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