Discrete-Time IIR Filter Design from Continuous-Time Filters Quote of the Day Experience is the name...
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Discrete-Time IIR Filter Design from Continuous-Time Filters
Quote of the DayExperience is the name everyone gives to their
mistakes. Oscar Wilde
Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer, and Buck, ©1999-2000 Prentice Hall Inc.
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 2
Filter Design Techniques• Any discrete-time system that modifies certain frequencies• Frequency-selective filters pass only certain frequencies• Filter Design Steps
– Specification• Problem or application specific
– Approximation of specification with a discrete-time system• Our focus is to go from spec to discrete-time system
– Implementation• Realization of discrete-time systems depends on target technology
• We already studied the use of discrete-time systems to implement a continuous-time system– If our specifications are given in continuous time we can use
D/Cxc(t) yr(t)C/D H(ej) nx ny
T/jHeH cj
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 3
Filter Specifications• Specifications
– Passband
– Stopband
• Parameters
• Specs in dB– Ideal passband gain =20log(1) = 0 dB– Max passband gain = 20log(1.01) = 0.086dB– Max stopband gain = 20log(0.001) = -60 dB
200020 01.1jH99.0 eff
30002 001.0jHeff
30002
20002
001.0
01.0
s
p
2
1
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 4
Butterworth Lowpass Filters• Passband is designed to be maximally flat• The magnitude-squared function is of the form
N2
c
2
cj/j1
1jH
N2c
2
cj/s1
1sH
1-0,1,...,2Nkfor ej1s 1Nk2N2/jcc
N2/1k
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 5
Chebyshev Filters• Equiripple in the passband and monotonic in the stopband• Or equiripple in the stopband and monotonic in the passband
xcosNcosxV /V1
1jH 1
Nc
2N
2
2
c
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 6
Filter Design by Impulse Invariance• Remember impulse invariance
– Mapping a continuous-time impulse response to discrete-time– Mapping a continuous-time frequency response to discrete-time
• If the continuous-time filter is bandlimited to
• If we start from discrete-time specifications Td cancels out– Start with discrete-time spec in terms of – Go to continuous-time = /T and design continuous-time filter– Use impulse invariance to map it back to discrete-time = T
• Works best for bandlimited filters due to possible aliasing
dcd nThTnh
k ddc
j kT2
jT
jHeH
T
jHeH
T/ 0jH
dc
j
dc
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 7
Impulse Invariance of System Functions • Develop impulse invariance relation between system functions• Partial fraction expansion of transfer function
• Corresponding impulse response
• Impulse response of discrete-time filter
• System function
• Pole s=sk in s-domain transform into pole at
N
1k k
kc ss
AsH
0t0
0teAth
N
1k
tsk
c
k
N
1k
nTskd
N
1k
nTskddcd nueATnueATnThTnh dkdk
N
1k1Ts
kd
ze1AT
zHdk
dkTse
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 8
Example• Impulse invariance applied to Butterworth
• Since sampling rate Td cancels out we can assume Td=1• Map spec to continuous time
• Butterworth filter is monotonic so spec will be satisfied if
• Determine N and c to satisfy these conditions
3.0 0.17783eH
2.00 1eH89125.0
j
j
3.0 0.17783jH
2.00 1jH89125.0
0.17783 3.0jH and 89125.0 2.0jH cc
N2
c
2
cj/j1
1jH
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 9
Example Cont’d• Satisfy both constrains
• Solve these equations to get
• N must be an integer so we round it up to meet the spec• Poles of transfer function
• The transfer function
• Mapping to z-domain
2N2
c
2N2
c 17783.013.0
1 and 89125.0
12.01
70474.0 and 68858.5N c
0,1,...,11kfor ej1s 11k212/jcc
12/1k
21
1
21
1
21
1
z257.0z9972.01z6303.08557.1
z3699.0z0691.11z1455.11428.2
z6949.0z2971.11z4466.02871.0
zH
4945.0s3585.1s4945.0s9945.0s4945.0s364.0s12093.0
sH 222
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 10
Example Cont’d
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 11
Filter Design by Bilinear Transformation• Get around the aliasing problem of impulse invariance• Map the entire s-plane onto the unit-circle in the z-plane
– Nonlinear transformation– Frequency response subject to warping
• Bilinear transformation
• Transformed system function
• Again Td cancels out so we can ignore it• We can solve the transformation for z as
• Maps the left-half s-plane into the inside of the unit-circle in z– Stable in one domain would stay in the other
1
1
d z1z1
T2
s
1
1
dc z1
z1T2
HzH
2/Tj2/T1
2/Tj2/T1s2/T1s2/T1
zdd
dd
d
d
js
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 12
Bilinear Transformation• On the unit circle the transform becomes
• To derive the relation between and
• Which yields
j
d
d e2/Tj12/Tj1
z
2tan
Tj2
2/cose22/sinje2
T2
je1e1
T2
sd
2/j
2/j
dj
j
d
2T
arctan2or 2
tanT2 d
d
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 13
Bilinear Transformation
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 14
Example• Bilinear transform applied to Butterworth
• Apply bilinear transformation to specifications
• We can assume Td=1 and apply the specifications to
• To get
3.0 0.17783eH
2.00 1eH89125.0
j
j
23.0
tanT2
0.17783jH
22.0
tanT2
0 1jH89125.0
d
d
N2
c
2
c/1
1jH
2N2
c
2N2
c 17783.0115.0tan2
1 and 89125.0
11.0tan21
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 15
Example Cont’d
• Solve N and c
• The resulting transfer function has the following poles
• Resulting in
• Applying the bilinear transform yields
6305.51.0tan15.0tanlog2
189125.0
11
17783.01
log
N
22
766.0c
0,1,...,11kfor ej1s 11k212/jcc
12/1k
5871.0s4802.1s5871.0s0836.1s5871.0s3996.0s20238.0
sH 222c
21
2121
61
z2155.0z9044.011
z3583.0z0106.11z7051.0z2686.11z10007378.0
zH
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 16
Example Cont’d