Plotting the convolution of two signals

I am given two functions x(t) =5[u(t+1)-u(t-1)] and h(t)=u(t-1)-u(t-7). I am asked to convolve these two signals and plot the result in the range -3 to 10. Here is the code that I wrote:

t=-3:0.1:10; t_c=-3:0.05:10; h_t=heaviside(t-1)-heaviside(t-7); x_t1=5.*(heaviside(t+1)-heaviside(t-1)); c_x_h=conv(x_t1,h_t); figure(1) plot(t_c,c_x_h)

However, since this is a simple convolution I verified it by hand and it does not look like the plot i get in MATLAB. The max value should be 10 but in MATLAB i get 100. Also, the duration of the function should be 8 but i get 4 with MATLAB. Please help.

Answer 1

[Rick Rosson]

Please try:

   t_c = -6:0.1:20;

and:

   dt = t(2) - t(1);
   c_x_h = dt*conv(x_t1,h_t);

[Dennis]

Thank you for the tip Rick. I tried that and it worked. I am unclear as to why the dt is needed though and why t_c needed to be extended to -6 to 20. Thanks again for your time.

[Baiyu Zhang]

What is dt = t(2)-t(1)? What does it mean?

[Celeste MacNeil]

What do you mean dt = t(2)-t(1)?

[Celeste MacNeil]

I assume you just mean the step?

Answer 2

[IKRAM jebali]

Hello! Thank you. I'had the same problem. Me too amnclear as to why the dt is needed though and why t_c needed to be extended to -6 to 20. I use it but i don't know why? There is any answer please. Thank you.

[Aswin Farzana Mohamed Ansar]

Because the length of the convolution of two matrices of ,length m and n will be m+n-1. inorder to get the same vector size, you need to do this.

Answer 3

[Paul]

Define the signals and plot them

syms t tau real
u(t) = heaviside(t);
x(t) = 5*(u(t+1)-u(t-1)); h(t) = u(t-1) - u(t-7);
figure
fplot([x(t),h(t)],[-5,10])

If y(t) is the convolution if x(t) and h(t), then we know that y(t) = 0 for t < 0 and t > 8 by adding the left and right edges of x(t) and h(t) respectively.

Compute the closed form expression of the convolution

ysym(t) = int(x(tau)*h(t-tau),tau,-inf,inf);

To compute the convolution numerically using conv, first define the the unit step function and the sampling period

u = @(t) heaviside(t);
dt = 0.05; % T

Define the time vectors to compute samples x(n*T) and h(n*T) where they are non-zero

tx = -1:dt:1; x = 5*(u(tx+1) - u(tx-1));
th =  1:dt:7; h = u(th-1) - u(th-7);

Now approximate the convolution integral using conv. We need to multiply by dt because conv computes the convolution sum, which is essentially dividing the integrand of the convolution integral into rectangles of width dt and summing their heights. We need to multiply the output of conv by dt to so that the result is the sum of their areas.

y = conv(x,h)*dt;

By the shift property of convolution, we know that the first element of y corresponds to tx(1) + th(1), we know the number of samples in y, and we know that the sampling period is dt. Hence, the time vector that corresponds to the elements of y is

ty = tx(1) + th(1) + (0:numel(y)-1)*dt;

Plot the closed form expression and overlay with the conv approximation, which only covers the finite duration of ysym

figure
fplot(ysym,[-3,10]);
hold on
plot(ty,y,'o'),axis padded

Note that y(1) is not exactly equal to zero

y(1)

ans = 0.0625

which is a consequence of approximating the integral via rectangular integration.

The original problem statement said to get the solution over the interval -3 <= t <= 10. We can take the solution above and pad y on the left and right with the appropriate number of zeros and then, of course, adjust ty as well. I think it's easier to start with the same time vector for both signals, making sure that it covers the non-zero portions of x and h and the desired duration of the output, and then extracting the part we want at the end

t = -2:dt:7;  % covers both x and h, output will cover -4 <= t <= 14, 
x = 5*(u(t+1) - u(t-1));
h = u(t-1) - u(t-7);
y = conv(x,h)*dt;
ty = 2*t(1) + (0:numel(y)-1)*dt;
ii = ty >= -3 & ty <= 10; % extract the part we care about
ty = ty(ii);
y = y(ii);
figure
fplot(ysym,[-3,10]);
hold on
plot(ty,y,'o'),axis padded