Determine the order of a Butterworth low-pass filter satisfying the following specifications: $f_p=0.10 \mathrm{~Hz}$ $f_s=0.15 \mathrm{~Hz}$ $\alpha_p=0.5 \mathrm{~dB}$ $\alpha_s=15 \mathrm{~dB} ; f=1 \mathrm{~Hz}$

Solution:

The type of transformation is not specified. Let us use bilinear transformation,

$ \frac{\Omega_2}{\Omega_1}=\frac{\frac{2}{T} \tan \frac{\omega_2}{2}}{\frac{2}{T} \tan \frac{\omega_1}{2}}=\frac{\frac{2}{1} \tan \frac{0.3 \pi}{2}}{\frac{2}{1} \tan \frac{0.2 \pi}{2}}=\frac{1.019}{0.649}=1.57\\ $

$ \begin{aligned} N & \geq \frac{1}{2} \frac{\log \left\{\left[\frac{1}{A_2^2}-1\right] /\left[\frac{1}{A_1^2}-1\right]\right\}}{\log \left(\frac{\Omega_2}{\Omega_1}\right)} \\\\ & \geq \frac{1}{2} \frac{\log \left\{\left[\frac{1}{0.177^2}-1\right] /\left[\frac{1}{0.944^2}-1\right]\right\}}{\log (1.57)} \\\\ & \geq 6.16 \approx 7\\ \end{aligned} $

So the order of the low-pass Butterworth filter is N = 7.