Chebyshev pole’s location and Network function
The Transfer function is given by
\[
|T_n(j \omega)|^2 = \frac{1}{1 + \varepsilon^2 C_n^2(\omega)}
\]
In s-domain, putting \( s = j \omega \) or \( \omega = s j \), substituting the values of \(\omega\), we get
\[
|T_n(s)|^2 = \frac{1}{1 + \varepsilon^2 C_n^2(s j)}
\]
The pole locations are given by
\[
1 + \varepsilon^2 C_n^2(s j) = 0
\]
\[
\varepsilon^2 C_n^2(s j) = -1
\]
\[
C_n^2(s j) = -\frac{1}{\varepsilon^2}
\]
\[
C_n(s j) = \pm j \frac{1}{\varepsilon}
\]
\[
C_n(s j) = 0 \pm j \frac{1}{\varepsilon}
\]
where,
\[
C_n(\omega) = \cos^n \left( \cos^{-1}(\omega) \right) \quad \text{for } \omega \leq 1
\]
For
\[
C_n(s j) = \cos^n \left( \cos^{-1}(s j) \right)
\]
Let
\[
C_n(s j) = x = u + j v \quad \dots \text{(i)}
\]
Then,
\[
C_n(s j) = \cos^n(u + j v)
\]
\[
C_n(s j) = \cos(n u) \cos(n j v) - \sin(n u) \sin(n j v)
\]
\[
C_n(s j) = \cos(n u) \cosh(n v) - \sin(n u) \, j \sinh(n v)
\]
\[
C_n(s j) = \cos(n u) \cosh(n v) - j \sin(n u) \sinh(n v) \quad \dots \text{(ii)}
\]
Comparing (i) and (ii), we get
\[
\cos(n u) \cosh(n v) = 0 \quad \dots \text{(iii)}
\]
\[
j \sin(n u) \sinh(n v) = \pm j \frac{1}{\varepsilon} \quad \dots \text{(iv)}
\]
From equation (iii),
\[
\cos(n u) \cosh(n v) = 0
\]
Since minimum value of \(\cosh(n v) = 1\), \(\cosh(n v) \neq 0\), so
\[
\cos(n u) = 0
\]
\[
n u = \cos^{-1}(0)
\]
\[
u_k = \frac{1}{n} \left( \frac{2k + 1}{2} \pi \right) \quad \text{for } k = 0, 1, 2, \ldots
\]
Similarly, from equation (iv), we have
\[
\sin(n u) \sinh(n v) = \frac{1}{\varepsilon}
\]
But from equation (iv),
\[
\sin(n u) = \sin \left( n \cdot \frac{1}{n} \cdot \frac{2k + 1}{2} \pi \right) = \sin \left( \frac{2k + 1}{2} \pi \right) = \pm 1
\]
We can say
\[
\sinh(n v) = \frac{1}{\varepsilon}
\]
\[
n v = \sinh^{-1} \left( \frac{1}{\varepsilon} \right)
\]
\[
v = \pm \frac{1}{n} \sinh^{-1} \left( \frac{1}{\varepsilon} \right) \quad \dots \text{(vi)}
\]
From (i) and (ii), the value of \(s\) can be evaluated as:
\[
\cos^{-1}(n) (s j) = x = u + j v
\]
\[
s = j \cdot \cos(u + j v)
\]
In general,
\[
s_k = j \left[ \cos(u_k) \cos(j v) - j \sin(u_k) \sin(j v) \right]
\]
\[
s_k = j \left[ \cos(u_k) \cosh(v) - j \sin(u_k) \sinh(v) \right]
\]
\[
s_k = j \cos(u_k) \cosh(v) - j^2 \sin(u_k) \sinh(v)
\]
\[
s_k = \sin(u_k) \sinh(v) + j \cos(u_k) \cosh(v) \quad \text{for } k = 0, 1, \ldots
\]
From equations (v) and (iv), we get
\[
s_k = \sigma_k + j \omega_k
\]
where
\[
\sigma_k = \sin \left( \frac{2k + 1}{2} \pi \right) \sinh(v)
\]
and
\[
\omega_k = \cos \left( \frac{2k + 1}{2} \pi \right) \cosh(v)
\]
Squaring both terms, we have
\[
\sigma_k^2 = \left( \sin \left( \frac{2k + 1}{2} \pi \right) \sinh(v) \right)^2
\]
\[
\left( \sigma_k \sin \left( \frac{2k + 1}{2} \pi \right) \right)^2 = (\sinh(v))^2 \quad \dots A
\]
\[
\omega_k^2 = \left( \cos \left( \frac{2k + 1}{2} \pi \right) \cosh(v) \right)^2
\]
\[
\left( \omega_k \cos \left( \frac{2k + 1}{2} \pi \right) \right)^2 = (\cosh(v))^2 \quad \dots B
\]
Adding A and B:
\[
\left( \sigma_k \sinh(v) \right)^2 + \left( \omega_k \cosh(v) \right)^2 = \left( \frac{2k + 1}{2} \pi \right)^2 + \left( \frac{2k + 1}{2} \pi \right)^2
\]
\[
\left( \sigma_k \sinh(v) \right)^2 + \left( \omega_k \cosh(v) \right)^2 = 1
\]
which is the general equation of an ellipse.
Therefore, we can say that the poles of the Chebyshev filter lie on an ellipse.
fig:Pole location of Chebyshev Response
Chebyshev Poles:
\[
\sigma_k \pm j \omega_k = s_k = \sin\left(\frac{(2k+1)\pi}{2n}\right) \sinh(v) + j \cos\left(\frac{(2k+1)\pi}{2n}\right) \cosh(v)
\quad \text{for } k = 0, 1, \ldots, 2n + 1
\]
where
\[
v = \pm \frac{1}{n} \sinh^{-1} \left( \frac{1}{\varepsilon} \right)
\]
---
k=0:
\[
s_0 = \sin\left(\frac{1 \cdot \pi}{2n}\right) \sinh(v) + j \cos\left(\frac{1 \cdot \pi}{2n}\right) \cosh(v)
\]
---
k=1:
\[
s_1 = \sin\left(\frac{3 \pi}{2n}\right) \sinh(v) + j \cos\left(\frac{3 \pi}{2n}\right) \cosh(v)
\]
---
k=2:
\[
s_2 = \sin\left(\frac{5 \pi}{2n}\right) \sinh(v) + j \cos\left(\frac{5 \pi}{2n}\right) \cosh(v)
\]