Lecture 23

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Step Functions and Discontinuous Forcing Functions Applications of Laplace Transforms
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MATLAB Quiz Tomorrow, March 13 Same Time As Your Section (APM B432) DON’T FORGET!!!! (NO MAKEUPS) Final Monday March 17, Here (Peterson 110), 3-6 PM Bring a Blue Book!
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On Monday Using the Laplace Transform, we changed Initial Value Problems into Algebra Problems Example: y 00 - 2 y 0 - 3 y = 0 y (0) = 3 y 0 (0) = 5 Becomes: L { y } = 2 1 s - 3 + 1 1 s + 1 Inverse Transform: y = 2 e 3 t + e - t
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Heavyside or “Step Function” u c ( t ) = 0 t < c 1 t c Think of it as “Flipping a switch” 2 0 2 4 6 8 10 1 0.5 0 0.5 1 1.5 2
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The Heavyside Function Can Be Used To Make Piecewise Continuous Functions 0 t < 1 cos ( t - 1 ) t 1 f ( t ) = 2 0 2 4 6 8 10 1 0.5 0 0.5 1 1.5 2 f ( t ) = u 1 ( t ) cos ( t - 1 )
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Laplace Transform With Discontinuous ODEs Models a system where “forces” change suddenly Need The Laplace Transform of L { u c ( t ) f ( t - c ) } u c ( t ) f ( t - c ) A y 00 + B y 0 + C y = u c ( t ) f ( t - c ) e - c s L { f } = Can Use This To Compute Inverse Transforms Too e - c s L { f } = y L { y } = u c ( t ) f ( t - c )
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Laplace Transform With Discontinuous ODEs L { u c ( t ) f ( t - c ) } e - c s L { f } = Examples e t 0 t < e t + sin ( t - ) t f = f = e t + u ( t ) sin ( t - ) L { f } = 1 s - 1 e - s 1 s 2 + 1 +
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Laplace Transform With
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