discrete-time iir filter design from continuous-time filters quote of the day experience is the name...

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



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



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



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



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



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



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



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



Example Cont’d



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



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



Bilinear Transformation



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



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



Example Cont’d

discrete-time iir filter design from continuous-time filters quote of the day experience is the name...

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