Dirac Delta Functions — Dirac Delta (Impulse) Functions (Intro) - Differential Equations

PREREQUISITES

Prerequisites for This Lesson

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PREREQUISITES FOR THIS LESSON

This lesson covers the Dirac delta function and its use with Laplace transforms, a standard part of an undergraduate course in differential equations, typically covered after an introduction to Laplace transforms (e.g., in a second-semester ODE or engineering mathematics course). To follow the derivations and worked example below, you should already have covered the prerequisite topics listed here.

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PREREQUISITES

To follow this lesson fully, you should already be comfortable with all of the following:

Integration, including improper integrals over (−∞, ∞)
Second-order differential equations and initial value problems
Laplace transforms and inverse Laplace transforms
The second shifting theorem
Partial fractions and algebra in the s-domain
The Heaviside (unit step) function

SECTION 1

Historical Context & Motivation

Imagine a baseball bat striking a ball. The entire force of the hit is delivered in a tiny fraction of a second, yet it completely changes the ball's speed and direction. In physics and engineering, events like these — a hammer strike, a lightning bolt, an electrical surge — happen so quickly that ordinary functions struggle to describe them. Mathematicians and physicists needed a new tool, one that could capture the idea of infinite intensity over zero duration while still producing a finite, measurable effect.

The Dirac delta function was developed to fill exactly this gap. Although its roots stretch back to the 1800s, it became a centerpiece of modern physics when Paul Dirac used it in quantum mechanics. Over time, mathematicians gave it a rigorous foundation, and today it appears everywhere from electrical engineering to differential equations.

1822 • Fourier's Early Hints

Joseph Fourier, while studying heat conduction, used integral expressions that behaved like what we now call the delta function, though he did not formally define them.

1893 • Heaviside's Impulse Ideas

Oliver Heaviside, a self-taught electrical engineer, introduced operational methods for circuits that implicitly used impulse-like objects to model sudden voltage changes.

1930 • Dirac Formalizes the Delta Function

Physicist Paul Dirac published The Principles of Quantum Mechanics, explicitly defining δ(t) as a function that is zero everywhere except at one point yet integrates to one.

1944–1950 • Schwartz's Distribution Theory

Mathematician Laurent Schwartz developed the theory of distributions, providing a rigorous mathematical framework that justified the delta function and earned him the Fields Medal.

The central question the Dirac delta answers is: How do we mathematically represent a force or input that acts instantaneously? Standard functions cannot be infinitely tall at a single point while having a finite integral, so a new kind of mathematical object was needed — and that is exactly what δ(t) provides.

SECTION 2

Core Principles & Definitions

The Dirac delta function, written δ(t), is not a function in the traditional sense. It is actually a generalized function (or distribution) that captures the concept of a perfect impulse — all of its effect concentrated at a single instant. To understand it, focus on these foundational ideas.

Zero Everywhere Except One Point

δ(t − a) equals zero for every value of t except t = a. All of the function's "action" is concentrated at that single point.

Infinite Height, Zero Width

At t = a, the delta function is thought of as having infinite height. It is infinitely tall but infinitely narrow, so it occupies no measurable interval of time.

Unit Area Under the Curve

Despite its extreme shape, the integral of δ(t − a) over all real numbers equals exactly 1. This means the total "impulse" delivered is one unit.

The Sifting Property

When you multiply any function f(t) by δ(t − a) and integrate, the result is simply f(a). The delta function "sifts out" or picks out the value of f at t = a.

Limit of Narrow Pulses

You can think of δ(t) as the limit of a sequence of increasingly tall, narrow pulses (rectangles, Gaussians, etc.) whose area is always 1 but whose width shrinks to zero.

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KEY TAKEAWAY

Think of the Dirac delta function like a perfectly aimed slingshot. A slingshot stores energy and releases it all in one explosive instant. The delta function does the same thing mathematically: it delivers a unit amount of input at one precise moment, with nothing before or after. If you graph it, you would see nothing but a single spike — an arrow shooting straight up at one point on the time axis.

