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?

Answer

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