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Digital Signal Processing

Objectives

To learn and understanddesign procedure of IIR filter.the frequency response, and its

properties of IIR filter.

Lecture

11

IIR FilterDesign

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IIR filter design technique

IIR filters have infinite-duration impulse responses, hence theycan be matched to analog filters , all of which generally haveinfinitely long impulse responses.

Therefore, the basic technique of IIR filter design is to transformanalog filters into digital filters using complex-valued mappings,which is called the analog-to-digital (A/D) filter transformation.

We need to apply frequency-band transformations to lowpassfilters in order to design other frequency-selective filters(highpass, bandpass, bandstop, etc.).

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IIR filter design technique contd..)

Two approaches of basic IIR filter design technique :

The main problem with these approaches is that we have no controlover the phase characteristics of the IIR filter. Hence IIR filter designswill be treated as magnitude-only designs.

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IIR filter design technique contd..)

IIR filter design technique we will follow the following steps:

Design analog lowpass filters.Study and apply filter transformations to obtain digitallowpass filters.Study and apply frequency-band transformations to obtain

other digital filters from digital lowpass filters.

There are three widely used prototype analog fi l ters :Butterworth filters

Chebyshev filtersChebyshev type-IChebychev type-II

Elliptic filters.

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Magnitude squared-response contd..)

The specifications are shown in Fig. below, from which we

observe that | H a( j )|2 must satisfy

sa

pa

A j H

j H

at,1

at

1

1

2

2

2

2

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Magnitude squared-response contd..)

The parameters and A are related to parameter R p and A s as

The ripples 1 and 2 are related to and A by

1101

1log10 10/210 p

R p R

20210 10

110 /log s A s A

A

A

1

12

1

1

1

2

1

1

1

1

2

1

1

2 111

A

A

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Butterworth lowpass filter

The magnitude-squared response for an N -th order Butterworth

lowpass filter is given by

where c is the cutoff frequency in rad/sec. The plot of themagnitude-squared response is shown below:

N

c

a j H 22

1

1

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Butterworth lowpass filter contd..)

From this plot we can observe the following properties:

At = 0, | H a( j0)|2 = 1 for all N .At = c, | H a( j )|2 = for all N , which implies a 3 dB attenuation at c.

| H a( j )|2 is a monotonically decreasing function of .

| H a( j )|2 approaches an ideal lowpass filter as N .

| H a( j )|2 is maximally flat at = 0 since the derivative of all order existand are equal to zero.

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Butterworth lowpass filter contd..)

The main properties of | H a( j )|2 is

To determine the system function

Finally,

with for odd N

for even N

j s

aaa j H s H s H /

2

N c N N

c N

c

j saaa j s

j

j s

j H s H s H 22

2

2/

2

1

1

k N c

a p s s H

po lesLHP

12,....,1,0,/ N k e p N jk ck

12,....,1,0,2

N k e p N k

N j

ck

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Example-1

Given that . Determine the analog filter

system function H a( s).

Solution: Given

Hence N = 3, c = 0.5.

since N is odd

62

6411

j H a

32326

2

5.01

1211

6411

j H a

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12,....,1,0,/ N k e p N jk ck

5050 300 .. / je p 433.025.05.0 3/11 je p j

433.025.05.0 3/22 je p j 5.05.0 3/33 je p

433.025.05.0 3/44 je p j 433.025.05.0 3/55 je p j

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Example-1 contd..)

25050501250

43302505043302501250

50

2

432

3

....

.....

.

.

polesLHP

s s s

j s s j s

p s p s p s p s s H k

N c

a

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Poles in the left-hand plane are:

The analog filter system functions:

-0.25 + j 0.433

-0.25 j 0.433-0.50

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Design equations

The analog lowpass filteris specified by the parameters p, RP , s,and As.

Therefore the essence ofthe design in the case of

Butterworth filter is toobtain the order N and thecutoff frequency c.

p N

c

p

pa p

R

R j H

210

2

10

1

1log10or

log10,at

s N

c

s

sa s

A

A j H

210

2

10

1

1log10or

log10,at

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Design equations contd..)

Solving for N , c and s , we have

To satisfy the specifications exactly at p,

or To satisfy the specifications exactly at s,

s p

A R s p

N /log2

110/110log

10

10/10/10

N R p

c p2 10/ 110

N A s

c s2 10/ 110

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Example-2

Design a lowpass Butterworth filter to satisfy:

p = 0.2 s = 0.3 R p = 7 dB A s = 16 dBSolution:

Filter order

To satisfy the specifications exactly at p,

397.23.0/2.0log2

110/110log

10

10/1610/710

N

4985.0

110

2.0

110 6 10/72 10/

N R

pc

p

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Example-2

To satisfy the specifications exactly at s,

We can choose any c between the above two numbers. Let us

choose c = 0.5.

