Convolution and Laplace Transform

Abstract

= e 7t t x=0 e −4x d x = e 7t · −1 4 e −4x t x=0 = −1 4 e 7t e −4t − −1 4 e 7t e −4·0 = −1 4 e 3t + 1 4 e 7t. Simplifying this slightly, we have f * g(t) = 1 4 e 7t − e 3t when f (t) = e 3t and g(t) = e 7t. It is common practice to also denote the convolution f * g(t) by f (t) * g(t) where, here, f (t) and g(t) denote the formulas for f and g. Thus, instead of writing f * g(t) = 1 4 e 7t − e 3t when f (t) = e 3t and g(t) = e 7t , we may just write e 3t * e 7t = 1 4 e 7t − e 3t. This simplifies notation a little, but be careful-t is being used for two different things in this equation: On the left side, t is used to describe f and g ; on the right side, t is the variable in the formula for the convolution. By convention, if we assign t a value, say, t = 2 , then we are setting t = 2 in the final formula for the convolution. That is, e 3t * e 7t with t = 2 means compute the convolution and replace the t in the resulting formula with 2 , which, by the above computations, is 1 4 e 7·2 − e 3·2 = 1 4 e 14 − e 6. It does NOT mean to compute e 3·2 * e 7·2 , which would give you a completely different result, namely, e 6 * e 14 = t x=0 e 6 e 14 dt = e 20 t. ! ◮ Example 27.2: Let us find 1 √ t * t 2 when t = 4. Here, f (t) = 1 √ t and g(t) = t 2. So f (x) = 1 √ x and g(t − x) = (t − x) 2 , and 1 √ t * t 2 = f * g(t) = t x=0 1 √ x (t − x) 2 = t x=0 x − 1 / 2 t 2 − 2t x + x 2 d x