C.T. signals convolution in Matlab
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Joshua Scicluna
on 15 Jan 2020
Commented: Star Strider
on 16 Jan 2020
Hi, I have 2 continues time signals (exp decay & step), is it possible to convolute them in MATLAB?
I am working with symbolic variables ‘s’ and ‘t’ since I have obtained a transfer function H(s) analyticlay then converted it to h(t) using ilapalce() function, hence now I need to obtain y(t) where y(t) = h(t)*x(t). x(t) = u(t) a step input and h(t) = exp(-2 t) 4 - 4 exp(-t)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/261977/image.png)
Thanks!
JS
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Accepted Answer
Star Strider
on 15 Jan 2020
One approach:
syms h(t) x(t) s t
Fcn1 = h(t) == exp(-2*t)*4 - 4*exp(-t);
Fcn2 = x(t) == heaviside(t);
convlap = laplace(Fcn1, t, s) * laplace(Fcn2, t, s);
Y(s) = simplify(rhs(convlap), 'Steps',250)
y(t) = ilaplace(Y, s, t)
Producing:
y(t) =
4*exp(-t) - 2*exp(-2*t) - 2
2 Comments
Star Strider
on 15 Jan 2020
Yes.
I did the convolution in the complex s-domain because (1) that is the only way it is possible to do it, and (2) I got the impression that was the process you described as desiring.
This:
syms h(t) x(t) s t T tau
h(t) = exp(-2*t)*4 - 4*exp(-t);
x(t) == heaviside(t);
y(t) = simplify(int(h(t)*x(t-tau), tau, -T, T), 'Steps',250)
produces:
y(t) =
-4*exp(-2*t)*(exp(t) - 1)*int(x(t - tau), tau, -T, T)
that appears to be the best result available. There is no specific convolution function in the Symbolic Math Toolbox. (I used symmetric integration limits because similar terms cancel each other, considerably simplifying the expression.)
Using asymmetric limits:
y(t) = simplify(int(h(t)*x(t-tau), tau, 0, T), 'Steps',250)
produces:
y(t) =
-4*exp(-2*t)*int(x(t - tau), tau, 0, T)*(exp(t) - 1)
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