Discrete-Time IIR Filter Design From Continuous-Time Filters

Discrete-Time IIR Filter Design from
Continuous-Time Filters
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Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer, and Buck, 1999-2000 Prentice Hall Inc.
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
xc(t)
C/D
xn
H(e )
j
H e j Hc j / T
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yn
D/C
yr(t)
351M Digital Signal Processing
Filter Specifications
Specifications
Passband
0.99 Heff j 1.01
0 2 2000
Stopband
Heff j 0.001
2 3000
Parameters
1 0.01
2 0.001
p 2 2000
s 2 3000
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
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Butterworth Lowpass Filters

Passband is designed to be maximally flat
The magnitude-squared function is of the form
Hc j
1 j / j c
sk 1
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1 / 2N
2N
jc ce j / 2N 2k N1
Hc s
1 s / j c
2N
for k 0,1,...,2N - 1
Chebyshev Filters
Equiripple in the passband and monotonic in the stopband
Or equiripple in the stopband and monotonic in the passband
Hc j
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1 V / c
2
2
N
VN x cos N cos 1 x
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
hn Tdhc nTd
He
Hc j
j
k
Td
k
Td
If the continuous-time filter is bandlimited to
Hc j 0
/ Td

H e j Hc j
Td
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
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Impulse Invariance of System Functions

Develop impulse invariance relation between system functions
Partial fraction expansion of transfer function
N
Ak
Hc s
k 1 s sk
Corresponding impulse response
N
Ak esk t t 0
hc t k 1
0
t0
Impulse response of discrete-time filter
hn Tdhc nTd
System function
TdAk e
k 1
H z
sknTd
k 1
TdAk
sk Td 1
z
k 1 1 e
N
Pole s=sk in s-domain transform into pole at
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un TdAk esk Td un
esk Td
Example
Impulse invariance applied to Butterworth

H e 0.17783
0.89125 H e j 1
0 0.2
0.3
Since sampling rate Td cancels out we can assume Td=1

Map spec to continuous time
0.89125 H j 1
0 0.2
H j 0.17783
0.3
Butterworth filter is monotonic so spec will be satisfied if

Hc j0.2 0.89125 and Hc j0.3 0.17783
Hc j
1 j / jc
2N
Determine N and c to satisfy these conditions
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Example Contd
Satisfy both constrains
0.2
1
c
2N
0.89125
and
0.3
1
c
2N
0.17783
Solve these equations to get

N 5.8858 6
and
c 0.70474
N must be an integer so we round it up to meet the spec

Poles of transfer function
1 / 12
jc ce j / 12 2k 11 for k 0,1,...,11
sk 1
The transfer function
H s
0.12093
s2 0.364s 0.4945 s2 0.9945s 0.4945 s2 1.3585s 0.4945
Mapping to z-domain
0.2871 0.4466z 1
2.1428 1.1455z 1
H z
1
2
1 1.2971z 0.6949z
1 1.0691z 1 0.3699z 2
1.8557 0.6303z 1
1 0.9972z 1 0.257z 2
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Example Contd
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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
2
s
Td
Transformed system function
1 z 1
1
1z
2
H z Hc
Td
1 z 1
1
1z
Again Td cancels out so we can ignore it

We can solve the transformation for z as
1 Td / 2 s 1 Td / 2 jTd / 2
1 Td / 2 s 1 Td / 2 jTd / 2
s j
Maps the left-half s-plane into the inside of the unit-circle in z

Stable in one domain would stay in the other
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Bilinear Transformation
On the unit circle the transform becomes
1 jTd / 2
e j
1 jTd / 2
To derive the relation between and
2
s
Td
1 e j
2 2e j / 2 j sin / 2
2j

tan

j
j / 2
Td 2e
Td
cos / 2
2
1e
Which yields
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2

tan
Td
2
or
Td
2 arctan
12
Bilinear Transformation
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13
Example
Bilinear transform applied to Butterworth

H e 0.17783
0.89125 H e j 1
j
0 0.2
0.3
Apply bilinear transformation to specifications

0.89125 H j 1
2
0.2
tan
Td
2
0.3

2
H j 0.17783
tan
Td
We can assume Td=1 and apply the specifications to

Hc j
1 / c
2N
To get
2 tan 0.1
1
c
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2N
0.89125
2 tan 0.15
and 1
c
2N
0.17783
14
Example Contd
Solve N and c
2

1
log
1

0.17783

N
2 log tan 0.15
0.89125
tan 0.1
5.305 6
c 0.766
The resulting transfer function has the following poles

1 / 12
jc ce j / 12 2k 11 for k 0,1,...,11
sk 1
Resulting in
Hc s
0.20238
s2 0.3996s 0.5871 s2 1.0836s 0.5871 s2 1.4802s 0.5871
Applying the bilinear transform yields

H z
1 1.2686z
1 0.9044z
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1
1
0.0007378 1 z 1
0.7051z 2 1 1.0106z 1 0.3583z 2
0.2155z 2
15
Example Contd
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Discrete-Time IIR Filter Design from

Filter Design Techniques

Approximation of specification with a discrete-time system

We already studied the use of discrete-time systems to

351M Digital Signal Processing

0.99 Heff j 1.01

351M Digital Signal Processing

Butterworth Lowpass Filters

351M Digital Signal Processing

351M Digital Signal Processing

Filter Design by Impulse Invariance

If the continuous-time filter is bandlimited to

If we start from discrete-time specifications Td cancels out

Works best for bandlimited filters due to possible aliasing

351M Digital Signal Processing

Impulse Invariance of System Functions

Pole s=sk in s-domain transform into pole at

351M Digital Signal Processing

Since sampling rate Td cancels out we can assume Td=1

Butterworth filter is monotonic so spec will be satisfied if

Determine N and c to satisfy these conditions

351M Digital Signal Processing

Solve these equations to get

N must be an integer so we round it up to meet the spec

351M Digital Signal Processing

351M Digital Signal Processing

Filter Design by Bilinear Transformation

Transformed system function

Again Td cancels out so we can ignore it

Maps the left-half s-plane into the inside of the unit-circle in z

351M Digital Signal Processing

To derive the relation between and

351M Digital Signal Processing

351M Digital Signal Processing

Apply bilinear transformation to specifications

We can assume Td=1 and apply the specifications to

351M Digital Signal Processing

The resulting transfer function has the following poles

Applying the bilinear transform yields

351M Digital Signal Processing

351M Digital Signal Processing

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