Unit Impulse and Dirac Delta Function

Introduction

The unit impulse function, also called the Dirac delta function $\delta(t)$, is arguably the most important signal in all of signals and systems theory. It's a peculiar mathematical object — infinitely tall, infinitely narrow, with unit area — that doesn't really exist as an ordinary function. Yet it's the key that unlocks the entire theory of LTI systems.

Here's why it matters: if you know how a system responds to an impulse (the impulse response $h(t)$), you can determine the system's response to ANY input through convolution. The impulse is the universal test signal.

Definition of the Continuous-Time Unit Impulse

The Dirac delta function $\delta(t)$ is defined by two properties:

$$\delta(t) = 0 \quad \text{for } t \neq 0$$

$$\int_{-\infty}^{\infty} \delta(t) \, dt = 1$$

It's zero everywhere except at the origin, where it's "infinitely concentrated" such that its total area equals one. Technically, $\delta(t)$ is not a function in the ordinary sense — it's a generalized function or distribution. No ordinary function can be zero everywhere yet have nonzero integral.

Visualization

Think of $\delta(t)$ as the limit of increasingly tall, narrow pulses that always maintain unit area:

As $\epsilon \to 0$, the pulse becomes infinitely narrow and tall, but the area remains exactly 1.

The Sifting Property

The most important operational property of $\delta(t)$ is the sifting (or sampling) property:

$$\int_{-\infty}^{\infty} x(t)\delta(t - t_0) \, dt = x(t_0)$$

This says: multiplying any signal $x(t)$ by a shifted impulse $\delta(t - t_0)$ and integrating extracts the value of $x$ at the point $t_0$.

Example: Find $\int_{-\infty}^{\infty} (3t^2 + 5)\delta(t - 2) \, dt$

Using the sifting property, substitute $t = 2$ into $x(t) = 3t^2 + 5$: $$= 3(2)^2 + 5 = 17$$

This property is why the impulse is so powerful — it "sifts out" (selects) exactly one value from a signal.

Relationship to the Unit Step Function

The unit impulse is the derivative of the unit step function:

$$\delta(t) = \frac{d}{dt}u(t)$$

Conversely, the unit step is the integral of the impulse:

$$u(t) = \int_{-\infty}^{t} \delta(\tau) \, d\tau$$

This makes intuitive sense: $u(t)$ is constant everywhere except at $t = 0$ where it jumps from 0 to 1. The "derivative" of that jump is infinitely large but infinitely brief — exactly what $\delta(t)$ represents.

Properties of the Dirac Delta

Scaling Property

$$\delta(at) = \frac{1}{|a|}\delta(t)$$

Compressing the impulse in time by factor $a$ scales its amplitude by $1/|a|$.

Product with a Function

$$x(t)\delta(t - t_0) = x(t_0)\delta(t - t_0)$$

Since $\delta(t-t_0)$ is nonzero only at $t = t_0$, we can replace $x(t)$ by its value at that point.

Derivative of the Delta (Doublet)

$$\int_{-\infty}^{\infty} x(t)\delta'(t) \, dt = -x'(0)$$

The doublet $\delta'(t)$ extracts the negative derivative of $x(t)$ at the origin.

Fourier Transform

$$\mathcal{F}\{\delta(t)\} = 1$$

The impulse contains ALL frequencies equally. Its spectrum is perfectly flat — this is why it's the ideal test signal. Hitting a system with an impulse excites every frequency simultaneously.

$$\mathcal{F}\{1\} = 2\pi\delta(\omega)$$

A constant signal (DC) has all its energy at zero frequency.

The Discrete-Time Unit Impulse

The discrete-time counterpart is much simpler — it's an ordinary sequence:

$$\delta[n] = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases}$$

Unlike the continuous-time version, $\delta[n]$ is a perfectly well-behaved function. It's simply the sequence $\{\ldots, 0, 0, \mathbf{1}, 0, 0, \ldots\}$ with the bold value at $n = 0$.

Discrete Sifting Property

$$\sum_{n=-\infty}^{\infty} x[n]\delta[n - n_0] = x[n_0]$$

Relationship to Unit Step

$$\delta[n] = u[n] - u[n-1]$$

$$u[n] = \sum_{k=-\infty}^{n} \delta[k] = \sum_{k=0}^{\infty} \delta[n-k]$$

The Impulse Response

The impulse response $h(t)$ is defined as the output of a system when the input is $\delta(t)$:

$$h(t) = T\{\delta(t)\}$$

For LTI systems, the impulse response completely characterizes the system. The output for any input $x(t)$ is:

$$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau$$

This is the convolution integral — and it works because any signal can be decomposed into shifted, scaled impulses.

Practical Approximations

In real life, we cannot create a true impulse (infinite amplitude, zero duration). Instead, we use approximations:

• A very short pulse of large amplitude
• A hammer strike on a mechanical system
• A spark gap discharge in an electrical circuit
• A hand clap in room acoustics (to measure room impulse response)

As long as the pulse duration is much shorter than the system's fastest response time, it effectively acts as an impulse.

Solved Examples

Example 1: Evaluate $\int_{-\infty}^{\infty} e^{-3t}\delta(t-2) \, dt$

By sifting property: $= e^{-3(2)} = e^{-6}$

Example 2: Evaluate $\int_{-\infty}^{\infty} \cos(2\pi t)\delta(3t) \, dt$

First apply scaling: $\delta(3t) = \frac{1}{3}\delta(t)$ Then: $= \frac{1}{3}\int_{-\infty}^{\infty} \cos(2\pi t)\delta(t) \, dt = \frac{1}{3}\cos(0) = \frac{1}{3}$

Example 3: Express $x(t) = 3\delta(t+1) - 2\delta(t) + \delta(t-2)$

This is a signal consisting of three impulses: amplitude 3 at $t = -1$, amplitude $-2$ at $t = 0$, and amplitude 1 at $t = 2$.

Key Takeaways

• $\delta(t)$ is zero everywhere except origin, with unit area — it's a distribution, not an ordinary function
• The sifting property $\int x(t)\delta(t-t_0)dt = x(t_0)$ is the most useful operational tool
• $\delta(t) = du(t)/dt$ — the impulse is the derivative of the step
• The Fourier transform of $\delta(t)$ is flat (all frequencies) — making it the ideal test signal
• The discrete-time $\delta[n]$ is much simpler — just the sequence with value 1 at $n=0$
• Knowing $h(t)$ (impulse response) completely characterizes any LTI system
• Any input's output can be found via convolution with $h(t)$