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Laplace transform combination

  1. Apr 3, 2010 #1
    1. The problem statement, all variables and given/known data


    Not a homework problem exactly, but in an EE textbook I saw something to the following effect:

    To obtain the time domain behavior for [tex]\frac{s}{s^2+\frac{\omega_0}{Q} + {\omega_0}^2}[/tex] the following Laplace transforms are combined to cancel the term in the numerator:

    [tex]\frac{1}{(s-\alpha)^2+b^2} = \frac{1}{b}e^{\alpha t}sin(bt)[/tex]

    and

    [tex]\frac{s+\alpha}{(s-a)^2 + b^2} = e^{\alpha t}cos(bt)[/tex]

    I may just be tired, but I'm having a lot of trouble seeing how those two Laplace transforms were algebraically combined to get "s" alone in the numerator. Any hint would be appreciated.
     
  2. jcsd
  3. Apr 3, 2010 #2

    gabbagabbahey

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    [tex]\frac{s}{(s-\alpha)^2 + b^2}=\frac{s+\alpha-\alpha}{(s-\alpha)^2 + b^2}[/tex] :wink:
     
  4. Apr 3, 2010 #3
    Aha! I see it now. Need to remember my "exponential shift" rule.

    So we get [tex]\frac{s+\alpha-\alpha}{(s-\alpha)^2 + b^2} = \frac{s-\alpha}{(s-\alpha)^2 + b^2} + \frac{\alpha}{(s-\alpha)^2 + b^2} = e^{\alpha t}L^{-1}(\frac{s}{s^2+b^2} + \alpha\frac{1}{s^2+b^2}) [/tex]

    which I believe is the Laplace transform of

    [tex]e^{\alpha t}*(cos(bt) + \frac{\alpha}{b}sin(bt))[/tex]

    Thanks!
     
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