Heaviside step function and Dirac delta function: Visual walkthrough

Prerequisites you may want open: Laplace Transform — Definition and Existence, First and Second Shifting Theorems.

Step 1 — Draw the switch (the step function)

WHAT. We draw the Heaviside step $u (t - a)$ : a graph that is glued to the floor (height $0$ ) for all time before $t = a$ , then jumps up to height $1$ and stays there forever.

WHY. Everything on this page is about change. Before we can talk about "how fast it changes", we must first see the thing that changes. The step is the simplest possible change: nothing, then click, then a flat plateau.

PICTURE.

Figure — Heaviside step function and Dirac delta function

Reading the figure term by term:

The horizontal axis is $t$ — time. Left is early, right is late.
The vertical axis is the height of the signal (its value).
The red mark at $t = a$ is the one instant where the whole story happens: the jump.
To the left of $a$ the curve sits at $0$ (switch off). To the right it sits at $1$ (switch on).

Step 2 — The jump is too sharp to differentiate, so we widen it

WHAT. We replace the instant jump by a ramp that climbs from $0$ to $1$ smoothly over a small width $ε$ (the Greek letter "epsilon", just a name for a tiny positive number). Call this widened switch $u_{ε}$ .

WHY. A derivative measures slope $= \frac{rise}{run}$ . On the true step the rise is $1$ but the run is $0$ , and $1/0$ is not a number. We cannot measure the slope of a vertical cliff. The cure: give the cliff a tiny but nonzero run $ε$ . Now the slope is an honest fraction. Later we shrink $ε \to 0$ and watch what the slope does.

PICTURE.

The red ramp rises over the interval $[a, a + ε]$ .
Its total rise is $1$ (bottom of the step to top).
Its horizontal run is $ε$ (the tiny width, marked with the double arrow).

Step 3 — Measure the slope of the ramp

WHAT. On the ramp, everywhere outside $[a, a + ε]$ the graph is flat, so its slope is $0$ . On the ramp itself the slope is constant: $slope = \frac{rise}{run} = \frac{1}{ε} .$

WHY. This is the derivative of the widened switch. We are literally computing $\frac{d}{d t} u_{ε} (t - a)$ . Where the graph is flat, the rate of change is nothing. Where it climbs, the rate of change is a single tall value $1/ ε$ .

PICTURE.

$\frac{1}{ε}$ appears each term as: numerator $1$ = the total height climbed; denominator $ε$ = the time taken to climb it.
The red block is the slope graph: a rectangle of height $1/ ε$ sitting over $[a, a + ε]$ , zero elsewhere.

Step 4 — Notice the box always has area 1

WHAT. The slope-box is a rectangle. Its area is height $\times$ width: $area = \frac{1}{ε} \times ε = 1.$

WHY. This is the pivot of the whole page. No matter how thin we make the ramp, the area under its slope-graph is locked at $1$ . Thin box $\Rightarrow$ tall box, but the product never changes. Area = the total amount the switch climbed = $1$ . That " $1$ " is the strength of the impulse we are about to build.

PICTURE.

Three boxes shown for shrinking $ε$ : wide-and-short, medium, thin-and-tall (the thinnest is red).
Each is labelled with its own $\frac{1}{ε} \times ε = 1$ : same area, different shape.

Step 5 — Squeeze the box to zero width

WHAT. Let $ε \to 0^{+}$ (epsilon shrinks toward zero from the positive side). The box becomes infinitely thin and infinitely tall, yet its area is still exactly $1$ . The limiting object is the Dirac delta: $δ (t - a) = lim_{ε \to 0^{+}} δ_{ε} (t - a) .$

WHY. This is the honest meaning of "the delta is the slope of the step." The true step (Step 1) is the limit of the ramps (Step 2); therefore its slope is the limit of the ramp-slopes (Step 3), and that limit is this infinite spike of area $1$ .

PICTURE.

The red spike (drawn as an arrow, because it is taller than any page) sits at $t = a$ .
The label "area $= 1$ " beside it reminds you: the height is meaningless ( $\infty$ ), only the area counts.

Step 6 — Feed a function through the box (the sifting property)

WHAT. Multiply any smooth function $f (t)$ by the box and integrate: $\int_{- \infty}^{\infty} f (t) δ_{ε} (t - a) d t = \frac{1}{ε} \int_{a}^{a + ε} f (t) d t .$

WHY. Outside $[a, a + ε]$ the box is $0$ , so it deletes all of $f$ except the sliver over that tiny interval. The right side is exactly the average value of $f$ on that sliver (total area of $f$ there, divided by the width $ε$ ). As the sliver shrinks onto the single point $a$ , the average of $f$ over it becomes just $f (a)$ .