SECTION 3

Visual Explanation

Building the Delta Function from Narrowing Pulses

One of the best ways to understand δ(t) is to watch what happens as a rectangular pulse gets taller and narrower while keeping its area equal to 1. The diagram below shows three such pulses. As the width ε shrinks toward zero, the height 1/ε grows without bound, yet the area (width × height = ε × 1/ε = 1) stays the same. The limiting "shape" is the Dirac delta function.

The violet rectangle is the widest pulse (ε = 1, height = 1). The cyan rectangle is narrower (ε = 0.5, height = 2). The pink rectangle is narrower still (ε = 0.2, height = 5). The amber arrow represents the limiting Dirac delta, an infinitely tall spike with unit area.

Notice that every rectangle has the same area of 1. The key insight is this: no matter how narrow the pulse becomes, the total "dose" of input stays constant. In the limit, we get δ(t) — something that is zero everywhere except at the origin, where it shoots up to infinity, while the integral over all time remains exactly 1. This is not a function you can graph in the usual sense; the arrow notation (the amber arrow in the diagram) is a conventional way to represent it.

SECTION 4

Mathematical Framework

Now let's state the key properties of the Dirac delta function in precise mathematical language and then connect it to Laplace transforms, the main tool for solving differential equations with impulse forcing.

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QUICK REMINDER: THE LAPLACE TRANSFORM

Recall that the Laplace transform converts a function of time, f(t), into a function of a complex variable s: ℒ{f(t)} = F(s) = ∫₀^∞

e^{−st}

f(t) dt. The inverse transform, written ℒ⁻¹{F(s)}, recovers f(t) from F(s). We use both operations freely throughout the rest of this lesson.

DEFINING PROPERTIES OF δ(t − a)

δ(t − a) = 0 for t ≠ a, and ∫₋∞^∞ δ(t − a) dt = 1

Here a is the time at which the impulse occurs. The function is zero at every instant except t = a, and the total integral is 1.

SIFTING (SAMPLING) PROPERTY

∫₋∞^∞ f(t) δ(t − a) dt = f(a)

When you integrate any continuous function f(t) multiplied by δ(t − a), the delta function "picks out" the value of f at t = a. This is the most powerful and frequently used property.

LAPLACE TRANSFORM OF δ(t − a)

ℒ{δ(t − a)} = e^(−as), a ≥ 0

This result follows directly from the sifting property. Since the Laplace transform is ∫₀^∞ $e^{−st}$ δ(t − a) dt, the delta function picks out $e^{−sa}$ . When a = 0, we get the especially clean result ℒ{δ(t)} = 1.

SPECIAL CASE: IMPULSE AT THE ORIGIN

ℒ{δ(t)} = 1

An impulse at t = 0 has the simplest possible Laplace transform — just the constant 1. This makes δ(t) an extremely convenient "test input" for analyzing systems.

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Why does ℒ{δ(t)} = 1 matter?

In systems theory, the output produced by a unit impulse input is called the impulse response. Because the Laplace transform of δ(t) is just 1, the Laplace transform of the output equals the system's transfer function directly. This means the delta function reveals the system's "personality" — its natural behavior — without any extra factors cluttering the algebra.

SECTION 5

Detailed Properties & Relationship to Step Functions

The Dirac delta function has a close relationship with the Heaviside step function u(t). Recall that u(t − a) equals 0 for t < a and 1 for t ≥ a — it "switches on" at time a. The derivative of this step function is the delta function: u′(t − a) = δ(t − a). This makes intuitive sense: the step function jumps instantaneously from 0 to 1 at t = a, and only an infinitely large instantaneous spike could produce such a jump.

