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Math Help - step Laplace

  1. #1
    Member pberardi's Avatar
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    step Laplace

    Hi,
    I am bit confused by the step function. I was under the impression that it is basically a plug and chug but I think I am doing something wrong and need some clarification and some guidance.

    Find the L[usub2(t)e^(-t)cos(3t)
    So I get e^(-2s)L[e^(-2t-4)cos(3t+6)]

    I am almost sure that this is wrong but really don't have any examples that are similar. Can someone please help me with this?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by pberardi View Post
    Hi,
    I am bit confused by the step function. I was under the impression that it is basically a plug and chug but I think I am doing something wrong and need some clarification and some guidance.

    Find the L[usub2(t)e^(-t)cos(3t)
    So I get e^(-2s)L[e^(-2t-4)cos(3t+6)]

    I am almost sure that this is wrong but really don't have any examples that are similar. Can someone please help me with this?
    \mathcal{L} u_2(t) e^{-t}\cos(3t) =\int_2^\infty e^{-t}\cos(3t) e^{-st} dt

    Now change the variable to \tau=t-2 then we have:

    \mathcal{L} u_2(t) e^{-t}\cos(3t) =\int_0^\infty e^{-\tau-2}\cos(3\tau + 6) e^{-s\tau-2s} d\tau

    Simplify:

    \mathcal{L} u_2(t) e^{-t}\cos(3t) =e^{-2s-2}\int_0^\infty e^{-\tau}\cos(3\tau + 6) e^{-s\tau} d\tau=e^{-2s-2}\left[\mathcal{L}e^{-t}\cos(3t + 6)\right]

    Or you may prefereint in the form:

    \mathcal{L} u_2(t) e^{-t}\cos(3t)=e^{-2s}\left[\mathcal{L}e^{-t-2}\cos(3t + 6)\right]

    Alternativly fiddle with the translation theorem to get the same result.

    CB
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  3. #3
    Member pberardi's Avatar
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    Well the step after that is actually the confusing part. How does the exponential with the t mesh with the exponential with the s? Also are you taking the Laplace of the exp with the t? How does that come out? Thank you for your assistance but it is from this part on that I get confused about.
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