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Unsure about Inverse Laplace Heaviside Function question

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the inverse Laplace transform of
    F(s)=5e^(-8s)/(s2+36)


    2. Relevant equations


    3. The attempt at a solution
    I know that to find the inverse Laplace transform of this function, I start by factoring out (e^(-8s)) to end up with 5/(s^2+36), and that my final answer will be step(t-8)f(t-8), where f(t-8) is the inverse of 5/(s^2+36). I've checked the Laplace table and I can't find the inverse of this function.
     
  2. jcsd
  3. Apr 11, 2014 #2
    It's because your understanding of the Heaviside Step Function is lacking definition. The fact that you have [itex]\int[/itex]e-8s([itex]\frac{5}{s^{2}+36}[/itex])ds is the same as saying you have a square wave with the function F(s) convoluted with it. When you integrate this, it's like integrating a Dirac-Delta function in the sense that if you do [itex]\int[/itex]δ(t-a)f(t)dt you get f(t-a). Now imagine you have a bunch of δ(s-c) "glued together". That is one basic way to define the Heaviside Step Function uc(t).

    Now if your integral was just [itex]\int[/itex]e-8sds you'd have δ(t-8). However, if you have [itex]\int[/itex][itex]\frac{e^{-8s}}{s}[/itex]ds, you get uc(t-8). This is because your integral is of the form [itex]\int[/itex]e-8sF(s)ds where your F(s) is [itex]\frac{1}{s}[/itex]. Now, the only difference between these two integrals is the F(s). What we are saying is we are performing the convolution between the Dirac Delta Function and a function F(s). The F(s) being [itex]\frac{1}{s}[/itex] though, when brought in the time domain, is just 1. So we are saying (once back in the time domain) that we have a 8 δ(t-a) "glued together", hence a square wave.

    Moving on from here, if we have a F(s) that is not [itex]\frac{1}{s}[/itex] , but say [itex]\frac{5}{s^{2}+36}[/itex], then we are saying we have a bunch of F(s) spanning the square wave's domain. So in the time domain this would be represented as uc(t)f(t-a). This says you have a function spanning the domain of the unit impulse uc(t).
     
  4. Apr 11, 2014 #3

    Ray Vickson

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    Homework Helper

    No: the inverse of ##5/(s^2 + 36)## is ##f(t)##, not ##f(t-8)##.

    To find the inverse, either consult a better table, or else do it manually, by writing ##1/(s^2 + 36) ## in partial fractions involving ##1/(s+ 6i)## and ##1/(s - 6i)##, ##i = \sqrt{-1}##. Now invert each term separately and combine the results.
     
  5. Apr 11, 2014 #4
    Isn't it 5u(t-8) sin(6(t-8))? Check a table of laplace transforms.

    Chet
     
  6. Apr 11, 2014 #5
    Yes Chet, it is, but not a 5...it's a 5/6 i believe. I was just trying not to give him f(t-a) :p

    So: [itex]\frac{5}{6}[/itex]uc(t)sin(6(t-8))
     
  7. Apr 11, 2014 #6
    Oh boy. You're right. I guess I should learn how to use the tables also.

    Chet
     
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