1

SUMMARY OF IIR FILTER THEORY

Yogananda Isukapalli

Analog Filter Specifications

w’p w’

s w’

Analog Frequency

AmplItude

ds

0

1+dp1

1-dp

|)(| w¢jH

Pass band Stop band

,for 1)(1 ppp jH wwdwd ¢£¢+£¢£-

,for )( ¥£¢£¢£¢ wwdw ssjH

,)(log20

,)1(log20

10

10

dBdB

ss

pp

da

da

-=

--=

where ap and as are the peak pass band ripple and minimum stop band attenuation in dB respectively

2

3

Analog Filterx(t) y(t)

åå==

+=M

kk

k

k

N

kk

k

k dttyd

bdttxd

aty00

)()()(

å

å

=

=

-= M

i

ii

N

i

ii

sb

sasH

0

0

1)(

Transfer function

0 0t

1

0

1

1

1

10

0)(

1)(

:)()(

³

<

-

=

-=

=

t

tbe

ba

th

sbasH

ExamplesHLth

Impulse response

4

Analog filters are stable when their poles are in the left half of the s-plane, where as digital filters are stable when their poles are confined within the unit circle.

z- plane Im(s)

Re(s)

S- plane Im(z)

Re(z)

Stable Stable

Analog Filter : Magnitude

Function :

Magnitude Square Function :

)()(|)()( www w ¢<¢==¢ ¢= jHjHsHjH js

ww ¢=-=¢ jssHsHjH |)()()( 2

Stability

5

Analog Butterworth Filters

Define : Lowpass Butterworth Filters (Nth order)

N

N

p

jH

or

jH

22

2

2

)(11|)(|

)(1

1|)(|

ww

www

¢+=¢

¢¢

+=¢

N > 50

N =1N =2

0

0.5

1.0

|H(j )|

w¢pw¢

w¢

( normalized prototype, =1)pw¢

Maximally flat response

6

Poles of the Lowpass Prototype (Normalized = 1)

Butterworth Filter

Magnitude Square Function :

N

jssN

N

ssHsH

sHsH

jHp

)(11)()(

11)()(

11|)(|

2

)(2

22

22

1

-+=-

¢+=-

¢+=¢

¢=-=¢

=¢

www

ww

w

pw¢

7

Roots :

b. N is even :

1 + s2N = 0 s2N = -1

Roots :

12,.........2,1,0.1.1

)()22(

-===Nkees N

kjN

kj

k

pp

12,.........2,1,0.1

)22(

-==

+

Nkes N

kj

k

pp

Poles: Solve: 1 + (-s2)N = 0

a. N is odd 1 - s2N = 0

s2N = 1

8

Poles of a normalized ( = 1) Butterworth (LPF) filter lie uniformly spaced on a unit circle in the s-plane at the following locations:

pw¢

NkNNkj

NNkes NNkj

k

,...,2,1 2

)12(sin

2)12(

cos2/)12(

=

úûù

êëé -+

+úûù

êëé -+

== -+ ppp

The poles occur in complex conjugate pairs and lie on the left-hand side of the s-plane.

9

Example: N = 3

s0 = 1, s1 = 1.ejp/3, s2 = 1.ej2p/3

s3 = 1.ejp, s4 = 1.ej4p/3, s5 = 1.ej5p/3

Im(s)

Re(s)

|sk|=1

1221)( 23 +++

=sss

sH

÷÷ø

öççè

梢

÷÷÷

ø

ö

ççç

è

æ

-

-

³

p

s

A

A

p

s

N

wwlog2

110

110log10

10

Butterworth Filter order N

10

Analog Chebyshev Filters

)(11|)(|

)/(1|)(|

:

222

222

wew

wwew

¢+=¢

¢¢+=¢

NLP

pN

CjH

CKjH

Define

p

Where :

CN( ) is the Nth order chebyshev polynomial.

