Problem in plotting the convolution

I can't plot the convolution of the two signals and I don't know why

dt= 0.0001;
t = 0:dt:10;

xt = 4*cos(2*pi*730*t) +3*cos(2*pi*440*t) + 5*cos(2*pi*490*t);
subplot(2,2,1)
plot(t,xt)
axis([ 0 0.02 -20 20 ])
title(' Ploting x(t)')
xlabel('time')
ylabel('x(t)')
grid on

ht= abs( (1/2*pi*j*t).*( exp(j*1460*pi*t) - exp(j*1450*pi*t)) );
subplot(2,2, 2)
plot(t,ht)
title(' plotting h(t)')
axis( [0 20 0 70] )
xlabel(' Time')
ylabel(' h(t)')
grid on

yt= conv(xt , ht);
subplot(2,2, 3)
plot(t,yt)
title(' plotting y(t)')
axis( [0 20 0 100] )
xlabel(' Time')
ylabel(' y(t)')
grid on

Answer 1

[Star Strider]

Use the 'same' argument so that the output is the same size as the input:

yt= conv(xt , ht, 'same');

This works —

dt= 0.0001;
t = 0:dt:10;

xt = 4*cos(2*pi*730*t) +3*cos(2*pi*440*t) + 5*cos(2*pi*490*t);
subplot(2,2,1)
plot(t,xt)
axis([ 0 0.02 -20 20 ])
title(' Ploting x(t)')
xlabel('time')
ylabel('x(t)')
grid on

ht= abs( (1/2*pi*j*t).*( exp(j*1460*pi*t) - exp(j*1450*pi*t)) );
subplot(2,2, 2)
plot(t,ht)
title(' plotting h(t)')
axis( [0 20 0 70] )
xlabel(' Time')
ylabel(' h(t)')
grid on

yt= conv(xt , ht, 'same');
subplot(2,2, 3)
plot(t,yt)
title(' plotting y(t)')
axis( [0 20 0 100] )
xlabel(' Time')
ylabel(' y(t)')
grid on

.

[Abdullah Alasfour]

Thank you man I realy appreciate it.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

[Abdullah Alasfour]

Sure and have a great day

[Star Strider]

Thank you!

Answer 2

[Paul]

Hi Abdullah,

Although not stated explicitly in the Question, I'm going to assume that

a) the objective is compute the convolution integral, i.e., xt and ht are continuous time signals, and

b) xt and ht are both zero for t < 0 adn t > 10

If those assumptions are correct, here's a similar problem that shows how to use conv to approximate the convolution integral. Maybe you can adapt it to your problem.

Let's use two simple functions for x(t) and h(t), both of which are zero outside the interval 0 <= t <= 3

syms t real
x(t) = exp(-0.2*t)*rectangularPulse(0,3,t);
h(t) = exp(-0.3*t)*rectangularPulse(0,3,t);

By defintion, the convolution integral for these signals is

syms tau real
y(t) = simplify(int(x(tau)*h(t-tau),tau,0,t))

y(t) = 

Plot x(t), h(t), and y(t). The duration of y(t) (6 seconds) is the sum of the durations of x(t) and h(t).

fplot([x(t) h(t) y(t)],[0 7])

Now use conv with numerical valuse of x(t) and h(t) sampled at 0.01 seconds

t = 0:.01:3;
x = exp(-0.2*t);
h =  exp(-0.3*t);

To get the correct answer, we need to use the full convolution and multiply the result by the sampling period (because both x(t) and h(t) are finite)

y = conv(x,h,'full')*0.01;  % 'full' is the default, so not necessary to specify explicitly

Of course, y will be longer than x and h, so we have to define a new time vector to plot it.

hold on
plot((0:numel(y)-1)*0.01,y)

For this problem, if we use the 'same' convolution, we get a y that has the same number of elements x and h and therefore covers 3 seconds. But that that's not sufficient to approximate the 6-second convolution integral. And we can see that plotting it would require a shift to the right by 1.5 seconds to match up with the "central" part of the convolution integral.

y = conv(x,h,'same')*0.01;
plot(t,y)
legend('x(t)','h(t)','y(t)','yconv-full','yconv-same')

[Image Analyst]

He's using Octave, not MATLAB.

[Paul]

I didn't notice the Octave tag because I don't look at the tags.

I'm also not sure about how to interpret this comment.

Are questions tagged with "octave" not supposed to be answered? Should they be closed?