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Laplace Transform Within a Domain

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the Laplace transform of

    f(t) = t [itex]\forall[/itex] 0≤t≤T, 0 otherwise

    2. Relevant equations



    3. The attempt at a solution
    I write the function as

    [itex]tu(t)-t*u(t-T)[/itex]

    That is turn on the function [itex]t[/itex] at [itex]t=0[/itex] and turn the function [itex]t[/itex] off at [itex]t=T[/itex]. It seems to be right to me.

    But now I struggle with trying to take the Laplace transform of this. I know that [itex]L(u(t)) = \frac{1}{s}[/itex]. I know that [itex]L(f(t-T)) = e^{-sT}F(s)[/itex]. So I know that [itex]L(u(t-T)) = e^{-sT}\frac{1}{s}[/itex], but I'm not sure how to evaluate [itex]L(t*u(t-T))[/itex] because of the extra [itex]t[/itex] term, hence I'm stuck.

    Thanks for any help.

    I seem to be confused because [itex]F(s) = ∫_{0^{-}}^{∞}f(t)e^{-st}dt[/itex]. So I don't see how you can take the Laplace transform over a domain other than over [itex]0^{-}≤t≤∞[/itex], which this question seems to be asking me to do.
     
    Last edited: Jan 20, 2014
  2. jcsd
  3. Jan 20, 2014 #2

    LCKurtz

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    You have a problem of the form ##\mathcal L(f(t)u(t-T))##.$$
    \mathcal L(f(t)u(t-T))=\int_0^\infty e^{-st}f(t)u(t-T)~dt =\int_T^\infty e^{-st}f(t)\cdot 1~dt$$Now make the substitution ##u = t - T##:$$
    =\int_0^\infty e^{-s(u+T)}f(u+T)~du=e^{-Ts}\int_0^\infty e^{-su}f(u+T)~du
    =e^{-Ts}\mathcal Lf(t+T)$$So for your problem$$
    \mathcal L(tu(t-T)) = e^{-Ts}\mathcal L(t+T) =~?$$
     
  4. Jan 21, 2014 #3
    So this problem becomes

    [itex]L(t*u(t) - t*u(t-T)) = \frac{1}{s^{2}} - L(t*u(t-T)) = \frac{1}{s^{2}} - \int^{∞}_{0^{-1}}e^{-st}t*u(t-T)dt = \frac{1}{s^{2}} - \int^{∞}_{0^{-}}e^{-st}tdt[/itex]

    Now I do [itex]u[/itex] substitution to evaluate the integral. [itex]u(t) =t, \frac{du}{dt} = 1[/itex]. [itex]\int dv = \int e^{-st}dt = -\frac{e^{-st}}{s}[/itex]

    Now I continue solving

    [itex]\frac{1}{s^{2}} - \int^{∞}_{0^{-}}e^{-st}tdt = \frac{1}{s^{2}} -(-\frac{te^{-st}}{s} - \int^{∞}_{0^{-}}e^{-st}dt) = \frac{1}{s^{2}} + \frac{te^{-st}}{s} - \frac{e^{-st}}{s}[/itex]

    The solutions has
    [itex]F(s) = \frac{1}{s^{2}} - \frac{1}{s^{2}}e^{-sT} - \frac{T}{s}e^{-sT}[/itex]. So I must be doing something but I don't see what.
     
  5. Jan 21, 2014 #4

    LCKurtz

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    You don't have to work out any integrals and re-do all the work. Just use the formula I gave you above. All you have to do is use that formula with ##f(t)=t##.
     
  6. Jan 22, 2014 #5
    Thanks, I was able to figure it out
     
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