Design a digital IIR low-pass filter with a passband edge at 1000 Hz and stopband edge at 1500 Hz for a sampling frequency of 5000 Hz. The filter is to have a passband ripple of 0.5 dB and a stopband ripple below 30 dB. Design a Butterworth filter using the bilinear transformation.

Solution:

Step 1:

Type of transformation. Bilinear transformation is to be used.

Step 2:

The ratio of analog filter edge frequencies,

$ \begin{aligned} & \Omega_1=\frac{2}{T} \tan \frac{\omega_1}{2}=2 \times 5000 \tan \frac{0.4 \pi}{2}=7265.425 \mathrm{rad} / \mathrm{s} \\ & \Omega_2=\frac{2}{T} \tan \frac{\omega_2}{2}=2 \times 5000 \tan \frac{0.6 \pi}{2}=13763.819 \mathrm{rad} / \mathrm{s} \\ & \frac{\Omega_2}{\Omega_1}=\frac{13763.819}{7265.425}=1.8944 \end{aligned} $

Step 3:

Order of the filter N,

$ \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(\Omega_2 / \Omega_1\right)} \\\\ & \geq \frac{1}{2} \frac{\log \left\{\left[\frac{1}{(0.0316)^2}-1\right] /\left[\frac{1}{(0.9446)^2}-1\right]\right\}}{\log (1.844)} \\\\ & \geq \frac{1}{2} \frac{\log \{1000.44 / 0.1207\}}{\log (1.844)} \\\\ & \geq 7.35=8 \end{aligned}\\ $

Step 4:

$ \Omega_c=\frac{\Omega_1}{\left[\frac{1}{A_1^2}-1\right]^{1 / 2 N}}=\frac{7265.425}{\left[\frac{1}{0.9446^2}-1\right]^{1 / 2 \times x 8}}=8292 \mathrm{rad} / \mathrm{s}\\ $

Step 5:

$ \begin{array}{ll} \therefore \quad & b_1=2 \sin \left[\frac{\pi}{16}\right]=0.390 \quad b_2=2 \sin \left[\frac{3 \pi}{16}\right]=1.111 \\\\ & b_3=2 \sin \left[\frac{5 \pi}{16}\right]=1.662 \quad b_4=2 \sin \left[\frac{7 \pi}{16}\right]=1.961 \\\\ H_a(s)= & \left(\frac{(8292)^2}{s^2+0.39 \times 8292 s+(8292)^2}\right)\left(\frac{(8292)^2}{s^2+1.111 \times 8292 s+(8292)^2}\right) \\\\ & \left(\frac{(8292)^2}{s^2+1.662 \times 8292 s+(8292)^2}\right)\left(\frac{(8292)^2}{s^2+1.961 \times 8292 s+(8292)^2}\right) \\\\ H_a(s)= & \prod_{k=1}^{N / 2} \frac{\Omega_c^2}{s^2+b_k \Omega_c s+\Omega_c^2}\\ \end{array} $

$ \text { where } b_k=2 \sin \left[\frac{(2 k-1) \pi}{2 N}\right]\\ $

Step 6:

$ H(z)=\left.H_a(s)\right|_{s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}=\left.H_a(s)\right|_{s=10000}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\ $

$ H(z)=\left.\left[\left(\frac{(8292)^2}{s^2+13781.3 s+(8292)^2}\right)\left(\frac{(8292)^2}{s^2+16260.6 s+(8292)^2}\right)\right]\right|_{s=10000\left(\frac{\left(1-z^{-1}\right.}{1+z^{-1}}\right)}\\ $