# Sampling the Dirac function

I would like to ask a theoretical question concerning the Dirac function. The Fourier Transform of the Dirac function is the value 1 (DC) for every frequency. If we consider the Sampling Theorem, we have to find a maximum frequency in the signal $\ f_{max}$, so that we can sample with $\ f_s \ge \ 2f_{max}$. But as we can see from its Fourier Transform, the Dirac function contains every frequency, so we cannot find an appropriate $f_s$ . My question is, from a theoretical point of view, can the Dirac function be sampled?

• In digital land the sequence x[n] = (1, n=0) (0, otherwise) does most of the jobs that the dirac distribution does in the analog world. It's the basis function for convolution, has a flat frequency response and is the impulse response of a "wire". That's actually one thing that is easier in digital – Hilmar May 24 '13 at 12:12
• personally, i think a more concise answer is "No, a dirac impulse, $\delta(t)$, cannot be sampled at $t=0$ because there is no value that the function (or distribution) takes at $t=0$." there is no dirac delta function in the physical world, only approximations to it. so there is nothing to sample. – robert bristow-johnson Jul 19 '14 at 18:43

Any signal can be sampled, independently of whether the sampling theorem holds or not. The sampling theorem tells you that, if the sampling rate is enough, then the samples represent the complete original signal.

Signals with discontinuities or, even worse, distributions such as $\delta(t)$, are not band-limited, so the sampling theorem's hypothesis will never hold.

Also note that the usual demonstration of the sampling theorem involves multiplying the signal by a pulse train. I believe this rules out signals being distributions altogether, because the products of distributions is not well defined.

In practice, imagine sampling $\delta(t)$ at $t=0$. This sample has an undefined value.

• "Any signal can be sampled" – well, a sampling algorithm can be applied to any signal, yes, but actually calling this process "sampling" might, depending on the context, already assert you expect to be able to reconstruct the signal from the result, i.e. that the preconditions for the sampling theorem are fulfilled. – leftaroundabout May 30 '13 at 17:17

I completely agree with Juancho's answer. I would just like to add a few things. I think the main problem is the misunderstanding which becomes evident in the last sentence of the question: "... can the Dirac function be sampled?" The Dirac impulse is NOT an ordinary function having definite values for every $t$, but it is a distribution (even though it is often called 'Dirac function'). One should therefore not try to 'evaluate' (or sample!) it. What is important about the Dirac impulse are its integral properties:

$$\int_{-\infty}^{\infty}\delta(t)dt = 1$$ and $$\int_{-\infty}^{\infty}\delta(t-t_0)f(t)dt = f(t_0)$$

As Juancho already pointed out, the square of a Dirac impulse $\delta^2(t)$ is not defined. So if you were to sample a Dirac impulse, you would get the undefined result

$$\sum_n\delta(t-nT)\delta(t)=\delta^2(t)$$

Dirac impulses are a convenient tool for analyzing linear time-invariant systems but they should be treated with care because common types of processing performed on ordinary signals (such as sampling) may lead to undefined and meaningless results when applied to Dirac impulses.

The information carried by a Dirac are its location and intensity. Vetterli et al. show how it possible to sample a signal given by the sum of N diracs:

$x(t)=\sum_{i=0}^{N-1}r_i\delta(t-t_i)$

Sampling $x(t)$ in this context means recovering $r_i$ and $t_i$ for $i=0,\ldots,N-1$. In short, this is done by low-pass filtering $x(t)$ and using standard spectral estimation techniques. For more details see:

Blu, Thierry, et al. "Sparse sampling of signal innovations." Signal Processing Magazine, IEEE 25.2 (2008): 31-40.