the Laplace transform
/ la-PLAHSS /
Imagine you have a tangled audio recording and a machine that, instead of replaying it, tells you how much of each pitch is hidden inside. The Laplace transform is a similar machine for functions of time. You feed it a function f(t) — say a voltage that changes as the seconds tick by — and it hands back a new function F(s) that repackages the very same information in a completely different, often far more convenient, form.
Concretely, the Laplace transform of f(t) is defined by the integral F(s) = integral from 0 to infinity of e^(-st) f(t) dt. You take your function f(t), multiply it by the decaying weight e^(-st), and add up (integrate) the result over all time from 0 to infinity. The answer no longer depends on t — that variable has been integrated away — but it does depend on the new variable s. We write this as F(s) = L{f(t)}. The factor e^(-st) is what makes the integral settle down to a finite number: for large enough s it crushes f(t) toward zero fast enough that the area is finite.
Why bother? Because under this transform the operations that are painful in the time world — differentiating, integrating — turn into ordinary algebra in the s world. That is the whole point: a differential equation for f(t) becomes an algebraic equation for F(s), which you solve with high-school manipulation and then translate back. The Laplace transform is the engine behind a clean, mechanical method for solving linear initial value problems.
Transform the constant function f(t) = 1. We compute integral from 0 to infinity of e^(-st) (1) dt = [ -e^(-st)/s ] from 0 to infinity = 1/s, valid for s > 0. So L{1} = 1/s: a constant in time becomes 1/s in the s-domain.
The simplest transform: the constant 1 becomes 1/s.
F(s) only exists for s large enough — there is a threshold below which the integral blows up. We almost always work where it converges and rarely fuss over the exact threshold, but it is real, and it is tied to how fast f(t) grows.