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Digital Signal Processing

1

Chapter 6IIR Filter

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Introduction

2

oDigital filters are widely used in almost all areas

of DSP.

oIf linear phase is not critical, infinite impulse

response (IIR) filters yield a much smaller filter

order for a given application.

oDigital filters process discrete-time signals.

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3

Low & constantly

decreasing cost

Freedom from

component

variations

Easy modification

of filter

characteristics

High accuracy

High noise

immunity

Advantages

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o Digital filters are therefore rapidly replacing

analog filters in many applications.

o Digital filtering is not only as a smoothing or

averaging operation, but also as any processing

of the input signal.

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Techniques of Digital Filter Design

5

oDigital filter design revolves around 2 differentapproaches.

oIf linear phase is not critical, IIR filters yield a

much smaller filter order.

oThe design starts with an analog lowpass

prototypebased on the given specifications.

oIt is then converted to the required digital filter,

by using appropriate mapping and spectral

transformation.

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oTo make the infinitely long symmetric impulse

response sequence of an IIR filter causal, we need

an infinite delay.

oSymmetric truncation (to preserve linear phase)

simply transforms the IIR filter into an FIR filter.

o

Only FIR filters can be designed with linearphase (no phase distortion).

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oTheir design is typically based on selecting a

symmetric impulse response sequence whose

length is chosen to meet design specifications.

oThis choice is often based on iterative techniques

(trial & error).

oFor given specifications, FIR filters requiremany more elements in their realization than do

IIR filters.

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IIR Filter Design

9

oThere are 2 related approaches:

Using the well-established analog filter design,

followed by a mapping that converts the analog

filter to the digital filter.

Directly designing the digital filter using

digital equivalents of analog approximations.

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Equivalence of Analog & Digital Systems

10

oThe impulse response h(t) of an analog system

may be approximated by

ots is the sampling interval corresponding to the

sampling rate S =1/ts.

oThe discrete-time impulse response hs[n]describes the samples h(nts) of h(t):

s

n

sss

n

sa nttnthtnttthtthth

~

knkhnthnhk

sss

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oThe Laplace transformHa(s) of is given by

oThe z-transformHd(z) of hs[n] is given by

oComparison suggests the equivalence

Ha(s) =t

sH

d(z) providedz-k= , or

k

skt

ssa

sekthtsHsH

k

ksd zkhzH

sskt

e

sstez

s

t

zs

ln

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oThese relations describe amappingbetween the

variableszands.

oSince s = + j, where is the continuous

frequency, the complex variable z can be

expressed as

where is the digital

frequency in radians/sample.

jttjttj eeeeez ssss

FS

fts

2

2

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Response Matching

13

oThe idea of response matching is to match the

time-domain analog & digital response.

oTypically the impulse response or step response.

o

Given the analog filter H(s) & the input x(t)whose invariance we seek, first find the analog

responsey(t) as the inverse transform ofH(s)X(s).

ox(t) & y(t) are then sampled at intervals ts toobtain their sampled versionsx[n] &y[n].

oFinally, H(z) = Y(z)/X(z) is computed to obtain

the digital filter.

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The concept of response invariance:

The quality of the approximation depends on the

choice of the sampling interval ts & a unique

correspondence is possible only if the sampling

rate S =1/tsis above Nyquist rate (avoid aliasing).

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oThis mapping is thus useful only for analog

systems such as lowpass & bandpass filters,

whose frequency response is essentially band-

limited.

o

The analog transfer function H(s) must bestrictly proper (with numerator degree less than

the denominator degree).

oThe response y(t) of H(s) matches the response

y[n] ofH(z) at the sampling instants t =nts.

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Example 6.1

Convert toH(z) by using:

I) Impulse invariance transformation.

II)Step invariance transformation.

Solution

I) Choosex(t) = (t). Then,X(s) = 1.

21

4

sssH

2

4

1

4

21

4

sssssXsHsY

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Taking the inverse Laplace Transform,

The sampled input and output are given by

x[n] = [n]

Their z-Transform gives

X(z) = 1

tuetuety tt 2

44

nuenueny ss ntnt 244

ss tt

ez

z

ez

zzY

2

44

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The ratio of their z-Transform gives the transfer

function of the digital filter,

II)Choosex(t) = u(t). Then,X(s) = 1/s.

ss ttI ez

z

ez

zzY

zX

zYzH

2

44

2

2

1

42

21

4

sssssssXsHsY

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Taking the inverse Laplace Transform,

The sampled input and output are given by

x[n] = u[n]

tuetuetuty tt 2

242

nuenuenuny ss ntnt 2242

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Their z-Transform gives

The ratio of their z-Transform gives the transfer

function of the digital filter,

1z

z

zX

ss tt ez

z

ez

z

z

zzY

2

24

1

2

ss ttS ez

zez

zzXzYzH

2

12142

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The Bilinear Transformation

21

oThe bilinear transformation is defined by

o

(6.9)where C = 2/tsoBy letting = 0, the complex variable z is

obtained in the form

1

1

z

zCs

sC

sCz

CjejC

jCz

1tan2

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oSince z = ej, where = 2F is the digital

frequency, then

(6.12)

oThis is a non-linear relation between the analog

frequency & the digital frequency .

oWhen = 0, = 0 &, .

oIt is a one-to-one mapping that nonlinearly

compressesanalog frequency range -

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It avoids the effect of aliasing at the expense of

distorting, compressing or warping the analog

frequencies as shown:

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oThe higher the frequency, the more severe is the

warping.

oWarping can be compensated but cannot be

eliminated!

oWarping can be compensated if the frequency

specifications is prewarped before designing theanalog system H(s) or by applying bilinear

transformation.

