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The Dirac Delta and the Instantaneous Kick

A hammer blow, a spark, a sudden tap — forces that act for almost no time at all. The Dirac delta is the idealisation that captures them, and through the Laplace transform it turns an impossible-looking input into the simplest one of all.

From a switch that stays on to a kick that is gone in an instant

The previous guide handed you the Heaviside step — a switch that flips a forcing term on at some time and leaves it on. That covers a circuit you plug in, a load you set down and leave. But nature also hands us forces that act and then vanish almost at once: a billiard ball striking another, a hammer tapping a tuning fork, a brief voltage spike. The total push is finite, yet it is delivered in a flash. How do you write down a forcing function for something that is over before you can blink?

Here is the trick of the trade. Picture a force that is a tall, thin rectangular pulse: zero everywhere, except on a short window of width e starting at t = 0, where it has height 1/e. Notice the area is exactly (1/e) times e = 1, no matter how small e is. Now shrink the window — make e go toward zero. The pulse grows taller and thinner, but it always encloses the same unit of area. The limit of this squeezing is the Dirac delta, written delta(t): an object of zero width, infinite height, and area exactly 1.

The one property that does all the work

Everything the delta does flows from a single rule, the sifting property. If you multiply delta(t - a) by any reasonable function g(t) and integrate over all time, the spike at t = a reaches in and plucks out the one value g(a): integral of g(t) delta(t - a) dt = g(a). That is the whole magic. The infinite spike, weighted against a smooth function, sifts out its height at the instant the spike lives. The shifted version delta(t - a) simply places the kick at time a instead of at the origin — exactly the t-shift idea you already met for the step.

Now watch the payoff in the s-domain. Take the Laplace transform of delta(t - a) straight from its definition: L{delta(t - a)} = integral from 0 to infinity of e^(-s t) delta(t - a) dt. But that integral is exactly the sifting property with g(t) = e^(-s t), so it equals g(a) = e^(-a s). For a kick at the origin (a = 0) this is even simpler: L{delta(t)} = 1. Read that again. The most violent, spiky input imaginable has the tamest transform of all — just the constant 1.

Sifting:        integral_0^infinity  g(t) delta(t - a) dt = g(a)         (a >= 0)

Transforms:     L{ delta(t) }     = 1
                L{ delta(t - a) } = e^(-a s)

Compare step:   L{ u(t - a) }     = e^(-a s) / s

Delta = derivative of the step:   delta(t) = u'(t)   (in the distribution sense)
The sifting property, the two delta transforms it produces, and how the delta sits one derivative below the Heaviside step.

There is a beautiful relationship hiding in that last line of the box. The Heaviside step jumps from 0 to 1; its rate of change is zero everywhere except at the jump, where it changes infinitely fast. That instantaneous, unit-area burst of slope is precisely the delta. So delta(t) is the derivative of the step, u'(t) — and conversely the step is the running integral of the delta. The transforms agree perfectly: differentiating multiplies the transform e^(-a s)/s by s, cancelling the 1/s to leave e^(-a s). The two new objects of this rung are not separate inventions; they are two views of the same edge.

Kicking a system: the impulse response

Now put the delta to work as a forcing term. Take a familiar mass-spring-damper, at rest, and strike it once: m y'' + b y' + k y = delta(t), with y(0) = 0 and y'(0) = 0. The system starts dead still, gets one sharp kick at t = 0, and then is left alone to ring down on its own. The motion that follows is so important it has its own name: the impulse response, usually written h(t). It is the fingerprint of the system — how this particular mass, spring, and damper answer a single unit jolt.

Solving it is almost embarrassingly quick, and this is where the work of the earlier guides pays off. Transform both sides. The derivative rule turns the left side into (m s^2 + b s + k) Y(s), with no leftover initial-condition terms because the system started from rest. The right side is L{delta(t)} = 1. So (m s^2 + b s + k) Y = 1, giving Y(s) = 1 / (m s^2 + b s + k) at once. No forcing integral to wrestle, no partial fractions on a messy right-hand side — the delta delivered a bare 1.

Why the kick is the master key

You might think the impulse response is a narrow curiosity — who really hits things with idealised infinite spikes? The answer is that the kick is the master key to every other input. Bring Y(s) home with the inverse transform for a lightly damped case and you get a decaying, oscillating h(t): the system's natural underdamped ringing, started by the blow and fading as the damper bleeds off energy. That single curve, it turns out, secretly encodes the response to any forcing at all.

Here is the intuition, made precise in the next guide. Any forcing function f(t) can be imagined as a dense parade of tiny kicks — at each instant a delta of strength f(t) dt. Because the equation is linear, the total response is just the sum of the responses to all those little kicks, each of which is a copy of h(t) started at its own moment and scaled by its strength. That superposition-of-kicks sum is exactly the convolution of f with h, the operation guide three is built around. The instantaneous kick is the atom from which every response is assembled.

And that same reciprocal 1 / (m s^2 + b s + k) you found above is the first sighting of the transfer function — the system's input-to-output gain in the s-domain, the star of guide four. Multiply any input's transform by it and you get the output's transform; its denominator is the characteristic polynomial, whose roots will decide whether the system settles or blows up. So this short guide quietly seeds the whole rest of the rung: the kick gives you h(t), h(t) builds convolution, and its transform is the transfer function.

Solving a kicked problem, start to finish

Let us walk one concrete example all the way through, so the abstraction lands. Take y'' + y = delta(t - pi), with y(0) = 0 and y'(0) = 0 — an undamped oscillator, at rest, struck once at time t = pi. We will ride the master algorithm: transform, solve for Y, invert. Watch how the delta makes step one trivial and the t-shift e^(-a s) does the bookkeeping for the delay.

  1. Transform both sides. With zero initial conditions L{y''} = s^2 Y, and L{delta(t - pi)} = e^(-pi s), so (s^2 + 1) Y = e^(-pi s).
  2. Solve for Y by simple division: Y(s) = e^(-pi s) / (s^2 + 1).
  3. Recognise the pieces: 1/(s^2 + 1) is the transform of sin(t), and the factor e^(-pi s) means 'delay by pi' — that is the second shifting theorem at work.
  4. Invert: y(t) = u(t - pi) sin(t - pi) — that is, y stays exactly 0 until t = pi, then begins a pure sine, the oscillator ringing forever because there is no damping.

Step back and admire how cleanly it ran. The system did absolutely nothing until the kick arrived at t = pi — causality is visibly built in by the step u(t - pi) the second shifting theorem supplied. After the kick it oscillates exactly as its impulse response prescribes. The delta, which looked like a monster you could never integrate, turned out to be the input that makes the algebra shortest of all. That is the recurring lesson of this rung: the rougher the input looks in the time domain, the friendlier it often is in the s-domain.