Plot convolution of continuous signal

I am trying to plot the convolution of 2 continuous signals. My figure has the correct shape, but the values are very large.

t= -2.5:.001:3.5;
he1 = (1-t).*(heaviside(t)-heaviside(t-1));
he2 = (t).*(heaviside(t+2)-heaviside(t-2));
m=conv(he1,he2);
subplot(224); plot(m); grid on;
xlabel('t'); ylabel('h_s(t), series (conv integral)');

When I try to plot

subplot(224); plot(t, m); grid on;

Error: Error using plot. Vectors must be the same length.

When I try to plot

subplot(224); plot(t, m(t)); grid on;

Error: Array indices must be positive integers or logical values.

When I try

t= -2.5:.001:3.5;
he1 = @(t) (1-t).*(heaviside(t)-heaviside(t-1));
he2 = @(t) (t).*(heaviside(t+2)-heaviside(t-2));
m=conv(he1(t),he2(t));
subplot(224); plot(m); grid on;
xlabel('t'); ylabel('h_s(t), series (conv integral)');

^ Define functions @t, I get the same plot as below

Problem: bottom right figure.

How do I get my convolution plot to have the same values as the original functions?

Answer 1

[Jiahao CHANG]

Hi Michelle,

Your code

t= -2.5:.001:3.5;
he1 = (1-t).*(heaviside(t)-heaviside(t-1));
he2 = (t).*(heaviside(t+2)-heaviside(t-2));
m=conv(he1,he2);
subplot(224); plot(m); grid on;
xlabel('t'); ylabel('h_s(t), series (conv integral)');

gives you the wrong value of 't' because you didn't define the x vector so Matlab will display the length of the variable 'm'.

subplot(224); plot(t, m); grid on;

goes wrong because t and m are not in the same length. (t is 1*6001 and m is 1*12001)

So i suggest you to use

t1= -2.5:6/length(m):3.5-6/length(m);

And the code should be like

t= -2.5:.001:3.5;
he1 = (1-t).*(heaviside(t)-heaviside(t-1));
he2 = (t).*(heaviside(t+2)-heaviside(t-2));
m=conv(he1,he2);
t1= -2.5:6/length(m):3.5-6/length(m);
subplot(224); plot(t1,m); grid on;
xlabel('t'); ylabel('h_s(t), series (conv integral)');

Try this and tell me whether it is want you want;)

[Michelle Watson]

Thanks for the information!

Yes the x-axis values are fixed. However, I'd like the y-axis to have the proper values as well. See the Desmos screencap below:

Also, could you elaborate on why exactly what your t1 variable does? Does it scale the length of m to fit the t values I set? If so, what is the purpose of the 6/length(m)? Why 6?

Is it possible to just have the convolution (m) already be scaled according to the signals that I input? (he1, he2)?

Why can he1 and he2 be plotted as a function of t? But when I try to plot m (which is a function of he1 which is a function of t), it gives these errors?

If this is not possible, how do I go about scaling m? Would I need to scale each value of me at each t value?

Thanks again!

[Jiahao CHANG]

For the y-axis value, if you want to achieve the same equation as you shown in this picture, setting variables by directly using these equations is easier.

Yes, t1 is a variable scale the length of m so that they could match. 6 is because your minimum time is -2.5 and max is 3.5 so the total time is 3.5-(-2.5)=6. In order to match the same length of m, you need to divide your time into length(m) pieces, each piece(or step) equals to 6/length(m). Because you already set the last number equals to 3.5, so maybe it is better to change that line as

t1= -2.5:6/(length(m)-1):3.5;

I'm not sure whether I understand this question or not. The length of convolution(m) equals to the length of he1+he2-1. You can see some examples of conv here.

he1 and he2 can plotted as a function of t is because for each t(i), you have and only have one he1(i) to match. However m has more values than he1 and t so that they can't match because the length are different. And Im not sure whether this is meaningful.

To understand this, you should know that the time you put

t= -2.5:.001:3.5;

is a discrete not continious.