Dsp Lab Report-2 by L Govind ec

8/3/2019 Dsp Lab Report-2 by L Govind ec

1/13

DSP LAB REPORT-2

1 | P a g e

NIT CALICUT

DSP LAB REPORT-2

DEPARTMENT OF ELECTRONICS ENGINEERING

NATIONAL INSTITUTE OF TECHNOLOGY CALICUT

NIT CAMPUS PO, CALICUT

KERALA, INDIA 673601

By

L Govind Reddy (M110117EC)


2/13


3/13

DSP LAB REPORT-2

3 | P a g e

NIT CALICUT

2. A unit impulse sequence ()is defined as

() = 1 when n = 0

0 when n 0

Unit step signal:1. Continuous time unit step signal u(t), is defined as

u(t) = 1 when t > 0= 0 when t < 0

2. Discrete time unit step signal u(n), is defined as

u(n) = 1 when n 0

0 when n < 0


4/13

DSP LAB REPORT-2

4 | P a g e

NIT CALICUT

Unit Ramp signal:1. A continuous time unit ramp signal r(t) is defined as

r(t) = t when t > 0= 0 when t < 0

2. A discrete time unit ramp signal r(n) is defined as

r(n) = n when n 0

= 0 when n < 0

1.3Even and odd signals?Even signal:

A signal () is said to be even signal if

() = ()

Example:

() =


5/13

DSP LAB REPORT-2

5 | P a g e

NIT CALICUT

Odd signal:

A signal () is said to be odd signal if

() = ()

Example:

() = sin ()

1.4Linear convolution. Circular convolution?Linear convolution:

1. Linear convolution of two signals () and () is defined as() = () () = ()( )

The output of any Linear Time Invariant system is the convolution of input signal withthe impulse response of the system.

2. Linear convolution of two sequences () and () is defined as() = () () = ()( )

The output of any Discrete Time Invariant system is the convolution of input signal with

the impulse response of the system.

Circular convolution:

The circular convolution of two sequences () and () is defined as

() = () () = ()[( ) N] .

Circular convolution is defined for finite length sequences (usually equal length sequences)

1.5 Causal versus non-causal systems?Causal system:

A system is said to be causal if the output () of the system at any time t = t0 depends onthe input () for t t0

I.e. the output of the system depends on present and/or past values of the input not on

future values of the inputExample:

() = ( 2)Non Causal system:

A system is said to be non-causal if its output depends on future values of the input.

Example:

() = ( + 3)

N


6/13

DSP LAB REPORT-2

6 | P a g e

NIT CALICUT

1.6System stability?Stable system:

A system is said to be stable if for any bounded value of input the output of system

should be bounded.

Example:

() = ()

2 Implement2.1 Generate two sinusoids with frequencies F0 = 1/8Hz and F1 = 7/8Hz. Sample both at Fs

= 1Hz. Generate stem plots for each, with a different color for each. Plot the original

continuous signals over the sampled one, again each color coded. What do you

observe? Determine the sampling rate at which you can avoid a problem here? Repeat

the procedure at this new sampling frequency. What do you observe?

Program:%sampling both with frequency 1 Hz

clc;

clear all;

f1=1/8;

f2=-7/8;

t1=0:.25:80/7;

t2=0:.05:80/7;

N1=0:1/1:80/7;

N2=0:1/1:80/7;

y1=sin(2*pi*f1*N1);

y2=sin(2*pi*f2*N2);y1c=sin(2*pi*f1*t1);

y2c=sin(2*pi*f2*t2);

stem(N1,y1,'red');

hold on;

plot(t1,y1c,'green');

hold on;

stem(N2,y2,'black');hold on;

plot(t2,y2c,'blue');

hold on;


7/13

DSP LAB REPORT-2

7 | P a g e

NIT CALICUT

Output:

Observation:

Here the problem is, signal one is completely reconstructed from its samples since

its frequency f1 is 1/8 and sampling frequency fs is 1 i.e. nearly 8 times signal frequency.

But for the signal we cannot reconstruct the full signal from the sampled one because signalfrequency f2 is 7/8 and sampling frequency fs is 1 which is less than the 2f2 .

So to avoid that problem we should take the sampling frequency at least twice the

frequency of second signal (=1.75 Hz).

