Residue Theorem for Laplace Transform

In summary, the Residue Theorem is a theorem that is used to calculate the inverse LT of a Laplace Transform. It requires the use of integral calculus and can be found in various articles on the internet.
  • #1
juan.
8
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:

[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)
 
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  • #2
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:

[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
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
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.
 

FAQ: Residue Theorem for Laplace Transform

What is the Residue Theorem for Laplace Transform?

The Residue Theorem for Laplace Transform is a mathematical concept used in complex analysis to evaluate integrals of complex functions. It states that the integral of a function along a closed path in the complex plane is equal to the sum of the residues of the function at its singularities inside the path.

What is a residue in the context of Laplace Transform?

In the context of Laplace Transform, a residue is the value of a complex function at a singularity, which is a point where the function is not defined or becomes infinite. Residues are used to calculate integrals using the Residue Theorem.

How is the Residue Theorem applied in Laplace Transform calculations?

The Residue Theorem is applied in Laplace Transform calculations by first identifying the singularities of the function and then finding the residues at those points. The integral can then be evaluated by summing the residues and multiplying by a factor of 2πi.

What are the benefits of using the Residue Theorem in Laplace Transform?

The Residue Theorem allows for the evaluation of complicated integrals using simple algebraic operations. It also provides a more efficient method for computing Laplace Transform compared to the traditional method of using partial fraction decomposition.

Are there any limitations to using the Residue Theorem in Laplace Transform?

The Residue Theorem can only be applied to functions that are analytic (smooth and continuous) in the complex plane. It also assumes that the path of integration encloses all the singularities of the function. If these conditions are not met, the Residue Theorem cannot be used.

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