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!

Laplace Transform Time Shift problem

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the Laplace transform:

    g(t) = 2*e[itex]^{-4t}[/itex]u(t-1)

    3. The attempt at a solution

    Essentially we're told for a time shift we multiply the Laplace transform pair of the function (without the delay) by e[itex]^{-as}[/itex]

    So here a = 1 (for the delay)

    The Laplace transform for e[itex]^{-4t}[/itex] is [itex]\frac{1}{s + 4}[/itex]

    Multiplying we should get e[itex]^{-s}[/itex]([itex]\frac{2}{s + 4}[/itex])

    However the answer is e[itex]^{-s}[/itex]*([itex]\frac{2}{s + 4}[/itex]) * [itex]\frac{1}{e^{4}}[/itex]
     
    Last edited: Apr 3, 2012
  2. jcsd
  3. Apr 4, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Shifting is tricky.

    You have used the formula L{f(t-a)U(t-a)} = e-asL{f(t)}.

    You need the formula for L{f(t)U(t-a)}.
     
  4. Apr 4, 2012 #3
    It's because you must be in the form
    [tex]e^{-4(t-a)}u(t-a)[/tex]
    Take a look at your exponential. It isn't time-shifted by a to the right. What you can do is use the fact that exponentials multiplied add their exponents and the fact that e^b/e^b = 1, so you can multiply by it without changing your values. So you choose e^b so that you can add its exponents and arrive to the needed shift. The e^b in the denominator is factored outside of the linear inverse Laplace operator and you go from there.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laplace Transform Time Shift problem
Loading...