1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Homework Helper
    Gold Member

    [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!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook