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28 laplace III convolution, partial fractions, complex numbers

28 laplace III convolution, partial fractions, complex numbers

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18.03 Lecture #28 Nov. 13, 2009: notes Four topics today. Topic zero is the formula L ( f * g ) = L ( f ) L ( g ) (Convolution formula) for the Laplace transform of a convolution. Topic one is computing inverse Laplace transforms using partial fractions, and particularly using complex numbers to help with this. Topic two is a review of generalized functions, Dirac delta functions, and their connection with Laplace transforms. Here is a proof of (Convolution formula). To make sure that all the integrals converge, you should assume that the functions f and g are of expnential type; I won’t worry about such things. The definition of f * g is ( f * g )( t ) = integraldisplay t 0 f ( τ ) g ( t - τ ) dτ. The Laplace transform of f * g is L ( f * g )( s ) = integraldisplay 0 ( f * g )( t ) e st dt = integraldisplay 0 integraldisplay t 0 e st f ( τ ) g ( t - τ ) dτdt. To make this look related to Laplace transforms for f and g , we factor e st = e e s ( t τ ) , and get L ( f * g )( s ) = integraldisplay 0 integraldisplay t 0 e f ( τ ) g ( t - τ ) e s ( t τ ) dτdt. The double integral is over the pie-shaped region 0 t < , 0 τ t in the ( t, τ ) plane. We now interchange the order of integration, integrating first with respect to t and then with respect to τ . In order to do this, we describe the same region as 0 τ < , t τ < , which gives L ( f * g )( s ) = integraldisplay 0 integraldisplay τ e f ( τ ) g ( t - τ ) e s ( t τ ) dtdτ. Now for the t integration we do the change of variables ξ = t - τ , = dt . As t runs from τ to , ξ runs from 0 to ; so we get L ( f * g )( s ) = integraldisplay 0 integraldisplay 0 e f ( τ ) g ( ξ ) e dτdξ = parenleftbiggintegraldisplay 0 e f ( τ ) parenrightbiggparenleftbiggintegraldisplay 0 g ( ξ ) e parenrightbigg = L ( f )( s ) · L ( g )( s ) .
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