This is purely theoretical question but can an actual delta-dirac function ever be sampled? To make it clear again, I'm not talking about an impulse response from the impulse function. Lets say we have a truly instantaneous impulse as defined with delta-Dirac that the impulse with infinite(or finite) height and no width, an instantaneous pulse. This pulse is to be detected by sampling, but to find this instant signal, wouldn't this require an infinitely fast sampling or infinite sampling rate? Thinking this in terms of frequency domain, I believe delta function is 1 or has infinite spectrum. So would delta function ever be sampled unless we have infinite sampling rate?
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A Dirac delta impulse cannot be sampled because it is a distribution and as such it does not make sense to talk about the value of $\delta(t)$ for e.g. $t=0$ because that value isn't defined. However, sampling extracts values of a function for certain values of $t$. The Dirac impulse is no function in the conventional sense. It's only under an integral that $\delta(t)$ makes sense.
Furthermore, the mathematical description of sampling is a multiplication with a Dirac comb. However, the expression $\delta^2(t)$ is undefined, so sampling a Dirac impulse can't be described mathematically in any meaningful way.
Note that the discrete-time unit impulse $\delta[n]$ can be sampled without any problem, because it has well-defined values for any $n$, unlike its continuous-time counterpart.
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$\begingroup$ Ok, that was a great explanation, thanks! $\endgroup$ Commented Jun 2, 2015 at 23:55
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Yeah, I think your conclusion is right. According to Nyquist' sampling theorem, the sampling rate should be at least twice as many as the maximum frequency of the target signal. And the maximum frequency of delta function is infinite. If sampled, aliasing must occur. In fact, the discrete delta function $\delta(n)$ is a sampled version of continuous delta function $\delta(t)$. The spectrum is overlapped while sampling, of course.
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1$\begingroup$ The reason why $\delta(t)$ can't be sampled has nothing to do with the sampling theorem. Neither is $\delta(n)$ a sampled version of $\delta(t)$. $\endgroup$– Matt L.Commented Jun 1, 2015 at 6:56
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$\begingroup$ @MattL. delta function is not a coventional function does not mean it can not be operated. And $\delta^2(t)$ is meaningful in some way. See this mathoverflow.net/questions/48067/… $\endgroup$– seven7eCommented Jun 1, 2015 at 10:42
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1$\begingroup$ Not sure what you mean by "cannot be operated", but I stick with my claim that $\delta^2(t)$ has no generally accepted and useful meaning, especially in the context of sampling the distribution $\delta(t)$. If you apply $\delta^2(t)$ to some test function which goes (at least quadratically) to zero for $t\rightarrow 0$, please use it if you can make sense out of it. But there's no way you can handle $\delta^2(t)$ in the context of the OP. Furthermore, $\delta[n]$ is definitely NOT a sampled version of $\delta(t)$. You might want to consider correcting or deleting your answer. $\endgroup$– Matt L.Commented Jun 1, 2015 at 11:51