Numerical 2
Passband extending from w = 0 rad/s to w=1000 rad/s, the attenuation shouldn’t exceed 0.25db
Stopband extending from w = 2500rad/s to w = infinity, the attenuation shouldn’t be less than 40db.
From Question:
$
\begin{aligned}
\textbf{Given:} \quad
\omega_p &= 1000 \, \text{rad/s}, \quad
\omega_s = 2500 \, \text{rad/s} \\
\alpha_{\text{max}} &= 0.25 \, \text{dB}, \quad
\alpha_{\text{min}} = 40 \, \text{dB}
\end{aligned}
$
Here,
$
n = \frac{ \cosh^{-1} \left( \sqrt{ \frac{10^{\frac{40}{10}} - 1}{10^{\frac{0.25}{10}} - 1} } \right) }
{ \cosh^{-1} \left( \frac{\omega_s}{\omega_p} \right) }
= \frac{ \cosh^{-1} \left( \sqrt{ \frac{10000 - 1}{1.059 - 1} } \right) }
{ \cosh^{-1} (2.5) }
\approx \frac{ \cosh^{-1}(464.2) }{ \cosh^{-1}(2.5) }
\approx \frac{6.14}{1.448} \approx 4.238 \Rightarrow n \approx 5
$
The Ripple Factor is given by:
$
\varepsilon = \sqrt{10^{\frac{0.25}{10}} - 1} = \sqrt{1.059 - 1} = \sqrt{0.059} \approx 0.2434
$
$
v = \pm \frac{1}{n} \sinh^{-1}\left( \frac{1}{\varepsilon} \right)
= \pm \frac{1}{5} \sinh^{-1}(4.108)
\approx \pm \frac{1}{5} \cdot 2.1206 \approx \pm 0.4241
$
Angle Components \( u_k \):
$
u_k = \frac{(2k+1)\pi}{2n}, \quad \text{for } k = 0,1,\dots,9
$
$
\begin{aligned}
u_0 &= \frac{\pi}{10}, \quad u_1 = \frac{3\pi}{10}, \quad u_2 = \frac{5\pi}{10}, \\
u_3 &= \frac{7\pi}{10}, \quad u_4 = \frac{9\pi}{10}, \quad u_5 = \frac{11\pi}{10}, \\
u_6 &= \frac{13\pi}{10}, \quad u_7 = \frac{15\pi}{10}, \quad u_8 = \frac{17\pi}{10}, \quad u_9 = \frac{19\pi}{10}
\end{aligned}
$
The Poles locations are given by:
$
\begin{aligned}
S_0 &= \phantom{-}0.135 + j\,1.0375 \\
S_1 &= \phantom{-}0.353 + j\,0.6414 \\
S_2 &= \phantom{-}0.4369 \\
S_3 &= \phantom{-}0.353 - j\,0.6414 \\
S_4 &= \phantom{-}0.135 - j\,1.0375 \\
S_5 &= -0.135 - j\,1.0375 \\
S_6 &= -0.353 - j\,0.6414 \\
S_7 &= -0.4369 \\
S_8 &= -0.353 + j\,0.6414 \\
S_9 &= -0.135 + j\,0.378
\end{aligned}
$
Selected Poles (Left Half s-Plane):
$
S_5 = -0.135 - j\,1.0375, \quad
S_6 = -0.353 - j\,0.6414, \quad
S_7 = -0.4369, \quad
S_8 = -0.353 + j\,0.6414, \quad
S_9 = -0.135 + j\,0.378
$
We only take values on the negative half of the s-plane, the poles are S5, S6, S7, S8, S9
The Transfer Functions
$
\begin{aligned}
T(s) &= \frac{1}
{(s - S_5)(s - S_6)(s - S_7)(s - S_8)(s - S_9)} \\
&= \frac{1}
{(s + 0.135 + j\,1.037)(s + 0.353 + j\,0.6414)(s + 0.4369)(s + 0.353 - j\,0.6414)(s + 0.135 - j\,0.378)}
\end{aligned}
$