Ch6.6:The ConvolutionIntegral装 Sometimes it is possible to write a Laplace transform H(s) asH(s) = F(s)G(s), where F(s) and G(s) are the transforms ofknown functions f and g, respectively. In this case we might expect H(s) to be the transform of the长product of f and g. That is, doesH(s) = F(s)G(s) = L(f}L(g) = L(f g)?* On the next slide we give an example that shows that thisequality does not hold, and hence the Laplace transformcannot in general be commuted with ordinary multiplication* In this section we examine the convolution off and g, whichcan be viewed as a generalized product, and one for which theLaplace transform does commute
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 LaplaceTransforms of f and g areL(0)-L(1-1, (g(0)-2(sm1-Thus米L((0g(0)=-L(sin )-+1and((ig(0)+)*Therefore for these functions it follows thatL(f(t)g(t)+ L( f(t) )L(g(t) )
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 L f t g t L t 1 1 , ( ) sin 1 ( ) 1 2 + = = = = s L g t L t s L f t L Lf (t)g(t) L f (t)Lg(t) ( 1) 1 ( ) ( ) 2 + = s s L f t L g t
Theorem 6.6.1Suppose F(s) = L(f(t)) and G(s) = L(g(t)) both exist for米s > a ≥ 0. Then H(s) = F(s)G(s) = L(h(t)) for s > a, whereh(t)= [' f(t-t)g(t)dt=, f(t)g(t-t)dtThe function h(t) is known as the convolution of f and g andthe integrals above are known as convolution integrals* Note that the equality of the two convolution integrals can beseen by making the substitution u = t - t.装The convolution integral defines a “generalized product"" andcan be written as h(t) = ( f *g)(t). See text for more details
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 - . The convolution integral defines a “generalized product” and can be written as h(t) = ( f *g)(t). See text for more details. = − = − t t h t f t g d f t g t d 0 0 ( ) ( ) ( ) ( ) ( )
Theorem 6.6.1 Proof OutlineF(s)G(s)-f e-" f(u)duJ" e-" g(t)dt-J。 g(t)dtJ。e(r) (u)duJ。g(t)dtf"e-"f(t-t)dt(t=t+u)-J。 J"e"g(t)f(t-t)ddtJ。J, e"f(t-T)g(t)dt dt-Je""f(t-t)g(t)dt di= L(h()taO
Theorem 6.6.1 Proof Outline ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 0 0 ( ) 0 0 L h t e f t g d dt e f t g d dt e g f t dtd g d e f t dt t u g d e f u du F s G s e f u du e g d t s t t s t s t s t s u s u s = = − = − = − = − = + = = − − − − − + − −
Example2* Find the Laplace Transform of the function h given belowh(t)=f (t-t)sin 2tdt* Solution: Note that f(t) = t and g(t) = sin2t, with1F(s) = L(f(t) = L(t) =S2G(s) = L(g(t)) = L(sin 2t) A二s2 +4*Thus by Theorem 6.6.1.2L(h(t)= H(s) = F(s)G(s) =32+4)
Example 2 Find the Laplace Transform of the function h given below. Solution: Note that f (t) = t and g(t) = sin2t, with Thus by Theorem 6.6.1, 4 2 ( ) { ( )} {sin 2 } 1 ( ) { ( )} { } 2 2 + = = = = = = s G s L g t L t s F s L f t L t = − t h t t d 0 ( ) ( )sin 2 ( 4) 2 ( ) ( ) ( ) ( ) 2 2 + = = = s s L h t H s F s G s