= cos(Ncos-1 ) 0£ £1

= cosh(Ncosh-1 ) > 1

e is the ripple parameter 0 < e < 1 sets the ripple amplitude in the ripple passband 0£ e £1

( normalized prototype, pw¢ =1)

w¢

CN( )w¢

CN( )w¢

w¢w¢

w¢w¢

11

The ripple amplitude in dB is given by :

÷øö

çèæ+

-= 210 11

log10e

g dB

( )210 1log10 e+=

Fig: Chebyshev prototype low pass response

12

w¢ w¢ w¢w¢CN+1( ) = 2 CN( ) - CN-1( ) Also,

÷÷ø

öççè

梢

÷÷÷

ø

ö

ççç

è

æ

-

-

³-

-

p

s

A

A

p

s

N

ww1

10

101

cosh2

110

110coshChebyshev Filter order N

13Fig: Prototype Chebyshev denominator polynomials

,...,2,1 ,2

)12(

);1

(sinh1

where

),sin()cosh()cos()sinh(

1

NkNNk

N

js

k

kkk

=-+

=

=

+=

-

pb

ea

baba

Poles of a normalized ( = 1) Chebyshev LPFpw¢

14

Mapping analog filters to digital domain:

• Mapping differentials:

Backward: Tnynyny

dttdy

nTt

a ]1[][]][[)( )1( --=Ñ=

=

Forward:T

nynynydttdy

nTt

a ][]1[]][[)( )1( -+=Ñ=

=

)1(Ñ Backward or forward difference operator

Also, ]]][[[]][[ )1()1()( nyny kk -ÑÑ=Ñ

with ][(.))0( ny=Ñ

15

å

å

=

== N

k

kk

M

k

kk

a

sc

sdsH

0

0)(å

å

=

-=

-

-

-

=N

k

kk

M

k

kk

Tzc

Tzd

zH

0

10

1

]1[

]1[)(

Analog Filters Digital Filters

Mapping Functions

Tzs11 --

=Tzs 1-

=

Backward Difference Mapping

Forward Difference Mapping

• Both the mapping functions may produce satisfactory results only for extremely small sampling times and are generally not recommended.

16

Impulse Invariance Mapping:

åå=

-= -

=«-

=N

kTpk

N

k k

ka ze

czHpscsH

k1

11 1

)( )(

• Evaluation – Pole Mapping:

(a) The impulse invariant mapping maps poles from the s-plane’s j axis to the z-plane’s unit circle.

(b) All s-plane poles with negative real parts map to z-plane poles inside the unit circle. In other words, stable analog poles are mapped to stable digital filter poles.

(c) All poles on the right half of the s-plane, i.e. with positive real parts, map to digital poles outside the unit circle.

The impulse invariant mapping, thus, preserves the stability ofthe filter.

w¢

17

• Deficiency of the impulse-invariance mapping:If s-plane poles have the same real parts and imaginary parts

that differ by some integer multiples of 2p/T, then there are an infinite number of s-plane poles that map to the same location in the z-plane. This will result in aliased poles in the z-plane.

• Aliasing in digital frequency response:

....)2(1)(1)( +±¢+¢=¢

TjjH

TjH

TeH aa

Tj pwww

Digital frequency response reproduces the analog frequency response every 2p/T.To avoid aliasing for lower sampling rates, the pole mapping is restricted to the primary strip

TTpwp

£¢£-Thus each horizontal strip of the s-plane of width 2p/T is mappedonto the entire z-plane.

18

s-plane

z-plane

Re(z)

Im(z)|z|=1

Im(s)

Re(s)Tp

Tp

-

To prevent significant aliasing:

TjHa /for 0)( pww >¢»¢

Band limit the analog filter.

19

• Example of the impulse-invariance mapping:

Third order Butterworth low-pass filter’s transfer function:

1) s 1)(s(s1 H(s) 2 +++

=

866.0-0.5p ,866.0-0.5p ,1p:are poles threeThe

)866.05.0(s0.577e

)866.05.0(s0.577e

1)(s1

:as tscoefficien thegives Algebra)866.05.0(s

C)866.05.0(s

C1)(s

C

j0.866)0.5j0.866)(s- 0.5 1)(s(s1 H(s)

:isexpansion fraction partial The

321

j2.62j2.62-

321

jj

jj

jj

-=+=-=

+++

-++

+=

+++

-++

+=

++++=

20.

