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Laplace Transform of a signal?

  1. Nov 17, 2011 #1
    Can someone explain the steps of this solution? In my linear systems class, we are doing Laplace transforms using transform tables and the properties. I can usually do the problems that closely resemble the table, but when they involve heavy algebraic manipulations, I don't know what to do.


    Why do the +1 and -1 appear? Where did the extra exponential come from? If someone can go line for line, I would appreciate it.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 17, 2011 #2


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    Gold Member

    [itex][/itex]well, that looks a little messy, but let's do this step by step:

    start from
    [tex] \mathcal{L} \{ (t-2)^2 e^{-5t} u(t-1) \} = \mathcal{L} \{ (t-1-1)^2 e^{-5t+1-1} u(t-1) \} [/tex]

    basically, [itex]t-2= t-1-1[/itex] and [itex] t = t + 0= t +1 -1 [/itex], straightforward enough I hope

    [tex] \mathcal{L} \{ (t-1-1)^2 e^{-5t+1-1} u(t-1) \} = \mathcal{L} \{ [ (t-1)^2 - 2 (t-1) +1 ] e ^{-5} e^{-5(t-1)} u(t-1) \} [/tex]

    the first part: [itex] (t-1-1)^2 = (t-1)^2 - 2 (t-1) +1 [/itex] is basically the same as expanding [itex] (x-1)^2 [/itex] seeing [itex]x=t-1[/itex]

    the second part: [itex] e^{-5t+1-1} = e ^{-5} e^{-5(t-1)} [/itex] is a property of exponents/logs [itex]e^a e^b = e^{a+b}[/itex]

    The rest is basically expanding the terms,
    things you need to know is [itex] e^{-5} [/itex] is just a constant, laplace transform is a linear operation so [itex] \mathcal{L} \{ e^{-5} f(t) \} = e^{-5} \mathcal{L} \{ f(t) \} [/itex]
    and [itex] \mathcal{L} \{ f(t-1) \} = e^{-s} \mathcal{L} \{ f(t) \} [/itex]

    thats basically it, feel free to ask any other questions you may have :)
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