chapter 8 design of infinite impulse response digital filter

Chapter 8Chapter 8

Design of infinite impulse Design of infinite impulse response digital filterresponse digital filter

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IIR filter– Recursive equation of IIR filter

– Transfer function of IIR filter

1. Basic property of IIR filters1. Basic property of IIR filters

1 20 1 2

1 21 2

0

1

( )1

1

NN

MM

Nk

kk

Mk

kk

a a z a z a zH z

b z b z b z

a z

b z

0

0 1

( ) ( ) ( )

( ) ( )

k

N M

k kk k

y n h k x n k

a x n k b y n k

3/47

– Transfer function of IIR filter• Factored form

1 2

1 2

( )( ) ( )( )

( )( ) ( )N

M

K z z z z z zH z

z p z p z p

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2. Design of IIR filter using analog 2. Design of IIR filter using analog filterfilter

Impulse invariant method– Identical impulse response of discrete filter to that of

analog filter

Analog Filter

Transfer Function

Digital Filter

Transfer Function

Impulse ResponseImpulse Response

Series

( )H s ( )H z

( )h t ( )h nT

Fig. 8-1.

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– Design a LPF using Impulse invariant method

( )a

H ss a

Fig. 8-2.

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• Inverse Laplace transform

• Sampling with T of inverse Laplace transform

• z-transform

( ) , 0ath t ae t

( ) anTnh h nT ae

0

0

1

0

1

( )

( )

1

z

z

z

nn

n

anT n

n

aT n

n

aT

H z h z

ae

a e

a

e

7/47

• Transfer function with single pole

• Inverse Laplace transform

• Sampling with T of inverse Laplace transform

1

( )N

i

i i

CH s

s p

1

( ) i

Np t

ii

h t C e

1

( ) i

Np nT

ii

h nT C e

8/47

• z-transform

• Using commutative law

0

0 1

( ) ( )

i

n

n

Np nT n

in i

H z h nT z

C e z

1 0

1

1 0

( )

( )

i

i

Np nT n

ii n

Np t n

ii n

H z C e z

C e z

9/47

• Using an infinite series

11

0

1( )

1i

i

p T np T

n

e ze z

11

( )1 i

Ni

p Ti

CH z

e z

10/47

– Repeated poles in designing filter• Repeated pole of l order

• z-transform

1 1

1 1

( 1)( )

( 1)! 1i

l li

l aT

a p

CH z

l a e z

( )( )

il

i

CH s

s p

11/47

– Complex number in designing filter (1)

• z-transform

– Complex number in designing filter (2)

• z-transform

( )( )( )

s aH s

s a jb s a jb

1

2 21

1 ( cos )( )

1 (2 cos ) z

aT

aTaT

e bT zH z

e bT z e

( )( )( )

bH s

s a jb s a jb

1

2 21

( sin )( )

1 (2 cos ) z

aT

aTaT

e bT zH z

e bT z e

12/47

– Example 8-1• second order Butterworth filter

• Partial fraction

• Impulse response function (T=1)

2

1( )

1 2H s

s s

2 2( )

( 1 ) 2 ( 1 ) 2

j jH s

s j s j

( 1 ) 2 1 ( 1 ) 2 1

1 2 1

1 2 1 2 2 2

2 2( )

1 1

2 sin(1 2)

1 2 cos(1 2)

j j

j jH z

e z e z

e z

e z e z

13/47

• Magnitude of impulse response function

– Summary of impulse invariance method(1)

(2) Multiply H(z) by T

1( ) ( )sT sz e

n

H z H s jmT

Fig. 8-3.

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Bilinear z transform– Replacing s in the transfer function depending on the

filter required

• Arranging to z variable

2 1

1

zs

T z

12

12

Ts

zT

s

15/47

• Replacing

• Considering frequency scaling

– Replacing ,

(1 )2 2

(1 )2 2

T Tj

zT T

j

s j

2 2

2 2

2 1

1

2

2tan

2

d

d

d d

d d

j T

j T

j T j T

j T j T

d

ej

T e

e e

T e eT

jT

s j dj Tz e

16/47

– Relationship between analog frequency and digital frequency

2tan

2dT

T

tan2 2

dTT

17/47

• Frequency warping

Fig. 8-4.

