signal and system basic matlab programs

7/29/2019 signal and system basic MATLAB Programs

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Assignment # 1


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Part A: Plotting Original Signal

t1= -5 : 5

x=[0,0,0,0,1,0,1,1,1,1,1]

plot(t1,x)

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time

x(t)

Question 1 Part (a) Original Signal


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Part b: Even Part of a Signal

t2=-fliplr(t1);

x1=[0 0 0 0 1 0 1 1 1 1 1];x2=fliplr(x1);

x= x1+x2;

y=0.5*x;

plot(t2,y)

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2Question 1 Part (b) Even Part of a Signal

time

x(t)


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Part (c): Odd Part of a signal

clc

t2=-fliplr(t1);

x1=[0 0 0 0 1 0 1 1 1 1 1];

x2=fliplr(x1);x= x1-x2;

y=0.5*x;

plot(t2,y)

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time

x(t)

Question 1 Part (c) Odd Part of a signal


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PART (a):

t1=-5:0.1:-4;

x1=zeros(1,11);

t2=-4:0.1:-3;

x2=ones(1,11);

t3=-3:0.1:-1;

x3=2*ones(1,21);

t4=-1:0.1:1;x4=3*ones(1,21);

x=[x1 x2 x3 x4];

t=[t1 t2 t3 t4];

negx=fliplr(x);

negt=-fliplr(t);

fx=[x negx];

ft=[t negt];

plot(ft,fx)

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4Question 2 Part (a)

time

x(t)


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PART (b):

t=-[1,0];

xt=[0,-1];

t1=[0,1];

xt1=[0,1];

t2=[1,2];

xt2=[1,1];

t3=[2,3];

xt3=[1,0];

x=[xt xt1 xt2 xt3];

t=[t t1 t2 t3];

plot (t,x)

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1.5Question 2 Part (b)

time

x(t)


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PART (c):

t0=[-3 -2];

t1=[-2,-1];

t2=[-1,0];t3=[0,1];

t4=[1,2];

t5=[2,3];

x0=[0,0];

x1=[1,1];

x2=[-1,-1];

x3=[-1,0];

x4=[1,1];

x5=[0,0];

x=[x0 x1 x2 x3 x4 x5];

t=[t0 t1 t2 t3 t4 t5];

plot (t,x)

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1.5Question 2 Part(c)

time

x(t)


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PART (a):

A=5Vp-p, F= 20Hz, T =

=

= 50ms.

Since Duty cycle is 0.6, which means that 60% of the time, signal remains positive and

remaining 40% the signal will be negative. In our case, the signal will be positive for

30msec and negative for 20msec.

MATLAB Code:

t1=-110:0.1:-100;t2=-100:0.1:-70;t3=-70:0.1:-50;t4=-50:0.1:-20;t5=-20:0.1:0;t6=0:0.1:30;t7=30:0.1:50;t8=50:0.1:70;t9=70:0.1:100;t10=100:0.1:110;x1=zeros(1,101);x2=5*ones(1,301);

x3=-5*ones(1,201);x4=5*ones(1,301);x5=-5*ones(1,201);x6=5*ones(1,301);x7=-5*ones(1,201);x8=5*ones(1,301);x9=-5*ones(1,201);x10=zeros(1,101);t=[t1 t2 t3 t4 t5 t6 t7 t8 t9 t10];y=[x1 x2 x3 x4 x5 x6 x7 x8 x9 x10];plot (t,y)

Note: I have calculated the size of x1 to x10 by getting the size of t1 to t10 respectively

by whos command at command window, the above code is written in M-File. Below is

the respective plot.


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-100

-80

-60

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40

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0

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60

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100

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Question3part(a)20HzSquarewa

veof0.6DutyCycle

time

x(t)


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PART (b):

A=5Vp-p,

MATLAB Code:

clct=[-100 -100 -75 -50 -50 -25 0 0 25 50 50 75 100 100];x=[ 0 -5 0 5 -5 0 5 -5 0 5 -5 0 5 0];plot (t,x)

-100 -80 -60 -40 -20 0 20 40 60 80 100-6

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6Question 3 Part(b) Sawtooth Wave of 20Hz

time

x(t)


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PART (a):

clc

y=0;for t=-15:0.1:15;

for n=-2:0.1:2;y=y+exp(-(t-3*n)^t);endplot(t,y)

end

As seen from the plotthe plot has no point in this time range, and as we cannot plot

the whole graph, by this small time range it is clear that the given signal is not periodic.

14 14.2 14.4 14.6 14.8 15 15.2 15.4 15.6 15.8 160

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1Question 4 Part (a)

time

x(t)


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PART (b):

clcsyms tfor n=-5:0.1:5;x=exp(-2*t)*exp(n)*heaviside(2*t-n);endezplot(x

It is clear from the graph that the signal is not periodic, because the constant

exponential signal is multiplying with the original signal

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(5221823812744079 heaviside(t - 5/2))/(35184372088832 exp(2 t))

t

x(t)


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PART (a): Plotting Original Signal

xt=[0 1 1 2 2 -1 0];t= [-1 0 1 1 2 2 3];plot(t,xt)

PART (a): Shifting Signal i.e. x(t+2)

st=(t-2)

plot(st,xt)

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2.5Question 5 Part (a) Original Signal x(t)

time

x(t)

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2.5Question 5 Part (b) Shifted Signal x(t+2)

time

x(t)


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PART (c): Scaling Signal i.e. x(t/3+2)

sct=st*3;

plot(sct,xt)

PART (d): Flipping the Signal i.e. x(2-t/3)

ft=-fliplr(sct);

fxt=fliplr(xt);

plot(ft,fxt)

PART (a):

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2.5Question 5 Part (c) Scaled Signal x(t/3+2)

time

x(t)

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2.5Question 5 Part (d) Flipped signal x(2-t/3)

time

x(t)


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MATLAB Code:

clear allclcsyms tsyms Tyt=exp(-2*t);magnitude = [abs(yt)]^2;energy= int(magnitude,0,inf)p= int(magnitude,-T,T)

power=limit((1/(2*T))*energy,T,inf)

Output in Command Window:

energy = 1/4

p = sinh(4*T)/2

power = 0


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PART (b):

MATLAB Code:

clear allclcsyms tsyms T

yt=1/2*cos(4*pi*t);magnitude = [(yt)]^2;energy = int(magnitude,-inf,inf)p= int(magnitude,-T,T)power = limit(1/(2*T) * p, T,inf)

Output in Command Window:

energy = Inf

p = T/4 + sin(8*pi*T)/(32*pi)

power = 1/8

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