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!

Inverse Laplace transform

  1. Jul 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the inverse laplace transform of (s+1)/ (s^2 + 4s + 5) + e^-2s / 3s^4

    2. Relevant equations



    3. The attempt at a solution

    For the first one

    (s+1) / (s^2 + 4s + 5), I completed the square for the denominator so

    (s+1) / [(s+1)^2 + 1]

    Now it gets confusing, how do I find out the inverse laplace of this.

    And for e^-2s / 3s^4 , is the inverse laplace 1/9 x 1/ (s+2)^4 = 1/9(s+2)^4
     
  2. jcsd
  3. Jul 14, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The same way you find the inverse Laplace transform of just about anything: look it up in a table.
    If you don't have one in your book, here's one:
    http://www.vibrationdata.com/Laplace.htm

    You will notice that they give an inverse Laplace transform for [itex]s/(s^2+ \alpha[/itex] as well as the inverse Laplace transform for [itex]F(x-\alpha)[/itex]. Use them together for [itex] (s+1)/((s+1)^2+ 2)[/itex].
     
  4. Jul 14, 2008 #3
    The problem is I dont get the LaTeX Code: s/(s^2+ \\alpha thing. I only get the f(x +a) thing.
     
  5. Jul 14, 2008 #4
    In the last line of Ivy you should recognize a form that can be found in the table. Ivy rewrote the equation in that form by taking out a square.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Inverse Laplace transform
Loading...