Attenuation in Chebyshev Filter
The attenuation in a Chebyshev filter is given by:
\[
\alpha = -20 \log\left( |T(j\omega)| \right)
\]
We know,
\[
|T_n(j\omega)|^2 = \frac{1}{1 + \varepsilon^2 C_n^2(\omega)}
\]
\[
|T_n(j\omega)| = \frac{1}{\sqrt{1 + \varepsilon^2 C_n^2(\omega)}}
\]
Then,
\[
\alpha = -20 \log\left( \frac{1}{\sqrt{1 + \varepsilon^2 C_n^2(\omega)}} \right)
\]
\[
\alpha = -20 \log\left( (1 + \varepsilon^2 C_n^2(\omega))^{-1/2} \right)
\]
\[
\alpha = -20 \cdot \left( -\frac{1}{2} \log(1 + \varepsilon^2 C_n^2(\omega)) \right)
\]
\[
\alpha = 10 \log(1 + \varepsilon^2 C_n^2(\omega))
\]
For \(0 < \omega < 1\), the maximum attenuation in the passband occurs when \(C_n(\omega) = 1\). Thus,
\[
\alpha_{\text{max}} = 10 \log(1 + \varepsilon^2)
\]
\[
\frac{\alpha_{\text{max}}}{10} = \log(1 + \varepsilon^2)
\]
\[
10^{\frac{\alpha_{\text{max}}}{10}} = 1 + \varepsilon^2
\]
\[
\varepsilon^2 = 10^{\frac{\alpha_{\text{max}}}{10}} - 1
\]
\[
\varepsilon = \sqrt{10^{\frac{\alpha_{\text{max}}}{10}} - 1} \qquad \text{(1)}
\]
If \(\varepsilon^2 C_n^2(\omega) = 1\), then \(\alpha = 3 \, \text{dB}\), which defines the half-power point.
\[
|T_n(j\omega)| = \frac{1}{\sqrt{1 + \varepsilon^2 C_n^2(\omega)}} = \frac{1}{\sqrt{2}} \approx 0.707
\]
\[
C_n^2(\omega) = \frac{1}{\varepsilon^2} \quad \Rightarrow \quad C_n(\omega) = \frac{1}{\varepsilon}
\]
Since \(\varepsilon < 1\), it follows that \(C_n(\omega) > 1\).
\[
C_n(\omega) = \cosh\left(n \cosh^{-1}(\omega_{\text{hp}})\right), \quad \omega_{\text{hp}} > 1
\]
\[
\cosh\left(n \cosh^{-1}(\omega_{\text{hp}})\right) = \frac{1}{\varepsilon}
\]
\[
n \cosh^{-1}(\omega_{\text{hp}}) = \cosh^{-1}\left(\frac{1}{\varepsilon}\right)
\]
\[
\cosh^{-1}(\omega_{\text{hp}}) = \frac{1}{n} \cosh^{-1}\left(\frac{1}{\varepsilon}\right)
\]
\[
\omega_{\text{hp}} = \cosh\left( \frac{1}{n} \cosh^{-1}\left( \frac{1}{\varepsilon} \right) \right)
\]
From Equation (1):
\[
\omega_{\text{hp}} = \cosh\left( \frac{1}{n} \cosh^{-1}\left( \frac{1}{\sqrt{10^{\frac{\alpha_{\text{max}}}{10}} - 1}} \right) \right)
\]