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Graph of versus for different values of C

compared to the linear relation = .

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oThe analog & digital frequencies always show a

match at the origin & at one other value dictated

by the choice of C.

oThe advantages of bilinear transformation are:

1.simple

2.stable

3.one-to-one mapping.

o It avoids problems caused by aliasing & hence

can be used for HP & BS filters.

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Using ilinear Transformation

27

oGiven an analog transfer function H(s) whose

response at analog frequency Ais to be matched

toH(z) at the digital frequency D, we can design

H(z) in one of 2 ways:

1)Take C by matching A & the prewarped

frequency D, & obtain H(z) from H(s) using the

Equation 6.9. From Equation 6.12,

DA C 5.0tan

DAC

5.0tan

11 zzCssHzH

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2)Take a convenient value for C, (say C= 1). This

matches the response at prewarped frequency x

given by

H(s) is frequency scaled to H1(s) = H(sA/x),

& then H(z) is obtained from H1(s) using the

transformations = (z- 1)/(z+ 1), where C= 1.

Dx 5.0tan

xAsssHsH /1

111

zzs

sHzH

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Example 6.2

Consider a Bessel filter described by

Design a digital filter whose magnitude at f0 = 3

kHz equals the magnitude of H(s) at A = 4 rad/s

if the sampling rate S= 12 kHz.

The digital frequency is given by = 2f0/S =

0.5.

33

32

ss

sH

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Solution

We can solve the problem by using either one of

the two methods discussed earlier:

Method 1

C is selected by choosing the prewarped

frequency equal to A = 4 rad/s.

A = 4 = Ctan(0.5)

4

5.05.0tan

4

5.0tan

4

C

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H(s) is transformed toH(z) using

1

14

1

1

z

z

z

z

Cs

114

zz

sHzH

3

1

1212

1

1216

3

31

143

1

14

3

2

22

z

z

z

zz

z

z

z

zzH

72631

13

2

2

zz

zzH

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Method 2

Choose C = 1.

Then,H(s) is frequency scaled to

H1(s) =H(sA/x) =H(s(4/1)) =H(4s)

15.05.0tan5.0tan Dx

31216

3

3434

3221

ssss

sH

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H(z) is obtained from H1(s) using the

transformations = (z- 1)/(z+ 1):

111

zzs

sHzH

3

1

1212

1

163216

3

31

112

1

116

3

2

22

z

z

z

zz

z

z

z

zzH

72631

132

2

zz

zzH

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The magnitude of H(s)= ats = j=j4 matches

the magnitude of H(z)atz = ej= ej/2=j.

33

3

2

sssH

12133

31216

3

3434

3

24 jjjjsH js

313

36

313

39

144169

3639

1213

1213

1213

3

4

jj

j

j

jsH js

1695.0313

36

313

39 22

4

jssH

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72631

13

2

2

zz

zzH

1695.01252

144

1252

156 22

jzzH

Spectral Transformations for IIR

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Spectral Transformations for IIR

Filters

36

oAs with analog filters, the design of IIR filters

usually starts with analog LP prototype, which is

converted to digital LP prototype by appropriate

mapping & transformed to the required filter type

by appropriate spectral transformation.

oFor bilinear mapping, we can perform the

mapping & spectral transformation (in a single

step) on the analog prototype itself.

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Digital-to-Digital Transformations

37

oIf a digital LP prototype has been designed, thedigital-to-digital transformations of Table 6.1 can

be used to convert it to the required filter.

oThe LP to BP transformations & LP to BStransformations yield a digital filter with twice the

order of the LP prototype.

oThe LP to LP transformation is actually a special

case of the more general allpass transformation.

oThe digital LP prototype cutoff frequency is D.

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oAll digital frequencies are normalized to

= 2f/S

Example 6.3

A LP filter operates at S = 8 kHz. Its cutoff

frequency is 2 kHz. UseH(z) to design a HP filter

with a cutoff frequency of 1 kHz.

