Dirac Delta Function | Unit Impulse and Its Laplace Transform

An explosion, a tennis ball struck by a racket, a concentrated load on a beam: each delivers a finite effect in a vanishingly short time. No ordinary function describes this, so we build one as a limit of tall narrow pulses of unit area. The limit is not a function at all, but its Laplace transform is the beautifully simple $e^{- a s}$ .

Quick Reference

Item	Statement
Pulse	P_h(t-a) = \begin{cases} \dfrac{1}{h} & a \leq t < a+h \\ 0 & \text{otherwise} \end{cases}
Unit area	$\int_{- \infty}^{\infty} P_{h} (t - a) d t = 1$ for every $h > 0$
Pulse in steps	$P_{h} (t - a) = \frac{1}{h} [u (t - a) - u (t - a - h)]$
Transform of the pulse	$ℒ {P_{h} (t - a)} = \frac{e^{- a s}}{s h} (1 - e^{- h s})$
Delta	$δ (t - a) = lim_{h \to 0} P_{h} (t - a)$
Transform of the delta	$ℒ {δ (t - a)} = e^{- a s}$
At the origin	$ℒ {δ (t)} = 1$
Relation to the step	\delta(t-a) = u'(t-a), in the sense of distributions
Status	A generalized function, or distribution, not a function

Motivation

In many engineering problems we must deal with a driving force acting over a very short period of time or length. Explosions, a tennis ball hit by a racket, and a concentrated force acting on a beam are all examples. These forces are modeled using the Dirac delta function, or unit impulse function.

What matters physically in each case is not the force at any instant, which may be enormous and is rarely known, but the total impulse delivered, the integral of the force over the brief interval. The construction below is built around preserving exactly that.

The Pulse Function

Let $h$ be a positive number and consider the function $P_{h} (t - a)$ defined by

P_h(t-a) = \left\{ \begin{aligned} &\frac{1}{h}, && \text{if } a \leq t < a+h \\ &0, && \text{otherwise} \end{aligned} \right.

Graph of a rectangular pulse of height one over h, extending from t = a to t = a + h and zero elsewhere. — The pulse $P_{h} (t - a)$ : a rectangle of width $h$ and height $\frac{1}{h}$ , so its area is exactly $1$ no matter how small $h$ becomes.

First notice that for every $h > 0$ ,

\int_{- \infty}^{\infty} P_{h} (t - a) d t = 1,

since the rectangle has base $h$ and height $\frac{1}{h}$ . The area is the invariant of the family. As $h$ shrinks, the pulse becomes taller and narrower, but it always delivers unit total impulse.

Now find its Laplace transform. Since the pulse is the difference of two steps,

P_{h} (t - a) = \frac{1}{h} [u (t - a) - u (t - a - h)],

we have, using $ℒ {u (t - c)} = \frac{e^{- c s}}{s}$ from Section 7.6,

\begin{aligned} \mathscr{L}\left\{P_h(t-a)\right\} &= \frac{1}{h}\left[\frac{e^{-as}}{s} - \frac{e^{-(a+h)s}}{s}\right] \\ &= \frac{e^{-as}}{sh}\left[1 - e^{-hs}\right] \end{aligned}

Taking the Limit

Now take the limit as $h \to 0$ :

\begin{aligned} \lim_{h\to 0}\mathscr{L}\left\{P_h(t-a)\right\} &= \lim_{h\to 0}\frac{e^{-as}}{sh}\left[1 - e^{-hs}\right] \\ &= \frac{e^{-as}}{s}\lim_{h\to 0}\frac{1 - e^{-hs}}{h} \qquad \left(\text{indeterminate } \tfrac{0}{0}\right) \\ &= \frac{e^{-as}}{s}\lim_{h\to 0}\frac{se^{-hs}}{1} \qquad \left(\text{l'H\^opital's rule in } h\right) \\ &= \frac{e^{-as}}{s}\cdot s = e^{-as} \end{aligned}

The differentiation in the third line is with respect to $h$ , treating $s$ as a constant, and the final step used $e^{- h s} \to 1$ as $h \to 0$ .

Denoting $lim_{h \to 0} P_{h} (t - a)$ by $δ (t - a)$ , that is,

δ (t - a) = lim_{h \to 0} P_{h} (t - a),

we can write

ℒ {δ (t - a)} = e^{- a s}, and in particular ℒ {δ (t)} = 1 .

$δ (t - a)$ , which is a limiting mathematical operation rather than a function, is called the Dirac delta function or the unit impulse function located at $t = a$ . It is infinite at $t = a$ and zero everywhere else:

\delta(t-a) = \left\{ \begin{aligned} &\infty, && \text{if } t = a \\ &0, && \text{otherwise} \end{aligned} \right.

