# What is the autocorrelation of a Dirac pulse?

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

Can you explain to me how to calculate it?

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.

• Perfect, thank you very much! – Kinka-Byo Oct 21 '19 at 19:00
• 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
• 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

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.

• 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
• 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
• 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 0:54