)866.0cos(2

))866.0cos(2(

)866.06/5cos(154.1

)866.06/5cos(154.1)866.0cos(2

H(z)

:as expressed and simplifiedfurther becan This

)e-(1)866.06/5cos(154.1z-

)e-(zz H(z)

get weg,Simplifyin)e-(1

1 )e-(1

1 )e-(1

1 H(z)

:as written becan function transfer theTherefore,

23

5.02

5.01

5.11

5.05.00

322

13

12

0

1j0.866)T(-0.5

5.02

T-

1j0.866)T(-0.51j0.866)T(-0.51T-

T

TT

TT

TT

TTT

T

eaTeeaTeea

TeebTeeTeb

whereazazaz

zbzb

zTe

zzz

-

--

--

--

---

-+

-

---+-

-=

+=

+-=

++=

+++-=

++++

=

+-+=

++=

p

p

p

21

11

2)(

-+

=zzTzH

ssH 1)( =

112

or 11

21

+-

=-+

=-

zz

Ts

zzTs

Bilinear Mapping:

• Evaluation – Pole Mapping:Stable analog filters are mapped from s-plane to z-plane as stable digital filters. Aliasing problem is eliminated.• Analog and Digital frequencies relationship for prewarping:

frequency Digital frequency Analog

®®¢

ww)2/tan( Tk ww =¢

22

Summary of Bilinear Transformation method1

23

Example 1:Low pass filter: Obtain the transfer function H(z) of the digital low pass filter to approximate the following transfer function:

121)(

2 ++=

sssH

Use Bilinear Transformation method and assume a 3 dB cut off frequency of 150Hz and a sampling frequency of 1.28kHz.

Soln. The critical frequency, and Fs =1/T = 1.28kHz, giving a prewarped critical frequency of

rad/s, 1502 ´= pw p

0.3857T/2)tan( ==¢ pp ww

24

Use of BZT and Classical Analog Filters

Analog Filters Review

N

p

jH2

2

)(1

1|)(|

www

¢¢

+=¢

• Low pass Butterworth filter of order N

Poles of a normalized ( = 1) Butterworth LPFpw¢

NkNNkj

NNkes NNkj

k

,...,2,1 2

)12(sin

2)12(

cos2/)12(

=

úûù

êëé -+

+úûù

êëé -+

== -+ ppp

÷÷ø

öççè

梢

÷÷÷

ø

ö

ççç

è

æ

-

-

³

p

s

A

A

p

s

N

wwlog2

110

110log10

10

Frequency response

Filter order N

(8.25)

(8.26)

(8.24)

25

Analog Filters Review contd

2. Low pass Chebyshev Type 1 Filter of order N

)/(1 |)(| 22

2

pNCKjH

wwew

¢¢+=¢Frequency

response

Poles of a normalized ( = 1) Chebyshev LPFpw¢

,...,2,1 ,2

)12(

);1

(sinh1

where

),sin()cosh()cos()sinh(

1

NkNNk

N

js

k

kkk

=-+

=

=

+=

-

pb

ea

baba

Filter order N

÷÷ø

öççè

梢

÷÷÷

ø

ö

ççç

è

æ

-

-

³-

-

p

s

A

A

p

s

N

ww1

10

101

cosh2

110

110cosh

(8.27)

(8.29)

(8.28)

26

Fig: Sketches of frequency response of some classical analog filters

(a) Butterworth response (b) Chebyshev type I © Chebyshev type II

(d) Elliptic

Analog Filters Review contd

3. Low pass Elliptic Filter of order N

)(1 |)(| 22

2

wew

¢+=¢

NGKjHFrequency

response(8.30)

27

Analog Frequency Transformations

1. Desired Low pass Low pass prototypeThe low pass to low pass transformation is:

p

ssw¢

=

Denote frequencies for

p

lpp

p

lpp jjww

www

w¢

=¢

= i.e.

lpw PrototypepwLPF and

From 8.21 (a)

(8.31)

28

2. Desired High pass Low pass prototypeThe low pass to high pass transformation is:

ss pw¢= From 8.21 (b)

Denote frequencies for hpw Prototype pwHPF and

hp

pp

hp

pp jjww

www

w¢

-=¢

-= i.e. (8.32)

29

30

Design Example

31

32

References

1. “Digital Signal Processing – A Practical Approach” -Emmanuel C. Ifeachor and Barrie W. Jervis Second Edition