The delta function was introduced by the English physicist Paul Dirac in 1927, though the mathematics justifying it was formalized only decades later.

Schematic of the Dirac delta: a single vertical arrow at t = a, with the function zero everywhere else. — The conventional picture of $δ (t - a)$ : an arrow at $t = a$ . The arrow's height is drawn to represent the *area*, which is $1$ , since the actual height is infinite and cannot be drawn.

Making the mathematics precise: distribution theory

Strictly speaking, $lim_{h \to 0} P_{h} (t - a)$ does not exist as a function. At $t = a$ the values $\frac{1}{h}$ grow without bound, and a function whose value is $\infty$ at one point and $0$ elsewhere has integral zero by any ordinary definition, contradicting the unit area we insisted on. So $ℒ {δ (t - a)}$ is not defined by the calculus developed so far.

We call $δ (t - a)$ a generalized function, or a distribution. Making the operations of this section rigorous requires distribution theory, in which $δ$ is defined not by its values but by what it does to other functions: the rule $\int_{- \infty}^{\infty} δ (t - a) ϕ (t) d t = ϕ (a)$ for every smooth $ϕ$ . That theory is beyond the scope of this book, but every formula here is a theorem within it.

The Delta as the Derivative of the Step

We defined $δ (t - a) = lim_{h \to 0} P_{h} (t - a)$ , and $P_{h} (t - a) = \frac{1}{h} [u (t - a) - u (t - a - h)]$ . Therefore

δ (t - a) = lim_{h \to 0} \frac{u (t - a) - u (t - a - h)}{h} .

The right-hand side is the limit definition of the derivative of $u (t - a)$ . To see this, set $k = - h$ and $f (t) = u (t - a)$ :

\begin{aligned} \delta(t-a) &= \lim_{k\to 0}\frac{f(t) - f(t+k)}{-k} \\ &= \lim_{k\to 0}\frac{f(t+k) - f(t)}{k} \\ &= f'(t) = u'(t-a) \end{aligned}

Notice that $u (t - a)$ is not continuous, so its derivative does not exist in ordinary calculus. The equation \delta = u' makes sense only within distribution theory. Nevertheless it is the right intuition: the step jumps by $1$ at $t = a$ , and the delta records that jump as an infinitely concentrated rate of change.

The relationship is consistent with the transforms already computed. Applying \mathscr{L}\left\{f'\right\} = sF(s) - f(0) formally to $f = u (t - a)$ with $a > 0$ , so that $f (0) = 0$ ,

\mathscr{L}\left\{u'(t-a)\right\} = s\cdot\frac{e^{-as}}{s} - 0 = e^{-as} = \mathscr{L}\left\{\delta(t-a)\right\} . \checkmark

Worked Examples

Solve y'' + 4y = \delta(t-\pi) with $y (0) = 0$ and y'(0) = 0, and interpret the result.

Solution

Physically: a mass on a spring sits at rest until it is struck by a hammer blow at $t = π$ .

Transforming, with both initial values zero,

s^{2} Y + 4 Y = e^{- π s} ⟹ Y (s) = \frac{e^{- π s}}{s^{2} + 4} .

Setting the exponential aside, $ℒ^{- 1} {\frac{1}{s^{2} + 4}} = \frac{1}{2} \sin 2 t$ . The factor $e^{- π s}$ delays this by $π$ :

y (t) = \frac{1}{2} \sin (2 (t - π)) u (t - π) = \frac{1}{2} \sin 2 t \cdot u (t - π),

using $\sin (2 t - 2 π) = \sin 2 t$ .

So $y (t) = 0$ for $t < π$ and $y (t) = \frac{1}{2} \sin 2 t$ afterward: the mass is motionless until struck, then oscillates forever with amplitude $\frac{1}{2}$ .

Note that $y$ is continuous at $t = π$ , both branches giving $0$ , but y' jumps from $0$ to $1$ there. An impulse produces a jump in velocity, not in position, which is exactly the physics of a hammer blow.

Solve y' + 2y = \delta(t-1) with $y (0) = 3$ .

Solution

Transforming, $s Y - 3 + 2 Y = e^{- s}$ , so

Y (s) = \frac{3}{s + 2} + \frac{e^{- s}}{s + 2} .

Inverting each piece,

y (t) = 3 e^{- 2 t} + e^{- 2 (t - 1)} u (t - 1) .

The solution decays from $3$ , then at $t = 1$ jumps up by exactly $1$ and resumes decaying. Here the jump is in $y$ itself, because the equation is first order: an impulse raises the order of the discontinuity by one relative to the highest derivative present.

Show that the pulse and its limit give consistent answers by solving y'' + 4y = P_h(t), y(0) = y'(0) = 0, and letting $h \to 0$ .

