The comb function can be used to sample a continuous signal. The comb function is defined as follows

$$ C_T(t) = \sum_{k=-\infty}^{\infty} \delta(t - kT) $$

where $\delta$ is the Dirac function, which is $\infty$ whenever its input is 0, and it is $0$ otherwise. So, when $t = kT$, then the Dirac delta is not zero. $T$ should be the "sampling period", which I am not sure what it is, but I think it is a way for us to specify when we are going to sample a value.

We could think of the comb function as multiple "impulses" (or multiple Dirac functions) at evenly spaced points (where the space between these single Dirac functions is $T$). We can see this from the following image

enter image description here

So, a discrete signal, $X_s(t)$ (where the subscript $s$ stands for "sampled") can be represented as follows

\begin{align} x_s(t) &= x(t) \cdot C_T(T)\\ &= x(t) \cdot \sum_{k=-\infty}^{\infty} \delta(t - kT)\\ &= \cdot \sum_{k=-\infty}^{\infty} x[t]\delta(t - kT) \end{align} where $x[t]$ is like a discrete version of $x(t)$ and $x(t)$ is the continuous signal associated with $x_s(t)$. However, as far as I understand, this assumes that we are sampling from $x(t)$ uniformly: more specifically, we are apparently sampling with frequency $F = \frac{1}{T}$ (why?).

Intuitively, why is the comb function the sampling function? The plot above of the comb function gives some intuition, but I think a simple step-by-step example of a sampling procedure would also help.

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    $\begingroup$ Your understanding breaks down right where you say "$\delta$ is the Dirac function, which is $\infty$ whenever its input is $0$, and it is $0$ otherwise." $\delta$ is not a function at all and you cannot ascribe it values as you have, no matter what your textbook or your instructor says. $\endgroup$ – Dilip Sarwate Mar 28 '19 at 13:56
  • $\begingroup$ @DilipSarwate Wikipedia states that Dirac function is a generalised function, which is a generalisation of a function, so it is a function. In general, you cannot ascribe the values of the function that you do not define (which is the case), but you can still pass different inputs to the function. I don't get your point. $\endgroup$ – nbro Mar 28 '19 at 14:18
  • $\begingroup$ The point is, it is sheer nonsense to say that "$\delta$ has value $\infty$ whenever its input is $0$ and value $0$ otherwise" regardless of whatever Wikipedia says about "$\delta$ is a generalized function" from which the only thing you seem to have taken away is that $\delta$ is a function. For dsp.SE purposes, $\delta$ is defined only in terms of how it behaves in definite integrals and ascribing any value or meaning to $x(t)\delta(t)$ is futile; the question is all about how $x(t)\delta(t)$ behaves when it is part of the integrand in a integral. $\endgroup$ – Dilip Sarwate Mar 28 '19 at 15:03
  • $\begingroup$ @Dilp Sarwate: I'm not a math guy but I understand what you mean by this. At the same time, at the bottom you say "in an integral". but you must mean "in an integral or summations" correct ? To OP: As Dilip said but I don't want to put words in his mouth: Don't think of the delta as a function unless you want to deal with generalized functions ( which are ugly for non-math people ). Think of it as an operator that is used in integrals and summations to "pick out" the element at the place where the associated delta's argument is zero. $\endgroup$ – mark leeds Mar 28 '19 at 19:31

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