1. Limited time only! Sign up for a free 30min personal 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!

Laplace Transform of this function

  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    We want to find the Laplace transform for

    f(t): 0 for t≤2 and (t-2)2 for t≥2

    2. Relevant equations
    I know that Lap{uc f(t-c)} = e-csLap{f(t)}=e-csF(s)
    I rewrite f(t)=0+g(t) where g(t) = 0 for 0≤t<2 and (t-2)2 for t≥2
    so we can write f(t)=g(t)= u2(t)*(t-2)2
    Lap{f(t)}= (e-2s/s) * Lap{(t-2)2}

    3. The attempt at a solution

    The answer in my bok is 2*(e-2s)/(s3)
    But Lap{(t-2)2} is (2/s3) - (4/s2)+(4/s)

    why ?!
  2. jcsd
  3. Oct 10, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    (t-2)^2 is not the same as your f(t) (especially in the range 0<t<2 :smile:) !
  4. Oct 10, 2016 #3
    TNX !

    Now I've got it!:smile:
  5. Oct 10, 2016 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Start with your formula ##{\cal L}[u(t-c) f(t-c)](s) = e^{-cs}{\cal L}[f(t)](s)##. So if ##F(s) = {\cal L}[t^2](s)##, then the answer you want is ##\text{answer} = e^{-2s} F(s)##.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Laplace Transform of this function