Solution

D= 2fC/S = 2(2)/8 = 0.5

C= 2fC/S = 2(1)/8 = 0.25

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Spectral transformation:

4142.0125.0cos

375.0cos

5.0cos

5.0cos

CD

CD

z

z

z

z

z

zz

4142.01

4142.0

4142.01

4142.0

1

74142.01

4142.026

4142.01

4142.031

14142.01

4142.03

72631

13

2

2

2

2

zz

zz

z

z

zz

zzHHP

0723.00476.0

128.0

2

2

zz

zzHHP

E l 6 4

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Example 6.4

A lowpass filter operates at

S= 8 kHz. Its cutoff frequency is 2 kHz. UseH(z)

to design a bandpass filter with band edges of 1

kHz & 3 kHz.Solution

D= 2fC/S = 2(2)/8 = 0.5

1= 2fC/S = 2(1)/8 = 0.252= 2fC/S = 2(3)/8 = 0.75

2- 1= 0.5

2+ 1=

72631

13

2

2

zz

zzH

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A1= 0,A2= 0

Spectral transformation:

125.0tan

25.0tan

5.0tan

5.0tan

12

DK

025.0cos

5.0cos

5.0cos

5.0cos

12

12

22

1

2

2

21

2

11

zz

zAzA

AzAzz

72631

13

72631

113

72631

13

72631

13

24

22

24

22

24

22

2

2

zz

z

zz

z

zz

z

zz

zzH

BP

f

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Direct Analog-to-Digital Transformations

for Bilinear Design

43

oAll stable transformations are free of aliasing &

introduce warping effects.

oOnly bilinear mapping offers a simple relation to

compensate for the warping.

oTable 6.2 shows the analog-to-digital

transformations of for bilinear design.

oThese can be used to convert a prewarped analog

lowpass prototype (with a cutoff frequency of 1

rad/s) directly to the required digital filter.

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Form Band Edges Mapping s Parameters

LP toLP

C C= tan(0.5C)

LP to

HP

C C= tan(0.5C)

LP toBP

1< 0 < 2 C= tan[0.5(2- 1)], = cos0or

LP to

BS

1< 0 < 2 C= tan[0.5(2- 1)], = cos0or

11

zCz

1

1

z

zC

112

2

2

zCzz

12

1

2

2

zz

zC

12

12

5.0cos

5.0cos

12

12

5.0cos

5.0cos

Bilinear Transformation for

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Bilinear Transformation for

Peaking & Notch Filters

45

oIf bilinear transformation is used to design a

second-order digital peaking (BP) filter with a 3

dB BW of & a center frequency of 0, westart with the LP analog prototype:

(6.20)

whose cutoff frequency is 1 rad/s.

11

ssH

Th l di i l LP BP f i i

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oThe analog-to-digital LP to BP transformation is

applied to obtain:

(6.21)

= cos0

(6.22)

C = tan(0.5) (6.23)

= 2 - 1

(6.24)

C

C

zCz

z

C

CzHBP

1

1

1

2

1

1 2

2

Si il l if bili f i i d

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oSimilarly, if bilinear transformation is used to

design a second-order digital notch (BS) filter

with a 3 dB notch BW of & a notch frequencyof 0, we start with LP analog prototype of

Equation 6.20.

oThe analog-to-digital LP to BS transformation is

applied to obtain:

(6.25)

C

Cz

Cz

zz

CzHBS

1

1

1

2

12

1

1

2

2

Th l f & C i b th E ti

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oThe value of & C are given by the Equation

6.22 & 6.23 respectively.

oIf either design requires for anAdB BW of ,the constant C is replaced byKC, where

(Ais in dB)

oThe LP prototype gain corresponds to an

attenuation ofA dB.

o0is used to determine the parameter .

110

1

1.0

AK

If only the band edges & are specified

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oIf only the band edges 1 & 2 are specified

instead of 0, can be found from the alternative

relation in terms of the given band edges.oHere, 0 can be found by means of geometric

symmetry of the prewarped frequencies:

oTo find the digital band edges of 1 & 2, we

can start from the expression:

210

5.0tan5.0tan5.0tan

5.0cos

5.0cos

5.0cos

5.0coscos

12

12

12

0

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From Eq. 6.24, we get 1= 2 - . Then,

Example 6.5

Design a peaking filter (BP filter) with a 3 dB BW

of 5 kHz & a center frequency of 6 kHz. The

sampling frequency is 25 kHz.

5.0coscos5.0cos012

5.0coscoscos5.00

1

12

5.0coscoscos5.00

1

22

5.0coscoscos5.00

1

2

Solution

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Solution

C =tan(0.5)= 0.7265

=cos0= 0.0628

Substituting these into Equation 6.21:

4.025

5223

S

fdB

48.0

25

6220

0

S

f

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Magnitude

spectrum of the

bandpass filter

7265.01

7265.01

7265.01

0628.02

1

7265.01

7265.0

2

2

zz

zzHBP

1584.00727.0

14208.0

2

2

zz

zzHBP

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The band edge 2 can be computed using

Equation 6.28:

2 = 2.1483

1 can be computed using Equation 6.24:

1= 2 - = 2.1483 - 0.4= 0.8917

4.05.0cos48.0coscos4.05.0 12

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As a result, these correspond to:

kHzS

f 55.32

10258917.0

2

3

1

1

kHzSf 55.8210251483.2

2

3

22