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Sign problem in laplace transform

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex] \int_{\pi/2}^{\infty}cos(t)e^{-st}dt [/tex]



    2. Relevant equations


    [tex] \int vdu=uv-\int udv [/tex]
    [itex] v=cos(t) , dv=-sin(t)dt [/itex], [itex] du=e^{-st} , u=\frac{-e^{-st}} {s}[/itex]

    3. The attempt at a solution
    [tex] \int_{\pi/2}^{\infty}cos(t)e^{-st}dt = \frac{-e^{-st}}{s}cos(t) -\int (-)\frac{-e^{-st}}{s}(-)sin(t)dt [/tex]
    the right side becomes:
    [tex] \frac{-e^{-st}}{s}cos(t) - \frac{1}{s} \int e^{-st}sin(t)dt [/tex]
    for the second integral on the right:[itex] v=sin(t) , dv=cos(t)dt [/itex] and [itex] du=e^{-st} , u=\frac{-e^{-st}} {s} [/itex]
    [tex] \int e^{-st}sin(t)dt =- \frac{1}{s}e^{-st}sin(t)dt - \int \frac{-e^{-st}}{s}cos(t) [/tex]
    now [tex] \int_{\pi/2}^{\infty}cos(t)e^{-st}dt = \frac{-e^{-st}}{s}cos(t) - \frac {1}{s} \left[ - \frac{1}{s}e^{-st}sin(t) + \int \frac{e^{-st}}{s}cos(t)dt \right] [/tex]
    thus
    [tex] \frac{-e^{-st}}{s}cos(t) + \frac {1}{s^2}e^{-st}sin(t) - \frac {1}{s^2} \int e^{-st}cos(t)dt [/tex]
    simplifying
    [tex] \int_{\pi/2}^{\infty}cos(t)e^{-st}dt \frac {s^2 +1} {s^2}= \frac {1}{s^2}e^{-st}sin(t) - \frac{e^{-st}}{s}cos(t) [/tex]
    going on :
    [tex] \int_{\pi/2}^{\infty}cos(t)e^{-st}dt = \frac {s^2}{s^2 +1} \left[ \frac {1}{s^2}e^{-st}sin(t) - \frac{e^{-st}}{s}cos(t) \right] [/tex]
    the right side is:
    [tex] = \frac {1}{s^2 +1}e^{-st}sin(t) + \frac{e^{-st}}{s^2+1}scos(t) [/tex]
    now when i evaluate the limits:
    [tex] =(0-0) - ( \frac{e^{-s\pi/2}}{s^2+1} - 0 ) [/tex]
    [tex] =- \frac{e^{-s\pi/2}}{s^2+1} [/tex]
    im sure this is wrong because of the sign (its supposed to be positive now where did i drop the sign? can u help? this was the function whose transform i was looking for
    [tex]
    f(x)=\left\{\begin{array}{cc}0,&\mbox{ if }0\leq x\leq \pi/2 \\
    cos(t), & \mbox{ if } x>\pi/2 \end{array}\right.
    [/tex]
     
    Last edited: Sep 17, 2009
  2. jcsd
  3. Sep 17, 2009 #2

    Dick

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    I get the same thing you got. Why would you think the answer should be positive? cos(t) is negative for pi/2<t<3pi/2. If you were trying to transform that step function f(x), where did the cos(t) come from?
     
  4. Sep 17, 2009 #3
    well typo on my part , let me edit it,
     
  5. Sep 17, 2009 #4
    Here's the culprit :
    [tex] L \left\{ U_{a}(t) f(t-a) \right\} = e^{-as} F(s) [/tex]
    which implies my result might be wrong
     
  6. Sep 17, 2009 #5

    Dick

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    That formula would be helpful for integrating cos(t-pi/2)*exp(-ts) from pi/2 to infinity. But that's not the same as your problem.
     
  7. Sep 17, 2009 #6
    So the answers for them would be different?
     
  8. Sep 17, 2009 #7
    So the answers for them would be different? As an aside how would you determine [tex]L(t^2u(t-1))[/tex] with the hint to write [tex]t^2u(t-1) [/tex] in the form f(t-a)u(t-a)
     
  9. Sep 17, 2009 #8

    Dick

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    Sure the answers to cos(t)e^(-st) and cos(t-pi/2)*e^(-st) integrated from pi/2 to infinity are different. If you want t^2 to be f(t-1) then I guess f(t) would have to be (t+1)^2, right?
     
  10. Sep 17, 2009 #9
    Thats sensible thanks,
     
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