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I just came from a class where the professor showed a slide with the definition of sampling:

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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?

$$\mathbf{1}_0(x) = \begin{cases} 1 &\text{if } x = 0, \\ 0 &\text{if } x \neq 0. \end{cases}$$

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that equation following "Conceptual scheme" is misleading. the $\implies$ arrow is not an equal sign. this is what is true:

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

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.

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  • $\begingroup$ I can't see why you may exchange $x(t)$ for $x(nT_s)$. $\endgroup$ – rmagno Nov 7 '18 at 19:19
  • $\begingroup$ because $\delta(t-nT_s) = 0$ for all $t \ne nT_s$. the only time when $\delta(t-nT_s) \ne 0$ is when $t = nT_s$. $\endgroup$ – robert bristow-johnson Nov 7 '18 at 19:23
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I think one important thing that you must understand is that a Dirac delta impulse is not an ordinary function that can be evaluated. So you do not multiply your signal with a value of $\infty$. What happens instead is that you weigh the individual shifted Dirac impulses with the corresponding values of the signal due to the following identity:

$$x(t)\delta(t-nT)=x(nT)\delta(t-nT)\tag{1}$$

which is true as long as $x(t)$ is continuous at $t=nT$.

Consequently, multiplying a signal with a Dirac impulse train results in a weighted impulse train, where the weights are the signal values at the sample instants. So what happens is that from a continuous signal $x(t)$ you only retain the sample values $x(nT)$, but you still have an expression that can be considered a continuous-time signal (in the sense that it can be integrated or convolved with another function).

Note that this is just a mathematical model. As pointed out in Carlos Danger's answer, what usually happens is that this impulse train is filtered, i.e., you get

$$\left(\sum_nx(nT)\delta(t-nT)\right)\star h(t)\tag{2}$$

where $h(t)$ is the impulse response of the filter. Since $\delta(t-nT)\star h(t)=h(t-nT)$, Eq. $(2)$ equals

$$\sum_nx(nT)h(t-nT)\tag{3}$$

which is now an ordinary function which can be evaluated for any $t$.

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In addition to the other answers, I would also like to point out that the "ideal" impulse sampler is usually considered to be followed by some filter of some kind, such as a zero-order hold. The infinite-amplitude spikes are "averaged out" by convolving with the filter impulse response.

Also, even without any filtering after the impulse sampling, the spectrum of the sampled signal creates a mathematical model for the aliased spectrum that is mathematically equivalent (up to a constant) with the periodic spectrum computed using the discrete-time Fourier transform (DTFT).

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