1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook