# Inverse Laplace Transform Control Systems

1. May 8, 2004

Hi I am currently working my way through a control systems textbook and I am bringing my understanding of the relevant mathematics up to speed. I have happened across the Laplace Transform which I am happy enough with. I am concerned however that given the fact that this is a real project I may find that the present calculations in the machine's software I am working on may not fit into something that can be easily inverse transformed using the standard models in the tables. Maybe I am worrying unnecessarily but I would like to be able to perform a complete Inverse Laplace Transform without the use of the tables. I am a competent mathematician, but using conventional methods I have not been able to perform this transform. Is it possible to do this, or this beyond the scope of anyone without a degree solely in mathematics? I have tried all sorts of substitutions and integration methods and so on, but I keep ending up with equations which cannot be manipulated any further, or just keep growing into even nastier equations. Can anyone give me a clue as to how this is actually done without the aid of tables or a computer?

I hope this has made some sense!!!

2. May 8, 2004

### enigma

Staff Emeritus

Welcome to the forums!

Doing inverse Laplaces is not an easy proposition. IIRC, you need to do a integration over a surface in state space.

FWIW, I've yet to see a problem which requires a direct inverse laplace to be performed.

Non-linearities are usually wrapped up into a 'D' disturbance function. This disturbance function contributes to steady state error, and extra poles at the origin of the controllers are added in to counter them.

3. May 9, 2004

Staff Emeritus
I have seen a Laplace transform inverted by setting up a "trick" contour and doing contour integration in the complex plane (no poles inside the contour). The LT turns out to be identically zero on part of the contour and hence the whole contour integral, which can be evaluated, equals the negative of its value on the other part, which was sought. Actually the zero part of the contour is the mirror image of the non-zero part in the imaginary axis, and the zero comes from the positivity of the LT.

4. May 9, 2004

Wow! I wasn't expecting advice as good as and as helpful as that. Thankyou both very much. The trick contour idea seems fascinating. I suspect that may well be a bit out of my depth at present, but what a great idea!! It's also a big relief to hear that it is unlikely I will have to perform an inverse LT for this project. I have to say however it is something out of shear curiousity I would love to be able to do one day.

The D(s) function was something I have not been introduced to before, but makes an awful lot of sense. Thankyou very much for that. Once I have some more data about the machine (likely this week) I will get back to you again.

5. Jul 15, 2004

### Feynman

The idea to use the contour complxe is exellent and u must pay attention how choose the contour
Then u have a complexe integral (or Cauchy's integral) and use Cauchy's integral theorem or The residue theorem.
Really i felicite THe Self adjoint "Operator"

6. Jul 16, 2004

### MiGUi

When you learn to integrate, you use tables that you learn to do hard integrals... obviously you can calculate it yourself, but is a very hard work...

With Laplace occurs the same, you are condemned to use tables xD

7. Jul 18, 2004

### Tyger

I seem to recall that the Schuam's Outline Series on the Laplace Transform gives a good explaination of the contour integral as the inverse transform.

One of the main reasons people use the Laplace Transform is that the tables and a few rules of thumb can be used to solve almost any real problem.

8. Jul 21, 2004

### turin

From what I remember, the contour must contain all of the poles. But from what selfadjoint has said, it can contain none of them? I am lost. When you do a Laplace transform, you should be implicitly aware of the domain of stability, which basically limits you to a region of the frequency domain to the left of some line. Then, you have to make sure to include this entire region inside your contour when you take the inverse Laplace transform. Look closely at the tables and you should see restrictions on the real part of s. For example, IIRC:

L{u(t)} = 1/s, where Re{s} > 0

This is directly related to the convergence of the kernel of the transform: e-st. If Re{s} < 0, then this kernel will "blow up" as the integral is taken to t -> infinity.