Numerical 1

Obtain 4th order network function of a low pass Chebyshev filter with αmax = 0.75db. Find half power point frequency.

$
\begin{aligned}
n &= 4 \\
\alpha_{\text{max}} &= 0.75 \, \text{dB} \\
\varepsilon &= \sqrt{10^{\frac{\alpha_{\text{max}}}{10}} - 1} 
= \sqrt{10^{0.075} - 1} 
= \sqrt{1.186 - 1} 
= \sqrt{0.186} 
\approx 0.434
\end{aligned}
$

The High-Pass Normalized Cutoff Frequency is given by:

$
\omega_p = \cosh\left( \frac{1}{n} \cosh^{-1}\left( \frac{1}{\varepsilon} \right) \right)
= \cosh\left( \frac{1}{4} \cosh^{-1}\left( \frac{1}{0.434} \right) \right)
\approx \cosh(0.582) \approx 1.0689 \, \text{rad/sec}
$

### \( u_k \) and \( v \):

$
u_k = \frac{(2k+1)\pi}{2n}, \quad \text{for } k = 0,1,\dots,7
$

$
\begin{aligned}
u_0 &= \frac{1\pi}{8}, \quad u_1 = \frac{3\pi}{8}, \quad u_2 = \frac{5\pi}{8}, \quad u_3 = \frac{7\pi}{8} \\
u_4 &= \frac{9\pi}{8}, \quad u_5 = \frac{11\pi}{8}, \quad u_6 = \frac{13\pi}{8}, \quad u_7 = \frac{15\pi}{8}
\end{aligned}
$

$
v = \pm \frac{1}{n} \sinh^{-1} \left( \frac{1}{\varepsilon} \right) 
= \pm \frac{1}{4} \sinh^{-1} \left( \frac{1}{0.434} \right) 
\approx \pm \frac{1}{4} \cdot 1.571 = \pm 0.393
$

The pole locations are given by

$
\begin{aligned}
S_0 &= 0.154 + j 0.996 \\
S_1 &= 0.3725 + j 0.4126 \\
S_2 &= 0.375 - j 0.4126 \\
S_3 &= 0.1547 - j 0.996 \\
S_4 &= -0.154 - j 0.413 \\
S_5 &= -0.373 - j 0.413 \\
S_6 &= -0.373 + j 0.413 \\
S_7 &= -0.154 + j 0.996
\end{aligned}
$

Only left-half plane poles are used:

$
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 S4, S5, S6, S7

The Transfer Functions

$
\begin{aligned}
T(s) &= \frac{1}{(s - S_4)(s - S_5)(s - S_6)(s - S_7)} \\
&= \frac{1}{(s + 0.154 + j0.996)(s + 0.373 + j0.413)(s + 0.154 - j0.996)(s + 0.373 - j0.413)} \\
&= \frac{1}{(s^2 + 0.308s + 1.0157)(s^2 + 0.7416s + 0.3096)}
\end{aligned}
$