Chapt 5 Design of Digital Filters_IIR

7/31/2019 Chapt 5 Design of Digital Filters_IIR

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2/2/2012Watumull Institute of Technology, Worli

Prepared by Chandrashekhar Padole, DSP-BE Computer, Mumbai Uni 1

Chapt 05

Design of Digital Filters ( IIR)B.E. Comps, Mumbai Uni

PrePrepared by Chandrashekhar Padole

Lecturer

Watumull Institute of Tech , Worli


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Chapt 05 Design of Digital Filters

Design of FIR filters

Design of IIR filters from analog filters

Frequency transformation

Design of digital filters based on least-squares method

Digital filters from analog filters

Properties of FIR filters

Design of FIR filters using windows

Comparison of IIR and FIR filters

Linear phase filters


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Need/Applications of filters

In wireless transmission such as FM transmission or mobilecommunication frequency selective filters are used to separate outthe one channel signal from another.

For noise removal

Speech Processing- Speech recognition, TTS

Biomedical signal processing to diagnosis

Signal analysis and synthesis

Data compression

DTMF( Dual Tone Multi-frequency Receivers)

Mobile receivers


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Common Filter Types

1

0.707

1 2

Low Pass Filter

Pass-band Transitionband

Stop-band

3-dB or cut-offfrequency

H(w)

1

0.707

12

High Pass Filter

Pass-bandTransitionband

Stop-band


H(w)

1

0.707

1 2

Band-pass Filter

P.B.T.B. S.B.


H(w)

S.B. T.B.

1

0.707

1 2

Band-stop Filter

P.B. T.B.


H(w)

S.B. T.B. P.B.


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Design of Digital Filter

1. Designing an analog prototype filter (s-domain) from a givenspecification

2. Transforming analog filters ( s-domain) to a digital filter( z-

domain)

3. Realization of z-domain transfer function

Step 1 is not only dependent on the problem specificationbut also on the transformation technique to be used in step2 to convert the analog filter into digital filter. Intransformation , analog frequencies are not mapped as it isin digital frequencies. This leads to a need of very important

step of modification of given specification while designingthe analog filter.


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Contd..

Filter

Specification

Modificationof

frequencies

Analog filter

Design

Transformation: analog to

digital

Digital Filter

z=f(s)s=f(z)

iableFrequencyDigital

iableFrequencyAna

var__

var_log_


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Designing Analog Filter

Depending on application requirements and availability , one has todecide which kind of response is suitable. ( Butterworth or Chebyshev orElliptic or any other type)Normally factors on which particular type of filter is selected , are

Order of filter ( trade-off between sharp cut-off and computations)

Allowed ripples in pass-band and/or stop-band


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Contd..

% order of the filtersN = 5;% cut-off normalized frequency

band = 0.5;% ripple in the pass-band (dB)

Rpass = 0.5;% ripple in the stop-band (dB)Rstop = 20; w = 0:pi/255:pi;[num, den] = butter(N, band, 'low');

butterfilter = abs(freqz(num, den ,w));

[num, den] = cheby1(N, Rpass, band);cheby1filter = abs(freqz(num, den ,w));[num, den] = cheby2(N, Rstop, band);cheby2filter = abs(freqz(num, den ,w));

[num, den] = ellip(N, Rpass, Rstop, band);

ellipfilter = abs(freqz(num, den ,w));F = [w/pi ; butterfilter; cheby1filter; cheby2filter; ellipfilter];F = F';


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Design of Butterworth type filter

The magnitude function of the Butterworth low pass filter is given by

( )[ ]

2/121

1)(

N

c

jH

+

=

where N is the order of filterc is cut-off frequency (3dB frequency)

At < c , |H(jw)| 1At > c , |H(jw)| decreases rapidlyAt = c , |H(jw)| =0.707 ( -3dB)

1

0.707

c

H(jw)

As order of filter increases , transition region becomes steeperand approaches ideal filter response as N , which is

practically impossible


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Contd..