Solution

Transforming with $ℒ {P_{h} (t)} = \frac{1 - e^{- h s}}{s h}$ ,

Y_{h} (s) = \frac{1 - e^{- h s}}{s h (s^{2} + 4)} .

As $h \to 0$ , the factor $\frac{1 - e^{- h s}}{h} \to s$ , exactly as in the l'Hôpital computation above, so

Y_{h} (s) \to \frac{s}{s (s^{2} + 4)} = \frac{1}{s^{2} + 4},

whose inverse is $\frac{1}{2} \sin 2 t$ , matching the delta calculation with $a = 0$ .

This is the honest justification of the whole section: the delta is shorthand for a limit taken at the very end. Every answer obtained with $δ$ is the limit of answers obtained with genuine pulses, so no physical conclusion depends on the fiction.

Exercises

Find $ℒ {3 δ (t - 2) + δ (t)}$ .

Solution

By linearity,

3 e^{- 2 s} + 1 .

Solve y'' + y = \delta(t-2\pi) with $y (0) = 1$ and y'(0) = 0.

Solution

Transforming, $s^{2} Y - s + Y = e^{- 2 π s}$ , so

Y (s) = \frac{s}{s^{2} + 1} + \frac{e^{- 2 π s}}{s^{2} + 1} .

Inverting,

y (t) = \cos t + \sin (t - 2 π) u (t - 2 π) = \cos t + \sin t \cdot u (t - 2 π),

using the $2 π$ -periodicity of the sine. So the mass oscillates as $\cos t$ , and after the blow at $t = 2 π$ it oscillates as $\cos t + \sin t = \sqrt{2} \sin (t + \frac{π}{4})$ , with the amplitude increased from $1$ to $\sqrt{2}$ .

Explain why $ℒ {δ (t)} = 1$ rather than $0$ , given that $δ (t) = 0$ for every $t \neq 0$ .

Solution

Because the delta is defined by its area, not by its values. The transform integral $\int_{0}^{\infty} e^{- s t} δ (t) d t$ picks out the value of $e^{- s t}$ at $t = 0$ , which is $1$ , exactly as the defining property $\int δ (t) ϕ (t) d t = ϕ (0)$ requires.

Trying to evaluate the integral pointwise, treating $δ$ as a genuine function that is zero almost everywhere, would give $0$ . This contradiction is precisely why $δ$ cannot be a function and why distribution theory is needed.

Frequently Asked Questions

What is the Dirac delta function?

The limit of rectangular pulses of unit area as their width shrinks to zero. It models an idealized instantaneous impulse: zero everywhere except at a single instant, yet delivering total effect $1$ .

Is it really a function?

No. A function that is zero except at one point has integral zero under any ordinary definition of the integral, contradicting the unit area. It is a generalized function, or distribution, defined by how it acts on other functions rather than by its values.

Why is

ℒ {δ (t - a)} = e^{- a s}

Because the transform of the approximating pulse is $\frac{e^{- a s}}{s h} (1 - e^{- h s})$ , and l'Hôpital's rule gives $\frac{1 - e^{- h s}}{h} \to s$ as $h \to 0$ , cancelling the $\frac{1}{s}$ . In distribution language, the transform integral simply evaluates $e^{- s t}$ at $t = a$ .

What is the physical meaning of an impulse?

Total effect delivered, not force applied. Striking a mass with $δ (t - a)$ imparts one unit of momentum instantaneously, so the velocity jumps by $1 / m$ while the position stays continuous. That is why the solution of y'' + 4y = \delta(t-\pi) has a corner rather than a jump at $t = π$ .

In what sense is the delta the derivative of the step?

In the sense of distributions. The step jumps by $1$ at $t = a$ , and its "rate of change" is infinitely concentrated there. The transforms confirm the relation: $s \cdot \frac{e^{- a s}}{s} = e^{- a s}$ . Classically $u$ has no derivative at $t = a$ , so the statement only lives in the generalized theory.

Can I trust results obtained using

δ

Yes, when they are understood as limits. Solving with a genuine narrow pulse and then letting the width go to zero gives the same answer, as the third worked example shows. Distribution theory makes this rigorous in general, and the results agree with physical measurement.

What is the impulse response, and why does it matter?

It is the solution of $L [y] = δ (t)$ with zero initial conditions. Since $ℒ {δ} = 1$ , its transform is $\frac{1}{p (s)}$ where $p$ is the characteristic polynomial, a quantity engineers call the transfer function. Knowing the impulse response determines the response to every input, by convolution, which is why so much of systems theory is organized around it.

Dirac Delta Function (Unit Impulse Function)

Quick Reference

Motivation

The Pulse Function

Taking the Limit

The Delta as the Derivative of the Step

Worked Examples

Exercises

Frequently Asked Questions

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