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Laplace transform combination

  • Thread starter bitrex
  • Start date
  • #1
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Homework Statement




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.
 

Answers and Replies

  • #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:
 
  • #3
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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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