Designing Butterworth filter has following steps

1. Determine the order of filter

2. Cut-off frequency

3. Determine transfer function in s-domain


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To determine the order of filter

In problem specification/requirements, two gains ( orattenuation) at two frequencies are given , which can be used

to determine the order of filter

1

Ap

As

1 2

H(w)

Since designed analog filter would undergo

through transformation by one of the technique( BLT, IIT etc) , modification of frequencies are

required.

For bilinear transformation

For impulse invariant method

2tan

2

T=

T=

Thus, now we have analog frequencyspecification1 and 2 ,


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Contd..

Now calculate the order of filter using this equation

p

s

A

A

p

s

N

log

1

1log

2

1 2

2

1

1

General formula

This equation is validonly if frequency

response is specified in

gains

p

s

P

S

N

log

1

1log

2

1

Normalized gains( A)

Gains in dB ()

Attenuation in dB()

where2

1

sAS = 2

1

pAP =

sS1.0

10

=

sS1.0

10=

pP1.0

10

=

pP1.0

10=


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Derivation for order of filter ( N )

For designing Butterworth filter , we will have analog frequencyspecificationp and s with corresponding gains Ap and As resp.

We have ,

( )[ ]2/121

1)(

N

c

jH

+

=

( ) 111

2

2

+

Np

c

pA

At p ,

( )[ ]2

2

11 sN A

c

s

+

At s ,

( ) 11 22

p

N

Acp ( ) 11

2

2

s

N

Acs


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Contd..

Considering equality from both equation and taking

ratio, we get

( )

=

11

11

2

22

p

sN

A

A

p

s

Taking log and solving we get,

p

s

A

A

p

s

N

log

1

1log

21

2

2

1

1Log can be of any

base but take same

base for num anddenum


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To determine cut-off frequency

Cut-off frequency,

N

p

p

c

A

21

112

=


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Derivation for cut-off frequency


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To determine the analog transfer function ( s-domain)

).().........)()((

1)(

1210 =

N

assssssss

sH

Where, each pole isgiven by

++

=

N

kj

ck es2

12

2


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Derivation for pole


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Butterworth polynomials for different orders N

Order ,n Factors of Polynomial Bn(s)

1 (s + 1)

2 s2 + 1.4142s + 1

3 (s + 1)(s2

+ s + 1)4 (s2 + 0.7654s + 1)(s2 + 1.8478s + 1)

5 (s + 1)(s2 + 0.6180s + 1)(s2 + 1.6180s + 1)

6 (s2 + 0.5176s + 1)(s2 + 1.4142s + 1)(s2 + 1.9319s + 1)

7 (s + 1)(s2 + 0.4450s + 1)(s2 + 1.2470s + 1)(s2 + 1.8019s + 1)

8 (s2

+ 0.3902s + 1)(s2

+ 1.1111s + 1)(s2

+ 1.6629s + 1)(s2

+ 1.9616s + 1)


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Design steps for Butterworth Filter ( summary)

Calculate the order of filter usingthis equation

p

s

A

A

p

s

N

log

1

1log

2

1 2

2

1

1 Cut-off

frequency, N

p

p

c

A

21

11

2

=

).().........)()((

1)(

1210 =

N

assssssss

sH

Where, each pole isgiven by

++

=

N

kj

ck es2

12

2

Transfer function


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Problem

Determine analog filter H(s) to design digital filter by IIT methodsatisfying following constraints

1)(707.0 jweH

2.0)( jweH

2/0

4/3

with T=1 s


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Solution

4/32/

2

1

=

=Given

2.0707.0

2

1

=

=

AA

Transformation technique is IIT

T

=

1

0.707

1 2

H(w)

1 2

2/11 ==T

with T=1 s 4/322 ==T

Order of filter

89.3

2/

4/3log

1

1log

2

1

log

11log

2

1 2

2

2

2

707.0

1

2.01

1

1

=

=

p

s

A

A

p

s

N

89.3N

4=NAs order filter should be integer


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Contd..

