The Convolution Operation

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1 The Convolution Operation Convolution is a very natural mathematical operation which occurs in both discrete and continuous modes of various kinds. We often encounter it in the course of doing other operations without recognizing it. Suppose, for example, we have two polynomials, one of degree n, the other of degree m: p(x) = a x n + a x n + + a n x + a n ; q(x) = b x m + b x m + + b m x + b m. The ordinary product of these two polynomials is a polynomial of degree n + m, i.e., p(x) q(x) = a b x m+n +(a b +a b ) x m+n + + (a b j +a b j + + a j b + a j b ) x m+n j + + (a n b m + a n b m ) x + a n b m. The coefficient of x m+n j is a b j + a b j + + a j b + a j b = j k= a k b j k, the sum of products a k b l for which k + l = j. It is instructive to arrange the coefficients of p(x) and q(x) in the following manner: a a a j a j a j+ b j+ b j b j b b from which we see that the sum j k= a k b j k is the sum of products of overlapping a k, b l in this diagram. The Matlab R operation conv(a, B), where A and B are vectors of dimension n and m, works in essentially this way.

2 Convolution of functions is defined in a very similar way. Let us suppose f(x) and g(x) are Laplace transformable, piecewise continuous functions defined on [, ). The convolution product of these two functions is again a function of x, denoted by (f g)(x), defined by f(ξ)g(x ξ) dξ. The integral involves products f(ξ)g(y) such that ξ + y = x and thus is directly analogous to the sum j k= a kb j k which occurs in multiplication of polynomials. Example Let f(x) = e ax, g(x) = e bx for constants a and b. Then for b a we have = ebx a b (e (a b)ξ ξ=x) ξ= e aξ e b(x ξ) dξ = e bx = ebx (e (a b)x ) a b e (a b)ξ dξ = eax e bx. a b A special case occurs with b = a, a, in which case we obtain the result that the convolution product of e ax with e ax is sinh ax. If b = a we have the integral a e aξ e a(x ξ) dξ = e ax dξ = x e ax. Theorem The convolution product is commutative: (f g) (x) (g f) (x). 2

3 Proof obtain For fixed x we let y = x ξ; then dξ = dy and we f(ξ) g(x ξ) dξ = y=x x y=x f(x y) g(y) dy = (since the name attached to the variable of integration is immaterial) = y=x y=x x g(y) f(x y) dy = g(ξ) f(x ξ) dξ = (g f) (x). The importance of both the Laplace transform and the convolution operation is very much due to the special relationship expressed by Theorem 2 Let f(x) and g(x) be Laplace transformable, piecewise continuous functions. Then (L (f g)) (s) = (Lf) (s) (Lg) (s), wherein the right hand side denotes the ordinary product of the two functions of s involved. Proof We have (L (f g)) (s) = = f(ξ) e sξ f(ξ) e sx ξ ξ f(ξ) g(x ξ) dξ dx e sx g(x ξ) dx dξ = e s(x ξ) g(x ξ) dx dξ = 3

4 ( with y = x ξ; dy = dx) = = ( e sξ f(ξ) dξ)( e sξ f(ξ) e sy g(y) dy dξ e sy g(y) dy) = (Lf) (s) (Lg) (s), again taking into account that the symbol attached to the variable of integration is immaterial. Example 2 Let s re-do Example in the context of this theorem. The ordinary product of the Laplace transforms of e ax and e bx is Since (s a)(s b) = ( ( L a b s a L ( s b (s a) 2 a b ( ), b a; (s a), b = a. 2 s a s b )) (x) = a b (eax e bx ); ) (x) = x e ax, we have the same result as before, but somewhat more quickly. Example 3 Let f(x) be arbitrary and let g(x), x. We compute f(ξ) g(x ξ) dξ = f(ξ) dξ. On the other hand, since (L) (s) = s, it must be true that (L (f g)) (s) = s (Lf) (s) 4

5 and so we conclude that ( L ) f(ξ) dξ (s) = s (Lf) (s). In the same way, since (Lx n ) (s) = n! s n+ ( L we can see that f(ξ) (x ξ) n ) dξ (s) = n! (Lf) (s). sn+ Example 4 For h > let g(x) be the pulse of duration h and amplitude A: g(x) = { A, < x < h;, otherwise. Then, for arbitrary Laplace transformable f(x) f(ξ) g(x ξ) dξ = A max{x h,} f(ξ) dξ. But, clearly, (Lg) (s) = A h (L (f g)) (s) = e sx dx = A ( s e sx ) h = A s ( e sh ) ( ) x L f(ξ) dξ (s) = A max{x h,} s ( e sh ) (Lf) (s). Now let g h (x) correspond to g(x), as defined, with A = h ; as we let h the function g h (x) approximates the distribution (as discussed in the section on distributions) δ, the Dirac delta function with support. On the other hand with A = 5

6 h we have A = and, as h, the function s sh sh ( e sh ) tends uniformly to the constant on any finite interval [, a), a >. With some further refinements in the analysis, which we do not detail here, this leads us to the conclusion that (L (f δ )) (s) = (Lf) (s) (f δ ) (x) f(x). We conclude that δ is the convolution identity; the convolution product of that distribution with f(x) simply returns f(x) again. In general, for a we have (Lδ a ) = e sa and preservation of the multiplication property of Laplace transforms of convolutions therefore requires that (L (δ a f)) (s) = e sa (Lf) (s) = (Lf a ) (s), so we must have (δ a f) (x) = f a (x) = u a (x) f(x a). The convolution multiplication relationship can be used to find inverse Laplace transforms. Example 5 Suppose ˆf(s) = s 2 (s+3) L ( s 2 ) (x) = x; L ( Therefore L s 2 (x) = (s + 3) e 3x ( ξ 3 e3ξ ) x e 3x 3 ; find f(x). We know that s + 3 ) (x) = e 3x. ξ e 3(x ξ) dξ = e 3x ξ e 3ξ dξ = e 3ξ dξ = e 3x 3 x e3x e 3x ( e 3x ) 9 = x e 3x 9. 6

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