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530_343lecture11_updated

530_343lecture11_updated - ME 530.343 Design and Analysis...

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ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 11 - Solving ODEs using the Laplace Transform Week of February 23, 2009 Today’s 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 ˙ ω + b eq ω = 0 Take Laplace transform: L [ J ˙ ω ] = J [ s Ω ( s ) - ω (0)] L [ b ω ] = b Ω ( s ) L [0] = 0 plug in ... J [ s Ω ( s ) - ω (0)] + b Ω ( s ) = 0 [ Js + b ] Ω ( s ) - J ω (0) = 0 Ω ( s ) = J ω (0) Js + b put in a form we can find in table no coe ffi cient in front of s: Ω ( s ) = ω (0) s + b J Looks like L [ e - at ] = 1 s + a , where a = b J L - 1 [ Ω ( s )] = ω (0) e - b J t ω ( t ) = ω (0) e - b J t What happens if we have applied torque? (Inhomogeneous) applied torque τ a ( t ) is a step function: τ a ( t ) = 0 for t 0 τ a ( t ) = T for t > 0 1
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J ˙ ω + b ω = τ a L [ TU ( t )] = T s J [ s Ω ( s ) - ω (0)] + b Ω ( s ) = T s let ω (0) = 0, zero initial condition ( Js + b ) Ω ( s ) = T s Ω ( s ) = T s ( Js + b ) Ω ( s ) = T J s ( s + b J ) Not in our chart! consider L [ t 0 f ( λ ) d λ ] = 1 s F ( s ) let F ( s ) = A s + a f ( t ) = Ae - at L - 1 A s ( s + a ) = t 0 Ae - a λ d λ = Ae - λ a - a | t 0 = A a (1 - e - at ) L - 1 A s ( s + a ) = A a (1 - e - at ) L - 1 [ Ω ( s )] = ω ( t ) = T J b J (1 - e - b J t ) ω ( t ) = T b (1 - e - b J t ) ω ( t ) = T b - T b e - b J t ω ( t ) =
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