Dirac delta in transform methods
/ dih-RAHK /
Some inputs are not switches but kicks: a hammer blow, a spark of charge, a collision — concentrated entirely at one instant. The Dirac delta is the idealized model of such an impulse: an input of zero duration but unit total strength, infinitely tall and infinitely thin, yet enclosing area one. In transform methods it is the cleanest way to model a sudden, instantaneous push.
The defining property is the sifting property: integral of delta(t - a) g(t) dt = g(a). It picks out the value of any function at the point a. Its Laplace transform follows immediately: L{delta(t - a)} = e^{-a s}, and the delta at the origin transforms to the constant 1. That a whole input transforms to the simple constant 1 is precisely why the impulse is the natural test signal — feed a system the input 1 and its output is the impulse response, the inverse transform of the transfer function.
The crucial honesty here is that the Dirac delta is not a function at all. No ordinary function is zero everywhere except one point yet integrates to one. It is a distribution (a generalized function), defined by how it acts inside an integral against test functions, and is rigorously the limit of ever-taller, ever-narrower bumps. Treating it as a genuine function leads to nonsense; treating it as a distribution makes everything — including its transform e^{-a s} and its role as the derivative of the Heaviside step — completely consistent.
For y'' + y = delta(t), y(0) = 0, y'(0) = 0: transforming gives (s^2 + 1) Y = 1, so Y = 1/(s^2 + 1) and y(t) = sin(t) — the system's impulse response.
Because L{delta(t)} = 1, the impulse-forced response is just the inverse transform of 1/(s^2 + 1) — the impulse response read straight off.
An impulse at t = 0 deposits its strength instantaneously, so it acts between 0- and 0+ and effectively jumps a derivative; this is why the same problem with delta(t) forcing differs from the homogeneous problem even though both have zero initial conditions.