Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Step Function (heaviside) laplace xform

  1. Feb 12, 2009 #1
    Hi, I currently have this problem to solve and I can't seem to figure it out

    it goes like this


    Find the Laplace transform of the given function:

    f(t) = { 0 t<2, (t-2)^2 t>=2

    I tried working it out and this is where i get stuck

    f(t) = (t-2)^2 * u(t-2)

    I am not sure if I got the write function for f(t), but if I did, I am not sure how to go on with solving this.
    Any help is appreciated, Thank YOu
    Chota
     
  2. jcsd
  3. Feb 12, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Did you consider integrating it directly?
    The laplace transform of function f(t) is defined as
    [tex]\int_0^\infty e^{-st}f(t)dt[/itex]
    and here
    [tex]\int_0^\infty e^{-st}(t-2)^2 u(t-2)dt= \int_2^\infty (t-2)^2e^{-st}dt[/tex]

    and that can be integrated using integration by parts, twice.
     
  4. Feb 15, 2009 #3
    You need to make use of a translation property of the Laplace transform. It states that:
    if F(s) is the Laplacetransform of f(t) then L{f(t-a)u(t-a)}=exp(-as)F(s)

    Can you apply this to your problem?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Step Function (heaviside) laplace xform
  1. Heaviside function (Replies: 3)

  2. Heaviside Functions (Replies: 1)

Loading...