This set of functions allows a user to numerically approximate an inverse Laplace transform for any function of "s". The function to convert can be passed in as an argument, along with the desired times at which the function should be evaluated. The output is the response of the system at the requested times.
For instance, consider a ramp function.
f = @(s) 2/s^2;
t = [1 2 3 4 5]';
The time response output is [2 4 6 8 10], as expected.
These methods can be used on problems of considerably more difficulty as well and are intended to approximate an inverse Laplace transform where an exact solution is unknown.
Two basic solvers (Euler and Talbot) are included, along with *symbolic* versions of those solvers. The symbolic solutions take substantially longer to calculate, but are capable of any desired accuracy. Also, the symbolic versions require the Symbolic Toolbox, whereas the basic versions do not.
Please see example_inversions.pdf or html/example_inversions.html to get started!
Thanks Tucker. This is really helpful.
Hi Li Zhang,
Thanks for the rating! This code was all based on the following work:
Abate, Joseph, and Ward Whitt. "A Unified Framework for Numerically Inverting Laplace Transforms." INFORMS Journal of Computing, vol. 18.4 (2006): 408-421. Print.
I can only take credit for coding it up in a MATLAB-appropriate way. Here's the MLA citation for the code. You'll need to change the last date to the date that you downloaded it.
McClure, Tucker. Numerical Inverse Laplace Transform. Computer software. Mathworks File Exchange. N.p., 4 Jan. 2013. Web. 8 Aug. 2016.
Hi Tucker, I used your package in my paper. How to cite it ?
Thank you. Would your inverse laplace result in a solution that can be used as a starting condition and still maintain the same solution?
Thank you very much for the help. I have just read your answer and I it helped me a lot.
Doesn't work very well with
periodic functions like:
F1 = @(s) 1 / (s * (1 + exp(-s)));
F1 = @(s) (1 - exp(-s) .* (s + 1)) ./ (s.^2 .* (1 - exp(-s)));
I would like to compute the inverse laplace transform of 1/(s-1i), which is exp(1i*t). But the programs are not working for this problem. Please help!
No, this is for continuous time only. However, Dr. Dan Ellis of Columbia University has an example of a numerical inverse z-transform written in MATLAB located here: http://www.ee.columbia.edu/~dpwe/e4810/matlab/s10/html/eval_z_transf.html
Note that this type of inversion is notoriously tricky to do numerically, as it requires very precise numbers. Working with the Symbolic Toolbox allows you to request arbitrary precision (e.g., 64 digits of precision).
Hope that helps!
Can we apply these functions to a discrete function?
Thank you :)
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