Cauer or Elliptic Response

• Butterworth: Flat at Passband
• Chebyshev: Equiripple at Passband and Flatter at Stopband
• Inverse Chebyshev: Equiripple at Stopband and Flatter at Passband

For same order of ‘n’, Chebyshev produces better response than Butterworth and has small ‘n’ than Butterworth at the cost of some ripples at Passband. Inverse Chebyshev produces a new approach by introducing zero of transfer function beyond s thus able to produce steeper response than Butterworth and Chebyshev on transition band. Elliptic Filter introduce a new response having equal ripple in both the passband and stopband. For the given Specification, the order will be less than that of Butterworth and Chebyshev such that no other filter achieves faster transition between passband and stopband for same order of ‘n’.

fig: Cauer Gain Response

fig: Cauer Attenuation Response

\[
R_n(s) = P(s) Q(s) \quad \text{(Quotient of Polynomials)}
\]

\[
|R_n(s)|^2 = P_s^2 Q_s^2 = P_s P(-s) Q_s Q(-s) = P(j\omega) P(-j\omega) Q(j\omega) Q(-j\omega)
\]

\[
\text{(i.e., separating odd and even terms)}
\]

\text{Where, for odd } n = 2k+1:
\[
R_n(s) = 1^2 - 2\cdot 2^2 - \dots \cdot (\omega_n^2 - 2) 1 - \omega_1^2 \cdot 2\cdot 1 - \omega_2^2 \cdot 2 \cdot \dots \cdot 1 - \omega_n^2 \cdot 2
\]

\text{Similarly, for even } n = 2k:
\[
R_n(s) = 1^2 - 2\cdot 2^2 - \dots \cdot (\omega_n^2 - 2) 1 - \omega_1^2 \cdot 2\cdot 1 - \omega_2^2 \cdot 2 \cdot \dots \cdot 1 - \omega_n^2 \cdot 2
\]

\[
\text{Then the transfer function is given by:}
\]

\[
|T_n(j\omega)|^2 = Q(j\omega) Q(-j\omega) + \varepsilon^2 \, [P(j\omega) P(-j\omega)]
\]

\[
|T_n(j\omega)|^2 = \frac{1}{1 + \varepsilon^2 R_n^2(\omega)} \quad \text{or} \quad \frac{1}{1 + \varepsilon^2 R_n^2(\omega, L)}
\]

\[
\text{where } R_n^2(\omega) \text{ is the Chebyshev Rational Function, and } L \text{ determines the ripple factor.}
\]

\[
\text{The attenuation is given by:}
\]

\[
\alpha = -20 \log_{10} |T(j\omega)|
\]

\[
\alpha = -20 \log_{10} \left( \frac{1}{\sqrt{1 + \varepsilon^2 R_n^2(\omega)}} \right)
\]

\[
\alpha = -20 \cdot (-0.5) \log_{10} \left( 1 + \varepsilon^2 R_n^2(\omega) \right)
\]

\[
\alpha = 10 \log_{10} \left( 1 + \varepsilon^2 R_n^2(\omega) \right)
\]

\[
\text{At the end of the passband, } \omega_p = 1 \text{ rad/sec, then } R_n(1) = 1
\]

\[
\alpha_p = \alpha_{\max} = 10 \log_{10} (1 + \varepsilon^2)
\]

\[
\text{Similarly, at the stopband:}
\]

\[
\alpha_s = \alpha_{\min} = 10 \log_{10} (1 + \varepsilon^2 L^2)
\]

The Poles and Zeros response is given by

fig: Poles of Cauer