Given:

Analog Pass-band edge $F_{p}=30 \mathrm{Hz}$

Analog Stop-band edge $F_{s}=75 \mathrm{Hz}$

Gain at Pass-band edge $A_{p}=0.89$

Stop band attenuation: $=0.20$

$\therefore$ Gain at Stop-band edge $A_{s}=0.2$

Let, Sampling period $(\mathrm{T})=1$

Step 1: Pre-warp Analog Frequency

Pass-band edge $\Omega_{p}=2 \pi F_{p}=2 \pi \times 30=60 \pi$

Stop-band edge $\Omega_{s}=2 \pi F_{s}=2 \pi \times 75=150 \pi$

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

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

$=\frac{\log \left(\frac{1}{0.2^{2}}-1\right)-\log \left(\frac{1}{0.89^{2}}-1\right)}{2(\log 150 \pi-\log 60 \pi)}$

$=2.4641$

since, $N \geq N_{1},$ let $\mathrm{N}=3$

$\therefore$ Order of Butterworth filter $=3$

Step 3: 3 dB Cut Off Analog Frequency

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

$=\frac{60 \pi}{\left(\frac{1}{0.89^{2}}-1\right)^{\frac{1}{6}}}$

$=235.57 \approx 236$

Step 4: T.F. $H(s)$ of the analog LPF

Since $\mathrm{N}=3$ is odd, T.F.

$H(s)=\frac{B_{0} \Omega_{c}}{s+c_{0} \Omega_{c}} \prod_{k=1}^{(N-1) / 2} \frac{B_{k} \Omega_{c}^{2}}{s^{2}+b_{k} \Omega_{c} s+c_{k} \Omega_{c}^{2}}$

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

$\therefore H(s)=\frac{1 \times 236}{s+1 \times 236} \prod_{k=1}^{1} \frac{1 \times 236}{s^2+2sin\left[ \frac{(2-1) \pi}{2 \times 3}\right] \times 236 s+1 \times 236^{2}}$

$\therefore H(s)=\frac{236}{s+236} \times \frac{55696}{s^{2}+236 s+55696} \cdots(1)$

Step 5: Digital Transfer Function

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

$\therefore H(z)=\frac{13144256}{\left[\frac{2 z-2}{z+1}+236\right]\left[\left(\frac{2 z-2}{z+1}\right)^{2}+236\left(\frac{2 z-2}{z+1}\right)+55696\right]}$

$=\frac{13144256(z+1)^{3}}{[2 z-2+236(z+1)] \times [(2 z-2)^{2}+236(2 z-2)(z+1)+55696(z+1)^{2}]}$

$=\frac{13144256(z+1)^{3}}{[238 z+234]\left[56172 z^{2}+111384 z+55228\right]}$

$=\frac{13144256(z+1)^{3}}{238[z+0.9832] \times 56172\left[z^{2}+1.9829 z+0.9832\right]}$

$=\frac{0.9832(z+1)^{3}}{(z+0.9832)\left(z^{2}+1.9829 z+0.9832\right)}$

$=\frac{0.9832(z+1)^{3}}{z^{3}+2.9661 z^{2}+2.9328 z+0.9667}$

$H(z)=\frac{0.9832(z+1)^{3}}{z^{3}+2.9661 z^{2}+2.9328 z+0.9667}$