Help with Inverse Laplace Transform Function

5 vues (au cours des 30 derniers jours)
Zachariah Ingram
Zachariah Ingram le 17 Oct 2017
If I run the following code, I get an answer.
n = 5;
ku = 1;
syms s;
HDs = (ku^n)/(s*(ku+s)^n);
sim_data = ilaplace(HDs,s,0.1);
sim_data = 1 - (265241*exp(-1/10))/240000;
However, if n is a decimal, such as 5.1, the inverse laplace function doesn't work.
n = 5.1;
ku = 1;
syms s
HDs = (ku^n)/(s*(ku+s)^n)
sim_data = ilaplace(HDs,s,0.1)
sim_data = ilaplace(1/(s*(s + 1)^(51/10)), s, 1/10)
I need to be able to calculate the inverse laplace transform for decimal values of n. Please help.

Réponses (2)

Aveek Podder
Aveek Podder le 26 Oct 2017
Hi,
The ilaplace function computes the analytic closed inverse Laplace form of a transfer function. It seems that mathematically a closed inverse Laplace form for this function cannot be found out, so ilaplace function is returning the input transfer function.
There is a community submission at MathWorks File Exchange which numerically approximates an inverse Laplace transform for any function of "s". I will encourage you to have a look at the submission. However, in case you have any query regarding the submission you have to get in touch with the owner of the submission.

Walter Roberson
Walter Roberson le 26 Oct 2017
For positive ku and n, and non-negative t, the inverse laplace apparently works out as
((-1-n)*igamma(1+n, ku*t)+ku^n*t^n*exp(-ku*t)*(1+n)+gamma(2+n))/gamma(2+n)

Catégories

En savoir plus sur Fluid Dynamics dans Help Center et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by