5 l lEC533: Digital Signal Processing

Lecture 9IIR Filter Design

9.1 – IIR Filter

Difference EquationNN

∑∑==

−+−=N

jj

N

ii jnybinxany

10)()()(

Transfer Function ∑ −N

i

ii za

0

∑ −

=

+= N

jj

i

zbzH 0

1)(

where; N is the filter’s order.ai,bj are the filter’s coefficients.

∑=j

j1

ai,bj are the filter s coefficients.

9.1 – IIR Filter – contd.• IIR filter have infinite-duration impulse responses, hence they can be

matched to analog filters, all of which generally have infinitely long impulse responses.

• The basic techniques of IIR filter design transform well-known analogfilters into digital filters using complex-valued mappings.

• The advantage of these techniques lie in the fact that both analog filter design (AFD) tables and the mappings are available extensively in the literature.

• The basic techniques are called the A/D filter transformation.

H th AFD t bl il bl l f l filt W l• However, the AFD tables are available only for lowpass filters. We also want design other frequency-selective filters (highpass, bandpass, bandstop, etc.)

• To do this, we need to apply frequency-band transformations to lowpass filters. These transformations are also complex-valued mappings, and they are also available in the literature.

9.2 – Design Methods

The following A/D filter transformation methods are used in calculating the coefficients of IIR filter:g ff f f

1. Impulse Invariant Method.2. Bilinear Z-transform Method.

to achieve a digital filter that has a certain specification equivalent to an g f p f qanalogue filter

NoteNote

We have no control over the phase characteristics of the IIR filter. Hence IIR filter designs will be treated as magnitude-onlydesigns.

9.3 – Impulse Invariant Method

• Here we require that the impulse response of the discrete system (digital filter) be the discrete version of the impulse response of the analogue system (filter), (H h l )(Hence the name: impulse invariant).

ZŁ-1 t = nTH(s) h(t) h(nT) X T H(z)

Steps:To remove the sampling effect:

(1/T) δ(t‐nT)Steps:1. Get H(s), continuous time transfer function of an analogue filter that satisfies the prescribed

magnitude response.2. Apply the inverse Laplace transform to get the impulse response h(t). h(n).3. Obtain a discrete version of h(t) by replacing t by nT h(nT).4. Apply the Z-transform to h(nT) to get H(z) and multiply by T.

h(t)h(nT)

h(n).

n ∞

n

Example - 1

Using the invariant impulse response method, design a digital filter that has the shown pole-zero distribution. jω

Solutionσ

x ω0

‐ a)]()][([)(

00 ωω jasjasassH

−++++

=

x - ω0

)])[()( 2

02 ω+++

=as

assH

)()cos()(1 thtesH at == −− ω L (1) )()cos()( 0 thtesH == ωL (1)

)cos( 0nTeh(nT) anT ω−= (2)aT 1)(1 T

zezTezTeh(nT)T aTaT

aT

×+−

−=× −−−−

−−

2210

10

)cos(21)cos(1

ωω

Z (3)

1aT

2210

10

))cos(2(1))cos(()( −−−−

−−

+−−

=zezTe

zTTeTzH aTaT

aT

ωω

Example - 1 – cont.

221

10

))cos(2(1))cos(()( −−−−

−−

+−

=zezTe

zTTeTzH aTaT

aT

ωω

0 ))cos(2(1 +− zezTe ωa0

Xx(n) y(n)

Xa1- b1

b

T

TX

- b2 TX

Tawhere, 0 =

aT

aT

Teb

TTea

Ta where,

01

0

)cos(2

)cos(−

−−=

ω

ω

aT1

eb

Teb 2

2

0 )cos(2−=

−= ω

9.4 – Bilinear Z-Transform (BZT) Methods t z

Impulse Invariant Method

ss1

z

Bilinear Z Transform Method

• It is the most important method of obtaining IIR filter coefficients.• In the BZT method, the basis operation is to convert an analogue filter H(s) into

i l di i l fil H( ) b i h bili i i

Bilinear Z‐Transform Method

an equivalent digital filter H(z) by using the bilinear approximation.

Ha(s) HD(z)

1 sTsT

...21

...21

2

2

+−

++=== − sT

sT

e

eez sT

sTsTQ 1st order bilinear

approximation...21 +e

21 sT+

212

sTz−

≅∴

9.4 – BZT Method – cont.

21 sTz

+≅ sTsTzz +=− 1

21 sTz−

≅

s2T1)z1-z

zz

+=

+=

(

21

2

2

12 −z112

+≅∴

zz

Ts

1

12)()( −=

= zsaD sHzH1+zT

s

9.4 – BZT Method – cont.

