Help with Inverse Laplace Transform Function
이전 댓글 표시
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.
답변 (2개)
Aveek Podder
2017년 10월 26일
0 개 추천
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
2017년 10월 26일
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)
카테고리
도움말 센터 및 File Exchange에서 Calculus에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!