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Find f(t) given an inverse Laplace transform

  1. Mar 22, 2009 #1

    JJBladester

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    1. The problem statement, all variables and given/known data

    Find [tex]f(t)[/tex].

    2. Relevant equations

    [tex]L^{-1}\left\{\frac{s}{s^{2}+4s+5}\right\}[/tex]

    3. The attempt at a solution

    I tried completing the square to get to the solution and I ended up with:

    [tex]L^{-1}\left\{\frac{s}{s^{2}+4s+5}\right\}[/tex] =

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

    Then, I used the inverse of a transform for cosine and the first translation theorum:

    [tex]coskt = L^{-1}\left\{\frac{s}{(s^{2}+k^{2})}\right\}[/tex]

    [tex]L\left\{e^{at}f(t)\right\} = F(s-a)}[/tex]

    with [tex]a[/tex] being -2 and [tex]k[/tex] being 1 to get an answer of:

    [tex]e^{-2t}cos(t)[/tex]

    However, I was wrong. The book had the following answer:

    [tex]e^{-2t}cos(t)-2e^{-2t}sin(t)[/tex]

    My question is where does the [tex]-2e^{-2t}sin(t)[/tex] come from?
     
  2. jcsd
  3. Mar 22, 2009 #2

    djeitnstine

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    Are you sure that is all of the Laplace transform?
     
  4. Mar 22, 2009 #3

    rock.freak667

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    [tex]\frac{s}{(s+2)^2+1}=\frac{s+2 -2}{(s+2)^2+1}[/tex]

    now when you split that into two fractions, the one with s+2 in the numerator will give the cos term and the other will give the sin term, assuming I remember laplace transform correctly.
     
  5. Mar 23, 2009 #4

    djeitnstine

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    Good call, I knew that was missing but for some reason failed to mention it =(
     
  6. Mar 23, 2009 #5

    JJBladester

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    Aaaaha! That looks like a Calc II trick coming back to light. The chapter in my Differential Equations book dealing with Laplace Transforms does talk a lot about fancy algebra techniques like partial fraction decomposition, rearranging constants, and completing the square. I just forgot this one in particular.

    Thanks for the explanation.
     
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