2

112

112

Tj

Tj

Tj

Tj eeT

zT

s Ω

Ω−

Ω

Ω

×−

=−

=Q211 TjTj

eeTzT Ω−Ω ++( )( )

2sin222 22 TjeeTjTj Ω

=−

=Ω−Ω

( )2cos222 TTeeT TjTj Ω=

+= Ω−Ω

( )t2 Tjj Ω( )2tan TT

jj aΩ=ω

( )2tan2 TΩ=ω Pre-Warping effect( )2tanTa =ω

⎠⎞⎜

⎝⎛=Ω −

2tan2 1 Taω Warping effect

p g ff

⎠⎜⎝ 2T

Digital frequency

p g ff

9.5 – Frequency Warping Effect (BZT)• The effect of the bilinear Z-transform is called

frequency warping: every feature appears in the f f th ti ti filt

ωaAnalogue frequency

frequency response of the continuous-time filter appears as it is in the frequency response of the discrete-time filter but at different frequency.

Ωa when

≤Ω≤−

∞≤≤∞−ππ

ω

σ

jω

S domain tan(θ)

Digital frequency

22ss

TT

Ω≤Ω≤

Ω−

≤Ω≤ σ

S1 domainΩs/2

θπ/2

‐ π/2jΩDigital frequency

• To compensate for this, every frequency specification the designer has a control over (ωc,ωs,…) has to be ‘prewarped’ by setting it with;

22- Ωs/2

( )2tan2 TT

Ω=ω 2

9.6 - IIR Filter Design Procedure Using BZTGiven specification in digital domain1 Given specification in digital domainConvert it into analog filter specification(prewarping)Design analog filter (Butterworth Chebyshov elliptic):H(s)

1

2

3 Design analog filter (Butterworth, Chebyshov, elliptic):H(s)Apply bilinear transform to get H(z) out of H(s)

3

4ω ω

sω32 ( )2tan2 T

TaΩ=ω

Ω|)(| ΩjeH

|)(| ωjHpω

1 1 1 1

( )2Ta

|)(| eH21 ε+

1

211ε+

A1

4 1

sΩpΩ πA1

1

1

112

)()(−

−

+

−⋅=

=zz

Ts

sHzHΩ

9.7 – IIR Filter Design Steps Using BZT

1. Prewarp any critical frequency in the digital filter specifications(ωc,ωp,ωs,…) using; ( ) TT 1t Ω

2. Use the digital filter specifications to find a suitable normalized prototype

( )sF

T; T2tan =Ω=ω

g f p f f p ypanalogue LPF, H(s), e.g., for butterworth;

• 1st order: 11)(+

=s

sH

• 2nd order:

1+s

121)(

2 ++=

sssH

1

For butterworth prototype filter

• 3rd order:

Where

Ap the passband ripple in dB

)1)(1(1)( 2 +++

=sss

sH

⎟⎠

⎞⎜⎜⎝

⎛ −A

A

p

s

1.0

1.0

10 110110log

p p ppAs the stopband attenuation in dB

⎟⎟⎠

⎞⎜⎜⎝

⎛⎠

⎜⎝ −≥

p

s

p

N

ωω

10log.2

110

9.7 – Design Steps – cont.

cs ωs

2. Denormalization according to filter type; (LPF, HPF, BPF, BSF).

• LPF

scω

ωω ss )( 20

2 +

s

s 210 where,; ωωω =2210 ωωω =

• HPF

• BPF

3 M f d i d i

)( 0

)( 20

2 ωω +sss 12

210

ωωω −=• BSF

3. Map from s-domain to z-domain;

11 −−=

zs

4 Realize the IIR filter

11 −+=

zs

4. Realize the IIR filter.

Example – 2

Design a 1st order Butterworth HPF with Ωc = 1 rad/sec, Fs = 1Hz.Solution 1)(H

1)(

+=

ssH scωs

ssH ==1)(

cc ss ωω ++1)(

⎞⎜⎛=⎞⎜⎛

Ω=1tantan Tcω⎠

⎜⎝

=⎠

⎜⎝

=2

tan2tan ccω

1)1( −− z

11

1

1

1

11

)1()1(

)(

)( −−

−

−

−

++−−

=−

+=zz

zz

zzHcc ωω

1)1()(

− ++ z

cccω

Example - 2 – cont.

1

11

11)(

−−=

zzH k 11

1111

)(−

+−

++ zc

cc

ωωω

c

kω+

=1

a0

Xx(n)

ab

y(n)

T

kX

10 =a where,X

a1- b1 TX

111

0

−−=

cb

a ω

1+=

c

c1b

ω

Example - 3

Design a 3rd order Butterworth LPF to have a cutoff frequency at 4 kHz using the BZT method & assuming a sampling frequency of 10 kHz.

1Solution secF

T & kHz fs

c41014 −===

rad Tc 513.2101042 43 =×××=Ω −πc

secrad T :prewarping c

c 0762.3)2tan( =Ω=ω

)1)(1(1)( 2 +++

=sss

sH

⎞⎜⎛

+⎞

⎜⎜⎛ −

+⎞

⎜⎜⎛ −⎞

⎜⎜⎛⎞

⎜⎜⎛

+⎞

⎜⎜⎛ −

=−−−

11111111

1)(12121 zzz

zH

⎟⎠

⎜⎜

⎝+⎠

⎜⎜⎝ +

+⎠

⎜⎜⎝ +⎠

⎜⎜⎝⎠

⎜⎜⎝

+⎠

⎜⎜⎝ + −−− 1

111

1 111 zzz ccc ωωω

Example - 3 – cont.

