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If the continuous-time unit impulse is the Dirac delta δ(t), its derivative is δ′(t), called the derivative of the Dirac delta or delta prime. This is a distribution, not an ordinary function with a finite value at each point. A common source of confusion: the derivative of the unit step is δ(t); the derivative of the impulse is δ′(t).
Unit step versus unit impulse
In continuous-time signals, the unit impulse usually means the Dirac delta distribution, δ(t). The unit step, written u(t), is a different signal. Their derivatives are:
| Signal | Derivative |
|---|---|
| Unit step, u(t) | δ(t) |
| Unit impulse, δ(t) | δ′(t) |
Thus, u′(t) = δ(t), while dδ(t)/dt = δ′(t). MIT’s signal-processing notes on the Dirac delta and Heaviside step distinguish these two objects.
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The Dirac delta is zero away from the origin and has unit total area:
δ(t) = 0 for t ≠ 0, and ∫−∞∞ δ(t) dt = 1.
Those statements are useful shorthand, but they do not define an ordinary function whose value at zero is a finite number. More precisely, the delta is a distribution: when integrated against a smooth test function φ(t), it selects that function’s value at the origin, ∫δ(t)φ(t)dt = φ(0). The University of Nebraska–Lincoln differential-equations text describes the delta as a generalized function and motivates it with narrow unit-area pulses.
How the derivative δ′ is defined
The derivative is defined by how it acts on a smooth test function:
∫−∞∞ δ′(t)φ(t) dt = −φ′(0).
The minus sign follows from integration by parts: the derivative transfers from the delta to the test function and changes sign, with the boundary term taken to vanish. This is the rigorous meaning of δ′(t); it is not obtained by taking an ordinary pointwise derivative of a graph of δ(t).
For an impulse shifted to t₀
If the impulse occurs at t = t₀, its signal is δ(t − t₀), and differentiation gives:
dδ(t − t₀)/dt = δ′(t − t₀).
Its action on a test function is ∫δ′(t − t₀)φ(t)dt = −φ′(t₀). The shifted-delta properties and impulse transforms are treated in the Penn State differential-equations material on impulse functions.
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Laplace and Fourier transforms
Laplace transform
For the one-sided engineering Laplace transform and the usual causal-distribution convention, ℒ{δ(t)} = 1 and ℒ{δ′(t)} = s. The derivative rule gives ℒ{δ′(t)} = sℒ{δ(t)} − δ(0⁻); for a causal impulse, the pre-origin term is zero. For t₀ ≥ 0, the shifted impulse has transform ℒ{δ(t − t₀)} = e−st₀. One-sided transform conventions require care for distributions at the origin; see the Nebraska–Lincoln discussion of Laplace transforms and impulses.
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Using the angular-frequency convention ℱ{x(t)} = ∫−∞∞x(t)e−jωtdt, the differentiation property gives ℱ{δ′(t)} = jω, since ℱ{δ(t)} = 1. If frequency is expressed as f in cycles per second, the corresponding factor is j2πf. Fourier-transform signs and scaling factors depend on the convention, so the convention should accompany the result.
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What δ′ looks like—and what a numerical plot can show
It is tempting to picture δ′ as a positive spike beside a negative spike, but that is only an informal image, not an ordinary graph of the exact distribution. The paired picture suggests a derivative-like change and zero net area; the test-function definition is the precise description.
A computer typically approximates an impulse with a narrow pulse of unit area. For example:
δε(t) = 1/(2ε) for |t| < ε, and 0 otherwise.
This rectangular pulse has area one. Its ordinary derivative is zero between the edges and has singular transitions at the edges; as ε shrinks, the pulse approximates the delta in the distributional sense, not by pointwise convergence. Numerical software therefore plots or computes a chosen approximation rather than an ideal delta derivative.
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In digital signal processing, the unit impulse may instead mean the unit sample δ[n], equal to one at n = 0 and zero at other integer indices. Discrete-time signals use differences rather than the continuous-time derivative. For the backward difference Δx[n] = x[n] − x[n − 1],
Δδ[n] = δ[n] − δ[n − 1].
For the forward difference Δfx[n] = x[n + 1] − x[n], the result is Δfδ[n] = δ[n + 1] − δ[n]. These are discrete sequences, not δ′(t).
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Common mistakes to avoid
- Answering δ(t): that is the derivative of the unit step, not the unit impulse.
- Calling the derivative zero: the delta is zero away from its impulse location, but its distributional derivative is not zero.
- Using “infinite at zero” as a definition: that is informal shorthand, not the mathematical definition of the delta.
- Dropping the minus sign: the test-function identity is −φ′(0).
- Mixing continuous and discrete time: discrete impulses are handled with a specified difference operator.
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