Inverse Laplace Transform -Exponential

Hello All: Does anyone know of a matlab code to obtain the inverse Laplace transform of an exponential? or hints

回答(2 个)

There does not appear to be any general form for all exponentials, but some exponential forms have simple transforms.
Perhaps you have a specific form you would like to consider?

3 个评论

Walter:
It is pretty general and pretty much the addition of two exponentials
A * exp(-lambda1 * x) + B * exp(-lambda2 * x)
any hints?
Under the assumptions that all the variables are real, and that lambda1 and lambda2 are positive (so you have a negative exponential), then the form for that is
A * Dirac(t-lambda1) + B * Dirac(t-lambda2)
However if lambda1 or lambda2 are complex or are negative, then you have a problem.
This actually solved my problem. In other words, assuming the "delay" parameter to be positive allows flawless inverse Laplace transform computation. Example below:
% Time and delay parameters
syms t, td real
% Laplace complex variable
syms s
F = exp(- td*s);
f = ilaplace(F)
Gives an unusable result:
f =
ilaplace(exp(-s*td), s, t)
But the explicit assumption of positive delay makes the trick and helps Matlab find the right solution. So, if the assumption is added:
assume(td > 0)
The output will be the expected one:
f =
dirac(t - td)
And this worked for me in a much more complex transfer function.

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Find the inverse Laplace transform of
F(s)=(100*(s+3))/(s+1)*(s+2)*(s^2+2*s+5)

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