1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating Inverse Laplace Using Convolutions

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data

    We are to use the convolution theorem to compute the inverse Laplace transform of the function

    [tex]L=\frac{1}{s^2 + 16}e^{-2s}[/tex]

    2. Relevant equations
    Using a table, I find that [tex]L^{-1}[\frac{1}{s^2 + 16}] = \frac{1}{4}sin(4t)[/tex]
    and
    [tex]L^{-1}[e^{-2s}] = \delta(t-2)[/tex]


    3. The attempt at a solution
    Using the convolution theorem,


    [tex]L^{-1}[\frac{1}{s^2 + 16}e^{-2s}] = \int_{0}^{t} \frac{1}{4}sin(4\tau)\delta(\tau-2)d\tau[/tex]

    My question is, how do I evaluate that integral? I know that when you have the delta function, e.g. [tex]\delta(t-2)[/tex] times a function f(x), and the limits are negative infinity to infinity, it is just f(2). But what if the limits are 0 to t?
     
  2. jcsd
  3. Feb 6, 2008 #2
    Can anyone help?
     
  4. Feb 6, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\int f(x)\delta(x-a) dx= f(a)[/tex]
    as long as the interval of integration includes x= a and is 0 if it doesn't. It doesn't have to be from [itex]-\infty[/itex] to [itex]\infty[/itex]

    In particular, with a variable upper limit (and assuming a> x0),
    [tex]\int_{x_0}^t f(x)\delta(x-a) dx[/tex]
    is the function that is 0 if t< a, f(a) if t[itex]\ge[/itex] a.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculating Inverse Laplace Using Convolutions
  1. Inverse Laplace (Replies: 4)

  2. Inverse Laplace? (Replies: 2)

Loading...