530_343lecture11_updated

530_343lecture11_updated - ME 530.343: Design and Analysis...

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Unformatted text preview: ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 11 - Solving ODEs using the Laplace Transform Week of February 23, 2009 Todays Objective Homogeneous and Inhomogeneous Solutions with the Laplace Transform Example: Solving a first order ODE with the Laplace Transform Example From Lab #1: Motor spin down test J eq + b eq = 0 Take Laplace transform: L [ J ] = J [ s ( s )- (0)] L [ b ] = b ( s ) L [0] = 0 plug in ... J [ s ( s )- (0)] + b ( s ) = 0 [ Js + b ]( s )- J (0) = 0 ( s ) = J (0) Js + b put in a form we can find in table no coefficient in front of s: ( s ) = (0) s + b J Looks like L [ e- at ] = 1 s + a , where a = b J L- 1 [( s )] = (0) e- b J t ( t ) = (0) e- b J t What happens if we have applied torque? (Inhomogeneous) applied torque a ( t ) is a step function: a ( t ) = 0 for t a ( t ) = T for t > 1 J + b = a L [ TU ( t )] = T s J [ s ( s )- (0)] + b ( s ) = T s let (0) = 0, zero initial condition ( Js + b )( s ) = T s ( s ) = T s ( Js + b ) ( s ) = T J s ( s + b J ) Not in our chart! consider L [ t f ( ) d ] = 1 s F ( s ) let F ( s ) = A s + a f ( t ) = Ae- at L- 1 A s ( s + a ) = t Ae- a d = Ae- a- a | t = A a (1- e- at ) L- 1 A s ( s + a ) = A a (1...
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530_343lecture11_updated - ME 530.343: Design and Analysis...

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