Residue Theorem for Laplace Transform

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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:

[tex]X(s).(s-1) = -Y(s)+5[/tex]

[tex]Y(s).(s-4) = 2.X(s)+7[/tex]

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:

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

but I need to know why and a general case (a Theorem, for example)
 

Answers and Replies

  • #2
SteamKing
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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:

[tex]X(s).(s-1) = -Y(s)+5[/tex]

[tex]Y(s).(s-4) = 2.X(s)+7[/tex]

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:

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

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" :)
 
  • #3
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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 [tex]\int_{C_1}{ } F(s).e^{s.t}ds[/tex] came from. If you look at the Inverse Transform of Laplace, you can see a [tex]\frac{1}{2j\pi}[/tex] and the limits of the integral are awful.

Thanks!
 
  • #4
SteamKing
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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.
 

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