left(-s_n\right)=\frac{1}{1+\epsilon^2 C_N^2\left(-j s_n\right)}\\ $

For the transfer function of the equation, we can determine $2 \mathrm{~N}$ poles which are given by the roots of the denominator polynomial.

It can be shown that the poles of the transfer function symmetrically lie on an ellipse in the s-plane.