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Math Help - step functions and Laplace Transforms

  1. #1
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    step functions and Laplace Transforms

    Can someone show me how to show that:

    f(t) = (t, 0 < t < 2
    (4-t, 2 < t < 4
    (0, t > 4

    is equal to

    f(t) = t - 2(t-2)U(t-2) + (t-4)U(t-4)

    and determine its Laplace transform.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Have you tried using the definition?:

    \mathcal{L}\left\{{f(t)}\right\}=\displaystyle\int  _0^{+\infty}e^{-st}f(t)\;dt=\displaystyle\int_0^{2}e^{-st}t\;dt+\ldots

    Regards.

    Fernando Revilla
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  3. #3
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    What ideas have you had for the first part so far?
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  4. #4
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    I would have put

    f(t) = tU(t-2) - 4-t-U(t-4)

    and not sure how it would be cleaned up.

    The laplace transform is a mystery.
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  5. #5
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    The way to think of this problem is that the Heaviside step functions (the U(t-a)'s) allow you to "switch on" certain portions of the function at different times. So, let's take your candidate. When t<2, both step functions are zero, because the arguments to both step functions are negative. You're left with f(t) = -4-t for t<2. That does not match up with the definition in the OP. So this can't be correct.

    Technically, the function in the OP is undefined for t<0, so you don't really need a step function U(t) multiplying the initial t.

    Keep in mind that the step functions are cumulative. As t marches past the various cutoffs, more and more step functions will be active. So, for example, when 2<t<4, you've got the initial t, but also the -2(t-2) multiplying the U(t-2). That coefficient of the step function is carefully chosen not only to cancel out the first piece (because it's still active), but also to get the desired 4-t.

    Does this help?

    As for the Laplace Transform, I would just use its linearity and the various properties and theorems.
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