dsp_iir

11

Infinite Impulse Response Filter (IIF)

= =

===

N

0k

M

0kkk

0k)kn(ya)kn(xb)kn(x)k(h)n(y

Realizable IIR is characterized by the following recursive equqtion:

=

=

+== M

0k

kk

N

0k

kk

za1

zb

)z(X)z(Y)z(H

X(n) y(n)Direct 1 form

bN

-a1

-a2

-aM

.

.

.

.

.

.

b0

b1

b2

z-1

z-1

z-1

z-1

z-1

z-1

y(n)X(n)Direct 2 form

z-1

z-1

z-1

b0

-a1

-a2

-aM

.

.

.

.

.

.

b1

b2

bN

22

Design Steps

1. Filter Specification

2. Coefficient Calculations

3. Realisation

4. Implementation

Pole-Zero PlacementImpulse InvariantBilinear Z-Transform (BZT)

Coefficient Calculations Methods

33

Pole-Zero PlacementBasic Concept:

H(f)=0Zero in the z- plane

H(f)=maxPole in the z- plane

x

xFs/4 3Fs/4

f

H(f)

Fs/4

Fs/4

3Fs/4

Fs/2 0

0

Fs/2

44

Pole-Zero Placement

Example: Design Bandpass Digital Filter to meet the following Specifications:

1. Comlete signal rejection: f=0 (dc) and f=250 Hz;2. Narrow passband centred: 125 Hz;3. At 3 dB bandwidth: Bw=10 Hz;4. Sampling frequency: Ft=500 Hz.

Solution:

Fs=0; 3Fs/4=250 Hz;

250Hz

x

xf

H(f)

00

125Hz

125Hz

125Hz0 250Hz

r

r=1-(Bw/Ft)*=0.937

55

Pole-Zero Placement

The difference equation:

y(n) = - 0.877969 y(n-2)+x(n)- x(n-2)1

z-1 z-1

z-1

z-1

x(n) y(n)-1

- 0.877969

2

2

2/j2/j

z877969.01z1

)rez)(rez()1z)(1z()z(H

+=

=+=

66

Impulse invariant Method

[ ] pt1 Ce)s(HL)t(h;ps

C)s(H ===

Basic Concept:

H(s) h(t)LT Sampling

h(nt)ZT

H(z)

Ilustration:

1ze1C)z(H

zCez)nT(h)z(H

pT

0n

npnT

0n

n

=

== =

=

77


0 50 100 150 200 250 300 350 400 450 500-25

-20

-15

-10

-5

0

M

a

g

n

i

t

u

d

e

(

d

B

)

Frequency (Hz)

FN=1000/2; % Nyquist frequency

fc=300; % cut offf frequency

N=5; % filter order

[z, p, k]=buttap(N);% create an analog filter

w=linspace(0, FN/fc, 1000); % plot the response of filter

h=freqs(k*poly(z), poly(p), w);

f=fc*w;

plot(f, 20*log10(abs(h))), grid;

ylabel('Magnitude (dB)')

xlabel('Frequency (Hz)')

Matlab-files for design analog filter

Example:Design a IIF filter with Butterworth characteristic to meetthe followingSpecifications:

88


0 50 100 150 200 250 300 350 400 450 500-18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

Frequency (Hz)

M

a

g

n

i

t

u

d

e

(

d

B

)

Fs=1000; % sampling frequency

fc=300; % cutoff frequency

WC=2*pi*fc; % cutoff frequency in radian

N=5; % filter order

[b,a]=butter(N,WC,'s'); % create an analog filter

[z, p, k]=butter(N, WC, 's');

[bz, az]=impinvar(b,a,Fs);% determine coeff. of filter

[h, f]=freqz(bz, az, 512,Fs);

plot(f, 20*log10(abs(h))), grid



Matlab-files for design digital Impulse invariant filter

99

IIF design using Impulse invariant Method

wp=0.25*pi; ws=0.55*piRp=0.5; Rs=15; s=0.1*Rs; e2=(10^s)-1;

v=0.1*Rp; e3=10^v-1; k=ws/wp; N=log10(e2/e3)/(2*log10(k)); N=ceil(N)

[z p k]=buttap(N) % Buttap - Butterworth analog lowpass filter

[num,den]=zp2tf(z,p,k); %[num,den]- numerator and denominator coefficients of analog filter

[bz,az]=impinvar(num,den) % [bz,az] numerator BZ and AZ denominator coefficients of digital filter

[h omega]=freqz(bz,az,512);%[h omega]-frequency response of digital filter

g=20*log10(abs(h));

plot(omega/pi,g);'grid'

xlabel('Normalized frequeny')

ylabel('Gain,dB')

[bz,az]=impinvar(num,den)

)wp/wslog(2110)110log(

N10/Rp

10/Rs

=

Design IIF with following specifications:

Passband edge frequency: wp=0.25*pi;Stopband edge frequency: ws=0.55*pi;Sampling frequency: T=1Passband ripple: Rp=-0.5;Stopband ripple: Rs=-15;Matlab Program:

