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Inverse Laplace Transform

  1. Feb 27, 2007 #1
    My book on signal processing says that:

    [tex] f(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} F(s) e^{st} ds = \lim_{\Delta s \to 0} \sum_{n = -\infty}^{\infty} \Big[ \frac{F(n\Delta s)\Delta s}{2\pi j} \Big] e^{n\Delta s t}[/tex]

    I don't get this. How/Why can you write a integration over a complex variable as the above sum?

    edit: I forgot a coefficient [tex]\frac{1}{2\pi j}[/tex] on the LHS. Sorry about that. Its fixed now.
     
    Last edited: Feb 28, 2007
  2. jcsd
  3. Feb 27, 2007 #2
    Isn't that definition of a Reimann integral?
     
  4. Feb 28, 2007 #3
    I understand how you could be confused! It would have to be explained what \Delta s is in the sum since the integral seems to have s lie on a vertical line c + jy in the complex plane (real y). So do we perhaps have \Delta s =
    j\Delta y???


    Supposing the above and not knowing anything else about F, using a Riemann sum, you should have limit as \Delta s -> 0 of sum over n of the following terms:
    F(c+n\Delta s)\exp((c+n\Delta s)t)\Delta s
    which is the limit as \Delta y-> 0 of the sum of terms

    F(c+n j \Delta y)\exp((c+ n j \Delta y)t)j\Delta y

    The 2\pi j in the denominator seems something like Cauchy's integral thm., but we need to know quite a bit more about F to get that. (Is F an entire fuction? Does it vanish very rapidly away from the y-axis in the complex plane?)

    Generally speaking, the independence of c on the RHS makes the formula dubious.

    Can you still get the money back for the book?
     
  5. Mar 4, 2007 #4
    Edited gammamcc's post to look nice...
     
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