written 5.6 years ago by | modified 2.6 years ago by |
second, to satisfy the following specifications.
0,707≤|H(ejw)|≤1.0:0≤w≤0.2π
H(ejw)|≤0.08:0.4π≤w≤π
written 5.6 years ago by | modified 2.6 years ago by |
second, to satisfy the following specifications.
0,707≤|H(ejw)|≤1.0:0≤w≤0.2π
H(ejw)|≤0.08:0.4π≤w≤π
written 5.6 years ago by |
Given:
Desired Pass-band edge ωp=0.2π
Desired Stop-band edge ωs=0.4π
Gain at Pass-band edge Ap=0.707
Gain at Stop-band edge As=0.08
T=1sec
Step 1: Pre-warp Analog Frequency:
By Bilinear Transformation (BLT)
Ω=2Ttanω2
∴Ωp=21tanωp2=2tan0.2π2=0.6498
Ωs=21tanωs2=2tan0.4π2=1.4531
Step 2: Order of the Filter (N)
N1=log(1A2s−1)−log(1A2p−1)2(logΩs−logΩp)
=log(10.082−1)−log(10.7072−1)2(log1.4531−log0.6498)=3.1340
since, N≥N1, let N=4,
∴ Order of Butterworth filter =4
Step 3: 3dB Cut Off Analog Frequency
Ωc=Ωp(1A2p−1)12N
=0.6498(10.7072−1)18
=0.6498
Step 4: T.F H(s) of the analog LPF
Since N=4 is even, Normalized T.F.
H(s)=∏N/2k=1BkΩ2cs2+bkΩcs+ckΩ2c
Here, bk=2sin[(2k−1)π2N],Bk=1 and ck=1,k=0,1,2,…
∴H(s)=∏2k=11×0.64982s2+2sin[(2k−1)π2×4]×0.6498s+1×0.64982
=0.4222s2+1.2996ssinπ8+0.4222×0.4222s2+1.2996ssin3π8+0.4222
=0.42222(s2+0.4973s+0.4222)(s2+1.2007s+0.4222)⋯(1)
Step 5: Digital Transfer Function
By BLT Method,
Put s=2(z−1)T(z+1)=2(z−1)1(z+1)=2z−2z+1 in (1)
∴ Digital Transfer Function
H(z)=0.1783[(2z−2z+1)2+0.4973(2z−2z+1)+0.4222]×[(2z−2z+1)2+1.2007(2z−2z+1)+0.4222]
=0.1783(z+1)4[(2z−2)2+0.4973(2z−2)(z+1)+0.4222(z+1)2]×[(2z−2)2+1.2007(2z−2)(z+1)+0.4222(z+1)2]
=0.1783(z+1)4[5.4168z2−7.1556z+3.4276][6.8236z2−7.1556z+2.0208]
=0.1783(z+1)45.4168[z2−1.3210z+0.6328]×6.8236[z2−1.0487z+0.2962]
=0.0048(z+1)4(z2−1.3210z+0.6328)(z2−1.0487z+0.2962)
∴ Digital Transfer Function
H(z)=0.0048(z+1)4(z2−1.3210z+0.6328)(z2−1.0487z+0.2962)