530_343lecture23

530_343lecture23 - ME 530.343: Design and Analysis of...

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Unformatted text preview: ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 23 - Solutions for Coupled ODEs Friday, April 24, 2009 (updated 4/27/2009) Today’s Objectives • Complete solution of coupled homogeneous ODEs • Two-rotor and Slinky demo Recall M ¨ x + B ˙ x + K x = f ( t ) x = x h + x p Assume x h ( t ) = u e pt This gives ( Mp 2 + Bp + K ) u = 0 Let B = 0 → ( p 2 + M- 1 K ) u = 0 Let λ =- p 2 → (- λI + M- 1 K ) u = 0 Let A = M- 1 K → (- λI + A ) u = 0 λ i = eigenvalues, u i = eigenvectors A u = λ u . Solving the eigenvalue problem The following are equivalent statements: 1. Find λ i which results in u i 6 = 0 ( i = 1 , 2 , . . . , n for n degrees of freedom) 2. Find λ i such that (- λI + A ) is singular, i.e. (- λI + A )- 1 does not exist 3. Find λ i such that det (- λI + A ) = 0 4. Find λ 1 , λ 2 , . . . , λ n , the eigenvalues of A For small systems of coupled equations ( n ≤ 3), method 3 works well for hand calculations For larger systems, use a computer. In Matlab, the command is eig(A) Note: A = A T ≥ 0, that is, A is a symmetric, positive (semi-)definite matrix. Therefore, it has all nonnegative eigenvalues and real eigenvectors....
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This note was uploaded on 03/02/2010 for the course MECH 530.343 taught by Professor Sun during the Spring '08 term at Johns Hopkins.

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530_343lecture23 - ME 530.343: Design and Analysis of...

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