Here in next program fs is taken as 3.5 Hz which equal to the four times of the second

signal frequency

Program:

%sampling frequency fs=3.5 Hz

clc;

clear all;

f1=1/8;

f2=-7/8;

t1=0:.25:80/7;


8/13

DSP LAB REPORT-2

8 | P a g e

NIT CALICUT

t2=0:.05:80/7;

N1=0:1/1:80/7;

N2=0:1/1.1:80/7;

y1=sin(2*pi*f1*N1);

y2=sin(2*pi*f2*N2);



stem(N1,y1,'red');

hold on;

plot(t1,y1c,'green');

hold on;

stem(N2,y2,'black');

hold on;

plot(t2,y2c,'blue');

hold on;

Output:

Here both signals can be reconstructed from there sampled ones.


9/13

DSP LAB REPORT-2

9 | P a g e

NIT CALICUT

2.2 Plot signal x (n) = [1 1 1 0 1 2 3 4], where x (0) = 0; Plot x (-n) and x (-n + 2). Obtain

linear convolution with system h (n) = [1 2 1 -1], h (0) = 2; Show the output.

Program:

clc;

clear all;

close all;

%x=[1 1 1 0 1 2 3 4];

%h=[1 2 1 -1 ];

x=input('enter x');

h=input('enter y');

[p q]=size(x);

[r s]=size(h);

p1=input('enter the indux range of x');

r1=input('enter the indux range of y');

%ploting x(n),x(-n),x(-n+2)

n=p1(1):1:p1(2);

subplot(2,2,1);stem(n,x);

title('x(n)');

subplot(2,2,2);

stem(-n,x);

title('x(-n)');

subplot(2,2,3);

stem(-n+2,x);

title('x(-n+2)');

%convolution....

for m=1:s+q-1if m


10/13

DSP LAB REPORT-2

10 | P a g e

NIT CALICUT

Giving inputs to above function through command prompt:Command window:

enter x[1 1 1 0 1 2 3 4]

enter y[1 2 1 -1]

enter range of x[-3 4] %since x(0)=0

enter range of y[-1 2] %since h(0)=2

Output:


11/13

DSP LAB REPORT-2

11 | P a g e

NIT CALICUT

2.3 Show stable and unstable conditions for the linear time invariant system h(n) = an

u(n).

Program:clc;

clear all;

close all;

a1=2;

a2=.5;

%n will have only positve values because y(n)=a^n u(n),u(n)=0

for n


12/13

DSP LAB REPORT-2

12 | P a g e

NIT CALICUT

3 Results and discussion3.1 Discuss sampling problem in implement 1. Can you give a theoretical explanation for

the particular frequencies that come up? Would you be able to give a F2 for which this

same sampling problem would occur at Fs?

The problem encountered in program one is with sampling frequency fs(1 Hz), fs is

greater than the 2f1(2*1/8 = 0.25 Hz) but it is not greater than 2f2 (2*7/8 = 1.75 Hz)so wecannot reconstruct the second signal from its sampled sequence.

For fs = 1 Hz any signal which is having frequency greater than 0.5 Hz will face the sameproblem during reconstructing original signal from the sampled sequence because fs should

be at least twice the signal frequency(from nyquist theorem).

3.2 How would the result in implement 2 change if we were to change the origin point of(a) x (n)

(b) h (n)1. For problem 2 initially

(Value indicated in bold is origin)x(n) = [1 1 1 0 1 2 3 4]

h(n) = [1 2 1 -1]And x(0) = 0 i.e. the index range of x is -3 to 4, h(0) = 2 i.e. index range of h is -1 to 2.

The convolution of x(n) and h(n) is

y(n) = [ 1 3 4 2 1 3 8 11 9 1 -4]

Here y(0) = 1, i.e. the index range of y is -4 to 6.

2. Now I will change the origin of x(n) and h(n)(Value indicated in bold is origin)

x(n) = [1 1 1 0 1 2 3 4]h(n) = [1 2 1 -1]

Here x(0) = 1 i.e. the index range of x is -4 to 3, h(0) = 1 i.e. index range of h is -2 to 1.The convolution of x(n) and h(n) is

y(n) = [ 1 3 4 2 1 3 8 11 9 1 -4]

Here y(0) = 8, i.e. the index range of y is -6 to 4.

So if we change the origin of any sequence same values will repeat with differentindexes.

3.3 How would the stability condition change if we were dealing with the systemh(n) = a

nn

0

= bn n < 01. If n 0 then

h(n) = an

is stable if 0 < a 1 and h(n) is unstable if 0 < b 1.


13/13

DSP LAB REPORT-2

13 | P a g e

NIT CALICUT

Dsp Lab Report-2 by L Govind ec

Documents