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

Can you explain to me how to calculate it?

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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.

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  • $\begingroup$ Perfect, thank you very much! $\endgroup$ – Kinka-Byo Oct 21 at 19:00
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    $\begingroup$ I am not sure mathematicians would agree. $\endgroup$ – Rodrigo de Azevedo Oct 21 at 19:13
  • $\begingroup$ At first, I was in line with your comment, but apparently this can be well defined $\endgroup$ – Laurent Duval Oct 21 at 20:32
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    $\begingroup$ The first equation $$\int_{-\infty}^\infty f(t)\delta(t-T)dt = f(T)$$ is valid only when $f(t)$ is an ordinary function that is continuous at $t=T$. So, applying this to conclude that $$\int_{-\infty}^\infty\delta(t)\delta(t+\tau)dt = \delta(-\tau)$$ is not very convincing. $\endgroup$ – Dilip Sarwate Oct 22 at 20:58
  • $\begingroup$ @DilipSarwate I fear you are right. $\endgroup$ – M529 Oct 23 at 18:09
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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:

Francois Treves, Topological Vector Spaces, Distributions and Kernels

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.

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    $\begingroup$ We do get into trouble for $\tau=0$ though, don't we? $\endgroup$ – Matt L. Oct 22 at 12:41
  • $\begingroup$ Intuitively yes, but the standard distribution theory seems to be OK with that. While the square of a Dirac is not defined. $\endgroup$ – Laurent Duval Oct 22 at 14:28
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    $\begingroup$ But for $\tau=0$ we get exactly that, the square of the Dirac impulse, which is indeed undefined, with or without integral, as far as I know. $\endgroup$ – Matt L. Oct 22 at 16:13
  • $\begingroup$ As far as I remember, operations on distributions are defined with respect to integration over a set of test functions. Don't mean to convince your with an analogy, yet it's a bit (more complicated) like the Cauchy principal value (a distribution), which can get sense over Schwartz functions $\endgroup$ – Laurent Duval Oct 22 at 21:25
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    $\begingroup$ Would you care to write a fuller explanation of how you came to change your mind from two years ago when you wrote that $(\delta(t))^2$ is undefined? $\endgroup$ – Dilip Sarwate Oct 23 at 0:54

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