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?

mathematicians and electrical engineers usually look at $\delta(t)$ differently. — robert bristow-johnson, Oct 22 '19 at 22:35

score 3 · Accepted Answer · answered Oct 21 '19 at 18:22

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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.

answered Oct 21 '19 at 18:22

M529

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Perfect, thank you very much! – Kinka-Byo Oct 21 '19 at 19:00
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I am not sure mathematicians would agree. – Rodrigo de Azevedo Oct 21 '19 at 19:13
At first, I was in line with your comment, but apparently this can be well defined – Laurent Duval Oct 21 '19 at 20:32
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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. – Dilip Sarwate Oct 22 '19 at 20:58
@DilipSarwate I fear you are right. – M529 Oct 23 '19 at 18:09
While your result is correct, the method of obtaining it isn't. You can easily fix it by taking the argument to the frequency domain. – Jazzmaniac Oct 24 '19 at 11:37
@Jazzmaniac Sadly, I cannot delete an accepted answer. Maybe you would like to elaborate on that in an additional answer, which then can hopefully be upvoted? – M529 Oct 25 '19 at 18:04

Laurent Duval · Answer 2 · 2019-10-22T22:28:14.200

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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:

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.

edited Oct 22 '19 at 22:28

answered Oct 21 '19 at 20:23

Laurent Duval

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We do get into trouble for $\tau=0$ though, don't we? – Matt L. Oct 22 '19 at 12:41
Intuitively yes, but the standard distribution theory seems to be OK with that. While the square of a Dirac is not defined. – Laurent Duval Oct 22 '19 at 14:28
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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. – Matt L. Oct 22 '19 at 16:13
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 – Laurent Duval Oct 22 '19 at 21:25
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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? – Dilip Sarwate Oct 23 '19 at 00:54
I really think I did not change my mind: $\delta^2(t)$ is (to me as a DSP guy) ill-defined for my purposes (or beyond my reach), but $\delta(t)\ast \delta(t)$ is OK (to me today), but remember this is only a shorthand notation for a linear operation over smooth dense function spaces. – Laurent Duval Oct 23 '19 at 21:37

What is the autocorrelation of a Dirac pulse?

2 Answers2