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Laplace transform of Diff. Eq containing sine Function

  • #1

Homework Statement


Find the Laplace transform of the pendulum equation.

Homework 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]

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
 

Answers and Replies

  • #2
1,341
3
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:
  • #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.
 
  • #4
1,341
3
What do you get if you take the Laplace transform of just a second deriavative?
 
  • #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.
 
  • #6
1,341
3
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.
 
  • #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:

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