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Beginner Laplace Transforms

  1. Aug 27, 2008 #1
    Hi all, 2 questions here

    1) I've been doing some questions on laplace transforms and have been running into some trouble getting my answers into the same form as the answers given with the questions.
    For example:

    f(t) = 1 - e^(-t)
    Using the linearity property i got 1/s - 1/(s-1) which is correct? But the answers give 1/s(s-1). Later in the lecture notes they show how to do the inverse laplace transform of the answer they gave and use partial fractions to get it into the form of the answer I got, i was just wondering what they have done to go from my answer to their answer, what steps am i missing?

    2) f(t) = sinhtcosht, does this involve convolution or is that only when doing the inverse laplace transform where L(f)L(g) = f(t)*g(t)?
  2. jcsd
  3. Aug 28, 2008 #2


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    Gold Member

    Its been a while since ive looked at Laplace transforms, but is the LT of e^-t not 1/(s+1) instead of 1/(s-1)?
  4. Aug 28, 2008 #3
    1) Try and simplify your expression using a common denominator of s and (s-1).

    2) Do you know the definition of sinh(t) and cosh(t) in terms of exponentials?
  5. Aug 29, 2008 #4
    oops yeah i did mean 1/(s+1) danago

    and Pere i wouldnt have a clue about sinh(t) and cosh(t) a exponentials, maybe i should have paid more attention in first year maths
  6. Aug 29, 2008 #5

    If you don't know about the definition of sinh(t) I don't see how you could calculate its LT..

    \sinh t =& \frac{e^t-e^{-t}}{2}\\
    \cosh t =& \frac{e^t+e^{-t}}{2}

    For some background reading you could have a look at wikipedia.
  7. Aug 29, 2008 #6


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    Homework Helper

    There should be a table of Laplace transforms which you can refer to check LT of sinh and cosh. But in any case, Pere Callahan's post is sufficient for you to do it, so long as you know the LT of e^t and e^-t.
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