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Inverse Laplace Transform

  1. May 15, 2005 #1
    I am asked to find the inverse laplace transform of the following function:

    [tex]
    \frac{ \left( s+3 \right) }{ \left( s+1 \right) \left( s+2 \right) }
    [/tex]

    Using tables, can anyone help me understand why the answer is:

    [tex]
    2e^{-t} - e^{-2t}
    [/tex]

    I'm completely loss on this one, and yet the book assumes this is easily determined. Any ideas?

    Note: I already realize that the bottom can be rewritten using partial fractions, but it seems to me that the book assumes that isn't even neccessary - which makes sense, since it doesn't seem to get me anywhere anyway.
     
    Last edited: May 15, 2005
  2. jcsd
  3. May 15, 2005 #2

    OlderDan

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    Look's like it might be this Heaviside formula for the ratio of polynomials. Have you encountered this before?

    http://www.plmsc.psu.edu/~www/matsc597/fourier/laplace/node10.html
     
  4. May 15, 2005 #3
    you have to do partial fractions thats all. [tex]\frac{\left( s+3 \right) }{ \left( s+1 \right) \left( s+2 \right) }= \frac{A}{s+1} + \frac{B}{s+2} [/tex]

    s+3= A(s+2) + B(s+1)= (A+B)s + (2A+B)
    A=2 B=-1

    so you get
    [tex] \frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{-1}{s+2} [/tex]

    from there you can see pretty simply from [tex]e^{at}= \frac {1}{s-a}[/tex] that you get the textbooks given answer.
     
  5. May 16, 2005 #4
    Good grief....

    That's totally correct Gale17 - thanks. I was trying to use partial fractions on just this part:

    [tex]
    \frac{ 1 }{ \left( s+1 \right) \left( s+2 \right) }
    [/tex]

    and then mutliply the answer by s+3. Obviously that was not getting me anywhere, because I still had an s term on the top to deal with.

    Thanks again! :biggrin:

    PS: OlderDan, thanks for the suggestion though. I don't think we have touched on that approach specifically yet, but based on your link, it does seem feasible.
     
  6. May 16, 2005 #5
    an interesting way to use partial fractions... :wink: also, when you have the right answer, try working backwards with it.

    you're welcome! (i like feeling useful anyways... sides, i have my own diff eq final coming up.. woo... )
     
  7. May 16, 2005 #6

    OlderDan

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    The partial fraction approach is definitely the way to go. I thought you were looking for an alternative. That other thing might come in handy for higher order polynomials.
     
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