Question: Design a Butterworth digital IIRIow pass filter using Bilinear transformation by taking $T=1$
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second, to satisfy the following specifications.

$0,707 \leq\left|H\left(e^{jw}) | \leq 1.0 : 0 \leq w \leq 0.2 \pi\right.\right.$

$\mathrm{H}\left(\mathrm{e}^{\mathrm{jw}}) | \leq 0.08 \quad : 0.4 \pi \leq \mathrm{w} \leq \pi\right.$

page butterworth filter • 34 views
 modified 4 weeks ago  • written 4 weeks ago by Ankit Pandey • 70
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Given:

Desired Pass-band edge $\omega_{p}=0.2 \pi$

Desired Stop-band edge $\omega_{s}=0.4 \pi$

Gain at Pass-band edge $A_{p}=0.707$

Gain at Stop-band edge $A_{s}=0.08$

$\mathrm{T}=1 \mathrm{sec}$

Step 1: Pre-warp Analog Frequency:

By Bilinear Transformation (BLT)

$\Omega=\frac{2}{T} \tan \frac{\omega}{2}$

$\therefore \Omega_{p}=\frac{2}{1} \tan \frac{\omega_{p}}{2}=2 \tan \frac{0.2 \pi}{2}=0.6498$

$\Omega_{s}=\frac{2}{1} \tan \frac{\omega_{s}}{2}=2 \tan \frac{0.4 \pi}{2}=1.4531$

Step 2: Order of the Filter $(\mathrm{N})$

$N_{1}=\frac{\log \left(\frac{1}{A_{s}^{2}}-1\right)-\log \left(\frac{1}{A_{p}^{2}}-1\right)}{2\left(\log \Omega_{s}-\log \Omega_{p}\right)}$

\begin{aligned} &=\frac{\log \left(\frac{1}{0.08^{2}}-1\right)-\log \left(\frac{1}{0.707^{2}}-1\right)}{2(\log 1.4531-\log 0.6498)} \\=& 3.1340 \end{aligned}

since, $N \geq N_{1},$ let $\mathrm{N}=4$,

$\therefore$ Order of Butterworth filter $=4$

Step 3: 3dB Cut Off Analog Frequency

$\Omega_{c}=\frac{\Omega_{p}}{\left(\frac{1}{A_{p}^{2}}-1\right)^{\frac{1}{2 N}}}$

$=\frac{0.6498}{\left(\frac{1}{0.707^{2}}-1\right)^{\frac{1}{8}}}$

$=0.6498$

Step 4: T.F $H(s)$ of the analog $\mathrm{LPF}$

Since $\mathrm{N}=4$ is even, Normalized T.F.

$H(s)=\prod_{k=1}^{N / 2} \frac{B_{k} \Omega_{c}^{2}}{s^{2}+b_{k} \Omega_{c} s+c_{k} \Omega_{c}^{2}}$

Here, $b_{k}=2 \sin \left[\frac{(2 k-1) \pi}{2 N}\right], B_{k}=1$ and $c_{k}=1, \mathrm{k}=0,1,2, \ldots$

$\therefore H(s)=\prod_{k=1}^{2} \frac{1 \times 0.6498^{2}}{s^{2}+2 \sin \left[\frac{(2 k-1) \pi}{2 \times 4}\right] \times 0.6498 s+1 \times 0.6498^{2}}$

$=\frac{0.4222}{s^{2}+1.2996 s \sin \frac{\pi}{8}+0.4222} \times \frac{0.4222}{s^{2}+1.2996 s \sin \frac{3 \pi}{8}+0.4222}$

$=\frac{0.4222^{2}}{\left(s^{2}+0.4973 s+0.4222\right)\left(s^{2}+1.2007 s+0.4222\right)} \cdots(1)$

Step 5: Digital Transfer Function

By BLT Method,

Put $s=\frac{2(z-1)}{T(z+1)}=\frac{2(z-1)}{1(z+1)}=\frac{2 z-2}{z+1}$ in $(1)$

$\therefore$ Digital Transfer Function

$H(z)=\frac{0.1783}{\left[\left(\frac{2 z-2}{z+1}\right)^{2}+0.4973\left(\frac{2 z-2}{z+1}\right)+0.4222\right] \times \left[\left(\frac{2 z-2}{z+1}\right)^{2}+1.2007\left(\frac{2 z-2}{z+1}\right)+0.4222\right]}$

$=\frac{0.1783(z+1)^{4}}{\left[(2 z-2)^{2}+0.4973(2 z-2)(z+1)+0.4222(z+1)^{2}\right] \times \left[(2 z-2)^{2}+1.2007(2 z-2)(z+1)+0.4222(z+1)^{2}\right]}$

$=\frac{0.1783(z+1)^{4}}{\left[5.4168 z^{2}-7.1556 z+3.4276\right]\left[6.8236 z^{2}-7.1556 z+2.0208\right]}$

$=\frac{0.1783(z+1)^{4}}{5.4168\left[z^{2}-1.3210 z+0.6328\right] \times 6.8236\left[z^{2}-1.0487 z+0.2962\right]}$

$=\frac{0.0048(z+1)^{4}}{\left(z^{2}-1.3210 z+0.6328\right)\left(z^{2}-1.0487 z+0.2962\right)}$

$\therefore$ Digital Transfer Function

$H(z)=\frac{0.0048(z+1)^{4}}{\left(z^{2}-1.3210 z+0.6328\right)\left(z^{2}-1.0487 z+0.2962\right)}$