My Dsp Lab Report

7/31/2019 My Dsp Lab Report

1/21

Lab Report

Of

Digital Signal Processing

Submitted in partial fulfillment of the requirement for the degree of

Master of TechnologyIn

VLSI Design

Submitted To

Ms. Sunita PrasadM. Tech (VLSI) Division,CDAC, NOIDA

Submitted By

Kumar Shekhar02511805211

M. TECH. (VLSI) IInd Semester

Centre for Development of Advanced Computing, NoidaAffiliated to

Guru Gobind Singh Indraprastha UniversityKashmere Gate, Delhi 110006


2/21

INDEX

SL.NO. EXPERIMENT NAME DATE REAMRK

1

2

3

4

5

6

7

8

9

10


3/21

Experiment -1

AIM :- Write a MATLAB program to generate the basic signal sequences andplot them.

(i) Unit Impulse Signal(ii) Unit Step Sequence(iii) Exponential Sequence(iv) Sinusoidal Sequence

Program:-

(i) x=[-3:1:5]y=[zeros(1,3),ones(1),zeros(1,5)]subplot(2,2,1)stem(x,y,'b')Xlabel('Time')

Ylabel('Unit Sample Output')

(ii) x=[-3:1:5]y=[zeros(1,3),ones(1),ones(1,5)]subplot(2,2,2)stem(x,y,'b')Xlabel('Time')Ylabel('Unit Step Output')

(iii) x=[-5:1:5]y=exp(x)subplot(2,2,3)stem(x,y,'b')Xlabel('Time')Ylabel('Exponential Output')

(iv) n=0:.2:2*pi;

x= sin(n);subplot(1,1,1)stem(n,x);Xlabel('Time')Ylabel('Sinusoidal Output')

Output at Command Window:-


4/21

(i)

-3 -2 -1 0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time

UnitSampleOutput

(ii)

-3 -2 -1 0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time

UnitStepOutput


5/21

(iii)

-5 -4 -3 -2 -1 0 1 2 3 4 50

50

100

150

Time

ExponentialO

utput

(iv)

0 1 2 3 4 5 6 7-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time

SinusoidalOutput


6/21

Experiment No 2

Aim :

1. Let x(n) = u(n)-u(n-10). Decompose x(n) into even and odd decompositiousing MATLAB.2. Determine computation y(n)=x(n)*h(n) of the following sequences

X[n]= [ 3, 11, 7, 0, -1, 4, 2] -3 n 3h[n]= [ 2, 3, 0, -5, 2, 1] -1 n 4

3. Given the following Difference Equation

y(n) - y(n-1) + 0.9 y(n-2) = x(n) ;

(a)Calculate and plot the impulse response at n=-20,100.(b)Calculate and plot the unit step response at n=-20,100.(c)Is the system specified by h(n) stable?

SOLUTION 1:

%Funtion for Even & odd parts

function[xe, xo, m]= evenodd(x,n)m = -fliplr(n);

m1 = min([m,n]);m2 = max([m,n]); m = m1:m2;nm = n(1)-m(1); n1 = 1:length(n);x1 = zeros(1,length(m));x1(n1+nm) = x; x = x1;xe = 0.5*(x+fliplr(x));xo = 0.5*(x-fliplr(x));

%Function Call

n=[0:10]; x=stepseq(0,0,10)-stepseq(10,0,10);[xe,xo,m]=evenodd(x,n);figure(1);clfsubplot(2,2,1); stem(n,x); title('Rectangular pulse')xlabel('n');ylabel('x(n)');axis([-10,10,0,1.2])subplot(2,2,2);stem(m,xe);title('even part')xlabel('n');ylabel('xe(n)');axis([-10,10,0,1.2])subplot(2,2,4);stem(m,xo);title('odd part')xlabel('n');ylabel('xo(n)');axis([-10,10,-0.6,0.6])


7/21

-10 -5 0 5 100

0.5

1

Rectangular pulse

n

x(n)

-10 -5 0 5 100

0.5

1

even part

n

xe(n)

-10 -5 0 5 10

-0.5

0

0.5

odd part

n

xo(n)


8/21

Solution 2

%Convolution of two sequences:

x=[3,11,7,0,-1,4,2,0];h=[0,0,2,3,0,-5,2,1];y=conv(x,h);subplot(2,2,1);stem(x); title('x(n)');subplot(2,2,2);stem(h); title('h(n)');subplot(2,2,3);stem(y); title('y(n)');

Output:

0 2 4 6 8-5

0

5

10

15x(n)

