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Laplace time differentiation property

  1. Jul 12, 2009 #1
    Hello,
    I'm trying to prove the time differentiation property of Laplace transform.
    dx(t)/dt = sX(s)


    http://img10.imageshack.us/img10/290/tlaplace.jpg [Broken] http://g.imageshack.us/img10/tlaplace.jpg/1/ [Broken]

    how do i continue from here ?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 12, 2009 #2

    Redbelly98

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    I'm not sure what the first step here is supposed to represent. Wouldn't the first thing to do be to write

    L[dx/dt] = ...

    And then use the definition of the Laplace transform?
     
  4. Jul 13, 2009 #3
    yes, you right
    the first term should be like you said
    but what i have to do with the first term of the last part?
    it should vanish but i don't know how? maybe by Final Value Theorem ?
     
  5. Jul 13, 2009 #4

    Redbelly98

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    In the final integral, t is the variable of integration.

    Take any term that is independent of t outside of the integral, and see what you get.

    p.s. your integration limits are wrong.
     
  6. Jul 13, 2009 #5
    the final integral is sX(s)
    but the one before should vanish somehow..., i think it's because of Final Value Theorem
    why do you think the integration limits are wrong?
     
  7. Jul 13, 2009 #6

    Redbelly98

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    Yes, good.

    It looks like there are two definitions of the Laplace transform:
    http://mathworld.wolfram.com/LaplaceTransform.html

    The integrations limits are usually taken from 0 to +∞, but maybe your class is using the -∞ to +∞ definition instead. I didn't know about that definition before.

    If the limits are 0 to +∞, you can just evaluate the x(t) e-st term at the limits. I'm not sure what to do if you're using the -∞ definition though.
     
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