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Unformatted text preview: 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 wont worry about such things. The definition of f * g is ( f * g )( t ) = integraldisplay t f ( ) g ( t- ) d. The Laplace transform of f * g is L ( f * g )( s ) = integraldisplay ( f * g )( t ) e st dt = integraldisplay integraldisplay t e st f ( ) g ( t- ) ddt. To make this look related to Laplace transforms for f and g , we factor e st = e s e s ( t ) , and get L ( f * g )( s ) = integraldisplay integraldisplay t e s f ( ) g ( t- ) e s ( t ) ddt. The double integral is over the pie-shaped region t < , t in the ( t, ) plane. We now interchange the order of integration, integrating first with respect to t and then with respect to . In order to do this, we describe the same region as < , t < , which gives L ( f * g )( s ) = integraldisplay integraldisplay e s f ( ) g ( t- ) e s ( t ) dtd. Now for the t integration we do the change of variables = t- , d = dt . As t runs from to , runs from 0 to ; so we get L ( f * g )( s ) = integraldisplay integraldisplay e s f ( ) g ( ) e s dd = parenleftbiggintegraldisplay e s f ( ) d parenrightbiggparenleftbiggintegraldisplay...
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Fall '09 term at MIT.
- Fall '09