Chebyshev and Inverse Chebyshev Relations
Obtain 4th order low pass Chebyshev filter with αmax =0.75db and relate to Inverse Chebyshev response.
From Previous Relation
Given
\[
n = 4, \quad \alpha_{\max} = 0.75 \, \text{dB}
\]
\[
\varepsilon = \sqrt{10^{\frac{\alpha_{\max}}{10}} - 1}
\]
\[
\varepsilon = \sqrt{10^{\frac{0.7}{10}} - 1} = 0.434
\]
\[
\omega_{hp} = \cosh\!\left(\tfrac{1}{n} \cosh^{-1}\!\left(\tfrac{1}{\varepsilon}\right)\right)
\]
\[
\omega_{hp} = \cosh\!\left(\tfrac{1}{4}\cosh^{-1}\!\left(\tfrac{1}{0.434}\right)\right) = 1.0689 \; \text{rad/sec}
\]
\[
u_k = \tfrac{1}{n} \cdot \tfrac{(2k+1)\pi}{2}, \quad k=1,2,\dots
\]
\[
u_0 = \tfrac{1}{4}\cdot \tfrac{(2\cdot0+1)\pi}{2} = \tfrac{\pi}{8}, \quad
u_1 = \tfrac{3\pi}{8}, \quad u_2 = \tfrac{5\pi}{8}, \quad
u_3 = \tfrac{7\pi}{8}, \quad u_4 = \tfrac{9\pi}{8}, \quad
u_5 = \tfrac{11\pi}{8}, \quad u_6 = \tfrac{13\pi}{8}, \quad u_7 = \tfrac{15\pi}{8}
\]
\[
v = \pm \tfrac{1}{n} \sinh^{-1}\!\left(\tfrac{1}{\varepsilon}\right)
\]
\[
v = \pm \tfrac{1}{4} \sinh^{-1}\!\left(\tfrac{1}{0.434}\right) = 0.393
\]
The pole locations are given by
\[
S_k = \sigma(k) \pm j \omega(k)
\]
Where,
\[
\sigma(k) = \sin\!\left(\tfrac{(2k+1)\pi}{2n}\right)\sinh(v),
\quad \omega(k) = \cos\!\left(\tfrac{(2k+1)\pi}{2n}\right)\cosh(v)
\]
for \(k = 0 \; \text{to} \; (2n+1)\)
\[
S_0 = \sin\!\left(\tfrac{\pi}{8}\right)\sinh(0.393) + j \cos\!\left(\tfrac{\pi}{8}\right)\cosh(0.393) = 0.154 + j0.996
\]
\[
S_1 = 0.3725 + j0.4126, \quad S_2 = 0.375 - j0.4126, \quad S_3 = 0.1547 - j0.996
\]
\[
S_4 = 0.154 - j0.413, \quad S_5 = -0.373 - j0.413, \quad S_6 = -0.373 + j0.413, \quad S_7 = -0.154 + j0.996
\]
We only take values on the negative half of the s-plane, the poles are
\[
S_4, \; S_5, \; S_6, \; S_7
\]
The Transfer Functions
\[
T(s) = \frac{1}{(s-s_4)(s-s_5)(s-s_6)(s-s_7)}
\]
\[
T(s) = \frac{1}{(s+0.154+j0.996)(s+0.373+j0.413)(s+0.154-j0.996)(s+0.373-j0.413)}
\]
\[
T(s) = \frac{1}{(s^2+0.308s+1.0157)(s^2+0.7416s+0.3096)}
\]
In terms of Inverse Chebyshev response.
For \(n = \text{odd}, \; T(0) = 1\)
For \(n = \text{even}, \; \)
\[
T(0) = \frac{1}{\sqrt{1 + \varepsilon^2}}
= \frac{1}{\sqrt{1 + (0.431)^2}} = \frac{1}{\sqrt{1.186}} \approx 0.92
\]
Here, General tansfer equation is:
\[
T(s) = \frac{k}{(s^2 + 0.308s + 1.0157)(s^2 + 0.7416s + 0.3096)}
\]
\[
\text{Set } s = 0:
\quad
T(0) = \frac{k}{(0 + 0.308 \cdot 0 + 1.0157)(0 + 0.7416 \cdot 0 + 0.3096)}
\]
Now solve for constant \( k \) in:
\[
\text{When } s = 0:
\quad
T(0) = \frac{k}{(1.0157)(0.3096)} = \frac{k}{0.3144}
\]
and on comparing,
\[
\frac{k}{(1.0157)(0.3096)} = \frac{1}{\sqrt{1 + \varepsilon^2}} = \frac{1}{\sqrt{1 + (0.431)^2}} = \frac{1}{\sqrt{1.186}} = 0.92
\]
\[
\Rightarrow k = 0.92 \cdot (1.0157)(0.3096) = 0.92 \cdot 0.3144 = 0.289
\]
So the final transfer function is:
\[
T_4(s) = 0.293 \cdot \frac{1}{(s^2 + 0.308s + 1.0157)(s^2 + 0.7416s + 0.3096)}
\]