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Unformatted text preview: Ch 6.6: The Convolution Integral Sometimes it is possible to write a Laplace transform H ( s ) as H ( s ) = F ( s ) G ( s ), where F ( s ) and G ( s ) are the transforms of known functions f and g , respectively. In this case we might expect H ( s ) to be the transform of the product of f and g . That is, does H ( s ) = F ( s ) G ( s ) = L { f } L { g } = L { f g }? On the next slide we give an example that shows that this equality does not hold, and hence the Laplace transform cannot in general be commuted with ordinary multiplication. In this section we examine the convolution of f and g , which can be viewed as a generalized product, and one for which the Laplace transform does commute. Example 1 Let f ( t ) = 1 and g ( t ) = sin( t ). Recall that the Laplace Transforms of f and g are Thus and Therefore for these functions it follows that { } { } 1 1 sin ) ( ) ( 2 + = = s t L t g t f L { } { } { } { } 1 1 sin ) ( , 1 1 ) ( 2 + = = = = s t L t g L s L t f L { } { } { } ) ( ) ( ) ( ) ( t g L t f L t g t f L { } { } ( 29 1 1 ) ( ) ( 2 + = s s t g L t f L Theorem 6.6.1 Suppose F ( s ) = L { f ( t )} and G ( s ) = L { g ( t )} both exist for s > a 0. Then H ( s ) = F ( s ) G ( s ) = L { h ( t )} for s > a , where The function h ( t ) is known as the convolution of f and g and the integrals above are known as convolution integrals . Note that the equality of the two convolution integrals can be seen by making the substitution u = t ....
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This note was uploaded on 08/13/2009 for the course DIFF 2343632 taught by Professor Csar during the Fall '09 term at Middle East Technical University.
 Fall '09
 Csar

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