18/47

– Example 8-2

• Using bilinear z transform

• in transfer function

1( )

11 ( )

1

H zk z

z

1( )

1 ( )c

H ss

k

11 1( )

2 2H z z

19/47

1) Impulse response of analog filter :

2) Frequency response of analog filter:

3) Impulse response of digital filter :

4) Frequency response of digital filter:

1/22

1

1 ( )c 1( )c - tan

cos2dT

1 sin-tan

1 cosd

d

T

T

cTce

Magnitude =

phase =

1 1, , 0, 0,

2 2

Magnitude =

phase =

20/47

5) Relationship between and

tan2d

c

T

dT

21/47

– Example 8-3• Specification of the desired filter

– Filter response : -3dB at 1000Hz

: -10dB at 3000Hz

– Sampling frequency : 10kHz

– Monotonic decrease in transition region(1000~3000Hz)

• Digital parameter from specification

–

–

–

1/100000[sec]T

2 1000[rad/sec], 0.2dp dpT

2 3000[rad/sec], 0.6dr drT

22/47

• Considering Frequency warp

– Prewarp

• Determining order of Butterworth filter

tan( ) 0.32492 2p dpT T

tan( ) 1.37642 2

drr TT

21.376410 log 1 ( ) 10

0.3249N

1N

23/47

• Using bilinear z transform

1( )

1 ( / )p

H ss

0.3949( )

10.3249

10.2452( 1)

0.5095

H zzz

z

z

2 1

1

zs

T z

Fig. 8-5.

24/47

Two transformation method– Impulse invariant method

– Bilinear z transform

3. Comparing two transformation 3. Comparing two transformation methodmethod

sTz e

2 1

1

zs

T z

25/47

– Example 8-4• Transfer function of analog filter

• Frequency response

• Using impulse invariant method (1)

1.333( )

1.333

aH s

s a s

1

( , 0.25 sec )0.75

a T

2 2

1.333( ) ( )

1.333aH j H j

1( ) tan1.333

j

1 1

1.333( )

1 1 0.716aT

aH z

e z z

26/47

• Frequency response with

• Using bilinear z transform (2)

cos( ) sin( )

cos(0.25 ) sin(0.25 )

j Tz e T j T

j

2 2

1.333( ) ( )

1 0.716cos(0.25 ) 0.716sin(0.25 )

j TIIH e H j

1 0.716sin(0.25 )( ) tan

1 0.716cos(0.25 )j

2 1

1

( ) ( )

( 2) ( 2)

0.333 0.333

2.333 1.667

zs

T z

H z H s

aTz aT

aT z aT

z

z

27/47

• Frequency response

2 22

2 2

( ) ( )

0.777cos (0.25 ) 0.222cos(0.25 ) 0.555 0.777sin(0.25 ) 1.332sin(0.25 )

2.333cos(0.25 ) 1.667 2.333sin(0.25 )

j TBH e H j

12 2

1.332sin(0.25 )( ) tan

0.777cos (0.25 ) 0.222cos(0.25 ) 0.555 0.777sin (0.25 )j

28/47

Analog

Constant Impulse Response

Bilinear z Transform

Fig. 8-6.

29/47

– Example 8-5

• Partial fraction of impulse response

• Inverse Laplace transform

2

1( )

2 1H s

s s

/ 2 / 2( )

1 1

2 2

j jH s

j js s

1

/ 2

( ) ( )

2 sin2

t

h t H s

te

L

30/47

• Frequency response

• Analog filter with -3db at

2

4

1( )

( ) 2( ) 1

1

1

H jj j

2

2 2

7

2 3 7

( ) ( )

2

3.948 10

8.886 10 3.948 10

c

c

c c

H s H s

s s

s s

2 1000[rad/sec]c

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• Impulse invariant method

– Partial fraction using sampling period ( ) 410 [sec]T

( 2 )

1 ( 2 ) 2( 2 )2

2 sin( 2)( )

2 cos( 2)

c

c c

Tc c

T Tc

e T zH z

z ze T e

0.20.4443

2 2cT

3

1 2

2.449 10( )

1.158 0.4112

zH z

z z

32/47

• Bilinear z transform

2 1

1

2

22

2 2 2

2 2 2 2 2

2

2

( ) ( )

2 1 2 12

1 1

( 1)

4( 1) 2 2 ( 1) ( 1)

0.064( 2 1)

1.168 0.424

zB sT z

c

c c

c

c c

H z H s

z zT z T z

T z

z T z T z

z z

z z

33/47

Design of various filters using frequency transformation

4. Frequency transformation4. Frequency transformation

Analog low pass filter

(normalization filter)

Analogfrequency transform Low pass Low pass High pass Band pass Band reject

Desired digital filter

Bilinear z transform

or

Impulse invariant method

Fig. 8-7.