Left: the Heaviside step function u(t − a) jumps from 0 to 1 at t = a. Right: its derivative, the Dirac delta function δ(t − a), is represented as an arrow at t = a since its "height" is infinite. The bottom boxes show the derivative-integral relationship between the two.

This relationship is extremely useful. It means that whenever you see a sudden switch or jump in a system's input, the corresponding "rate of change" input involves a delta function. Conversely, integrating a delta function gives you a step. The Laplace transforms of both are closely linked: ℒ{u(t − a)} = $e^{−as}$ /s, while ℒ{δ(t − a)} = $e^{−as}$ . Notice that dividing by s in the Laplace domain corresponds to integration in the time domain — a pattern you will use often.

Property	Formula	Meaning
Definition	δ(t − a) = 0 for t ≠ a; ∫ δ dt = 1	Zero everywhere except at t = a; total area is 1
Sifting Property	∫ f(t)δ(t − a) dt = f(a)	"Picks out" the value of f at t = a
Laplace Transform	ℒ{δ(t − a)} = $e^{−as}$	An exponential shift in the s-domain
Derivative of Step	u′(t − a) = δ(t − a)	The impulse is the rate of change of a sudden jump
Scaling	δ(ct) = (1/\|c\|) δ(t)	Compressing the time axis rescales the delta

Essential Properties of the Dirac Delta Function

SECTION 6

Worked Example: Solving a Differential Equation with an Impulse

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BEFORE YOU BEGIN

This example uses the Laplace transform, the second shifting theorem, and the Heaviside step function, all introduced above. Keep the properties table from Section 5 nearby as a reference while you work through the steps.

Let's solve a differential equation that models a system being hit by a sudden impulse. Consider a spring-mass system at rest that receives a unit impulse at t = 3.

PROBLEM

y″ + 4y = δ(t − 3), y(0) = 0, y′(0) = 0

This represents a spring system with natural frequency 2 (since ω² = 4), initially at rest, struck by a unit impulse at t = 3.

Solving y″ + 4y = δ(t − 3) Using Laplace Transforms

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Step 1 — Take the Laplace Transform of Both Sides

Apply ℒ to the entire equation. Using the derivative properties ℒ{y″} = s²Y(s) − sy(0) − y′(0), and the initial conditions y(0) = 0, y′(0) = 0, the left side simplifies. On the right side, ℒ{δ(t − 3)} = $e^{−3s}$ . We get:

s²Y(s) + 4Y(s) = $e^{−3s}$
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Step 2 — Solve for Y(s)

Factor Y(s) from the left side: Y(s)(s² + 4) = $e^{−3s}$ . Now divide both sides by (s² + 4).

Y(s) = $e^{−3s}$ / (s² + 4)
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Step 3 — Recognize the Inverse Transform Pattern

We know that ℒ⁻¹{1/(s² + 4)} = (1/2) sin(2t). The factor $e^{−3s}$ indicates a time shift by 3 units, which introduces the Heaviside step function u(t − 3). By the second shifting theorem: ℒ⁻¹{e^(−as) F(s)} = f(t − a) u(t − a).

ℒ⁻¹{1/(s² + 4)} = (1/2) sin(2t)
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Step 4 — Apply the Inverse Laplace Transform

Combine the time-shift result with the inverse transform. Replace t with (t − 3) in the sine function and multiply by u(t − 3) to enforce that the response only begins at t = 3.

y(t) = (1/2) sin(2(t − 3)) · u(t − 3)
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Step 5 — Interpret the Result

For t < 3, y(t) = 0: the system sits at rest, undisturbed. At the instant t = 3, the impulse strikes and the system begins oscillating as a sine wave with amplitude 1/2 and angular frequency 2. The step function u(t − 3) guarantees that the oscillation only appears after the impulse.

No motion before t = 3; sine oscillation with amplitude 1/2 for t ≥ 3.

