# Why is dirac delta used in continuous signal sampling?

Why this concept is the most widely accepted model of signal sampling? By multiplying the continuous signal value with the dirac delta we get an infinite value. However if we perform convolution of our signal and the dirac comb, we get our signal again - how is it useful?

The answer to a similar question has been given here, but it's still not 100% clear to me.

Sampling of a continuous function: Kronecker's or Dirac's delta?

• The Dirac delta is a distribution. Distributions can't be taken outside the integral, which means you don't multiply with them, and thus can't get any value (such as infinite) from said multiplication. – hotpaw2 Jan 5 '14 at 18:12
• If you must multiply by something, try RBJ's Planck delta (around 1e-44 wide and 1e44 high, which is less than infinite), as nothing narrower in time can ever be observed, according to Quantum Mechanics – hotpaw2 Jan 5 '14 at 18:15
• $f_s(t)=\sum_{k=-\infty}^{\infty} f(t) \delta(t-kT)$ - so for example here there is no multiplication by the dirac delta? – user107986 Jan 5 '14 at 18:22
• wow @hotpaw2 ! i didn't expect to see anything of mine migrate from comp.dsp to here. user107986 in fact, some texts do multiply $x(t)$ by $f_s(t) = \sum_{k=-\infty}^{+\infty} \delta(t-kT)$ – robert bristow-johnson Jan 5 '14 at 18:29
• my issue with the Dirac delta being a function vs. distribution is that this differentiation of concepts is not particularly useful in engineering and most physics. but if you take a Real Analysis course, you will learn Lebesgue integration and also that the integral of a function that is zero "almost everywhere" is zero. but the Dirac delta is zero almost everywhere and we're saying its integral is 1. – robert bristow-johnson Jan 5 '14 at 18:32

The best explanation of this that I've seen is from the "Digital Signal Processing Handbook" by Madisetti. Essentially the multiplication by the delta function is equivalent to sampling because the Fourier transforms are the same. So although the result of $s(t)\delta(t-nT)$ may not make much sense it's Fourier transform does exist because we are taking the integral over the delta function.

I'm cutting and pasting his text and redoing the equations. Note CT=continuous time, DT= discrete time. I have not included figures 1.1 and 1.2.

The relationship between the CT and the DT domains is characterized by the operations of sampling and reconstruction. If $s_a(t)$ denotes a signal $s(t)$ that has been uniformly sampled every T seconds, then the mathematical representation of $s_a(t)$ is given by

$s_a(t) =\sum_{n=- \infty}^{\infty} s(t)\delta(t-nT) \qquad$ (1.1)

where $\delta(t)$ is a CT impulse function defined to be zero for all $t\not=0$, undefined at t=0, and has unit area when integrated fromt $t=-\infty$ to $t=\infty$. Because the only places at which the product $s(t)\delta(t−nT)$ is not identically equal to zero are at the sampling instances, $s(t)$ in (1.1) can be replaced with $s(nT)$ without changing the overall meaning of the expression. Hence, an alternate expression for $s_a(t)$ that is often useful in Fourier analysis is given by

$s_a(t) =\sum_{n=- \infty}^{\infty} s(nT)\delta(t-nT) \qquad$ (1.2)

The CT sampling model $s_a(t)$ consists of a sequence of CT impulse functions uniformly spaced at intervals of $T$ seconds and weighted by the values of the signal $s(t)$ at the sampling instants, as depicted in Fig.1.1. Note that $s_a(t)$ is not defined at the sampling instants because the CT impulse function itself is not defined at $t$. However, the values of $s(t)$ at the sampling instants are imbedded as “area under the curve” of $s_a(t)$, and as such represent a useful mathematical model of the sampling process. In the DT domain the sampling model is simply the sequence defined by taking the values of $s(t)$ at the sampling instants, i.e.,

$s[n]=s(t)|_{t=nT} \qquad$(1.3)

In contrast to $s_a(t)$, which is not defined at the sampling instants, $s[n]$ is well defined at the sampling instants, as illustrated in Fig.1.2. Thus, it is now clear that $s_a(t)$ and $s[n]$ are different but equivalent models of the sampling process in the CT and DT domains, respectively. They are both useful for signal analysis in their corresponding domains. Their equivalence is established by the fact that they have equal spectra in the Fourier domain, and that the underlying CT signal from which $s_a(t)$ and $s[n]$ are derived can be recovered from either sampling representation, provided a sufficiently large sampling rate is used in the sampling operation (see below).

Cheers,

David

• "Essentially the multiplication by the delta function is equivalent to sampling because the Fourier transforms are the same." boy, no matter how i parse that statement, i don't think it's correct. – robert bristow-johnson Jan 6 '14 at 17:47
• it is true that multiplication by a delta function is equivalent to sampling the function at the instant of the delta function. this is because the delta function is equal to zero everywhere else than at the sampling instance, so it throws away all of that other information, except at the sampling instance. – robert bristow-johnson Jan 6 '14 at 17:49
• Nice explanation. As I've found another interesting material, here's the link: springer.com/cda/content/document/cda_downloaddocument/… – user107986 Jan 6 '14 at 21:35

The answer to your question is that multiplying by the dirac comb thus:

\begin{align} x(t) \cdot \sum_{n=-\infty}^{+\infty} \delta(t - nT) & = \sum_{n=-\infty}^{+\infty} x(t) \delta(t - nT) \\ & = \sum_{n=-\infty}^{+\infty} x(nT) \delta(t - nT) \\ \end{align}

shows explicitly that $x(t)$ is sampled at the uniformly-spaced sample times $nT$. all information between the samples is discarded because multiplying by zero discards the information.

the the Sampling Theorem shows that, if $x(t)$ is sufficiently bandlimited, then the entire $x(t)$ can be recovered from the sampled signal above and thus from the samples $x(nT)$.