The Dirac Delta Function

The Dirac delta is simply a model of a sudden whack, like a hammer hitting something: all of its push happens in a single instant. It's named after the physicist Paul Dirac, and it's called a "function" even though, strictly, it isn't an ordinary function at all (more on that below). It's good for modeling strikes, kicks and sudden doses. The formula $\int_{-\infty}^{\infty}\delta(t - c)\,f(t)\,dt = f(c)$ looks strange, but it says something simple: $\delta(t - c)$ (a kick at the instant $c$), multiplied by $f(t)$ (any function) and added up, just gives back $f(c)$ (the value of the function at the moment of the kick). And $\mathcal L\{\delta(t - c)\} = e^{-cs}$ means the kick's transform is only the delay tag, with nothing else attached. It's an odd object, but bear with it; it behaves very nicely.

From pulses to a kick. In the last lesson you handled inputs that stay on for a while. Now squeeze one down. Take the pulse $\frac1\varepsilon\left(u_c(t) - u_{c + \varepsilon}(t)\right)$: it's on for a time $\varepsilon$ with height $\frac1\varepsilon$, so its area is always 1. As $\varepsilon$ shrinks, it gets taller and thinner, and the limit is the delta $\delta(t - c)$. A hammer strike works like that: a huge force for a tiny moment, and what matters is the total push.

The delta isn't a function in the usual sense, since no ordinary function is zero everywhere but one point and still has area 1. But its effect under an integral is precise: it picks out the value at $t = c$, which is called the sifting property. It makes sense once you picture the thin pulse. Over that tiny interval, $f(t)$ is basically the constant $f(c)$, and the pulse has area 1, so the integral is $f(c)$.

The transform follows straight from sifting: $\mathcal L\{\delta(t - c)\} = \int_0^\infty e^{-st}\delta(t - c)\,dt = e^{-cs}$. That's the same delay factor you saw for steps, with nothing else attached.

A common mistake. It's tempting to think a hammer strike is a step input, because both start suddenly at a given time. It isn't: a step stays on forever after it switches, while a delta is over the instant it happens. A step pushes the spring to a new resting position; a delta gives it a sudden velocity and then lets it go.

Impulse response. Solving with forcing $\delta(t)$ gives the system's response to a single kick. With convolution (next lesson), that one response determines the reaction to any input.

Engineers find a structure's impulse response by striking it with a hammer and recording the ringing; audio engineers capture a room's impulse response to reproduce its reverb.