( )212211

211

)1()1(32490)1(1050)6751032491()1()1()( −−−−

−−

++++++

=z zzH ( )212211 )1()1(3249.0)1(105.0)675103249.1( ++−+−+ zz z z .

211 )21()1( −−− +++ zzz( )211 78060789143051)509501(3249.1

)21()1(−−− +++

+++=

z. z.. z . zzz

( )21

21

1

1

5457.02506.114305.1)21(

)509501(3249.1)1(

−−

−−

−

−

++++

×++

= z z

zzz .

z

( )21

21

1

1

54570250611)21(

)509501()1(52763.0 −−

−−

−

− ++×

+×=

zzz( )211 5457.02506.11)509501( +++ zzz.

Example - 3 – cont.

Filter Realization (Cascaded Realization)

0 52763

x(n)2- 1.2506

y(n)

T

0.52763

X- 0.5095

TX

TX

T

TX

- 0.5457

X

Example - 4

Using BZT method, design a digital filter meeting the specifications given by the following tolerance structure, |H(f)|

assume a Butterworth characteristic for thisfilter. 0.707

1

Solutionf

0.1

(1) From the specifications, the prewarped frequencies are: 0.5 2 4 kHz

19891205002tan =⎞⎜⎛ ×

=πω 198912.0

80002tan =

⎠⎜⎝ ×

=ω p

120002tan =⎞⎜⎛ ×

=πω 1

80002tan =

⎠⎜⎝ ×

=ωs

(2) D t i th d f th t t l B tt th LPF i th f ll i

Example - 4 – contd.(2) Determine the order of the prototype analogue Butterworth LPF using the following

relation,

⎞⎛⎟⎞

⎜⎜⎛ −

Aδ11

2

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−⎟⎟

⎠⎜⎜⎜

⎝−

−A

A

p

s

p

s

Nδ

δ

110

110log1)1(

1log

10

10

2

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎠⎝=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎠⎝≥

p

s

p

s

pN

ωω

ωω log2

110

log2

)(

⎠⎝⎠⎝ pp

995635.1199log1121

)1(1991

01.0111

22 =⎟⎠⎞

⎜⎝⎛⇒=−=−

−=−=− ,

ps δδ

402678.1198912.0

1log2log2

)(

=⎟⎠⎞

⎜⎝⎛×=⎟

⎟⎠

⎞⎜⎜⎝

⎛×

⎠⎝

p

s

ps

ωω

2.423.1402678.1995635.1

==≥

⎠⎝⎠⎝

N Let N

p

Example - 4 – contd.(3) A 2nd d B tt th l LPF h th l t f f ti H( ) i b(3) A 2nd order Butterworth analogue LPF has the s-plane transfer function, H(s), given by,

121)(

2 ++=

sssH

(4) For a denormalized LPF transfer function, substitute ps ωs

22

2

2 21)( psH

ωωω

++=

⎞⎛=

212

pp

pp

ssss ωω

ωω

++++⎟

⎟⎠

⎞⎜⎜⎝

⎛

(5) Applying the BZT,

2121

2

11

11)()(

1

1 p

zzs zz

sHzHω

⎞⎜⎛ −⎞

⎜⎛ −

==−−⎟

⎟⎠

⎞⎜⎜⎝

⎛

+−

= −

−

211

1

112

11

pp

z

zz

zz ωω +⎟

⎠

⎞⎜⎜⎝

⎛+

+⎟⎠

⎞⎜⎜⎝

⎛+ −−

⎠⎝ +

( )212 1 −+ zω ( )( ) ( )( ) ( )2121121 11121

1−−−− +++−+−

+=

zzz z

z

pp

p

ωω

ω

Example - 4 – contd.

( )212( )( ) ( ) ( )212221

212

21122121

)(−−−−−

−−

+++−++−

++=

zzzzzzz

zHpp

p

ωωω

( ) ( ) ( )pp

( )22122

212 21)(

−− ++=

zzzH pω

22122 )21()1(2)21()(

−− +−+−+++ zz ppppp ωωωωω

,parameterstheofvalues the ngSubstituti

96043.0132087.12122

22 −=−=++ ppp ,

,pfg

ωωω

0395659.07582858.021 22 ==+− ppp , ωωω

( )21210395659.0)(−− ++ zzH ( )

21 7582858.092086.132087.1)( −− +−=

z z zH

Example - 4 – contd.2 32087.1)21( 2 =++ pp by buttom& top Dividing ωω

( )212102995.0)(−− ++ zzH ( )

21 57408.04542.112102995.0)( −− +−

++=

z z zzzH

(6) Realizationa0x(n) y(n)k

(6) Realization,

X

Xx(n)

a1- b1

y(n)

TX

X

X

Xa2- b2 T

X

X

12102995.0

210 ====

a ,a ,a k where,

57408.04542.1210

=−= 21 b , b ,,