the convolution of two functions
/ kon-vo-LOO-shun /
Convolution is a way of blending two functions into a third — a 'smeared overlap' that asks, at each moment, how much of one function's past is still echoing now, weighted by the other. It sounds abstract, but it is exactly how a present output accumulates the effects of every past input: today's reading is a blend of everything that came before, faded by how long ago it happened.
The convolution of f and g, written f * g, is defined by (f * g)(t) = integral from 0 to t of f(tau) g(t - tau) d tau. Picture it as a sliding, flipping, summing operation: hold f(tau) fixed, take g flipped and shifted so its 'origin' sits at the current time t, multiply them point by point, and add up the overlap. The variable tau runs over all past instants; g(t - tau) weights the input from tau ago by the system's memory of it. Convolution is commutative (f * g = g * f), so you may flip whichever function is more convenient.
Convolution earns its keep because of the convolution theorem: in the s-domain a messy convolution integral becomes a plain product, L{f * g} = F(s) G(s). That single fact lets you invert a product of transforms (which you often cannot split by partial fractions) by writing it as a convolution in time. It is also the language in which a linear system's output is 'input convolved with impulse response' — the mathematical shape of cause echoing through a system.
The convolution of 1 and e^(-t) is integral from 0 to t of 1 · e^(-(t - tau)) d tau = 1 - e^(-t); equivalently, L{1 * e^(-t)} = (1/s)·(1/(s+1)) = 1/(s(s+1)), whose inverse is also 1 - e^(-t).
Same answer two ways: a convolution in time equals a product of transforms in s.
Convolution is NOT pointwise multiplication: (f * g)(t) is generally not f(t) g(t). The product happens inside the integral over a shifted argument, not at the single time t — that integral is the whole point.