Definition of sampling using delta or indicator function? I just came from a class where the professor showed a slide with the definition of sampling:



But I do not understand how we can multiply a signal $x(t)$ with the delta function $\delta(t)$, as the $\delta(t)$ is infinite at $x=0$. So this multiplication would amount to multiply a real-valued function with $\infty$ which is obviously not what is meant here. I think.

So would it make the formula formally correct if we would swap the $\delta(t)$ with the indicator function?

$$\mathbf{1}_0(x) =
\begin{cases}
1 &amp;\text{if } x = 0, \\
0 &amp;\text{if } x \neq 0.
\end{cases}$$
that equation following "Conceptual scheme" is misleading.  the $\implies$ arrow is not an equal sign.  this is what is true:

$$\begin{align}
x(t) \times \sum\limits_{n=-\infty}^{\infty}\delta(t-nT_s) &amp;= \sum\limits_{n=-\infty}^{\infty}x(t)\times\delta(t-nT_s) \\
&amp;= \sum\limits_{n=-\infty}^{\infty}x(nT_s)\times\delta(t-nT_s) \\
&amp;= \sum\limits_{n=-\infty}^{\infty}x[n]\times\delta(t-nT_s) \\
\end{align}$$

because we simply define $x[n] \triangleq x(nT_s)$.

the difference between the dirac delta and the indicator function is that one integrates to an area of 1 and the other integrates to an area of 0.