Poles and Zeros

We know that,

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

\[
T(s)T(-s) = Z(s)Z(-s) \, P(s)P(-s)
\]

Where,

\[
Z(s)Z(-s) \;\; \text{for } s=j\omega = \varepsilon^2 C_n^2\!\left(\tfrac{1}{\omega}\right)
\]

And

\[
P(s)P(-s) \;\; \text{for } s=j\omega = 1 + \varepsilon^2 C_n^2\!\left(\tfrac{1}{\omega}\right)
\]

For Zero’s locations:

\[
\varepsilon^2 C_n^2\!\left(\tfrac{1}{\omega}\right) = 0
\]

Since \(\varepsilon \neq 0\), we can say

\[
C_n^2\!\left(\tfrac{1}{\omega}\right) = 0
\]

\[
\cos\left(n \cos^{-1}\!\left(\tfrac{1}{\omega_k}\right)\right) = 0
\]

\[
\cos\left(n \cos^{-1}\!\left(\tfrac{1}{\omega_k}\right)\right) = \cos\!\left(\tfrac{k\pi}{2}\right), \quad k = 1,3,5,\dots \;\text{(odd values)}
\]

\[
n \cos^{-1}\!\left(\tfrac{1}{\omega_k}\right) = \tfrac{k\pi}{2}
\]

\[
\tfrac{1}{\omega_k} = \cos\!\left(\tfrac{k\pi}{2n}\right)
\]

\[
\omega_k = \sec\!\left(\tfrac{k\pi}{2n}\right), \quad k = 1,3,5,\dots
\]

Similarly for poles:

\[
1 + \varepsilon^2 C_n^2\!\left(\tfrac{1}{\omega}\right) = 0
\]

\[
\varepsilon^2 C_n^2\!\left(\tfrac{1}{\omega}\right) = -1
\]

The poles location of Inverse Chebyshev is similar to Chebyshev. Simply \(\omega_k\) is replaced by \(\tfrac{1}{\omega_k}\).

If poles in Chebyshev are \(p_i\),

Then poles in Inverse Chebyshev are \(\tfrac{1}{p_i}\).