34/47

Analog

low pass filter

(normalization)

Digital

low pass filter

(normalization)

Analog

Low pass

High pass

Band pass

Band reject

Digital

Low pass

High pass

Band pass

Band reject

Bilinear transform

Bilinear transform

Analog frequency transform

Digital frequency transform

Fig. 8-8.

35/47

– Low pass filter• -3dB at

– High pass filter• By replacing to

• -3dB at

2

1( )

2 1H s

s s

2

1( )

( / ) 2( / ) 1c c

H ss s

1[rad/sec]c

( )1

sH s

s

( )c

sH s

s

s1

s

1[rad/sec]c

36/47

– Band pass filter• -3dB at

– Band reject filter• -3dB at

2 20s

ss

2 20

ss

s

1[rad/sec]c

1[rad/sec]c

Table. 8-1 Analog Frequency Transform

Low pass filter (cutoff frequency )

Low pass filter (cutoff frequency : )

High pass filter (cutoff frequency : )

Band pass filter(Upper cufoff frequency : ,Lower cufoff

frequency: , Band pass frequency : )

Band reject filter (Upper cufoff frequency : ,Lower cufoff frequency :

, Band reject frequency : )

1[rad sec]c ( )H s

c

ss

c

css

20 ,h l h l

h l2 2

0

( )h l

ss

s

20 ,h l h l

h l2 2

0

( )h lss

s

c

38/47

– Example 8-6• Specification of filter design

– -3dB at 10Hz

– Sampling frequency ( )

– Bilinear z transform

– Transfer function :

• Considering prewarp

2

1( )

1.414 1H s

s s

'tan tan 0.325

2 10c

c

T

100[Hz]sf

39/47

• Transfer function of analog LPF

• Bilinear z transform

2

2

1( )

1.414 10.325 0.325

0.105

0.46 0.105

H ss s

s s

11

2

2

2

1 2

1 2

0.105( ) ( )

1 10.46 0.105

1 1

0.105 0.21 0.105

1.56 1.79 0.645

0.067 0.135 0.067

1 1.147 0.413

zzs

H z H sz zz z

z z

z z

z z

z z

40/47

• For computational efficiency

• For accurate frequency

'2tan

2c

c

T

T

2 1

1

zs

T z

'tan

2c

c

T

1

1

zs

z

'2tan

2

2tan 65[rad / sec]

0.01 10

cc

T

T

41/47

Fig. 8-9.

42/47

– Example 8-7• Specification of filter design

– cutoff frequency ( )

– Sampling frequency ( )

– Bilinear z transform

– Transfer function :

• Cutoff frequency using prewarp

1( )

1LH ss

'tan tan 0.7265

2 5c

c

T

150[Hz]sf

30[Hz]sf

43/47

• Analog HPF using table 8-1

• Bilinear z transform

( ) ( )

1

0.72651

cs

L s

c

H s H s

s

ss

11

1

1

( ) ( )

0.5792 0.5792

1 0.1584

zzs

H z H s

z

z

44/47

Fig. 8-10.

45/47

– Example 8-8• Specification of filter design

– Band pass frequency ( )

– Sampling frequency ( )

– Order of filter : 2

– Bilinear z transform

• Using table 8-1

200 ~ 300[Hz]sf

2[kHz]sf

2 20

( )h l

ss

s

46/47

• Analog bandpass filter

1( )

1LH ss

20

'tan tan 0.3249

2 10

' 3tan tan 0.5095

2 20

0.1655

0.1846

hh

ll

h l

h l

T

T

2 20

2

( ) ( )

0.1846

0.1846 0.0274

s

s

Ls

H s H s

s

s s

47/47

• Bilinear z transform

11

2

1 2

( ) ( )

0.1367 0.1367

1 1.2362 0.7265

zzs

H z H s

z

z z