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Inverse Laplace Transform -Exponential

4 vues (au cours des 30 derniers jours)
Melany
Melany le 6 Mar 2012
Commenté : Walter Roberson le 17 Fév 2021
Hello All: Does anyone know of a matlab code to obtain the inverse Laplace transform of an exponential? or hints

Réponses (2)

Walter Roberson
Walter Roberson le 6 Mar 2012
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 commentaires
Walter Roberson
Walter Roberson le 6 Mar 2012
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.
Giuseppe Maria D'Aucelli
Giuseppe Maria D'Aucelli le 20 Jan 2016
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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MANOHAR POKA
MANOHAR POKA le 17 Fév 2021
Find the inverse Laplace transform of
F(s)=(100*(s+3))/(s+1)*(s+2)*(s^2+2*s+5)
  1 commentaire
Walter Roberson
Walter Roberson le 17 Fév 2021
ilaplace((100*(s+3))/(s+1)*(s+2)*(s^2+2*s+5))

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