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Laplace transform: where am I messing up?

  1. Apr 16, 2004 #1
    Ok, using the definition of Laplace transforms to find [itex]\L\{f(t)\}[/itex]

    Given:
    [tex]f(t)=\{^{\sin{t}, 0\le{t}<{\pi}}_{0, t\ge{\pi}}[/tex]

    So, this is what I did:

    [tex]\L\{\sin t\}=\int^{\pi}_{0} e^{-st}\sin t dt+\int^{\infty}_{\pi} e^{-st}(0)dt[/tex]

    [tex]=\int^{\pi}_{0} e^{-st}\sin t dt[/tex]

    [tex]=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}\int^{\pi}_{0} e^{-st}\cos t dt[/tex]

    [tex]=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}(\frac{-e^{-st}\cos{t}}{s}]^{\pi}_{0}-\frac{1}{s}\int^{\pi}_{0} e^{-st}\sin t dt)[/tex]

    [tex]=\frac{-se^{-s\pi}}{s^2}+\frac{1}{s}(\frac{-1}{s}-\frac{1}{s}\L\{\sin t\})[/tex]

    [tex]\L\{\sin t\}(\frac {s^2+1}{s^2})=\frac{-se^{-s\pi}}{s^2}-\frac{1}{s^2}[/tex]

    [tex]\L\{\sin t\}=\frac {-se^{-s\pi}-1}{s^2+1}[/tex]

    Which I know is wrong because the Laplace for sin t should be:

    [tex]\L\{\sin t\}=\frac {1}{s^2+1}[/tex]

    I know my limits of integration will affect the problem but the restricted limits should only add a term to the numerator [itex]e^{-s\pi}[/itex] I believe.

    Where did I mess up?

    Thanks a lot.
     
  2. jcsd
  3. Apr 16, 2004 #2

    arildno

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    Your mistake is in the 5.line, in the evaluation of cos(t) at the limits indicated.
     
    Last edited: Apr 16, 2004
  4. Apr 16, 2004 #3

    arildno

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    Also, the sin(t) evaluation in the 5.line should be zero.
     
  5. Apr 16, 2004 #4
    Got it. Thanks. How silly of me.
     
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