2/

1707.0

1

2/

11

81

21

22

=

=

=N

p

p

c

A

Cut-off frequency

Transfer function).().........)()((

1)(

1210 =

N

assssssss

sH

++

=

N

kj

ck es2

12

2

Ha(s) poles

++

=

Nkj

k es2

122

We use normalized poles ( with c =1 ) as indigital transformation , actual c would beconsidered


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Contd..

k ( 0N-1) = kth pole

0 -0.382+j0.92

1 -0.92+j0.382

2 -0.92-j0.382

3 -0.382-j0.92

++

=

N

kj

k es2

12

2

+

=82

0

j

es

+

=8

3

2

1

j

es

+

=8

5

2

2

jes

+

=8

7

2

3

j

es

=8

5

0

j

es

=8

7

1

j

es

=8

9

2

j

es

=8

11

3

j

es


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Contd..

))()()((1)(

3210 sssssssssHa

=

))92.0382.0())(382.092.0())(382.092.0())(92.0382.0((

1

jsjsjsjs ++=

)184.1)(1765.0(

1)(

22++

=ssss

sHa


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Problem

Design a low pass filter ( Butterworth) with following specificationsPass-band 0-500HzStop-band 2-4kHzPass-band ripple 3dB

Stop-band attenuation 20dBSampling Frequency 8kHzBilinear transformation to be used

Ans- Frequency warping2

tan2

T=

Order of filter N=2

Cut-off frequency c = ???

Transfer function12

1)(

2+

=ss

sHa


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Design of Chebyshev type filter


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Contd..

Chebyshev low-pass filter has magnitude response , given by

2/1

221

)(

+

=

c

NC

AjwH

where A is filter gain 11

2

1

=A

and c is cut-off frequency and

The chebyshev polynomial of Nth order CN(x) is given by

=

)coshcosh(

)coscos()(

1

1

xN

xNxCN

1||

1||

xfor

xfor


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Design Steps

Determine the order of filter

1

21

2

2

1

cosh

111coshA

N

Transfer function

11

2

1

=

A

Poles are calculated as :

N/12 11

++=

2

12

11

+=r

2

12

12

=r

Now, the kth normalized pole is given by

kkk jrrs sincos 12 += 1,......1,0.. = Nkfor

whereN

kk

2

)12(

2

++=

Actual poles can be given assk=sk c


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Problem

Determine analog chebyshev filter for following specifications ofdigital filter by using bilinear transformation technique

1)(8.0 jweH

2.0)( jweH

2.00

6.0

with T=1 s


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Solution

6.02.0

2

1

==Given

2.08.0

2

1

==

AA

Transformation technique is bilinear

1

0.707

1 2

H(w)

1 2

with T=1 s

Order of filter

20.1N 2=NAs order filter should be integer

2tan

2

T=

65.02

tan2 1

1 ==T

75.22

tan2 2

2 ==T

1

21

2

2

1

cosh

111

coshA

N

75.018.0111 22

1

===A

( ) 20.112.2

563.2

23.4cosh

)53.6(cosh

65.0

75.2cosh

12.0

1

75.0

1cosh

1

1

1

2

1

===

N

C d


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Contd..

2nd order transfer function

Now, the kth normalized pole is given by

kkk jrrs sincos 12+=

1,......1,0..=

Nkfor

whereN

kk

2

)12(

2

++=

2

12

11

+

=r

2

12

12

=rand

where

75.018.0

11

122

1

===A

N/12 11

++=

where

C td


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Contd..

73.175.0

175.01

2/1

2=

++=

75.0

73.1*2

173.165.0

2

1 22

11 =+

=+

=

r 375.0

73.1*2

173.165.0

2

1 22

12 =

=+

=

r

4

3

420 =+=

N

kk

2

)12(

2

++=

4

5

4

3

21

=+=

kkk jrrs sincos 12 +=

53.0265.04

3sin75.0

4

3cos375.00 jjs +=

+

=

53.0265.04

5sin75.0

4

5cos375.01 jjs =

+

=

C td


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Contd..

