The Convolution Operation
Convolution is a very natural mathematical operation which oc-curs 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) = a0xn + a1xn−1 + · · · + an−1x + an; q(x) = b0xm + b1xm−1 + · · · + bm−1x + bm.
The ordinary product of these two polynomials is a polynomial of degree n+m, i.e., p(x)·q(x) = a0b0xm+n+(a0b1+a1b0)xm+n−1+· · ·+ (a0bj+a1bj−1+ · · ·+aj−1b1+ajb0)xm+n−j +· · ·+ (an−1bm +anbm−1)x+anbm. The coefficient of xm+n−j is a0bj +a1bj−1 + · · · + aj−1b1 + ajb0 = j X k=0 akbj−k,
the sum of productsakb` for whichk +` = j. It is instructive to
arrange the coefficients ofp(x) andq(x) in the following manner:
a0 a1 · · · aj−1 aj aj+1 · · ·
· · · bj+1 bj bj−1 · · · b1 b0
from which we see that the sum Pjk=0 akbj−k is the sum of
prod-ucts of overlapping ak, b` in this diagram. The MatlabR
Convolution of functions is defined in a very similar way. Let us suppose f(x) and g(x) are Laplace transformable, piece-wise continuous functions defined on [0,∞). The convolution product of these two functions is again a function of x, denoted by (f ∗g)(x), defined by
(f ∗g) (x) = Z x
0 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 Pjk=0 akbj−k which
occurs in multiplication of polynomials.
Example 1 Let f(x) = eax, g(x) = ebx for constants a and b. Then for b 6= a we have
(f ∗g) (x) = Z x 0 e aξ eb(x−ξ)dξ = ebx Z x 0 e (a−b)ξ dξ = e bx a−b e (a−b)ξ ξ=x ξ=0 ! = e bx(e(a−b)x−1) a−b = eax−ebx a−b .
A special case occurs with b = −a, a 6= 0, in which case we obtain the result that the convolution product of eax with e−ax
is 1a sinhax. If b = a we have the integral
Z x 0 e aξ ea(x−ξ)dξ = eax Z x 0 1dξ = x e ax.
Theorem 1 The convolution product is commutative: (f ∗g) (x) ≡ (g∗f) (x).
Proof For fixedx we let y = x − ξ; then dξ = −dy and we obtain (f ∗g) (x) = Z x 0 f(ξ)g(x−ξ)dξ = − Z y=x−x y=x−0 f(x−y)g(y)dy = (since the name attached to the variable of integration is imma-terial)
= Z y=x−0
y=x−x g(y)f(x−y)dy = Z x
0 g(ξ)f(x−ξ)dξ = (g ∗f) (x).
The importance of both the Laplace transform and the con-volution operation is very much due to the special relationship expressed by
Theorem 2 Let f(x) and g(x) be Laplace transformable, piece-wise 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) = Z ∞ 0 e −sx Z x 0 f(ξ)g(x−ξ)dξ dx = Z ∞ 0 f(ξ) Z ∞ ξ e −sxg(x−ξ)dx dξ = Z ∞ e−sξf(ξ) Z ∞ e−s(x−ξ)g(x−ξ)dx dξ =
( with y = x−ξ; dy = dx) = Z ∞ 0 e −sξ f(ξ) Z ∞ 0 e −sy g(y)dy dξ = (Z ∞ 0 e −sξ f(ξ)dξ)(Z ∞ 0 e −syg(y)dy) = (Lf) (s)·(Lg) (s),
again taking into account that the symbol attached to the vari-able of integration is immaterial.
Example 2 Let’s re-do Example 1 in the context of this the-orem. The ordinary product of the Laplace transforms of eax
and ebx is 1 (s−a)(s−b) = 1 a−b( 1 s−a − 1 s−b), b 6= a; 1 (s−a)2, b = a. Since L−1 1 a−b 1 s−a − 1 s−b !! (x) = 1 a−b(e ax−ebx); L−1 1 (s−a)2 ! (x) = x eax,
we have the same result as before, but somewhat more quickly.
Example 3 Let f(x) be arbitrary and let g(x) ≡ 1, x ≥ 0. We compute
(f ∗g) (x) = Z x
0 f(ξ)g(x−ξ)dξ =
Z x
0 f(ξ)dξ. On the other hand, since (L1) (s) = 1s, it must be true that
(L(f ∗g)) (s) = 1
and so we conclude that LZ x 0 f(ξ)dξ (s) = 1 s (Lf) (s).
In the same way, since (Lxn) (s) = n!
sn+1 we can see that
LZ x
0 f(ξ) (x−ξ)
ndξ(s) = n!
sn+1(Lf) (s).
Example 4 For h > 0 let g(x) be the pulse of duration h
and amplitude A:
g(x) =
(
A, 0< x < h; 0, otherwise.
Then, for arbitrary Laplace transformable f(x) (f ∗g) (x) = Z x 0 f(ξ)g(x−ξ)dξ = A Z x max{x−h,0} f(ξ)dξ. But, clearly, (Lg) (s) = A Z h 0 e −sxdx = −A 1 s e −sx ! h 0 = A s (1 − e −sh ) ⇒ (L(f ∗g)) (s) = LZ x max{x−h,0} f(ξ)dξ ! (s) = A s (1−e −sh) (Lf) (s).
Now let gh(x) correspond to g(x), as defined, with A = 1h; as we let h → 0 the function gh(x) approximates the distribution
1
h we have A
1
s =
1
sh and, as h → ∞, the function
1
sh (1 − e−sh) tends uniformly to the constant 1 on any finite interval [0, a), a > 0. With some further refinements in the analysis, which we do not detail here, this leads us to the conclusion that
(L(f ∗δ0)) (s) = (Lf) (s) ⇒ (f ∗δ0) (x) ≡ f(x).
We conclude that δ0 is the convolution identity; the convolution product of that distribution withf(x) simply returnsf(x) again. In general, for a ≥ 0 we have (Lδa) = e−sa and preservation
of the multiplication property of Laplace transforms of convolu-tions therefore requires that (L(δa∗f)) (s) = e−sa(Lf) (s) =
(Lfa) (s), so we must have
(δa∗f) (x) = fa(x) = ua(x)f(x−a).
The convolution – multiplication relationship can be used to find inverse Laplace transforms.
Example 5 Suppose ˆf(s) = s2(s1+3); findf(x). We know that
L−1 1 s2 ! (x) = x; L−1 1 s+ 3 ! (x) = e−3x. Therefore L−1 1 s2(s+ 3) (x) = Z x 0 ξ e −3(x−ξ)dξ = e−3xZ x 0 ξ e 3ξ dξ = e−3x ξ 3 e 3ξ ! x 0 − e−3x 3 Z x 0 1·e 3ξdξ = e −3x 3 x e 3x − e −3x 9 e3x−1 = x 3 − 1 9 + e−3x 9 .