0 2 4 6 8-5

0

5h(n)

0 5 10 15-100

-50

0

50y(n)


9/21

Solution 3a)

clear all;close all;clc;disp('Difference Equation of a digital system');N=input('Desired Impulse response length = ');

b=input('Coefficients of x[n] terms = ');a=input('Coefficients of y[n] terms = ');h=impz(b,a,N);disp('Impulse response of the system is h = ');disp(h);n=-20:1:100;figure(1);stem(n,h);xlabel('time index');ylabel('h[n]');title('Impulse response');

Output:Difference Equation of a digital systemDesired Impulse response length = 100Coefficients of x[n] terms = 1Coefficients of y[n] terms = [1 -1 0.9]'Impulse response of the system is h =

-20 0 20 40 60 80 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

time index

h[n]

Impulse response


10/21

b)clear all;close all;clc;disp('Difference Equation of a digital system');N=input('Desired step response length = ');b=input('Coefficients of x[n] terms = ');a=input('Coefficients of y[n] terms = ');u=stepz(b,a,N);disp('Step response of the system is U = ');disp(u);n=-20:1:100;figure(1);stem(n,u);xlabel('time index');ylabel('U[n]');title('Step response');

Output:Difference Equation of a digital systemDesired step response length = 121Coefficients of x[n] terms = 1Coefficients of y[n] terms = [1 -1 0.9]Step response of the system is U =

-20 0 20 40 60 80 1000

0.5

1

1.5

2

2.5

time index

U[n]

Step response


11/21

c)clc;clear all;close all;

b=[0.9 -1 1];k=poly2rc(b);knew=fliplr(k);s = all(abs(knew));if(s==1)

disp(' "Given system is a Stable System" ');else

disp(' "Given system is an Unstable System" ');

end

Output:

Given system is a Stable System


12/21

Experiment No 3

Aim :

Point sequence x(n) defined as

x(n) =[ 1 2 3 4 5 6 6 5 4 3 2 1]

a) Determine DFT X(k) of x(n). Plot (using stem) its magnitude and phase ancalculate manually. Show both are same.b) Plot the magnitude and phase of DTFT X(e^jw) of x(n) using MATLABc)Verify that the above DFT is the sampled version of X(e^jw). It might behelpful to combine the above two plots in a graph using hold function.Solution:a)DFT:::::

xn= [1 2 3 4 5 6 6 5 4 3 2 1];N = 12;n = [0:1:N-1]; k = [0:1:N-1];WN = exp(-j*2*pi/N); nk = n'*k;WNnk = WN .^nk; Xk= xn*WNnk;magXk = abs(Xk);angXk =angle(Xk);figure;subplot(2,2,1);stem(magXk);gridxlabel('frequency in units of pi');ylabel('|X|')subplot(2,1,2);stem(angXk);gridxlabel('frequency in units of pi');ylabel('radians/pi')


13/21

output:

0 5 10 150

20

40

60

frequency in units of pi

|X

|

0 2 4 6 8 10 12-3

-2

-1

0

1

2


radians/pi

To calculate manually we have to compute DFT Coefficents one by one.

X(k)=

[42 -13.9 -3.73i 0 -1- 1i 0 0.0718-0.2679i 0 -0.0718+0.2679i 0 -1+1i -13.9282+3.7321i ]

From above coefficients we analyze that results are similar to our calculatedresults.


14/21

b)N=12; x = [1 2 3 4 5 6 6 5 4 3 2 1];n = [0:1:N-1]; k = [0:1:N-1];w = (pi/12)*k;X = x * (exp(-j*pi/12)) .^ (n'*k);magX = abs(X); angX =angle(X);subplot(2,2,1); plot(w/pi,magX);gridxlabel('frequency in units of pi');ylabel('|X|')subplot(2,1,2); plot(w/pi,angX/pi);gridxlabel('frequency in units of pi');ylabel('radians/pi')

0 0.5 10

20

40

60


|X|

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


radians/pi


15/21

c)using hold function we compare the graphs and find that DTFT sam-ples are DFT coefficients themselves

0 0 . 5 10

2 0

4 0

6 0

f r e q u e n c y i n u n i t s o f p i

|X

|

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


radians

/pi

0 5 1 0 1 50

2 0

4 0

6 0


|X

|

0 2 4 6 8 1 0 1 2- 3

- 2

- 1

0

1

2


radians/pi


16/21

Experiment No 4

Aim :

Given a causal system y(n) = 0.9y(n-1) + x(n)

a)Find H(z) and sketch the pole zero plot in MATLAB

b)Plot |H(e^jw)| & H(e^jw)Solution

a) Difference equation can be put in the formy(n) - 0.9y(n-1) = x(n)

The system transfer fn from the difference equation is given as

H(z) = 1/(1 0.9 z-1); |z|>0.9

There is one pole at 0.9 and a zero at origin. We will use zplane fn to

represent it in MATLAB

b = [ 1 , 0 ]; a = [ 1 , - 0.9 ];zplane(b,a)

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Part

ImaginaryPart


17/21

b) Now for determining the frequency response we use freqz fn in MAT-LAB.

