Transfer Function of Coupled Diff Eq

  • #1
I have two coupled harmonic oscillators:

[tex]\ddot{x}_{1} = -2kx_{1} + kx_{2} + f(t)[/tex]
[tex]\ddot{x}_{2} = kx_{1} - kx_{2}[/tex]

Mass 1 is at position [tex]x_{1}[/tex] and subject to force [tex]f(t)[/tex].

I take the Laplacian of the first equation and solve for [tex]X_{1}[/tex] to get

[tex]X_{1} = \frac{ F(p) + k X_{2} }{ p^{2} + 2k }[/tex]

I then do the same for the second to get

[tex]X_{2} = \frac{ k X_{1} }{ p^{2} + 2k }[/tex]

I then substitute [tex]X_{1}[/tex] into [tex]X_{2}[/tex], divide out [tex]F(p)[/tex], and then wind up with the transfer function

[tex]T(e^{jwt}) = \frac{ z^{-2} + 2kz^{-4} }{ 1 + 4kz^{-2} + 3k^{2}z^{-4} }[/tex]

My question:

Does this method work for finding the transfer function of coupled differential equations?
 

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
Sure, why not? All you're shooting for is [itex]T(p)=X_1(p)/F(p)[/itex], right? So if you can get there by this method (which you clearly can), then it must work!
 

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