[EE3101] Laboratory 2 Report

8/2/2019 [EE3101] Laboratory 2 Report

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EE3101 Digital Signal Processing Laboratory ReportAng Zhi Ping U066463H

EE3101 Laboratory 2: IIR Digital Filter Design

Name: Ang Zhi PingMatric No.: U066463H

Date: 12th November 2008

Q1.

432 6130.24140.36130.211)(

sssssHa

++++=

The commands executed are:

>> analog_pole = roots([1 2.6130 3.4140 2.6130 1])

analog_pole =

-0.3826 + 0.9239i

-0.3826 - 0.9239i

-0.9239 + 0.3827i

-0.9239 - 0.3827i

Since the real part of all poles of Ha(s) is negative (-0.3826 and -0.9239), all poles of the

filter lies in the open left half plane. This implies that the analog filter is stable.

Q2.

The commands executed are:

>> [H,w] = freqs([1],[1 2.6130 3.4140 2.6130 1]);

>> plot(w,20*log10(abs(H)),'k'),title('Magnitude Response of

H_a(s)'),xlabel('Frequency (rad s^-^1)'),ylabel('Magnitude

(dB)'),xlim([0 10]),ylim([-80 10]),grid on,hold;>> plot([0:10],-3*ones(length([0:10])),'--k');

>> plot([1 1],[-80,10],'--k');

Refer to Graph A.

The filter is a low pass filter with 3 dB cutoff frequency at 1 rad/s It has unity DC gain


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% repeated convolutionsb0 = [1]; % 1

b1 = conv(b0,a); % sb2 = conv(b1,a); % s^2b3 = conv(b2,a); % s^3b4 = conv(b3,a); % s^4b0(length(b4)) = 0; % Pad coefficients to fit the longest vector b4b1(length(b4)) = 0;b2(length(b4)) = 0;b3(length(b4)) = 0;B = [1];A = 1*b0 + 2.6130*b1 + 3.4140*b2 + 2.6130*b3 + 1*b4;[H,w] = freqz(B,A,8192);plot(w,20*log10(abs(H))),title('Frequency Response for IIR Filter Using

Derivative Approximation Method for T=0.05'),xlim([0

pi]),xlabel('Frequency (rad/sample)'),ylabel('Magnitude (dB)'),grid on;

Refer to Graph B.

Q4.

The script which is executed is:

% EE3101 Laboratory 2 Question 4% Ang Zhi Ping% U066463H

% The MATLAB function impinvar() is used which uses the impulse% invariance method to perform analog to digital filter conversionanalog_b = [1];analog_a = [1 2.6130 3.4140 2.6130 1];

% T=0.05[bz1,az1] = impinvar(analog_b,analog_a,1/0.05);

% T=0.25

[bz2,az2] = impinvar(analog_b,analog_a,1/0.25);

% T=0.45[bz3,az3] = impinvar(analog_b,analog_a,1/0.45);

% Computing digital frequency response[h1 w1] = freqz(bz1 az1 8192);


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(rad/sample)'),ylabel('Magnitude (dB)'),xlim([0 pi]),ylim([-80

10]),grid;

Refer to Graph C.

Q5.

The script which is executed is:


% The MATLAB function bilinear() is used which uses the bilinear% transformation to perform analog to digital filter conversionanalog_b = [1];analog_a = [1 2.6130 3.4140 2.6130 1];

% T=0.05[bz1,az1] = bilinear(analog_b,analog_a,1/0.05);



% Computing digital frequency response[h1,w1] = freqz(bz1,az1,8192);[h2,w2] = freqz(bz2,az2,8192);[h3,w3] = freqz(bz3,az3,8192);

% Plotting all magnitude responses on the same graphsubplot(3,1,1),plot(w1,20*log10(abs(h1))),title( 'Magnitude Response of

