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Laplace transform of unit step function

  1. Aug 22, 2013 #1
    1. The problem statement, all variables and given/known data

    dut86FX.jpg

    3. The attempt at a solution

    I know that u(t) is a unit step function and holds a value of either 0 or 1. In laplace transform, when we integrate f(t) from 0 to infinity, we take u(t) to be 1.

    In this case, since u(t) is u(-t), does this mean it holds a value of 0? Does not make sense to me that both answers are 0.

    Thanks in advance for the help. :) Any links to notes on similar laplace transforms would be very much appreciated as well! :)
     
  2. jcsd
  3. Aug 23, 2013 #2

    HallsofIvy

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    I don't understand what you want to do. The Laplace transform is defined only for functions on the positive real numbers. u(t) is 1 for [tex]0\le t\le 1[/tex], 0 otherwise. So u(-t) is 1 for [tex]-1\le t\le 0[/tex]. Since both of your x(t) is 0 for all positive t, their Laplace transform is 0.
     
  4. Aug 23, 2013 #3
    Thank you so much for your reply Halls of Ivy! It's much appreciated.

    The exact question is actually just "Find the Laplace transform of:". I apologize for leaving it out.

    So just to confirm, the answers for both questions are actually 0? Seems a little weird that they've give us two questions on the same "concept".

    Could it be possible that instead of a unit step function, it could be some sort of shift?

    Once again, thank you so much! :)
     
  5. Aug 23, 2013 #4

    Ray Vickson

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    Careful: the usual definition of the unit step function u(t) is
    [tex] u(t) = \left\{ \begin{array}{cl}1 & \text{ if } t \geq 0\\
    0 & \text{ if } t < 0.
    \end{array} \right. [/tex]
    So, as said already, your functions are multiples of u(-t), so are 0 for t >= 0. Thus, their Laplace transforms are zero.
     
  6. Aug 23, 2013 #5

    LCKurtz

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    I agree. I would almost bet there is something missing or misunderstood about notation or transcription here. Have you copied the complete problem exactly, word for word?
     
  7. Aug 23, 2013 #6
    Yups, that's all the question says. Nothing before that and no hints after that. :( I guess I'll just stick to 0 then.

    Is it possible though that it could be a shifting problem instead of a unit step function?
     
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