C.T. signals convolution in Matlab

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Joshua Scicluna
Joshua Scicluna 2020-1-15
评论： Star Strider 2020-1-16
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)
Thanks!
JS

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Star Strider
Star Strider 2020-1-15
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 个评论
Joshua Scicluna
Joshua Scicluna 2020-1-15
are you using Y(s)=H(s)X(s)?
I need to do time domain convoltion using the equation i mentiond befor.
Thanks!
Star Strider
Star Strider 2020-1-15
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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更多回答（1 个）

Joshua Scicluna
Joshua Scicluna 2020-1-16
Agreed, Thanks for your help!

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