LABEX 4 Xử lý tín hiệu số

HBP,1z = transfer function of the IIR bandpass transfer function HBP,2z = by < Insert MATLAB figures here.. > of a stable IIR bandstop filter as HBSz = The plot of the gain response of

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June 20, 1998

Name:Tr n Qu c L m 1063739ầ ố ắ

Nguyễn Minh Đẵng 1063723

Laboratory Exercise 4 LINEAR, TIME-INVARIANT DISCRETE-TIME SYSTEMS:

FREQUENCY-DOMAIN REPRESENTATIONS 4.1 TRANSFER FUNCTION AND FREQUENCY RESPONSE

Project 4.1 Transfer Function Analysis

Answers:

Q4.1

The modified Program P3_1 to compute and plot the magnitude and phase spectra of a

% Program P3_1

% Evaluation of the DTFT

clf;

% Compute the frequency samples

of the DTFT

w = 0:8*pi/511:2*pi;

m = 5;

num =ones(1,m)/m;

h = freqz(num, 1, w);

% Plot the DTFT

subplot(2,1,1)

plot(w/pi,abs(h));grid

subplot(2,1,2) plot(w/pi,angle(h));grid

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0

0.5

1

Magnitude Spectrum |H(e j ω )|

ω / π

-2 0 2 4

Phase Spectrum arg[H(e j ω )]

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The types of symmetries exhibited by the magnitude and phase spectra are due to

The results of Question Q2.1 can now be explained as follows

-

0

0.5

1

-2

-1

0

1

2

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.5

1

Magnitude Spectrum |H(e )|

ω / π

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-4

-2

0

2

4

Phase Spectrum arg[H(e j ω )]

ω / π

nhau

t t h n ố ơ

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

2

4

6

8

Magnitude Spectrum |H(ejω )|

ω / π

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1

-0.5

0

0.5

1

Phase Spectrum arg[H(ejω )]

ω / π

Questions 4.2 and 4.3 obtained using the program developed in Question Q3.50

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0 10 20 30 40 50 60 70 80 90 100

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

impule

ω / π

zplane are shown below:

4

0 10 20 30 40 50 60 70 80 90 100

-4

-3

-2

-1

0

1

2

3

4

5

6x 10

ω / π

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-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

4.2 TYPES OF TRANSFER FUNCTIONS

Project 4.2 Filters

% Program P4_1

% Impulse Response of Truncated

Ideal Lowpass Filter

clf;

fc = 0.25;

n = [-6.5:1:6.5];

y = 2*fc*sinc(2*fc*n);k = n+6.5;

13 -0.2 0.6]);

Answers:

0 2 4 6 8 10 12

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

N = 13

Time index n

[-6.5:1:6.5];

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 -2

-1.5 -1 -0.5 0 0.5 1 1.5 2

Real Part

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The parameter controlling the cutoff frequency is – Fc = 0.25

response of the FIR lowpass filter of Project 4.2 with a length of 20 and a cutoff

% Program P4_1

% Impulse Response of

Truncated Ideal Lowpass Filter

clf;

fc = 0.45;

n = [-10:1:10];

y = 2*fc*sinc(2*fc*n);k = n+10;

Program P4_1 to compute and plot the impulse response of the FIR lowpass filter of Project 4.2 with a length of 15 and a cutoff frequency of ωc = 0.65 are as

% Program P4_1

% Impulse Response of

Truncated Ideal Lowpass Filter

clf;

fc = 0.65;

n = [-7.5:1:7.5];

y = 2*fc*sinc(2*fc*n);k = n+7.5;

6

0 2 4 6 8 10 12 14 16 18 20 -0.2

0 0.2 0.4 0.6 0.8

N = 20

Time index n

-0 2 -0 1 0 0.1 0.2 0.3 0.4 0.5 0.6

N = 1 5

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Q4.10 The MATLAB program to compute and plot the amplitude response of the FIR

% Program P4_1

% Impulse Response of Truncated

Ideal Lowpass Filter

clf;

fc = 0.25;

n = [-6.5:1:6.5];

y = 2*fc*sinc(2*fc*n);k =

n+6.5;

lowpass filter for several

-0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6

N = 13

Time index n

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0 2 4 6 8 10 12

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

N = 15

Time index n

% Program P4_2

% Gain Response of a Moving

Average Lowpass Filter

clf;

M = 2;

num = ones(1,M)/M;

[g,w] = gain(num,1);

plot(w/pi,g);grid axis([0 1 -50 0.5])

Answers:

Q4.11 A plot of the gain response of a

length-2 moving average filter obtained using Program P4_2 is

8

-0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6

Time index n

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

ω / π

M = 2

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From the plot it can be seen that the 3-dB cutoff frequency is at – 0.5

Q4.12 The required modifications to Program P4_2 to compute and plot the gain response

% Program P4_2

clf;

M = 2;

num =[1 1]/M;

The plot of the gain response for a cascade of 3 sections obtained using the

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

M = 3

Q4.13 The required modifications to Program P4_2 to compute and plot the gain response

% Program P4_2

clf;

M = 5;

num =[1 -1 1 -1 1]/M;

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The plot of the gain response for M = 5 obtained using the modified program is

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

M = 5

Q4.14 From Eq (4.16) for a 3-dB cutoff frequency ωc at 0.45 π we obtain α = 0.0787

function of the first-order IIR lowpass and highpass filters, respectively, given by

