Half Power Point
We know for Chebyshev Response
\[
C_n(\omega) = \cos\left(n \cos^{-1}(\omega)\right), \quad 0 < \omega < 1
\]
For Inverse Chebyshev
\[
C_n\left(\tfrac{1}{\omega}\right) = \cos\left(n \cos^{-1}\!\left(\tfrac{1}{\omega}\right)\right), \quad 0 < \tfrac{1}{\omega} < 1
\]
For,
\[
\cos\left(n \cos^{-1}\!\left(\tfrac{1}{\omega_k}\right)\right) = 0
\]
\[
\cos\left(n \cos^{-1}\!\left(\tfrac{1}{\omega_k}\right)\right) = \cos\left(\tfrac{k\pi}{2}\right) \quad \text{where } k \text{ is odd } (1,3,5,\dots)
\]
and for
\[
\cos\left(n \cos^{-1}\!\left(\tfrac{1}{\omega_k}\right)\right) = 1
\]
\[
\cos\left(n \cos^{-1}\!\left(\tfrac{1}{\omega_k}\right)\right) = \cos\left(\tfrac{k\pi}{2}\right) \quad \text{where } k \text{ is even } (2,4,6,\dots)
\]
The frequency for maxima and minima can be given as
\[
\cos^{-1}\!\left(\tfrac{1}{\omega_k}\right) = \tfrac{k\pi}{2}
\]
For odd values of \(k\),
\[
C_n\!\left(\tfrac{1}{\omega_k}\right) = 0, \quad |T_n(j\omega)|^2 = \frac{\varepsilon^2 C_n^2}{1+\varepsilon^2 C_n^2} = 0
\]
and for even values of \(k\),
\[
C_n\!\left(\tfrac{1}{\omega_k}\right) = 0, \quad |T_n(j\omega)|^2 = \frac{\varepsilon^2 C_n^2}{1+\varepsilon^2 C_n^2} = \frac{\varepsilon^2}{1+\varepsilon^2}
\]
In terms of attenuations:
\[
\alpha = \infty \quad \text{for } k = \text{odd}, \quad
\alpha = \alpha_{\min} \quad \text{for } k = \text{even and } \omega = 1
\]
---
Half Power Frequency
The point when \(|T_n(j\omega)| = \tfrac{1}{\sqrt{2}}\) is known as half power point and the frequency is known as half power frequency.
At Half power point:
\[
|T_n(j\omega)| = \tfrac{1}{\sqrt{2}}
\]
The transfer function is given as
\[
|T_{ic,n}(j\omega)|^2 = \frac{1}{1 + \tfrac{1}{\varepsilon^2 C_n^2(1/\omega)}}
\]
For Half power point:
\[
\varepsilon^2 C_n^2\!\left(\tfrac{1}{\omega}\right) = 1
\]
\[
C_n^2\!\left(\tfrac{1}{\omega}\right) = \tfrac{1}{\varepsilon^2}
\]
\[
C_n\!\left(\tfrac{1}{\omega}\right) = \pm \tfrac{1}{\varepsilon}
\]
Since \(\varepsilon < 1\), we can say \(\tfrac{1}{\omega} > 1\) such that
\[
\cosh\left(n \cosh^{-1}\!\left(\tfrac{1}{\omega_{hp}}\right)\right) = \tfrac{1}{\varepsilon}
\]
\[
n \cosh^{-1}\!\left(\tfrac{1}{\omega_{hp}}\right) = \cosh^{-1}\!\left(\tfrac{1}{\varepsilon}\right)
\]
\[
\cosh^{-1}\!\left(\tfrac{1}{\omega_{hp}}\right) = \tfrac{1}{n} \cosh^{-1}\!\left(\tfrac{1}{\varepsilon}\right)
\]
\[
\tfrac{1}{\omega_{hp}} = \cosh\!\left(\tfrac{1}{n} \cosh^{-1}\!\left(\tfrac{1}{\varepsilon}\right)\right)
\]
\[
\omega_{hp} = \frac{1}{\cosh\!\left(\tfrac{1}{n} \cosh^{-1}\!\left(\tfrac{1}{\varepsilon}\right)\right)}
\]
Which gives the relation for Half power Point.