))((1)(

10 sssssHa

=

))53.0265.0())(53.0265.0((

1)(

jsjssHa

+=

35.053.0

1)(

2++

=ss

sHa

R t P bl f h b h


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Repeat Problem for chebyshev

Determine analog filter H(s)( chebyshev) to design digital filter byIIT method satisfying following constraints

1)(707.0 jweH

2.0)( jweH

2/0

4/3

with T=1 s

Solution


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Solution

4/32/

2

1

=

=Given

2.0707.0

2

1

=

=

AA

Transformation technique is IIT

1

0.707

1 2

H(w)

1 2

with T=1 s

Order of filter

358.2N 3=NAs order filter should be integer

1

21

2

2

1

cosh

111

coshA

N

11707.0111 22

1

===A

358.296.0

27.2

2/

4/3cosh

12.0

1cosh

1

2

1

==

N

2/11 ==T

4/322 ==T

T=

Compare with Butterworth .

Order of chebyshev is

smaller but at the cost ofripples in pass band

Contd


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Contd..

Transfer function for 3rd

order chebyshev

Analog to digital Transformation( Step II)


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Analog to digital Transformation( Step II)

Transforming analog filter in to digital filter ( s-domain z-domain)

From first step , we get the analog filter transfer function in s-domain with a low pass characteristics always.

If requirement is of other than low pass filter, transfer functionwould be transformed for desired characteristic filter ( high pass, band

pass ,band reject etc).

This transformation can be done either in analog domain itselfor in digital domain.

Cont


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Cont

Normalized LP-Analog filter( s-domain)

Transformation in s-

domain for HP/BP/BR

Analog to digitaltransformation

s z

Analog to digitaltransformation sz

( BLT, IIT etc)

Digital FilterLP

Transformation ins-domain forHP/BP/BR

Digital FilterHP/BP/BR

Digital FilterHP/BP/BR

De-normalization

Ss/

Frequency transformation in s-domain


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Frequency transformation in s-domain

Normalized LPanalog filter in s-

domain

Frequencytransformation in

s-domain

Desired Filter in S-domain

Desired Filter Frequency transformation

Low pass filter

High pass filter

Band-pass filter

Band-stop filter

c

sS

sS c

s

sQS

0

2

0

2 )(

+ 21

2

0 =

12

0

=Q

)(2

0

2

0

+

sQ

sS

Problem


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Problem

From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters

cs

BcsHa

+=)(

Solution


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Solution

A ) High Pass filter

cs

BcsH

LP

a+

=)(

ss

LP

a

HP

acsHsH /)()(

=

sc

Bcs

c

Bc

csc +

=+

=

c

HP

acs

BssH

+

=)(

Contd


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Contd..

B) Band pass filter

cs

sQ

BcsH

BP

a

+

+=

0

2

0

2 )()(

csQQs

sBc

.02

0

2

0

++

=

( )( ) 20

20

0

++=

ss

sBc

Qc

Q

Contd..


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Contd..

C) Band reject filter

csQ

s

BcsH

BR

a

++

=

)(

)(

2

0

2

0

)(

)(.2

0

2

0

2

0

2

++

+=

sQcs

sQBc

2

002

2

0

2 )(

+

+

+=

sQ

s

sB

c

20

02

2

0

20

02

2

++

+

++

=

sQ

s

B

sQ

s

Bs

cc

Analog to Digital Transformation


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Analog to Digital Transformation

Mapping of s-plane to z-plane is done such that

1. Points in left-hand plane of s-plane mapped into correspondingpoints

2. Points in right-hand plane of s-plane mapped into corresponding

points outside unit circle of z-plane3. j axis is mapped along the perimeter of unit circle of z-plane

j

s-planez-plane

1

Different Approaches


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pp

Following are different methods for obtaining digital filter from ananalog filter ( mapping of s-plane to z-plane)

1. Approximation of derivatives2. Impulse-invariant transformation

3. Bilinear Transformation4. Matched z-transformation

Approximation of Derivatives


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pp

Differentiation is given by

nT-T nT

y(t)

T

TnTynTy

dt

tdy

nTt

)()()( =

=

)1()( = nyny

where T is sampling interval

Differentiation in s-domain,

Differentiatory(t)

dt

tdy )(

ssH = )(

Differentiation in z-domain,

Differentiatory(n)

TzzH

1

1)(

=

T

nyny )1()(

Contd..


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Thus , if we replace s by 1-z

Problem


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Problem


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Problem


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Problem


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Problem


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Problem


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Problem


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End of Chapter 05


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Chapt 5 Design of Digital Filters_IIR