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Having difficulty finding the inverse laplace transform

  1. Aug 14, 2004 #1
    Having difficulty finding the inverse laplace transform!!

    Hello everyone, I am really stuck on finding the inverse Laplace transform for this:

    [tex]f(s)=\frac{5se^{-3s} - e^{-3s}}{s^{2}-4s+17}[/tex]

    Heres my reasoning: I feel that I should rewrite the denominator in some kind of form such as (s-2)^2 + 13, and note the similarity with some of the problems ive been doing before, however its that 13 that is bothering me! Its not something you can take the squareroot of--- and also in addtion, I tried factoring out e^-3s on top and splitting this into two equations, but its this denominator that I absolutely despise.

    Any help with finding the right method would be greatly appreciated thank you!!!
     
    Last edited: Aug 14, 2004
  2. jcsd
  3. Aug 14, 2004 #2

    arildno

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    I've forgotten just about every transform formula I knew (and I'm not in the mood rederiving them), but you may write [tex]13=(\sqrt{13})^{2}[/tex]

    if that helps..
     
  4. Aug 14, 2004 #3
    hmm i still am having difficulty---
     
  5. Aug 14, 2004 #4

    arildno

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    Well, what about rewriting:
    [tex]e^{-3s}=e^{-6}e^{-3(s-2)}[/tex]

    Then you would get one expression on the form:
    [tex]k\frac{e^{-3w}}{w^{2}+a^{2}},k=-e^{-6},w=s-2,a=\sqrt{13}[/tex]

    Is this a familiar transform in w?
     
  6. Aug 14, 2004 #5

    Tom Mattson

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    The exponentials are due to shifts in the t-domain:

    L{f(t-T)}=e-sTF(s).

    Just find the inverse transform of the rational functions of s, and then let the t-domain functions be delayed by the appropriate amount, given by the coefficient in the exponents of the exp functions.
     
  7. Aug 14, 2004 #6
    interesting let me see if i can get anywhere with that...
     
  8. Aug 14, 2004 #7
    Uh oh, you see, I cannot split the denominator into two linear factors like the book seems to be doing with most of the problems---- and because I cant, i dont know how to proceed like the examples do... Im sorry if im a bit slow but we just started Laplace and this is a challenge problem id like to know to prepare.
     
    Last edited: Aug 14, 2004
  9. Aug 15, 2004 #8
    Well I cannot solve it but thanks for your help anyway...
     
  10. Aug 16, 2004 #9

    Tom Mattson

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    You could factor the denominator into 2 linear factors, but the roots of that polynomial are complex. It would be better instead to complete the square in the denominator. The solution will be t-shifted, exponentially damped sines and cosines.
     
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