Dirac delta function

The Dirac delta function $\delta (t)$ is an idealized “function” that’s zero everywhere except at $t=0$, where it’s infinite, with the property that its integral equals $1$. It models an instantaneous unit impulse — an infinite force applied for zero duration, with finite total impulse.

Image: Schematic of the Dirac delta function (a line surmounted by an arrow), CC BY-SA 3.0

Strictly speaking, $\delta$ isn’t a function in the ordinary sense — it’s a distribution (or generalized function). But it behaves like a function under integration and is used freely in physics and engineering.

Definition (operational)

There is one defining property — the sifting property — and one heuristic picture that helps you remember why it’s true.

Sifting property (the definition):

${\int }_{-\infty }^{\infty }f(t)\delta (t)dt=f(0)$

for every test function $f$ that is continuous at $0$. Equivalently, $\delta$ is the linear functional that maps $f\mapsto f(0)$.

Heuristic picture (not a definition):

$\delta (t)\approx \{\begin{array}{ll}0& t\ne 0\\ \text{very large}& t=0\end{array}\text{with}{\int }_{-\infty }^{\infty }\delta (t)dt=1$

It is tempting to write $\delta (0)=+\infty$ and call this the definition. Don’t. There is no ordinary function that is zero everywhere except at one point and integrates to $1$ — Lebesgue integration of any pointwise-defined function tells you the integral is $0$. The “spike of infinite height, zero width, area one” picture is a useful intuition for visualizing approximations (see “How it arises” below), but the actual mathematical object is the sifting functional. Anything you want to prove about $\delta$ should be derived from the sifting property, not from the heuristic.

Shifted version

Shifted sifting (this is the substantive statement):

${\int }_{-\infty }^{\infty }f(t)\delta (t-a)dt=f(a)$

Heuristically, $\delta (t-a)$ is “concentrated at $t=a$” instead of $t=0$.

How it arises

Consider a physical impulse: a force $F(t)$ applied during a short interval $[t_0-\tau ,t_0+\tau ]$. The total impulse is

$I(\tau )={\int }_{t_0-\tau }^{t_0+\tau }F(t)dt$

If we want to keep $I=1$ but shrink $\tau \to 0$, the force has to grow $\to \infty$. In the limit, the force becomes the delta function $\delta (t-t_0)$:

${\lim }_{\tau \to 0}{F}_{\tau }(t)=\{\begin{array}{ll}0& t\ne t_0\\ \infty & t=t_0\end{array},{\int }_{-\infty }^{\infty }{F}_{\tau }(t)dt=1$

This is the rigorous way to think of $\delta$ — a limit of increasingly tall, narrow pulses of constant area.

Laplace transform

$\mathcal{L}\{\delta (t-a)\}(s)=e^{-as}$

For $a=0$: $\mathcal{L}\{\delta (t)\}=1$.

That’s a clean transform — the delta becomes a constant (or pure exponential) in the s-domain.

Use case: impulsive forcing

Solve $y^{\prime \prime }+9y=3\delta (t-\pi )$, $y(0)=1$, $y^{\prime }(0)=0$.

This models an undamped spring at rest at position $y_0=1$, then hit with an impulse $I=3$ at $t=\pi$.

Laplace. Using $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)$ with $y(0)=1$, $y^{\prime }(0)=0$:

$\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-s\cdot 1-0=s^{2}Y-s$

The ODE becomes $(s^{2}Y-s)+9Y=3e^{-\pi s}$, i.e.

$(s^2+9)Y(s)=s+3e^{-\pi s}$

The $+s$ on the right comes from the $-sy(0)$ initial-condition term moved to the right side; the $3e^{-\pi s}$ is $\mathcal{L}\{3\delta (t-\pi )\}$.

Solve:

$Y(s)=\frac{s}{s^2+9}+\frac{3e^{-\pi s}}{s^2+9}$

Inverse:

${\mathcal{L}}^{-1}\{s/(s^2+9)\}=\cos (3t)$.

${\mathcal{L}}^{-1}\{3e^{-\pi s}/(s^2+9)\}$: by the second shifting theorem, this is $u(t-\pi )\cdot {\mathcal{L}}^{-1}\{3/(s^2+9)\}(t-\pi )=u(t-\pi )\sin (3(t-\pi ))$.

Final:

$y(t)=\{\begin{array}{ll}\cos (3t)& t<\pi \\ \cos (3t)+\sin (3(t-\pi ))& t>\pi \end{array}$

Before $t=\pi$, the system oscillates as a simple cosine. At $t=\pi$, the impulse “kicks” it, adding a sine term that shifts the trajectory. After $\pi$, the system continues oscillating but with new amplitude/phase.

Properties

Beyond the sifting property:

1. Identity for convolution: $f\ast \delta =f$.
2. Derivative of step: in a distributional sense, $\frac{d}{dt}u(t)=\delta (t)$, where $u$ is the Heaviside step function. Differentiating a jump gives an infinite spike.
3. Scaling: $\delta (at)=\delta (t)/|a|$ for $a\ne 0$.

Why it’s not really a function

Mathematicians built up distribution theory (Schwartz, 1940s) to give $\delta$ a rigorous foundation. A distribution is a continuous linear functional on a space of test functions. The delta corresponds to “evaluation at a point” — a perfectly well-defined operation, even though no ordinary function realizes it.

For engineering use, you don’t need the rigor — just remember the sifting property and the Laplace transform formula. They’re enough to handle impulsive forcing in ODEs.

Three properties used constantly in Continuous-Time Signals and Systems

The unit impulse has three operational properties used over and over in signals-and-systems work, all derivable from the limit-of-rectangles picture above:

• Equivalence: $g(t)\delta (t-t_0)=g(t_0)\delta (t-t_0)$ — see Equivalence property of the impulse.
• Sampling (sifting): $\int g(t)\delta (t-t_0)dt=g(t_0)$ — covered above as the defining property.
• Scaling: $\delta (a(t-t_0))=\delta (t-t_0)/|a|$ — see Scaling property of the impulse.

In a typical signal-processing derivation, you spot one of these being needed (most often the sampling property in convolutions), apply it, and the integral collapses.

Building blocks made from the impulse

A periodic impulse train ${\delta }_T(t)={\sum }_n\delta (t-nT)$ is the central object in Sampling: multiplying a continuous signal by an impulse train models the sampling operation.

For the partner concept (a unit step, the “antiderivative” of the delta), see Heaviside step function. For applying delta-driven inputs to systems, see Impulse response.