Solution:

$0 \cdot 8 \leq\left|H\left(e^{j \omega}\right)\right| \leq 1 \quad 0 \leq \omega \leq 0 \cdot 2 \pi$

$\left|H\left(e^{j \omega)} | \leq 0.2 \quad 0.6 \pi \leq \omega \leq \pi\right.\right.$

Given:

$\delta p=0.8 \quad \omega_{p}=0.2 \pi$

$\delta_{S}=0.2 \quad \omega_{S}=0.6 \pi$

Step 1: Obtain analog specifications.

$\Omega p=\frac{2}{T} \tan \left(\frac{\omega P}{2}\right)$

$=\frac{2}{0.1} \tan \left(\frac{0.2 \pi}{2}\right)$

$=20 \tan \left(\frac{\pi}{10}\right)$

$\Omega p = 6.4383 \ rad/sec$

$\Omega_{S}=\frac{2}{T} \tan \left(\frac{0.6 \pi}{2}\right)$

$=20 \tan \left(\frac{3 \pi}{10}\right)$

$\Omega s = 27.5276 \ rad/sec$

Step 2: Obtain the order N.

$N \geq \frac{\frac{1}{2} \log _{10}\left[\frac{\frac{1}{\delta_{5}^{2}}-1}{\frac{1}{8 p^{2}}-1}\right]}{\log _{10}\left[\frac{\Omega s}{\Omega p}\right]}$

$N \geq \frac{1}{2} \log_{10} \frac{ \left[\frac{\frac{1}{0.2}-1}{\frac{1}{0.8^{2}}-1}\right]}{log_{10} [-] \frac{27.5276}{6.4983}}$

$N \geq \frac{1}{2} \frac{ \log _{10}\left[\frac{24}{0 \cdot 5625}\right]} {log_{10} [4.2361]}$

$N \geq \frac{1}{2} \quad \frac{1 \cdot 6300}{0 \cdot 6269}$

$N = \ \geq 1.300$

N = 2

Step 3: Cutoff frequency $\Omega c$

$\Omega c=\frac{\Omega p}{\left[\frac{1}{\delta^2 p}-1\right]^{1 / 2 N}}$

$\Omega_{C}=\frac{6.4983}{\left[\frac{1}{0.8^{2}}-1\right]^{1 / 4}}$

$\Omega_{C}= \ 7.5035 \ rad/sec$

Step 4: Transfer Function H(S)

N is even.

$H(s)=\frac{\Omega_{p}^{N}}{\prod_{k=1}^{N / 2} s^{2}+\Omega_{e} b_{k}S+\Omega_{C}^{2}}$

$H(s)=\frac{(7.5035)^{2}}{\pi_{k = 1} \ s^{2}+7.5035 b_{R} S+(5.5035)^{2}}$

$=\frac{56.3025}{S^{2}+7.5035 b_{1} S+56.3025}$

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

$=2 \sin \left[\frac{\pi}{4}\right]$

$b_1 = 1.4142$

$H(s)=\frac{56.3025}{s^{2}+10.61143+56.3025}$

Step 5: Transfer function H(z) By using BLT.

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

$H(z)=\frac{56.3025}{\left[\frac{2}{T}\left(\frac{1-\overline{z}^{1}}{1+2^{-1}}\right)\right]^{2}+10 \cdot 6114\left(\frac{2}{T}\left(\frac{1-\overline{z}^{1}}{1+\overline{z}^{1}}\right)\right)+56 \cdot 3025}$

$\therefore$ $H(z)=\frac{0.0841+0.168 \ z^{-1} +0.0841 \ z^{-2}}{1-1 \cdot 0284 \ z^{-1}+0.3650 \ z^{-2}}$