Incorrect continuous signal convolution values in Matlab

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I am trying to convolve the following signals in Matlab:
and
for a=0.5. The result, say z(t) is:
So here is my Matlab code:
%Define initial signals
alpha = .5;
Ts = 0.01;
t_x = -5:Ts:5;
x = heaviside(t_x);
t_y = -5:Ts:5;
y = heaviside(t_y).*(alpha.^abs(t_y));
%Plot initial signals
figure(1)
plot(t_x,x);
xlabel('t');
ylabel('x(t)');
axis([min(t_x)-1 max(t_x)+1 min(x)-1 max(x)+1]);
figure(2)
plot(t_y,y);
xlabel('t');
ylabel('y(t)');
axis([min(t_y)-1 max(t_y)+1 min(y)-1 max(y)+1]);
%Convolution
z = Ts*conv(x,y);
t_z = min(t_x)+min(t_y):Ts:max(t_x)+max(t_y);
figure(3)
plot(t_z,z);
xlabel('t');
ylabel('z(t)');
axis([min(t_z)-1 max(t_z)+1 min(z)-1 max(z)+1]);
The problem is that the final signal that Matlab calculates has incorrect values for values of t greater than 5. I have checked this in detail and noticed that the signal is wrong for t > max(t_x). Here is the final plot:
What am I doing wrong?

采纳的回答

Star Strider
Star Strider 2016-6-27
The convolution is actually defined only for your ‘t_x’ vector (that must be the same as ‘t_y’).
For example, if you have the Symbolic Math Toolbox, run this:
syms a t x y
assume(t>0)
assume(a>0)
x(t) = 1;
y(t) = a^t;
X = laplace(x);
Y = laplace(y);
Z = X*Y;
z = ilaplace(Z)
za = subs(z, a, 0.5);
figure(1)
fplot(za, [-5 5])
  6 个评论
Konstantinos
Konstantinos 2016-6-29
编辑:Konstantinos 2016-6-29
Anyway, I found the answer on some university notes. The problem is that I am trying to convolve two signals of infinite duration. In this case, the convolution is only valid in the interval that the first signal is defined at (this is a Matlab convention about the conv(...) function). You can see more inside the PDF. If you don't know Greek, you can check the code :) Thank you!

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