H(z) Using Bilinear Transformation Method (T=0.05)'),xlabel('Frequency

(rad/sample)'),ylabel('Magnitude (dB)'),xlim([0 pi]),ylim([-30010]),grid;subplot(3,1,2),plot(w2,20*log10(abs(h2))),title( 'Magnitude Response of

H(z) Using Bilinear Transformation Method (T=0.25)'),xlabel('Frequency


10]),grid;subplot(3,1,3),plot(w3,20*log10(abs(h3))),title( 'Magnitude Response of

( ) i ili f i h d ( 0 45)') l b l('


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Zooming in to see the 3 dB cutoff frequency, the T = 0.05 filter has a 3 dB cutoff at

exactly 0.05 rad (given = 1 rad/s). For T = 0.25, the 3 dB cutoff has shifted leftwards to

0.248 rad instead of 0.25 rad, and for T = 0.45, the 3 dB cutoff has shifted further to 0.442rad instead of 0.45 rad. This is a result of the non-linear mapping of the analog frequency

axis to the digital frequency unit circle. As the mapping function from analog to digital

frequency is given by the arctangent function, the mapping is approximately linear for

small values ofT but large analog frequencies will be mapped to smaller-than-expected

values of digital frequencies if the linear property is used as an estimate.

Refer to Graph E for the nonlinear frequency mapping observation.

Q6.

The standard form of third order Butterworth filters is:

)1)(1(

1

)( 2ssssH

a+++

=

Given 1,c = 0.1 rad/s and 2,c = 0.2 rad/s, the required transformed Butterworth filters are:

)2551)(51(

1)(

)100101)(101(

1)(

2,2

2,1

ssssH

ssssH

a

a

+++=

+++=

The following script is executed:


[h1,w1]=freqs([1],conv([100 10 1],[10 1]));[h2,w2]=freqs([1],conv([25 5 1],[5 1]));

% Plotting magnitude response of Butterworth filtersubplot(2,1,1),plot(w1,20*log10(abs(h1))),title( 'Magnitude Response of

Butterworth Filter (N=3,\Omega_c=0.1 rad s^-^1)'),xlabel('Frequency (rad


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Q7.

For frequency transformation method, the transformation equation is:

=

+

=

+

+

+

++

+

=

2tan

2cot,

2cos

2cos

,

1

1

1

21

1

2

1

1

12

12

12

21

21

1

k

Zk

kZ

k

k

ZZk

k

k

k

z

The following script is executed for both matched-Z and frequency transformation

methods:


analog_b1 = [1];

analog_a1 = conv([100 10 1],[10 1]);

analog_b2 = [1];analog_a2 = conv([25 5 1],[5 1]);

% The MATLAB function c2d() is used which converts an analog filter to% discrete filter using the 'matched' option which does the matched-Z% transformationsys1 = tf(analog_b1,analog_a1);sys2 = tf(analog_b2,analog_a2);

% Perform matched-Z transformation using T=1sysd1 = c2d(sys1,1,'matched');sysd2 = c2d(sys2,1,'matched');

% Extracting digital coefficientsnum1 = get(sysd1,'num');den1 = get(sysd1,'den');num2 = get(sysd2,'num');den2 = get(sysd2,'den');

b1 = num1{1};a1 = den1{1};b2 = num2{1};2 d 2{1}


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% The MATLAB function iirlp2bp() is used to perform lowpass to

bandpass% frequency transformation[b3,a3] = iirlp2bp(b1,a1,0.1/pi,[0.1 0.2]);[b4,a4] = iirlp2bp(b2,a2,0.2/pi,[0.1 0.2]);

% Calculating frequency response of frequency transformed bandpass% filters[h3,w3] = freqz(b3,a3,8192);[h4,w4] = freqz(b4,a4,8192);

% Plotting magnitude response of frequency transformed bandpass

filtersfigure(2),subplot(2,1,1),plot(w3,20*log10(abs(h3))),title( 'Magnitude

Response of Frequency Transformed Butterworth Filter (N=3,\Omega_c=0.1

rad s^-^1,\phi_1=0.1\pi rad,\phi_2=0.2\pi rad)'),xlabel('Frequency


10]),grid;subplot(2,1,2),plot(w4,20*log10(abs(h4))),title( 'Magnitude Response of

Frequency Transformed Butterworth Filter (N=3,\Omega_c=0.2 rad s^-^1,\phi_1=0.1\pi rad,\phi_2=0.2\pi rad)'),xlabel('Frequency


10]),grid;

Refer to Graph G (matched-Z transformation) and Graph H (frequency transformation).


