Residue Theorem for Laplace Transform

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)

• I just Googled "inverse laplace residue" and there were heaps of hits. – David Feb 7 '15 at 5:27
• The inverse laplace transform IS by definition the residue of the function F(s)*e^(st). Here's what I would do: Open three windows. Window one shows the inverse laplace transform forumla, window two shows the cauchy integral equation, window three shows the definition of the residue (all from wikipedia that is fine). See the correspondence between the three formula! – Carlos - the Mongoose - Danger Feb 7 '15 at 5:32
• There is actually also a website explaining the detailed steps: lmgtfy.com/?q=inverse+laplace+residue. – Fabian Feb 7 '15 at 6:24
• @David Thanks for your comment. I have done that before I made this thread, but I couldn't find what I was asking, that's why I made this thread :) Please, if you can, help me! Thanks! – Unnamed Feb 8 '15 at 8:00
• @IllegalImmigrant Thanks for your comment. I have done that before I made this thread, but I couldn't find what I was asking, that's why I made this thread :) Please, if you can, help me! Thanks! – Unnamed Feb 8 '15 at 8:01

The use of the residue theorem comes from the definition of the ILT:

$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F(s) e^{s t}$$

where the line $\operatorname{Re}{s}=c$ is to the right of the rightmost pole of $F$. To evaluate this integral when $F$ has simple poles, one closes a contour to the left of the line of integration for $t \gt 0$; the contour then contains all of the poles of $F$. The ILT is then the sum of the residues of the poles of $F$. To evaluate for $t \lt 0$, one closes the contour to the right; since there are no poles there, the ILT for $t \lt 0$ is zero.

With that in mind, let's solve the above system for $X$ and $Y$:

\begin{align}(s-1) X + Y &= 5 \\ -2 X+(s-4) Y &= 7 \\ \end{align}

Solving, I get

$$X(s) = \frac{5 s-27}{(s-2)(s-3)}$$ $$Y(s) = \frac{7 s+3}{(s-2)(s-3)}$$

So, for $t \gt 0$, $x(t)$ is equal to the sum of the residues of $X(s) e^{s t}$ at $s=2$ and $s=3$, and zero otherwise. This is then

$$x(t) = \left (17 e^{2 t}- 12 e^{3 t} \right ) \theta(t)$$

where $\theta$ is the Heaviside step function, zero for $t \lt 0$, one for $t \gt 0$. Similarly,

$$y(t) = \left (-17 e^{2 t}+24 e^{3 t} \right ) \theta(t)$$