the Laplace transform
/ lah-PLAHSS /
Solving a differential equation in time is hard because derivatives couple neighbouring instants. The Laplace transform trades the whole problem for an easier one: it repackages a function of time t as a function of a new complex variable s, and in doing so turns calculus (differentiation, integration) into algebra (multiplication, division). You solve the easy algebra problem, then transform back.
The Laplace transform of f(t) is F(s) = the integral from 0 to infinity of e^(-s t) f(t) dt, defined for complex s in a right half-plane where the integral converges. Its defining virtue is the derivative rule: the transform of f'(t) is s F(s) - f(0), and of f''(t) is s^2 F(s) - s f(0) - f'(0) -- the initial conditions are built in automatically. A linear constant-coefficient ODE therefore becomes a polynomial equation for F(s); one solves algebraically and inverts, usually by partial fractions or the Bromwich contour integral (which is where the residue theorem re-enters).
It is the everyday tool for initial-value problems and transients: RLC circuits, damped oscillators, control systems, and the step and impulse response all yield to it. The poles of F(s) in the complex s-plane are the system's natural frequencies and decay rates -- a pole with positive real part means a growing mode and instability. The one-sided integral (from 0) is what makes it, unlike the Fourier transform, naturally suited to causal switch-on-at-t=0 problems.
For the decay f(t) = e^(-a t), F(s) = the integral of e^(-(s + a) t) dt = 1/(s + a). The single pole at s = -a is exactly the decay rate a of the original signal.
A decaying exponential transforms to a single pole at minus its rate.
Convergence needs a region of s (a half-plane Re(s) greater than some abscissa); a function like e^(t^2) grows too fast and has no Laplace transform. The transform existing is not automatic.