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Easy Laplace Transform

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Apply the definition in (1) to find directly the Laplace transforms of the functions described (by formula or graph).
    1) f(t)=t


    2. Relevant equations



    3. The attempt at a solution
    Seems pretty easy... Question is, I don't understand the directions exactly.. Am I allowed to use the tables or do I have to solve the transform? Anyway this is how I solved, but i got stuck..
    $$\mathcal{L}(f(t))=\int_0^{\infty}e^{-st}f(t)dt=\int_0^{\infty}te^{-st}dt$$
    Then using an integration by parts, u=t and dv=e-stdt, I get:
    $$\frac{-te^{st}}{s}\bigg|_0^{\infty}-\int_0^{\infty}\frac{-e^{st}}{s}dt=\left(\frac{-te^{st}}{s}-\frac{e^{s}t^{s}}{s^{2}}\right)\bigg|_0^{\infty}$$
    Is that right up to there and what do I do after?
     
  2. jcsd
  3. Nov 18, 2013 #2

    Dick

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    Yes, you solve it by integrating by parts and you are starting with right parts. But it goes badly after that. How did ##e^{-st}## turn into ##e^{st}##? And I have no idea how you got some of the other stuff.
     
  4. Nov 18, 2013 #3
    Yes I messed up, it should be u= t, du=dt; ##dv=e^{-st}dt##, ##v=\frac{-e^-st}{s}## Which leads to:
    $$\frac{-te^{-st}}{s}-\int\frac{-e^{-st}}{s}dt=\frac{-te^{-st}}{s}-\frac{e^{-st}}{s^{2}}$$

    Right?
     
  5. Nov 18, 2013 #4

    LCKurtz

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    On your last line you have dropped the minus sign in the ##e^{-st}##. And check that antiderivative in red.

    [Edit] Took too long to type it, others beat me to it.
     
  6. Nov 18, 2013 #5

    Dick

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    That looks much better. Now you evaluate it between 0 and infinity. It's going to involve thinking about limits. Take s>0.
     
  7. Nov 18, 2013 #6
    $$\lim_{t \to \infty} \frac{-te^{-st}}{s}-\frac{e^{-st}}{s^{2}}$$
    It seems like both would go to 0... Maybe I'm not seeing something correctly
     
  8. Nov 18, 2013 #7

    Dick

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    Yes, they both go to 0. That ##te^{-st}## goes to zero might take a little justification, but if you know why, that's fine. Now you have to take the limit as t->0 and take the difference.
     
  9. Nov 18, 2013 #8
    ##te^{-st}## can be written as ##\frac{t}{e^{st}}## and then you can use L'Hopitals since you get ##\frac{\infty}{\infty}## and end up with ##\frac{1}{se^{st}}=0##

    AHHH now I see, I forgot you have to evaluate at 0 also...
    So the first fraction goes to 0 and the second one will be e-s(0)=1 so you end up with: ##\frac{1}{s^{2}}##

    Thanks for the help!
     
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