10

10

bz =-0.0000 0.0828 0.1657 0.0226 0

az =1.0000 -1.5587 1.2283 -0.4722 0.0733

Transfer function:

4321

321

z07.0z4.0z2.1z5.11z02.0z16.0z08.0)z(H

++++=

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-40

-35

-30

-25

-20

-15

-10

-5

0

5

Normalized frequeny

G

a

i

n

,

d

B

11

11

Bilinear Z-Transform (BZT)

Basic Concept:

H(s) H(z)

S-plane

x

x

x

x

Im

Rex

x

Im

Re

z-plane

x

xx

x xx

x

x

x

T2or1k;

1z1zks =+

=

12

12

Warping Effect

2T

tank p'p=

- analog frequency- digital frequency

T-sampling period

2Ttank' =

(analog)

(digital)

2Ttank;

1e1ekj 'Tj

Tj' =+

=

Relationship between and is nonlinear (tangensoidal) leading to a distortyion or warping of

digital frequency.Passband for the analog filter are with

regular intervals, whereas the passband for the digital

filterare somewhat squashed up. This effect is

Compensated for by prewarping the analog filter

before applying the BZT.

p- specified cutoff frequency p- prewarped frequency

'js;ez Tj == T2or1k;

1z1zks =+

=

13

13

Summary of the BZT

'p2

'p1

'p2

'p1

20

20

2s

s

20

2

'p

'p

W;

BSF);to(LPFs

WsBPF);to(LPFWss

HPF);to(LPFs

sLPF);to(LPFss

==+=

+=

==

1. Use the DF specifications to find a suitable analog LPF, H(s)2. Determine and prewrap bandedge frequencies

;2T

ktan;2T

ktan

2T

ktan

p2'P2

p1'P1

p'P

==

= For LPF and HPF

For Band pass and stopband filters

p- desired cutoff frequency ; p- prewarped critical frequency3. Denormalized analog filter or H(s) by replac s one of the following transformations

4. Apply BZT

1z1zks +

=

14

14

0 50 100 150 200 250 300 350 400 450 500-250

-200

-150

-100

-50

0

50

M

a

g

n

i

t

u

d

e

(

d

B

)

Frequency (Hz)

0 50 100 150 200 250 300 350 400 450 500-18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

Frequency (Hz)

M

a

g

n

i

t

u

d

e

(

d

B

)

Matlab-files using BZTFs=1000; % Sampling frequency

FN=Fs/2; % Nyquist frequency

fc=300; % cut offf frequency

N=5; % filter order

[z, p, k]=butter(N, fc/FN);

[h, f]=freqz(k*poly(z), poly(p), 512, Fs);





BZT

Fs=1000; % sampling frequency

fc=300; % cutoff frequency

WC=2*pi*fc; % cutoff frequency in radian

N=5; % filter order

[b,a]=butter(N,WC,'s'); % create an analog filter

[z, p, k]=butter(N, WC, 's');

[bz, az]=impinvar(b,a,Fs);% determine coeffs of IIR filter

[h, f]=freqz(bz, az, 512,Fs);




15

15

Matlab-files using BZT

Fs=2000; % Sampling frequency

FN=Fs/2; fc1=200/FN;

fc2=300/FN;

[b,a]=butter(4,[fc1, fc2]); % Create and digit. analogue filter.

[z,p,k]=butter(4, [fc1, fc2]);

subplot(2,1,1) % Plot magnitude freq. response

[H, f]=freqz(b, a, 512, Fs);

plot(f, abs(H))


ylabel('Magnitude Response ')

subplot(2,1,2) % Plot pole-zero diagram

zplane(b, a)

0 100 200 300 400 500 600 700 800 900 10000

0.2

0.4

0.6

0.8

1

Frequency (Hz)

M

a

g

n

i

t

u

d

e

R

e

s

p

o

n

s

e

-3 -2 -1 0 1 2 3-1

-0.5

0

0.5

1

4 4

Real Part

I

m

a

g

i

n

a

r

y

P

a

r

t

Problem: Design Bandpass filter with Butterworth characteristics rhat meets the following specifications:

Passband 200-300 HzSampling frequency 2000 HzFilter order 8

16

16

Matlab-files using BZTProblem: Design Bandpass filter with Butterworth characteristics rhat meets the following specifications:

Lower passband edge frequency 200 HzUppere passband edge frequency 300 HzLower stopband edge frequency 50 HzUppere stopband edge frequency 450 HzPassband ripple 3 dBStopband attenuation 20 dBSampling frequency 1 kHz

Determine the order N of filter; b) poles, zeros, gain, coefficients and transfer functon of DF

