Numerical related 1

Q.N What are the characteristics of Butterworth filter? Derive an expression to estimate the order ‘n’ of low pass Butterworth approximation. Use that formula to estimate that order of Butterwoth filter with following specification

\[
W_p = 1000 \text{ rad/sec}, \quad \alpha_{\max} = 0.5 \text{ dB}
\]

\[
W_s = 2000 \text{ rad/sec}, \quad \alpha_{\min} = 20 \text{ dB}
\]

\[
\text{Calculate the cutoff frequency } \omega_0 \text{ and the pole locations.}
\]

We have,

\[
\begin{aligned}
n &= \frac{\log_{10} \left( \frac{10^{\alpha_{\min}/10} - 1}{10^{\alpha_{\max}/10} - 1} \right)}{2 \log_{10} \left( \frac{W_s}{W_p} \right)} \\[8pt]
n &= \frac{\log_{10} \left( \frac{10^{20/10} - 1}{10^{0.5/10} - 1} \right)}{2 \log_{10} \left( \frac{2000}{1000} \right)} \\[8pt]
n &= \frac{\log_{10} \left( \frac{100 - 1}{1.122 - 1} \right)}{2 \log_{10} (2)} \\[8pt]
n &= \frac{\log_{10} (99 / 0.122)}{2 \times 0.301} = \frac{\log_{10} (811.48)}{0.602} \\[8pt]
n &= \frac{2.909}{0.602} = 4.83 \approx 5
\end{aligned}
\]

\[
\therefore \text{Order } n = 5
\]

We Know,

\[
\omega_0 = W_p \times \left(10^{\alpha_{\max}/10} - 1\right)^{-\frac{1}{2n}}
\]

\[
\omega_0 = 1000 \times (1.122 - 1)^{-\frac{1}{10}} = 1000 \times (0.122)^{-0.1}
\]

Using logarithms:

\[
(0.122)^{-0.1} = e^{-0.1 \ln 0.122} = e^{0.214} = 1.238
\]

\[
\Rightarrow \omega_0 = 1000 \times 1.238 = 1238.1 \text{ rad/sec}
\]

\[
\text{Pole locations for } n=5:
\]

\[
\text{Poles are at angles: } 0^\circ, \pm 36^\circ, \pm 72^\circ
\]

\[
B_5 = (s + 1) \times (s^2 + 2 \cos 36^\circ \, s + 1) \times (s^2 + 2 \cos 72^\circ \, s + 1)
\]

\[
B_5 = (s + 1) \times (s^2 + 1.62 s + 1) \times (s^2 + 0.62 s + 1)
\]

\[
\text{The transfer function is:}
\]

\[
T_5(s) = \frac{1}{B_5} = \frac{1}{(s + 1)(s^2 + 1.62 s + 1)(s^2 + 0.62 s + 1)}
\]