# 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?

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

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) &= \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.

• I can't see why you may exchange $x(t)$ for $x(nT_s)$. – rmagno Nov 7 '18 at 19:19
• because $\delta(t-nT_s) = 0$ for all $t \ne nT_s$. the only time when $\delta(t-nT_s) \ne 0$ is when $t = nT_s$. – robert bristow-johnson Nov 7 '18 at 19:23

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 $$t$$.

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).