Numerical 3

Passband extending from f = 0 Hertz to f= 3.2 Kilo Hertz, the attenuation shouldn’t exceed 0.4db

Stopband extending from f= 9.8 Kilo Hertz to f = infinity, the attenuation shouldn’t be less than 52db.

\[
\begin{aligned}
& \text{Given specifications:} \\
& f_p = 3.2\, \text{kHz}, \quad f_s = 9.8\, \text{kHz} \\
& \alpha_{\max} = 0.4\, \text{dB}, \quad \alpha_{\min} = 52\, \text{dB} \\
\\
& \text{Convert to angular frequencies:} \\
& \omega_p = 2 \pi f_p = 2 \pi \times 3200 = 20106.2\, \text{rad/s} \\
& \omega_s = 2 \pi f_s = 2 \pi \times 9800 = 61575.6\, \text{rad/s} \\
\\
& \text{Ripple factor } \varepsilon: \\
& \varepsilon = \sqrt{10^{\frac{\alpha_{\max}}{10}} - 1} = \sqrt{10^{0.04} - 1} = \sqrt{1.096 - 1} = \sqrt{0.096} \approx 0.3098 \\
\\
& \text{Filter order } n: \\
& n = \frac{
\cosh^{-1} \left( \sqrt{\frac{10^{\frac{\alpha_{\min}}{10}} - 1}{\varepsilon^2}} \right)
}{
\cosh^{-1} \left(\frac{\omega_s}{\omega_p}\right)
} \\
& 10^{\frac{52}{10}} - 1 = 10^{5.2} - 1 \approx 158489 - 1 = 158488 \\
& \varepsilon^2 = (0.3098)^2 = 0.096 \\
& \sqrt{\frac{158488}{0.096}} = \sqrt{1,651,000} \approx 1285.7 \\
& \cosh^{-1}(1285.7) = \ln(1285.7 + \sqrt{1285.7^2 - 1}) \approx \ln(2571.4) \approx 7.85 \\
& \frac{\omega_s}{\omega_p} = \frac{61575.6}{20106.2} \approx 3.06 \\
& \cosh^{-1}(3.06) = \ln(3.06 + \sqrt{3.06^2 - 1}) = \ln(5.93) \approx 1.78 \\
& n = \frac{7.85}{1.78} = 4.41 \implies n = 5 \quad (\text{round up}) \\
\\
& \text{Calculate } v: \\
& v = \frac{1}{n} \sinh^{-1} \left(\frac{1}{\varepsilon}\right) \\
& \frac{1}{\varepsilon} = \frac{1}{0.3098} = 3.23 \\
& \sinh^{-1}(3.23) = \ln(3.23 + \sqrt{3.23^2 + 1}) = \ln(6.6) \approx 1.887 \\
& v = \frac{1}{5} \times 1.887 = 0.3774 \\
& \sinh(v) \approx 0.386, \quad \cosh(v) \approx 1.073 \\
\\
& \text{Calculate poles } s_k \text{ for } k=1,2,...,5: \\
& \theta_k = \frac{(2k - 1) \pi}{2n} = \frac{(2k - 1) \pi}{10} \\
\\
& \begin{array}{c|c|c|c}
k & \theta_k (\text{rad}) & \sin(\theta_k) & \cos(\theta_k) \\
\hline
1 & \frac{\pi}{10} \approx 0.314 & 0.309 & 0.951 \\
2 & \frac{3\pi}{10} \approx 0.942 & 0.809 & 0.588 \\
3 & \frac{5\pi}{10} = \frac{\pi}{2} = 1.571 & 1.000 & 0.000 \\
4 & \frac{7\pi}{10} \approx 2.199 & 0.809 & -0.588 \\
5 & \frac{9\pi}{10} \approx 2.827 & 0.309 & -0.951 \\
\end{array} \\
\\
& s_k = -\omega_p \sinh(v) \sin(\theta_k) + j \omega_p \cosh(v) \cos(\theta_k) \\
\\
& \begin{cases}
s_1 = -20106.2 \times 0.386 \times 0.309 + j\, 20106.2 \times 1.073 \times 0.951 \approx -2397 + j 20510 \\
s_2 = -20106.2 \times 0.386 \times 0.809 + j\, 20106.2 \times 1.073 \times 0.588 \approx -6275 + j 12696 \\
s_3 = -20106.2 \times 0.386 \times 1.000 + j\, 20106.2 \times 1.073 \times 0.000 \approx -7761 + j 0 \\
s_4 = -20106.2 \times 0.386 \times 0.809 + j\, 20106.2 \times 1.073 \times (-0.588) \approx -6275 - j 12696 \\
s_5 = -20106.2 \times 0.386 \times 0.309 + j\, 20106.2 \times 1.073 \times (-0.951) \approx -2397 - j 20510 \\
\end{cases} \\
\\
& \text{Transfer function } T(s): \\
& T(s) = \frac{1}{(s - s_1)(s - s_2)(s - s_3)(s - s_4)(s - s_5)} \\
& \quad = \frac{1}{(s^2 - 2 \text{Re}(s_1) s + |s_1|^2)(s^2 - 2 \text{Re}(s_2) s + |s_2|^2)(s - s_3)} \\
\\
& |s_1|^2 = (-2397)^2 + (20510)^2 = 5.75 \times 10^{6} + 4.20 \times 10^{8} \approx 4.26 \times 10^{8} \\
& |s_2|^2 = (-6275)^2 + (12696)^2 = 3.94 \times 10^{7} + 1.61 \times 10^{8} \approx 2.01 \times 10^{8} \\
\\
& \therefore \quad T(s) = \frac{1}{
(s^2 + 4794 s + 4.26 \times 10^{8})(s^2 + 12550 s + 2.01 \times 10^{8})(s + 7761)
}
\end{aligned}
\]


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