0.707≤|H(w) |≤1 ; for 0<ω<0.3π

|H(w) |≤0.2 ; for 0.75π<ω<π

Step-1: Identify the filters specification

$A_p=0.707 ; A_s=0.2 ; ω_p=0.3π ; ω_s=0.75π ; T=1 sec $

Now,

$Ω_p=\frac{ω_p}{T}=0.942 rad/sec$

$Ω_s=\frac{ω_s}{T}=39.25 rad/sec$

Step-2: Calculation of order of filter

The order of filter is given by

$N≥\frac{1}{2} \frac{log⁡[\frac{(1/(As^2 )-1)}{(1/(Ap^2 )-1)}]}{log⁡(\frac{Ω_s}{Ω_p} }$

$N≥1.73≅2$

Step-3: Calculation of cut off frequency

$Ω_c=\frac{Ω_p}{(\frac{1}{(Ap^2-1)})^{\frac{1}{2N}}}$

$Ω_c=0.941 rad/sec$

Step-4: Calculation of poles

$P_k=Ω_c=e^{j(N+2k+1} \frac{π}{2N}$

Now,

when k=0;

$∴P_o=-0.665+j0.665$

when k=1;

$∴P_1=-0.665-j0.665$

Step-5: Calculation of Transfer function H(s)

$H(s)=\frac{(Ω_c)^N}{((s-P_o )(s-P_1))}$

$=\frac{0.885}{((s+0.665-j0.665)(s+0.665+j0.665))}$

$∴H(s)=\frac{0.885}{((s+0.665)^2+(0.665)^2 )}$

Step-6: Conversion of analog Transfer function to digital Transfer function

We know that,

$\frac{b}{((s+a^2 )+b^2 )}=\frac{(e^{-aTs} [sin⁡bTs ] z^{-1})}{(1-2e^{-aTs} [cos⁡bTs ] z^{-1}+e^{-2aTs} z^{-2} )}$

$∴H(z)=\frac{(e^{-0.665Ts} [sin⁡0.665Ts ] z^{-1})}{(1-2e^{-0.665Ts} [cos⁡0.665Ts ] z^{-1}+e^{-1.33Ts} z^{-2})}$