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0 1 2 3 4 5 6 7 8 9 10-80

-70

-60

-50

-40

-30

-20

-10

0

10

Magnitude Response of Ha(s)

Frequency (rad s-1

)

Magnitude(dB)

7

Graph A


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0 0.5 1 1.5 2 2.5 3-150

-100

-50

0Frequency Response for IIR Filter Using Derivative Approximation Method for T=0.05

Frequency (rad/sample)

Magnitude(dB)

0 0.5 1 1.5 2 2.5 3-80

-60

-40

-20



Magnitude(dB)

0 0.5 1 1.5 2 2.5 3

-60

-40

-20



Magnitude(dB)

8

Graph B


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0 0.5 1 1.5 2 2.5 3

-100

-50

0

Magnitude Response of H(z) Using Impulse Invariance Method (T=0.05)


Magnitude(dB)

0 0.5 1 1.5 2 2.5 3-100

-80

-60

-40

-20

0



M

agnitude(dB)

0 0.5 1 1.5 2 2.5 3

-80

-60

-40

-20

0



Magnitude(dB)

9

Graph C


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0 0.5 1 1.5 2 2.5 3-300

-200

-100

0

Magnitude Response of H(z) Using Bilinear Transformation Method (T=0.05)


Magnitude(dB)

0 0.5 1 1.5 2 2.5 3-300

-200

-100

0



M

agnitude(dB)

0 0.5 1 1.5 2 2.5 3

-300

-200

-100

0



Magnitude(dB)

10

Graph D


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0.04 0.045 0.05 0.055 0.06

-4

-3.5

-3

-2.5

-2



Magnitude

(dB)

0.24 0.245 0.25 0.255 0.26

-4

-3.5

-3

-2.5

-2



M

agnitude(dB)

0.44 0.445 0.45 0.455 0.46

-4.5

-4

-3.5

-3

-2.5



Magnitude(dB)

11

Graph E


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-70

-60

-50

-40

-30

-20

-10

0

10

Magnitude Response of Butterworth Filter (N=3,c=0.1 rad s

-1)

Frequency (rad s-1

)

Magnitude(dB)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-70

-60

-50

-40

-30

-20

-10

0

10

Magnitude Response of Butterworth Filter (N=3,c

=0.2 rad s-1

)

Frequency (rad s-1

)

Magnitude(dB)

12

Graph F


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0 0.5 1 1.5 2 2.5 3-200

-150

-100

-50

0

Magnitude Response of Matched-Z Transformed Butterworth Filter (N=3,c=0.1 rad s

-1)


Magnitude(dB)

0 0.5 1 1.5 2 2.5 3

-200

-150

-100

-50

0

Magnitude Response of Matched-Z Transformed Butterworth Filter (N=3,c

=0.2 rad s-1

)


Magnitu

de(dB)

13

Graph G


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0 0.5 1 1.5 2 2.5 3-200

-150

-100

-50

0

Magnitude Response of Frequency Transformed Butterworth Filter (N=3,c=0.1 rad s

-1,

1=0.1 rad,

2=0.2 rad)


Magnitude(dB)

0 0.5 1 1.5 2 2.5 3

-200

-150

-100

-50

0

Magnitude Response of Frequency Transformed Butterworth Filter (N=3,c

=0.2 rad s-1

,1

=0.1 rad,2

=0.2 rad)


Magnitu

de(dB)

14

Graph H

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[EE3101] Laboratory 2 Report