Given: Pass band attenuation: $-2 \mathrm{dB}$jm

Gain at Pass-band edge $: A_{p}=10^{-2 / 20}=0.7943 $

Stop band attenuation $: \geq-10 \mathrm{dB}$

Gain at Stop-band edge $:A_{s}=10^{-10 / 20}=0.3162$

Step 1: Pre-warp Analog Frequency

Since the given frequencies are in rad/sec, they must be Pre-warp analog frequencies

Pass-band edge $\Omega_{p}=20$

Stop-band edge $\Omega_{s}=30$

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

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

$=\frac{\log \left(\frac{1}{0.3162^{2}}-1\right)-\log \left(\frac{1}{0.7943^{2}}-1\right)}{2(\log 30-\log 20)}$

$=3.37$

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

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

Step 3: 3 dB Cut Off Analog Frequency

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

$=\frac{20}{\left(\frac{1}{0.7943^{2}}-1\right)^{\frac{1}{8}}}$

$=21.3863$

$\approx 21.39 \mathrm{rad} / \mathrm{sec}$

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

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

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

$b_{k}=2 \sin \left[\frac{(2 k-1) \pi}{2 N}\right]=2 \sin \left[\frac{(2 k-1) \pi}{8}\right] \& c_{k}=1, \mathrm{k}=0,1,2, \ldots$

$\therefore H(s)=\prod_{k=1}^{2} \frac{1 \times 21.39^{2}}{s^{2}+2 \sin \left[\frac{(2 k-1) \pi}{8}\right] \times 21.39 s+1 \times 21.39^{2}}$

$=\frac{457.53}{s^{2}+42.78 s \sin \frac{\pi}{8}+457.53} \times \frac{457.53}{s^{2}+42.78 s \sin \frac{3 \pi}{8}+457.53}$

$=\frac{457.53}{s^{2}+16.37 s+457.53} \times \frac{457.53}{s^{2}+39.52 s+457.53} \cdots(1)$

Step 5

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

$\therefore$ Digital Transfer Function

$=\frac{457.53}{\left(\frac{2 z-2}{z+1}\right)^{2}+16.37\left(\frac{2 z-2}{z+1}\right)+457.53} \times \frac{457.53}{\left(\frac{2 z-2}{z+1}\right)^{2}+39.52\left(\frac{2 z-2}{z+1}\right)+457.53}$

$=\frac{457.53(z+1)^{2}}{(2 z-2)^{2}+16.37(2 z-2)(z+1)+457.53(z+1)^{2}} \times \frac{457.53(z+1)^{2}}{(2 z-2)^{2}+39.52(2 z-2)(z+1)+457.53(z+1)^{2}}$

$=\frac{457.53(z+1)^{2}}{494.27 z^{2}+907.06 z+428.79} \times \frac{457.53(z+1)^{2}}{540.57 z^{2}+907.06 z+382.49}$

$=\frac{457.53^{2}(z+1)^{4}}{494.27\left(z^{2}+1.8352 z+0.8675\right) \cdot 540.57\left(z^{2}+1.6780 z+0.7076\right)}$

$H(z)=\frac{0.7835(z+1)^{4}}{\left(z^{2}+1.8352 z+0.8675\right)\left(z^{2}+1.6780 z+0.7076\right)}$