What is the autocorrelation of a Dirac pulse?

Question

What is the autocorrelation of $x(t) = \delta(t)$?

Can you explain to me how to calculate it?

Answer 1

Well, by definition of the $\delta$ distribution, you have:

$\int_{-\infty}^{\infty} f(t) \delta(t-T)\, \textrm{d}t = f(T)$

The autocorrelation of a function $g(t)$ can be computed via:

$\int_{-\infty}^{\infty} g^{*}(t)g(t + \tau)\, \textrm{d}t$, with $g^*$ as the complex conjugate of $g$. Since $\delta(t)$ is real-valued, this is conjugation can be skipped. So you are left with:

$\int_{-\infty}^{\infty} \delta^{*}(t)\delta(t + \tau)\, \textrm{d}t = \int_{-\infty}^{\infty} \delta(t)\delta(t + \tau)\, \textrm{d}t = \delta(-\tau)$.

The first = sign comes from the autocorrelation of the real-valued $\delta$, the second from the definition of the $\delta$-distribution.

So, the autocorrelation function of the $\delta$-distribution is the distribution itself. A eigenfunction of the autocorrelation function, so to say ;)

Think about it, this does make sense: the only perfect match is achieved with no time shift, ie at $\tau = 0$. All other shifts would end up with one of the arguments of the $\delta$ being different from 0, hence with the $\delta$-function being 0 there.

BTW: $\delta(-\tau) = \delta(\tau)$, since the function/distribution is symmetric.

Answer 2

Looking at documents like Lecture notes on Distributions, Hasse Carlsson, or Convolution dans l'espace $\mathcal{D}'_+(\mathbb{R})$, convolution of distributions can be defined under some technical conditions. However, when one of the operand has a compact support, as $\delta(t)$ does, the convolution is well-defined. From Wikipedia:Distribution-Convolution:

Distribution of compact support: It is also possible to define the convolution of two distributions S and T on $\mathbb{R}^n$, provided one of them has compact support.

If you prefer classical books, there is: Francois Trèves, Topological Vector Spaces, Distributions and Kernels, 1967:

Adding that $\delta(t)$ is the neutral element, then $\delta(t)$ is its own autocorrelation.

This lazy version avoids to write cumbersome Dirac products under integral signs. Nota: I do admit that I initially thought it was way less simple.