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Laplace transform

  1. Feb 5, 2016 #1
    for the right part of the notes, why the integral of (e^-su)f(u) from 0 to T will become integral of (e^-st)f(t) from 0 to T suddenly ? why not integral of (e^-s (t-nT) )f(t-nT) from 0 to T ? as we can see, u = t +nTd given/known data


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    3. The attempt at a solution
     

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  3. Feb 5, 2016 #2

    Samy_A

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    Do you ask about this step?
    laplace.jpg
    If so, that's just renaming the (dummy) integration variable from u to t.
     
  4. Feb 5, 2016 #3
    yes , why thgere is no need to turn u into t -nT ? since it is given at the left part of the notes
     
  5. Feb 5, 2016 #4

    Samy_A

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    It's just a name. It doesn't matter whether the integration variable is called u or t (or something else).

    Example: the two following integrals are equal
    ##\displaystyle \int_a^b e^x dx = \int_a^b e^y dy##
     
    Last edited: Feb 5, 2016
  6. Feb 7, 2016 #5

    vela

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    The author was a little sloppy. The earlier ##t## (the one related to ##u##) and the ##t## in the last integral aren't the same ##t##.
     
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