Impulse Function

Discussion in 'Homework Help' started by thisonedude, Sep 11, 2014.

  1. thisonedude

    Thread Starter Member

    Apr 20, 2014
    Do i'm studying for an exam and i needed help with some clarification. How is it that the integral of the impulse function multiplied with e^-st is equal to the integral of just the impulse function? What happened to e^-st. According to the sifting property, doesn't e^-st become f(a) = e^-sa, since it's continuous at a. And so then shouldn't he final integral be f(a) and not 1? Is anyone able to help? I've attached the function: 1.JPG
    Last edited: Sep 11, 2014
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    The Laplace tranform shown is for the delta function at t=0, not at t=a.
  3. blah2222

    Well-Known Member

    May 3, 2010
    As t_n_k mentioned, the last equation that you are referring to the time offset 'a' is equal to zero, making e^-s*t = e^s*0 = 1