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WHY THIS MATTERS

Notice the elegant pattern: the delta function on the right side becomes a simple exponential

e^{−3s}

in the s-domain. This turns what could be a nightmare of piecewise analysis into a clean algebraic problem — factor, divide, look up the inverse transform, done. That is the superpower of Laplace transforms combined with the Dirac delta.

SECTION 7

Delta Function vs. Other Common Inputs

In differential equations, you encounter several types of forcing functions — the inputs that "drive" a system. Understanding how the delta function compares to other common inputs helps you choose the right model for a given situation.

Input Type	Laplace Transform	Physical Interpretation
Dirac Delta δ(t − a)	$e^{−as}$	An instantaneous impulse at t = a (hammer strike, spike of current)
Step Function u(t − a)	$e^{−as}$ / s	A sudden switch that stays on forever after t = a (flipping a light switch)
Constant c	c / s	A steady, unchanging force applied for all time (gravity)
Sine sin(ωt)	ω / (s² + ω²)	A smooth, periodic oscillation (vibrating motor, AC voltage)
Exponential $e^{at}$	1 / (s − a)	An input that grows (or decays) exponentially (charging capacitor)

Common Forcing Functions and Their Laplace Transforms

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KEY TAKEAWAY

The delta function is the simplest possible forcing function in the Laplace domain: its transform is just an exponential (or even just 1 when a = 0). This is why engineers love it — feeding an impulse into a system reveals the system's transfer function directly, with no extra factors of s to untangle. It is the "clean test signal" for any linear system.

SECTION 8

Connection to Advanced Theory

The Dirac delta function is your entry point into a much larger world. In more advanced courses, you will encounter the theory of distributions (also called generalized functions), where objects like δ(t) are defined rigorously not by their values at each point but by how they act inside integrals. This framework, developed by Laurent Schwartz, places the delta function on solid mathematical ground.

This Introductory Course	Advanced Theory
δ(t) is an "infinitely tall spike" with unit area	δ is a linear functional that maps test functions to their values at the origin
Laplace transform of δ(t − a) is $e^{−as}$	Generalized Fourier/Laplace transforms extend to distributions
Solve ODEs with impulse inputs	Green's functions use δ to solve PDEs and boundary value problems
Single impulse at one point in time	Multidimensional delta functions δ³(r) in 3D space for point sources
Derivative of the step function	Distributional derivatives allow differentiation of any locally integrable function

From Introductory to Advanced: How the Delta Function Grows

For now, the key ideas to carry forward are: the Dirac delta lets you model instantaneous events cleanly; its Laplace transform is elegantly simple; and the sifting property makes it a powerful computational tool. As you move into courses on partial differential equations or signal processing, you will see δ(t) appearing as the foundation for Green's functions and impulse response analysis — powerful techniques for understanding how systems respond to arbitrary inputs.

SECTION 9

Practice Problems

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PRACTICE NOTES

These problems apply the Laplace transform techniques from this lesson, ranging from conceptual reasoning to a full applied differential equation. Refer back to the properties table in Section 5 and the worked example in Section 6 as needed.

Question 1: CONCEPTUAL

Explain in your own words why the Dirac delta function δ(t) is not a "regular" function. What two properties does it have that no ordinary function can satisfy simultaneously?

Show answer

An ordinary function that equals zero everywhere except at one point must have an integral of zero — the single point contributes no area. The delta function, however, is defined to be zero for all t ≠ 0 and yet its integral over all real numbers equals 1. No standard function can be zero at every point except one and still produce a nonzero integral. This is why δ(t) is called a generalized function or distribution. The two contradictory properties are (1) δ(t) = 0 for t ≠ 0 and (2) ∫δ(t) dt = 1.

Question 2: BASIC CALCULATION

Evaluate the integral: ∫₀^∞ (t² + 5t) δ(t − 2) dt.

Show answer

By the sifting property, ∫ f(t) δ(t − a) dt = f(a). Here f(t) = t² + 5t and a = 2. Since a = 2 lies within the interval [0, ∞), we simply evaluate f(2) = (2)² + 5(2) = 4 + 10 = 14. Therefore the integral equals 14. The delta function "picks out" the value of the integrand at t = 2 and ignores everything else.

