530_343lecture23

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

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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 = 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. The eigenvectors, u i , are called modes of the system. Say we have solved for λ 1 , λ 2 , . . . , λ n and u 1 , u 2 , . . . , u n The poles p are related to the eigenvalues via ± p = ± i λ

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