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Inverse Laplace Transform -Exponential

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Melany
Melany 2012 年 3 月 6 日
コメント済み: Walter Roberson 2021 年 2 月 17 日
Hello All: Does anyone know of a matlab code to obtain the inverse Laplace transform of an exponential? or hints

回答 (2 件)

Walter Roberson
Walter Roberson 2012 年 3 月 6 日
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 件のコメント
Walter Roberson
Walter Roberson 2012 年 3 月 6 日
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 2016 年 1 月 20 日
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 2021 年 2 月 17 日
Find the inverse Laplace transform of
F(s)=(100*(s+3))/(s+1)*(s+2)*(s^2+2*s+5)
  1 件のコメント
Walter Roberson
Walter Roberson 2021 年 2 月 17 日
ilaplace((100*(s+3))/(s+1)*(s+2)*(s^2+2*s+5))

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