Therefore, we have to design a Butterworth filter with N = 3 andc = 0.5, which we did in example-1.

Hence 25.05.05.0125.0

2 s s s s H a

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5122.0

1103.0

110 6 10/162 10/

N A

sc

s

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Chebyshev lowpass filter

There are two types of Chebyshev filters:Chebyshev-I: These filters have equiripple response in the passband .

Chebyshev-II: These filters have equiripple response in the stopband.

Butterworth filters have monotonic response in both bands.

We noted that by choosing a filter that has an equiripple ratherthan a monotonic behavior, we can obtain a lower-order filter.

Therefore, Chebyshev filters provide lower order thanButterworth filters for the same specifications.

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Chebyshev lowpass filter contd..)

The magnitude-squared response of a Chebyshev-I filter is

where N is the order of the filter, E is the passband ripple factor,which is related to R p , and T N ( x) is the N th-order Chebyshev

polynomial given by

c N

a

T

j H 22

2

1

1

c N x

x

x

x N

x N xT where

1

10

,coshcosh

,coscos1

1

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Chebyshev lowpass filter contd..) ECR 305_L11

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MATLAB implementation

MATLABExamples

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MATLAB implementation

MATLAB provides a function called [z, p, k] =buttap (N) todesign a normalized (i.e. c = 1) Butterworth analog prototypefilter of order N, which returns zeros in z array, poles in p array,and the gain value k .

However, we need an unnormalized Butterworth filter witharbitrary c. But we have to scale the array p of the normalizedfilter by c, and the gain k by c N.

In the following function, called U_buttap (N, Omegac) , wedesign the unnormalized Butterworth analog prototype filter.

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u_buttap function

function [b,a] = u_buttap(N, Omegac);

% Unnormalized Butterworth lowpass filter prototype% --------------------------------------------------

% b = numerator polynomial coefficients of Ha(s)% a = denominator polynomial coefficients of Ha(s)% N = Butterworth filter order% Omegac = cutoff frequency in radian per sec

[z,p,k] = buttap(N); p = p*Omegac;k = k*Omegac^N;B = real(poly(z));

b0 = k; b = k*B;a = real(poly(p));

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Example 8.1

Given that . Determine the analog filter

system function H a( s).

Solution: Given

Hence N = 3, c = 0.5.

62

6411

j H a

32326

2

5.01

1211

6411

j H a

% File name ex8p1.m% Definition

N = 3;Omegac = 0.5;

% Calculation[z,p,k] = buttap(N);

p = p*Omegack = k*Omegac^N

p =

-0.2500 + 0.4330i-0.2500 - 0.4330i

-0.5000

k =

0.1250

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Example 8.1 contd..)

25.05.05.0125.0

433.025.05.0433.025.0125.0

5.0

2

432

3

po lesLHP

s s s

j s s j s

p s p s p s

p s s H

k

N c

a

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Example 8.3

Design a lowpass Butterworth filter to satisfy:

p = 0.2 s = 0.3 R p = 7 dB A s = 16 dBSolution: Filter order

To satisfy the specifications exactly at p,

To satisfy the specifications exactly at s,

397.2

3.0/2.0log2110/110log

10

10/1610/710

N

4985.0

110

2.0

110 6 10/72 10/

N R

pc

p

5122.0

110

3.0

110 6 10/162 10/

N A

sc

s

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fd b f i

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afd_butt function

% File name: afd_butt.m

function [b,a] = afd_butt(Wp,Ws,Rp,As);

% Analog Lowpass Filter Design: Butterworth

% ------------------------------------------------% b = numerator coefficients of Ha(s)% a = denominator coefficients of Ha(s)% Wp = Passband edge frequency in rad/sec; Wp>0

% Ws = Stopband edge frequency in rad/sec; Ws>Wp>0% Rp = Passband ripple in +dB; Rp>0% As = Stopband ettenuation in +dB; As>0% ------------------------------------------------

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afd_butt function contd..)

if Wp

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Example 8.3 contd..)

% File name: ex8p3.m% DefinitionWp = 0.2*pi; Ws = 0.3*pi; Rp = 7; As = 16;

% Analog filter design[b,a] = afd_butt(Wp,Ws,Rp,As)

% Frequency responsewmax = 0.5*pi; w = [0:500]*wmax/500;H = freqs(b,a,w);mag = abs(H); pha = angle(H);db = 20*log10((mag+eps)/max(mag));

% Plotting plot(w/pi,db); gridxlabel('Frequency in pi units','fontsize',15);ylabel('Decibel','fontsize',15);

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Example 8.3 contd..)

N = 3

b = 0.1238

a = 1.0000 0.9969 0.4969 0.1238

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References

1. Vinay K. Ingle, and John G. Proakis, Di gital SignalProcessing using M ATL AB , Thomson LearningBookware Companion Series, 2007. (pp. 313 385)

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Next class..

Filter Bank

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