Fs=1000; % Sampling frequencyAp=3;As=20;Wp=[200/500, 300/500]; % Band edge frequenciesWs=[50/500, 450/500];[N, Wc]=buttord(Wp, Ws, Ap, As); % Determine filter order[zz, pz, kz]=butter(N,Wp); % Digitise filter[b, a]=butter(N, Wp);subplot(2,1,1) % Plot magnitude freq. response[H, f]=freqz(b, a, 512, Fs);plot(f, abs(H))xlabel('Frequency (Hz)')ylabel('Magnitude Response ')subplot(2,1,2) % Plot pole-zero diagramzplane(b, a)

0 50 100 150 200 250 300 350 400 450 5000

0.2

0.4

0.6

0.8

1

Frequency (Hz)

M

a

g

n

i

t

u

d

e

R

e

s

p

o

n

s

e

-3 -2 -1 0 1 2 3-1

-0.5

0

0.5

1

22

Real PartI

m

a

g

i

n

a

r

y

P

a

r

t

17

17

N=2

pz= [-0.1884 0.7791j ; 0.1884 0.7791j]zz= [1 1 -1 -1]

kz=0.0675

b= [0.0675 0 -0.1349 0 0.0675]

a= [1.0000 -0.0000 1.1430 -0.0000 0.4128]

42

42

z4128.0z1430.11z0675.0z1349.00675.0)z(H

+++=

y(n)= 0.0675x(n)- 0.1349x(n-2) +0.0675x(n-4) -1.1430y(n-2)- 0.4128y(n-4)

x(n)

0.0675

z-1 z-1

z-1

z-1

y(n)0.0675

- 0.4128

z-1z-1

-0.1349 z-1

z-1

- 1.1430

18

18

Realization of IIR

NM)kn(ya)kn(xb)n(yN

0k

M

1kkk =

= =

z-1

z-1

X(n) y(n)b0

b1

b2

-a1

-a2

w(n)

Direct 2 2nd order

z-1

z-1

z-1

z-1

-a1

-a2

b1

b2

b0X(n) y(n)

Direct 12nd order

= =

=2

0k

2

1kkk )kn(ya)kn(xb)n(y

=

=

=

=2

0kk

2

1kk

)kn(wb)n(y

)kn(wa)n(x)n(w

22

11

22

110

zaza1zbzbb)z(H

++++=

19

19

Cascade Realization of IIR

==

=

++++=

2/N

1k k

k2/N

1k2

21

1

22

110

)z(D)z(N

zaza1zbzbb)z(H 2

21

1k

22

110k

zaza1)z(D

zbzbb)z(N

++=++=

)z(D)z(N

)z(D)z(N)z(H;

)z(D)z(N

)z(D)z(N)z(H

)z(D)z(N

)z(D)z(N)z(H;

)z(D)z(N

)z(D)z(N)z(H

1

1

2

2

2

1

1

2

1

2

2

1

2

2

1

1

==

==

X(n)

z-1

z-1

b01

b11

b21

-a11

-a21

z-1

z-1

y(n)b0n

b1n

b2n

-a1n

-a2n

X(n)

z-1

b01

b11-a11

z-1

z-1

y(n)b0n

b1n

b2n

-a1n

-a2nCascade realization of a 3rd order IIR

20

20

Parallel Realization of IIR

=

+=2/N

1kk )z(HC)z(H

z-1

z-1

b01

b11-a11

-a21

z-1

z-1

b0n

b1n-a1n

-a2n

X(n)

y(n)

.

.

.

C

1k1

1k1

1k1k0

N

N

zaza1zbb)z(H,

abC

+++==

=

++++=

n

1k1

k11

k1

1k1k0

zaza1zbbC)z(H

21

21

Develop Cascade and Parallel Realization

)z0165.0z3419.0z1801.01)zz3z31(1432.0)z(H 321

321

++++=

1. For casacade realization H(z) is expressed in factored form:

1

1

21

21

z0490.01z1

z3355.0z1307.01zz211432.0)z(H

+

+++=

z-1z-1

z-1

y(n)

0.1307

-0.33557 1

2

10.1432

0.0490 1

1x(n)

1. For parallel realization H(z) is expressed usingpartial fraction expansion as the sum second- and first order sections:

7107.8z049.01

1764.10z3355.0z131.01

z08407.02916.1)z(H 1211

++=

z-1

z-1X(n) y(n)

z-1

-0.3355

0.1310 -0.0841

1.2916

-8.7107

0.0490

10.1764

22

22

Finite Wordlength Effects

The main errors in IIR:

1. ADC quantizing noise, with resulths from representing the samplesof x(n) by small number of bits

2. Coefficients ak and bk quantizing errors, caused by representing ak and bk by a finite number of bits

3. Overflow errors, which result from arithmetic operations and storage the results in a limited register length

One of reserves to reduce errors in IIF is broken H(z) into smaller first and second order bloks which are then

cconnected up in cascade or in parallel

Application of IIR1. Telecommunication- in Transmitter as antialiasing filter and in Receiver anti imaging filter

2. Digital Telephony-digital dual tone multifrequency touch-tone receiver

3. Clock recover in Data communications etc

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