HLP(z) =

HHP(z) =

10

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

ω / π

M = 5

A plot of the magnitude response of the sum HLP(z) + HHP(z) obtained using

< Insert MATLAB figure(s) here Copy from figure window(s) and paste >

Q4.15 From Eq (4.24), we get substituting K = 10, B = 1.86

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

ω / π

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Using this value of α in Eq (4.22) we arrive at the transfer function of the cascade

of 10 IIR lowpass filters as

1 – α z 1









10

= (7.85+7.85z-1)/(2+13.7z-1)

IIR lowpass filter

HLP,1(z) = 1 – α

1 – α z 1 = (7.7+7.7z-1)/(2+13.4z-1)

< Insert

Q4.16 Substituting ωo = 0.61 π in Eq (4.19) we get β = cos(0.61 π ) =

Substituting ∆ω3dB = 0.15 π in Eq (4.20) we get (1 + α2)cos(0.15 π ) − 2 α = 0,

transfer function of the IIR bandpass transfer function

12

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -50

-45 -40 -35 -30 -25 -20 -15 -10 -5 0

M = 21

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -50

-45 -40 -35 -30 -25 -20 -15 -10 -5 0

HHP

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HBP,1(z) =

transfer function of the IIR bandpass transfer function

HBP,2(z) =

by

of a stable IIR bandstop filter as

HBS(z) =

The plot of the gain response of the transfer function HBS(z) obtained using MATLAB is shown below:

From these plots we observe that the designed filters do/do not meet the

A plot of the magnitude response of the sum HBP(z) + HBS(z) obtained using

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Q4.17 The transfer function of a comb filter derived from the prototype FIR lowpass filter

of Eq (4.38) is given by

G(z) = H0(zL) =

Plots of the magnitude response of the above comb filter for the following values of

= _and _ peaks at ωk = _ = _, where k = 0, 1, , _.

Q4.18 The transfer function of a comb filter derived from the prototype FIR highpass filter

of Eq (4.41) with M = 2 is given by

G(z) = H1(zL) =

Plots of the magnitude response of the above comb filter for the following values of

From these plots we observe that the comb filter has _ notches at ωk = _

Q4.19 A copy of Program P4_3 is given below:

% Program P4_3

% Zero Locations of Linear Phase

FIR Filters

clf;

b = [1 -8.5 30.5 -63];

num1 = [b 81 fliplr(b)];

num2 = [b 81 81 fliplr(b)];

num3 = [b 0 -fliplr(b)];

num4 = [b 81 -81 -fliplr(b)];

n1 = 0:length(num1)-1;

subplot(2,2,1); stem(n1,num1);

pause subplot(2,2,1); zplane(num1,1);

subplot(2,2,2); zplane(num2,1);

are');

14

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are');

disp(roots(num2));

are');

disp(roots(num3));

are');

disp(roots(num4));

- Các tín hiệu kế tiếp nhau cho ta thấy các giá trị M của tín hiệu là khác nhau : tín

hiệu 1 và 3 thì giá trị M chẵn và tín hiệu 2 và 4 thì giá trị M lẽ.

- Tín hiệu 1 có M=8 và a = 1.979

- Tín hiệu 2 có M= 9 và a = 1

- Tín hiệu 4 có M = 9 và a = 1

The plots of the impulse responses of the four FIR filters generated by running Program

From the plots we make the following observations:

-100 -50 0 50 100

Time index n

Type 1 FIR Filter

-100 -50 0 50 100

Time index n

Type 2 FIR Filter

-100 -50 0 50 100

Time index n

Type 3 FIR Filter

-100 -50 0 50 100

Time index n

Type 4 FIR Filter

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Filter #4 is of length with a impulse response and is

Plots of the phase response of each of these filters obtained using MATLAB are

Q4.20 The plots of the impulse responses of the four FIR filters generated by running

% Program P4_3

% Zero Locations of Linear Phase

FIR Filters

clf;

b = [1 -8.5 30.5 -63];

num1 = [b 81 fliplr(b)];

num2 = [b 81 81 fliplr(b)];

num3 = [b 0 -fliplr(b)];

num4 = [b 81 -81 -fliplr(b)];

pause subplot(2,2,1); zplane(num1,1);

are');

disp(roots(num1));

are');

disp(roots(num2));

are');

16

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are');

disp(roots(num4));

- Các tín hiệu kế tiếp nhau cho ta thấy các giá trị M của tín hiệu là khác nhau : tín

hiệu 1 và 3 thì giá trị M chẵn và tín hiệu 2 và 4 thì giá trị M lẽ.

- Tín hiệu 1 có M=8 và a = 1.979

- Tín hiệu 4 có M = 9 và a = 1

Plots of the phase response of each of these filters obtained using MATLAB are

-100 -50 0 50 100

Time index n

Type 1 FIR Filter

-100 -50 0 50 100

Time index n

Type 2 FIR Filter

-100 -50 0 50 100

Time index n

Type 3 FIR Filter

-100 -50 0 50 100

Time index n

Type 4 FIR Filter

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Answers:

Q4.21 A plot of the magnitude response of H1(z) obtained using MATLAB is shown below:

H1(z) by _ and arrive at a bounded-real transfer function

H2(z) =

Q4.22 A plot of the magnitude response of G1(z) obtained using MATLAB is shown below:

G1(z) by _ and arrive at a bounded-real transfer function

G2(z) =

18

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4.3 STABILITY TEST

A copy of Program P4_4 is given below:

< Insert program code here Copy from m-file(s) and paste >

Answers:

Q4.23 The pole-zero plots of H1(z) and H2(z) obtained using zplane are shown below:

Q4.24 Using Program P4_4 we tested the stability of H1(z) and arrive at the following

Q4.25 Using Program P4_4 we tested the root locations of D(z) and arrive at the following

circle.

Q4.26 Using Program P4_4 we tested the root locations of D(z) and arrive at the following

circle.

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