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While learning sampling theory - I noticed that examples of continuous signal sampling always achieved the goal via multiplying the signal with a "Dirac Comb". I was intrigued by the requirement to use a Dirac Comb instead of a "Ones Comb" - I.E: multiply the signal by the value '1' at the sampling times.

The explanation I got: "That's necessary because multiplying by a Dirac function is equivalent to Convolution in the frequency domain which maintains the spectrum of the original sampled signal".

This made a lot of sense to me...But than, when I proceeded to learn about discrete domain sampling - I was surprised to find out that examples used a "Ones Comb" to sample a signal (not the "Dirac Comb").

Why the difference ? How does discrete time sampling get away with multiplying by a train of discrete ones ?

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the integral

$$ \int\limits_{-a}^{a} \delta(x) \, \mathrm{d}x = 1 \qquad \forall a>0$$

but this integral

$$ \int\limits_{-a}^{a} g(x) \, \mathrm{d}x = 0 \qquad \forall a>0$$

where

$$ g(x) = \begin{cases} 1 \qquad & \text{for } x=0 \\ 0 \qquad & \text{for } x\ne0 \\ \end{cases} $$

this is why we need this dirac delta thing for the sampling or sifting property:

the integral

$$ \int\limits_{-\infty}^{\infty} f(x) \, \delta(x-x_0) \, \mathrm{d}x = f(x_0) $$

but this integral

$$ \int\limits_{-\infty}^{\infty} f(x) \, g(x-x_0) \, \mathrm{d}x = 0 $$

can you see what's wrong with using $g(x)$ for sampling?

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  • $\begingroup$ Well, I was already convinced that δ(x) (and not g(x)) is the correct way to do it... The question is why why this isn't also required for discrete time signals? Why do discrete time signals CAN "get away with using g(x)" while continues signals MUST use δ(x) ? $\endgroup$ – shaiko Mar 29 at 21:18
  • $\begingroup$ well, a difference equation is similar, but not the same as a differential equation. summations are similar, but not the same as integrals. Discrete-time signals, $f[n]$, are only defined for integer indices. if $\delta[n]$ is the Kronecker delta, then $$ \sum\limits_{n=-\infty}^{\infty} f[n] \delta[n-n_0] = f[n_0] $$ so it works for discrete-time signals. $\endgroup$ – robert bristow-johnson Mar 29 at 21:30
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Dirac combs are for sampling in the domain of continuous time while retaining a continuous time viewpoint. That's a bit tricky since as a windowing function they are not supposed to look at any values except at integral arguments, but retaining a continuous time viewpoint also means that you want to work on the results using integral transforms (like Fourier transforms). And a single point does not make a difference for an integral. So a dirac function is more or less defined by its behavior under integration, making the integral "magically" jump by 1 when including the location of the dirac pulse in the integration interval. So there is a whole new way of looking at function-like objects in the context of their extended behavior under integration.

That needs to change some of the definitions, turning functions into distributions, and Riemann integrals into Lebesgue integrals, often in the form of Stieltjes integrals which arise in integral substitutions. So while the fundamentals when working with distributions rather than functions are very much different, a whole lot of the mathematics built on top of it works the same.

Now the purpose of a dirac comb is extracting a finite number of values from a "function" and giving them an existence that is visible even under integration in the continuous domain. If you already are in a discrete domain, there is no such thing as integration and Lebesgue measures and distributions. A function (in this case a sequence) that selects all discrete points in a discrete domain is just 1.

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When a signal is expressed in discrete time notation, it is equivalent to a continuous time signal comprised of an infinite series of Dirac impulses scaled by the signal values. A Ones Comb in discrete time is equivalent to a Dirac Comb in continuous time.

So you would use a Dirac to sample a continuous signal, and a Ones to resample a discrete signal. Again though, they mean the same thing.

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