It is easily checked that this so called convolution

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The convolution can be thought of a some kind of product between two functions. It is easily checked that this so-called convolution product satisfies some of the properties of ordinary multipulcation. For example, for any functions f , g , amd h that are piecewise continuous over every [0 , T ] we have g f = f g h ( f + g ) = h f + h g h ( g f ) = ( h g ) f commutative law , distributive law , associative law . The commutative law is proved by introducing τ = t τ as a new variable of integration, whereby one sees that ( g f )( t ) = integraldisplay t 0 g ( t τ ) f ( τ ) d τ = integraldisplay t 0 g ( τ ) f ( t τ ) d τ = ( f g )( t ) . Verification of the distributive and associative laws is left to you. The convolution differs from ordinary multipulcation in some respects too. For exam- ple, it is not generaly true that f 1 = f or that f f 0. Indeed, one sees that (1 1)( t ) = integraldisplay t 0 1 · 1 d τ = t negationslash = 1 , and that (sin sin)( t ) = integraldisplay t 0 sin( t τ ) sin( τ ) d τ = sin( t ) integraldisplay t 0 cos( τ ) sin( τ ) d τ + cos( t ) integraldisplay t 0 sin( τ ) 2 d τ = 1 2 sin( t ) 3 + 1 2 t cos( t ) 1 2 sin( t ) cos( t ) 2 negationslash≥ 0 for every t > 0 . The main result of this section is that the Laplace transform of a convolution of two functions is the ordinary product of their Laplace transforms. In other words, the Laplace transform maps convolutions to multiplication. Convolution Theorem. Let f ( t ) and g ( t ) be piecewise continuous over every [0 , T ] of exponential order α as t → ∞ . Then L [ f g ]( s ) is defined for every s > α with L [ f g ]( s ) = F ( s ) G ( s ) , where F ( s ) = L [ f ]( s ) and G ( s ) = L [ g ]( s ) . (8.16) Proof. For every T > 0 definition (8.15) of convolution implies that integraldisplay T 0 e st ( f g )( t ) d t = integraldisplay T 0 e st integraldisplay t 0 f ( t τ ) g ( τ ) d τ d t = integraldisplay T 0 integraldisplay t 0 e st f ( t τ ) g ( τ ) d τ d t .
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22 We now exchange the order of the definite integrals over τ and t on the right-hand side. As you recall from Calculus III, this should be done carefully because the upper endpoint of the inner integral depends on the variable of integration t of the outer integral. When viewed in the ( τ, t )-plane, the domain over which the double integral is being taken is the triangle given by 0 τ t T . In general, when the order of definite integrals is exchanged over this domain we have integraldisplay T 0 integraldisplay t 0 d τ d t = integraldisplay T 0 integraldisplay T τ d t d τ , where denotes any appropriate integrand. We thereby obtain integraldisplay T 0 e st ( f g )( t ) d t = integraldisplay T 0 integraldisplay T τ e st f ( t τ ) g ( τ ) d t d τ . We now factor e st as e st = e s ( t τ ) e , and group the factor e s ( t τ ) with f ( t τ ) and the factor e with g ( τ ), whereby integraldisplay T 0 e st ( f g )( t ) d t = integraldisplay T 0 integraldisplay T τ e s ( t τ ) f ( t τ ) e g ( τ ) d t d τ = integraldisplay T 0 e g ( τ ) integraldisplay T τ e s ( t τ ) f ( t τ ) d t d τ .
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