[H,w] = freqz(b,a,100);magH = abs(H) ; phaH = angle(H);

subplot(2,1,1); plot(w/pi,magH);gridxlabel(frequency in pi units); ylabel(Magnitude);title(Magnitude Response)subplot(2,1,2); plot(w/pi,phaH/pi); gridxlabel(frequency in pi units); ylabel(Phase in pi units);title(Phase Response)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

frequency in pi units

Magnitud

e

Magnitude Response

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

-0.3

-0.2

-0.1

0

frequency in pi units

Phaseinpiunits

Phase Response

Experiment No 5


18/21

Aim :

A causal LTI system is described by the following differential equation.y(n) = 0.81y(n-2) + x(n) x(n-2)Determinea)System fn H(z)b)Unit impulse response h(n).

c) Unit step response v(n),that is, response to the unit step u(n), andd)the frequency response function H(e^jw), and plot its magnitude and

phase over 0 w

Solution:a)

Its a causal system , the ROC will be outside of a circle with radiusequal to largest pole magnitude.

Taking the z-transform of both sides and then solving for Y(z)/X(z) we get

H(z) = (1 z-2)/(1 0.81z-2) = (1 z-2)/(1+0.9z-1)*(1-0.9z-1)b)

Using MATLAB for partial fn expansionb = [1, 0, -1]; a = [1, 0, -0.81];[R, p, C] = residue(b,a)

Output:R =

-0.1173-0.1173

p =

-0.9000

0.9000

C =

1.2346So the system function is given by


19/21

H(z) = 1.2346 0.1173 * 1/(1 + 0.9z-1) - 0. 1173 * 1/(1 - 0.9z-1), |z| >0.9

Taking inverse Z transform of above we geth(n) = 1.2346 (n) 0.1173{1 + (-1)^n}(0.9^n)u(n)

c) we know [ u(n) ] = 1/ (1-z-1), |z|>1

V(z) = H(z)U(z) =[(1 + z-1)(1 - z-1)/ (1 + 0.9z-1)(1 - 0.9z-1)]* [1/ (1-z-1)]

= (1 + z-1)/ (1 + 0.9z-1)(1 - 0.9z-1) , |z| > 0.9

V(z) = 1.0556* 1/(1 - 0.9z-1) - 0.0556*(1 + 0.9z-1) , |z| > 0.9

Using inverse Z transform we getv(n) = [1.0556(0.9)^n - 0.0556 (-0.9)^n ] u(n)

d) Substituting z= e^jw in H(z) we getH(z) = (1 e^-j2w)/(1 0.81e^-j2w)

Using MATLABw = [0:1:500]*pi/500;

H = freqz(b,a,w);magH = abs (H); phaH = angle (H);subplot(2,1,1); plot(w/pi, magH); gridxlabel(frequency in pi units); ylabel(Magnitude);title(Magnitude Response)subplot(2,1,2); plot(w/pi,phaH/pi); gridxlabel(frequency in pi units); ylabel(Phase in pi units);title(Phase Response)


20/21

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 10

0 . 2

0 . 4

0 . 6

0 . 8

1

1 . 2

1 . 4


|X

|

M a g n i t u d e P a r t

0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1- 0 . 5

0

0 . 5


radians/pi

A n g l e P a r t

Experiment No 5


21/21

Aim :

Given the following Difference equation .

y(n) = x(n) + 2 x(n-1) 0.2 y(n-2)

Implement in simulink

Solution:

To create a new model:1. If Simulink is not running, enter simulink in the MATLAB Command Win-

dow to open the Simulink Library Browser.2. Select File > New > Model in the Simulink Library Browser to create a

new model.The software opens an empty model window.To construct a model, we first copy blocks from the Simulink Library Brows-er to the model window. To create the required model , we need 8 blocks .These are 2 gain blocks, sum, output(scope),input and three unit delays.

By connecting as shown in figure below we realize our required system.

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