Help with Inverse Laplace Transform Function

Asked by Zachariah Ingram

Zachariah Ingram (view profile)

on 17 Oct 2017
Latest activity Answered by Walter Roberson

Walter Roberson (view profile)

on 26 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.

Aveek Podder (view profile)

2017 年 10 月 26 日

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 (view profile)

2017 年 10 月 26 日

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)