Question 3: INTERMEDIATE

Find the Laplace transform of 3δ(t − 5) + 7δ(t − 1). Write your answer as a function of s.

Show answer

The Laplace transform is linear, so we handle each term separately. Using ℒ{δ(t − a)} =

e^{−as}

, we get ℒ{3δ(t − 5)} =

3e^{−5s}

and ℒ{7δ(t − 1)} =

7e^{−s}

. Adding these gives the total Laplace transform:

3e^{−5s}

7e^{−s}

. This result represents a system that receives a pulse of magnitude 7 at t = 1 and another pulse of magnitude 3 at t = 5.

Question 4: APPLIED

A spring-mass system is modeled by y″ + 9y = 2δ(t − π), with y(0) = 0 and y′(0) = 0. Find y(t) using Laplace transforms.

Show answer

Take the Laplace transform of both sides: s²Y(s) + 9Y(s) =

2e^{−πs}

, so Y(s) =

2e^{−πs}

/(s² + 9). We know ℒ⁻¹{1/(s² + 9)} = (1/3)sin(3t). The factor

e^{−πs}

introduces a time shift of π using the second shifting theorem. Therefore y(t) = 2 × (1/3) sin(3(t − π)) × u(t − π) = (2/3) sin(3(t − π)) u(t − π). Since sin(3(t − π)) = sin(3t)cos(3π) − cos(3t)sin(3π) = −sin(3t), we can write y(t) = −(2/3) sin(3t) · u(t − π). The system is at rest until t = π, then oscillates with amplitude 2/3.

Question 5: CRITICAL THINKING

Suppose a system has transfer function H(s) = 1/(s + 2). If the input is δ(t), explain why the output equals the inverse Laplace transform of H(s) directly. Then find the impulse response h(t) and explain what it tells you about the system's behavior.

Show answer

When the input is δ(t), its Laplace transform is ℒ{δ(t)} = 1. The output in the s-domain is Y(s) = H(s) × Input(s) = H(s) × 1 = H(s). Therefore Y(s) = H(s), which means the output's Laplace transform IS the transfer function. Taking the inverse transform: h(t) = ℒ⁻¹{1/(s + 2)} =

e^{−2t}

for t ≥ 0. This impulse response tells us the system decays exponentially after being "kicked" by the impulse. The decay rate of 2 means the response halves roughly every 0.35 seconds. This is exactly why engineers use impulse inputs for testing: the response reveals the system's natural dynamics — in this case, pure exponential decay with no oscillation, indicating an overdamped or first-order system.

SUMMARY

Lesson Summary

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LOOKING AHEAD

Together, the Dirac delta function and Laplace transforms give you a compact, algebraic method for handling impulse forcing in differential equations — a technique used throughout engineering, physics, and applied mathematics coursework.

Summary

The Dirac delta function δ(t − a) is a generalized function that models an instantaneous impulse at time t = a. It is defined by two key properties: it equals zero for all t ≠ a, and its integral over all real numbers equals 1. Its most powerful feature is the sifting property — integrating f(t)δ(t − a) yields f(a), effectively "picking out" the function's value at the impulse location. The delta function is the derivative of the Heaviside step function, connecting sudden jumps to instantaneous spikes.

In the context of Laplace transforms, the delta function has the elegant transform ℒ{δ(t − a)} = $e^{−as}$ , which simplifies to just 1 when a = 0. This makes solving differential equations with impulse forcing remarkably clean: take the Laplace transform, solve the resulting algebraic equation for Y(s), and apply the second shifting theorem to find y(t). The impulse response of a system — its output when driven by δ(t) — directly equals the inverse Laplace transform of the transfer function, making the delta function the ultimate diagnostic tool for linear systems.