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Laplace transform of Diff. Eq containing sine Function

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the Laplace transform of the pendulum equation.

    2. Relevant equations
    The pendulum equation:
    [tex]
    \frac{d^{2}\theta}{dt^{2}} + \alpha \frac{d \theta}{dt} + g \sin(\theta) = 0
    [/tex]

    [tex]
    s = \sigma + i \omega
    [/tex]

    3. The attempt at a solution
    Taking the laplace transform I get

    [tex]
    s^{2} + \alpha s + g \frac{\omega}{s^{2}+\omega^{2}} = 0
    [/tex]

    I usually drop the [tex] \sigma [/tex] term in [tex]s = \sigma + i \omega[/tex], which leaves

    [tex]
    s^{2} + \alpha s + g \frac{\omega}{-\omega^{2}+\omega^{2}} = 0
    [/tex]

    The [tex]\frac{\omega}{-\omega^{2}+\omega^{2}}[/tex] is problematic, but I'm not sure what to do. I don't see the usefulness of the [tex] \sigma [/tex] variable, and I'm not sure that I'm doing this correctly to begin with
     
  2. jcsd
  3. Apr 14, 2009 #2
    You're not doing the transform correctly. For laplace you're just taking the real part of z in [tex]exp^{-zt}, z = \sigma + i \omega[/tex].
     
    Last edited: Apr 14, 2009
  4. Apr 15, 2009 #3
    I'm not sure what you're saying but I had a thought. I think the Laplace of the sine function is
    [tex]
    \mathcal{L} \left\{\sin(t)\right\} = \frac{1}{s^{2}+1}
    [/tex]

    which would make the transform

    [tex]
    s^{2} + \alpha s + g \frac{1}{s^{2}+1} = 0
    [/tex]

    This is more sensible, but I still need to verify it some how.
     
  5. Apr 17, 2009 #4
    What do you get if you take the Laplace transform of just a second deriavative?
     
  6. Apr 19, 2009 #5
    I think it is something like

    [tex]\mathcal{L} \left\{y(t)\right\} = s^{2} Y(s)[/tex]

    presuming that
    [tex]\dot y(0) = 0[/tex]
    and
    [tex]\ddot y(0) = 0[/tex]

    I think the sine term would make this a nonlinear differential equation, and I'm now I'm thinking that maybe the Laplace transform doesn't work on nonlinear differential equations.
     
  7. Apr 20, 2009 #6
    Sure it does!

    Why do you keep dropping the Y(s) terms? That's what you need to take the inverse laplace transform of! Solve for Y(s) and take the inverse transform.
     
  8. Apr 26, 2009 #7
    I'm trying to get a transfer function of:

    [tex]
    \frac{d^{2}\theta}{dt^{2}} + \alpha \frac{d \theta}{dt} + g \sin( \theta ) = f(t)
    [/tex]

    With the Y(s) terms I get:

    [tex]
    s^{2} Y(s) + \alpha s Y(s) + g \frac{1}{s^{2}+1}} = F(s)
    [/tex]

    For the transfer function, I want to Y(s) and F(s) on the same side like this:

    [tex]
    \frac{Y(s)}{F(s)}
    [/tex]

    but I don't see and way to factor things out that way.
     
    Last edited: Apr 26, 2009
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