PICTURE.

The black curve is $f (t)$ .
The red strip is where the box lives; only the piece of $f$ inside it survives.
The arrow shows the strip collapsing onto $a$ , where $f$ takes the single value $f (a)$ (red dot).

Term-by-term of the limit:

$\frac{1}{ε} \int_{a}^{a + ε} f d t$ — the " $\frac{1}{ε}$ " and the "width- $ε$ integral" cancel in size, leaving an average.
As $ε \to 0$ , the average of $f$ over a vanishing interval is the value at the point: $f (a)$ .

Step 7 — The same trick gives the Laplace transform

WHAT. Choose the special function $f (t) = e^{- s t}$ (the ingredient inside every Laplace transform) and sift: $L {δ (t - a)} = \int_{0}^{\infty} e^{- s t} δ (t - a) d t = e^{- a s} .$

WHY. The Laplace transform of anything is $\int_{0}^{\infty} e^{- s t} (\cdot) d t$ . When the " $\cdot$ " is the delta, sifting just reads off the value of $e^{- s t}$ at $t = a$ . No new work — the same picture from Step 6 with a specific curve.

PICTURE.

The black curve is $e^{- s t}$ , sliding downward as $t$ grows.
The red spike at $a$ picks off its height there: $e^{- a s}$ (red dot on the curve).

Reading $e^{- a s}$ : the " $- a$ " says how long we waited (the delay), the " $s$ " is the transform variable. Together $e^{- a s}$ is the delay stamp — the same stamp that appears in the First and Second Shifting Theorems.

Recall Check the special case

Set $a = 0$ . Then $L {δ (t)} ::: e^{0} = 1$ — the delta is the input whose transform is exactly $1$ , which is why its response is the system's impulse response.

Step 8 — Edge case: what if $a = 0$ or the kick lands off-screen?

WHAT. Two degenerate situations:

$a = 0$ — the spike sits at the very start of the Laplace window $[0, \infty)$ .
$a < 0$ — the spike would land before $t = 0$ , outside the window.

WHY. The Laplace integral starts at $t = 0$ . If the whole spike lives at or after $0$ it is captured; if it lives strictly before $0$ the integral over $[0, \infty)$ never sees it, and the transform is $0$ . This is the boundary case a solver must never trip over.

PICTURE.

Left: spike exactly at $t = 0$ (red) — half-caught by convention, transform $= 1$ (we take $a \geq 0$ ).
Right: spike at $a < 0$ (red, greyed window) — the integration window starts at $0$ , so nothing is collected, transform $= 0$ .

The one-picture summary

One image, the whole chain:

Top row: the step $u (t - a)$ softened into a ramp of width $ε$ (red), then sharpened back to a cliff.
Bottom row: its slope — a box of height $1/ ε$ and area $1$ — squeezed into the red spike $δ (t - a)$ .
The vertical arrow between rows is the operation " $\frac{d}{d t}$ " going down, and " $\int d t$ " going up. Differentiate the switch, get the kick; integrate the kick, rebuild the switch.

Recall Feynman retelling (explain the whole walkthrough in plain words)

Picture a light switch. Off, then click, then on — that jump from dark to bright is the step function, and it happens at one exact moment, $t = a$ . Now, "how fast did it brighten?" is a fair question, but the honest answer for a real switch is: super fast, over a tiny sliver of time $ε$ . During that sliver the brightness climbs from $0$ to $1$ , so the speed of brightening is $1$ divided by that tiny time — a big number, $1/ ε$ . Draw that speed and you get a thin tall box. Here's the magic: the box is $\frac{1}{ε}$ tall and $ε$ wide, so its area is always $1$ , no matter how quick the click. Make the switch flip faster ( $ε \to 0$ ) and the box gets taller and thinner but its area stays glued at $1$ . In the limit it's an infinitely tall, infinitely thin needle of area $1$ — the delta function, the "click" itself. Because it's a needle, asking for its height is silly (it's infinite). Instead you ask what it does: slide it under any curve $f$ and it grabs exactly one value, $f (a)$ , ignoring everything else — that's sifting. Slide it under the Laplace ingredient $e^{- s t}$ and it grabs $e^{- a s}$ , the delay stamp. And the one sentence that ties it together: the kick is the speed of the switch flipping — the delta is the derivative of the step.

See also: Solving IVPs with Laplace Transforms, Piecewise and Periodic Forcing Functions, Distributions and Generalized Functions.