FILTER DESIGN TECHNIQUESFILTER DESIGN TECHNIQUES
FILTER DESIGN TECHNIQUES Filter:
Any discrete-time system that modifies certain frequencies Frequency-selective filters pass only certain frequency components
and totally reject all others 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
Desired filter is generally implemented with digital computation It is used to filter a sampled and digitized continuous-time signal
DSP
FILTER DESIGN TECHNIQUES We have 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:
It can be noted that the specifications for filters are typically given in frequency domain
DSP
D/Cxc(t) yr(t)C/D H(ej) nx ny /j
cH e H j T
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
DSP
0.99 1.01 0 2 2000effH j
0.001 2 3000effH j
1
2
0.01
0.001
2 2000
2 3000
p
s
DISCRETE SIGNAL PROCESSING
Discrete-Time IIR Filter Design from Continuous-Time Filters
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
In this design procedure: Start with discrete-time spec in terms of (Td has no role in the process) Go to continuous-time = /T and design continuous-time filter Use impulse invariance to map it back to discrete-time = T
Works best for practical filters due to possible aliasing
DSP
2jd c d c
k d d
h n T h nT H e H j j kT T
0 /
c d
jc
d
H j T
H e H jT
IMPULSE INVARIANCE OF SYSTEM FUNCTIONS Develop impulse invariance relation between system functions Partial fraction expansion of transfer function of continuous-time
filter:
Corresponding impulse response:
Impulse response of discrete-time filter:
System function:
Pole s=sk in s-plane transforms into a pole at in the z-plane
DSP
1
Nk
ck k
AH s
s s
1
0
0 0
k
Ns t
kkc
A e th t
t
1 1
k d k d
N N ns nT s Td c d d k d k
k k
h n T h nT T A e u n T A e u n
11 1 k d
Nd ks T
k
T AH z
e z
k ds Te
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
DSP
0.89125 1 0 0.2
0.17783 0.3
j
j
H e
H e
0.89125 1 0 0.2
0.17783 0.3
H j
H j
0.17783 3.0jH and 89125.0 2.0jH cc
N2
c
2
cj/j1
1jH
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
DSP
2 22 20.2 1 0.3 1
1 and 10.89125 0.17783
N N
c c
5.8858 6 and 0.70474cN
1/12 /12 2 111 for k 0,1,...,11j kk c cs j e
1 1
1 2 1 2
1
1 2
0.2871 0.4466 2.1428 1.1455
1 1.2971 0.6949 1 1.0691 0.3699
1.8557 0.6303
1 0.9972 0.257
z zH z
z z z z
z
z z
2 2 2
0.12093
0.364 0.4945 0.9945 0.4945 1.3585 0.4945H s
s s s s s s
EXAMPLE CONT’D
10
DSP
FILTER DESIGN BY BILINEAR TRANSFORMATION Avoids the aliasing problem of impulse invariance Maps 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
DSP
1
1
2 1
1d
zsT z
1
1
2 1
1cd
zH z H
T z
1 / 2 1 / 2 / 2
1 / 2 1 / 2 / 2d d d
d d d
T s T j Tz
T s T j T
s j
BILINEAR TRANSFORMATION On the unit circle the transform becomes
To derive the relation between and
Which yields
DSP
1 / 2
1 / 2jd
d
j Tz e
j T
/ 2
/ 2
2 sin / 22 1 2 2tan
1 2 cos / 2 2
jj
j jd d d
e je js jT e T e T
2tan or 2arctan
2 2d
d
T
T
BILINEAR TRANSFORMATION DSP
EXAMPLE Bilinear transform applied to Butterworth
Apply bilinear transformation to specifications
We can assume Td=1 and apply the specifications to
To get
DSP
0.89125 1 0 0.2
0.17783 0.3
j
j
H e
H e
2 0.20.89125 1 0 tan
2
2 0.30.17783 tan
2
d
d
H jT
H jT
2
2
1
1 /c N
c
H j
2 22 22 tan 0.1 1 2 tan 0.15 1
1 and 10.89125 0.17783
N N
c c
EXAMPLE CONT’D Solve N and c
The resulting transfer function has the following poles
Resulting in
Applying the bilinear transform yields
DSP
2 21 1
log 1 10.17783 0.89125
5.305 62log tan 0.15 tan 0.1
N
766.0c
1/12 /12 2 111 for k 0,1,...,11j kk c cs j e
2 2 2
0.20238
0.3996 0.5871 1.0836 0.5871 1.4802 0.5871cH s
s s s s s s
61
1 2 1 2
1 2
0.0007378 1
1 1.2686 0.7051 1 1.0106 0.3583
1
1 0.9044 0.2155
zH z
z z z z
z z
EXAMPLE CONT’D DSP
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