Chebyshev II

• It has maximally flat responses in the passband and equi ripple in the stop band.

• In terms of same order, Chebyshev Response is found better than Butterworth response, so it could be more efficient if we have ripples at stopband and have transmission zeros as well which can be defined as Inverse Chebyshev.

fig: Chebyshev II response

The transfer function of Chebyshev low pass filter is given by

\(|T_n(j\omega)| = \frac{1}{\sqrt{1 + \varepsilon^2 C_n^2(\omega)}}\)

\(|T_n(j\omega)|^2 = \frac{1}{1 + \varepsilon^2 C_n^2(\omega)} \quad \text{(1)}\)

Inverse of the function can be obtained by subtracting Chebyshev Response from 1 and replacing by 1/w  (Inverting w).

\(\text{Inverse is: } 1 - |T_n(j\omega)|^2 = 1 - \frac{1}{1 + \varepsilon^2 C_n^2(\omega)}\)

\(= \frac{\varepsilon^2 C_n^2(\omega)}{1 + \varepsilon^2 C_n^2(\omega)}\)

\(\text{Replacing } \omega \text{ by } 1:\)

\(1 - |T_n(j\omega)|^2 = \frac{\varepsilon^2 C_n^2(1)}{1 + \varepsilon^2 C_n^2(1)}\)

\(= \frac{\varepsilon^2 C_n^2(1)}{\varepsilon^2 C_n^2(1)} \cdot \left(1 + \frac{1}{\varepsilon^2 C_n^2(1)}\right)\)

\(\Rightarrow |T_{ic,n}(j\omega)|^2 = \frac{1}{1 + \frac{1}{\varepsilon^2 C_n^2(1)}}\)

The magnitude squared Response of Inverse Chebyshev low pass filter can be represented as

\(\text{For } 0 < \omega < 1,\quad C_n(\omega) = \cos(n \cos^{-1} \omega)\)

\(\text{So } C_n(1) = \cos(n \cos^{-1} 1) = \cos(n \cdot 0) = 1\)

\(\Rightarrow |T_{ic,n}(j\omega)|^2 = \frac{\varepsilon^2}{1 + \varepsilon^2}\)

\(|T_{ic,n}(j\omega)| = \sqrt{\frac{\varepsilon^2}{1 + \varepsilon^2}} \quad \text{(2)}\)

The ripple in passband of Chebyshev ( \( \omega = 0 \) to \( \omega = 1 \) ) corresponds to \( \omega = 1 \) to \( \omega = \infty \) in Inverse Chebyshev. The maximum value of attenuation in passband for Chebyshev when \( C_n(\omega) = 1 \) corresponds to the minimum value of attenuation in stopband for Inverse Chebyshev response ( \( C_n(1) = 1 \) ), and can be termed as \( \alpha_{\min} \).

\(\text{Chebyshev passband ripple (} 0 \leq \omega \leq 1\text{)} \Rightarrow\) corresponds to inverse Chebyshev stopband ripple.

\(\text{Define minimum attenuation in stopband as: } \alpha_{\min} = -20 \log_{10}(|T(j\omega)|) \quad \text{(3)}\)

\(|T(j\omega)| = \sqrt{\frac{\varepsilon^2}{1 + \varepsilon^2}}\)

\(\Rightarrow \alpha_{\min} = -20 \log_{10} \left( \sqrt{\frac{\varepsilon^2}{1 + \varepsilon^2}} \right)\)

\(= -20 \cdot \frac{1}{2} \log_{10} \left( \frac{\varepsilon^2}{1 + \varepsilon^2} \right)\)

\(= -10 \log_{10} \left( \frac{\varepsilon^2}{1 + \varepsilon^2} \right)\)

\(= 10 \log_{10} \left( \frac{1 + \varepsilon^2}{\varepsilon^2} \right)\)

\(\Rightarrow \frac{\alpha_{\min}}{10} = \log_{10} \left( \frac{1 + \varepsilon^2}{\varepsilon^2} \right)\)

\(\Rightarrow 10^{\frac{\alpha_{\min}}{10}} = \frac{1 + \varepsilon^2}{\varepsilon^2}\)

\(\Rightarrow \frac{1}{\varepsilon^2} = 10^{\frac{\alpha_{\min}}{10}} - 1\)

\(\Rightarrow \varepsilon^2 = \frac{1}{10^{\frac{\alpha_{\min}}{10}} - 1}\)

\(\Rightarrow \varepsilon = \sqrt{ \frac{1}{10^{\frac{\alpha_{\min}}{10}} - 1} } \quad \text{(4)}\)

fig: Response Comparisions