written 6.5 years ago by | • modified 5.5 years ago |
0.707≤|H(w) |≤1 ; for 0<ω<0.3π
|H(w) |≤0.2 ; for 0.75π<ω<π
written 6.5 years ago by | • modified 5.5 years ago |
0.707≤|H(w) |≤1 ; for 0<ω<0.3π
|H(w) |≤0.2 ; for 0.75π<ω<π
written 6.5 years ago by |
Step-1: Identify the filters specification
Ap=0.707;As=0.2;ωp=0.3π;ωs=0.75π;T=1sec
Now,
Ωp=ωpT=0.942rad/sec
Ωs=ωsT=39.25rad/sec
Step-2: Calculation of order of filter
The order of filter is given by
N≥12log[(1/(As2)−1)(1/(Ap2)−1)]log(ΩsΩp
N≥1.73≅2
Step-3: Calculation of cut off frequency
Ωc=Ωp(1(Ap2−1))12N
Ωc=0.941rad/sec
Step-4: Calculation of poles
Pk=Ωc=ej(N+2k+1π2N
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} [sinbTs ] z^{-1})}{(1-2e^{-aTs} [cosbTs ] z^{-1}+e^{-2aTs} z^{-2} )}
∴H(z)=\frac{(e^{-0.665Ts} [sin0.665Ts ] z^{-1})}{(1-2e^{-0.665Ts} [cos0.665Ts ] z^{-1}+e^{-1.33Ts} z^{-2})}