Residue Theorem for Laplace Transform

Click For Summary

Discussion Overview

The discussion centers around the application of the Residue Theorem in the context of Laplace Transforms, specifically regarding the inverse transformation of functions represented in the Laplace domain. Participants explore the theoretical underpinnings and practical applications of using residues to compute inverse Laplace Transforms, as well as the challenges associated with this method.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on the Residue Theorem as it applies to Laplace Transforms, expressing a need for a general theorem or explanation.
  • Another participant mentions that residues can be useful for calculating inverse Laplace Transforms when one does not want to rely on tables, highlighting the necessity of evaluating complex integrals.
  • A participant expresses confusion regarding the use of a specific contour in the integral and questions the origin of the integral expression used for the inverse transform.
  • One participant suggests that many prefer using tables of Laplace Transforms due to the complexity involved in understanding the necessary concepts from complex variables.
  • Concerns are raised about the readability of the article referenced, particularly regarding the clarity of certain mathematical expressions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to using the Residue Theorem for Laplace Transforms, with some favoring traditional methods and others advocating for the residue approach. Confusion and differing levels of familiarity with the concepts are evident.

Contextual Notes

Participants express uncertainty about the specific contours used in integrals and the derivation of certain integral expressions. There is a noted reliance on external articles for clarification, which may not fully address all participants' questions.

juan.
Messages
8
Reaction score
0
I need to know what's the Residue Theorem for a Laplace Transform. Does anyone know the name or something, so I can search it? I couldn't find anything.

For example, if I have this two equations:

X(s).(s-1) = -Y(s)+5

Y(s).(s-4) = 2.X(s)+7

I know how to solve them using Simple Fractions, but I need to know how to solve that using Residue Theorem.

Oh, I forgot to mention that I'm looking for the Inverse Transform of Y(s) and X(s)
Thanks!

EDIT:

I know that, for example, for y(t) I'm going to have this:

y(t) = Res[Y(s).e^{st}, 2] + Res[Y(s).e^{st}, 3]

but I need to know why and a general case (a Theorem, for example)
 
Physics news on Phys.org
juan. said:
I need to know what's the Residue Theorem for a Laplace Transform. Does anyone know the name or something, so I can search it? I couldn't find anything.

For example, if I have this two equations:

X(s).(s-1) = -Y(s)+5

Y(s).(s-4) = 2.X(s)+7

I know how to solve them using Simple Fractions, but I need to know how to solve that using Residue Theorem.

Oh, I forgot to mention that I'm looking for the Inverse Transform of Y(s) and X(s)
Thanks!

EDIT:

I know that, for example, for y(t) I'm going to have this:

y(t) = Res[Y(s).e^{st}, 2] + Res[Y(s).e^{st}, 3]

but I need to know why and a general case (a Theorem, for example)
Most LT can be calculated using integral calculus. If you don't want to use a table of LT to calculate the inverse, then residues come in handy, since you need to evaluate a complex integral.

The attached article shows how to use residues to compute an inverse LT:

http://www.staff.city.ac.uk/~george1/laplace_residue.pdf

There are other articles which can be found if you Google "Inverse laplace transform by residue theorem" :)
 
Perfect! But I have 2 problems:
- I never used a contour like this: D. It was always with the line in the x axis
- Then, I don't understand where \int_{C_1}{ } F(s).e^{s.t}ds came from. If you look at the Inverse Transform of Laplace, you can see a \frac{1}{2j\pi} and the limits of the integral are awful.

Thanks!
 
That's why most people use a table of LTs and manipulate them to obtain the inverse. It requires a lot less knowledge of complex variables in order to obtain the inverse of the LT.

The article I linked was apparently written for electrical engineers, who use j2 = -1 for the complex constant i.

As far as the integral used for the inverse LT, see p. 1 of the article. It comes from taking the Fourier transform of the LT and then taking the inverse Fourier Transform.

Some of the printing on this page is hard to read unless magnified. E.g., in some terms involving e-jωt, the negative sign is almost illegible unless you zoom in on the page.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K