0.6≤|H(ω) |≤1 ; 0≤ω≤0.35π

|H(ω) |≤0.1 ; 0.7π≤ω≤π

Step-1: Identification of filters specification

$A_p=0.6 ; A_p=0.1 ; ω_p=0.35π ; ω_s=0.7π ; T=0.1 sec$

Now,

$Ω_p=\frac{2}{T} tan⁡(\frac{ω_p}{2})=12.25 rad/sec$

$Ω_s=\frac{2}{T} tan⁡(\frac{ω_s}{2})=39.25 rad/sec$

Step-2: Calculation of order of filter

The order of filter is given by

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

N≥1.72≅2

Step-3: Calculation of cut off frequency

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

$Ω_c=10.60 rad/sec$

Step-4: Calculation of poles

$P_k=Ω_c e^{j(N+2k+1) \frac{π}{2N}}$

when k=0;

$∴P_o=-7.49+j7.49$

when k=1;

$∴P_1=-7.49-j7.49$

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

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

$=\frac{(10.60)^2}{((s+749-j7.49)(s+7.49+j7.49))}$

$∴H(s)=\frac{112.36}{((s+7.49)^2+(7.49)^2 )}$

Conversion of analog Transfer function to digital Transfer function:

$H(z)=H(s)_(s=\frac{2}{T} \frac{(z-1)}{(z+1)}$

$H(z)=\frac{112.36}{\frac{(20((z-1)}{(z+1)}}+(7.49)^2+(7.49)^2$