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I have a band-limited signal and sample it at sample rate of $f_\mathrm{s}$. By taking the continuous time fourier transform, I will see copies of the same signal at $n\cdot f_\mathrm{s}$ where $n \in \mathbb{Z}$. I have attached a spectrum of the signal before and after sampling as shown below.

enter image description here

My question is, intuitively, I would expect that the sampled signal ought to still have the same energy as the original signal; yet, as the equation and picture show, the sampled signal now has infinite spectrum; hence a power signal and infinite energy. Given that the original signal is band-limited, it must be an energy signal, hence finite energy. So they are not equal.

What is my misunderstanding?

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  • $\begingroup$ This answers should also answer your question. $\endgroup$
    – Matt L.
    Oct 24 at 6:22
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The spectrum that you give in your question corresponds to the "train of impulses" sampling process $$x(k) = \sum_{k=-\infty}^\infty x(t) \delta(t - kT_s). \tag1$$

This model is a very good one for tying together the Laplace and $z$ transforms, and for expressing the four kinds of Fourier transform in a unified framework.

However, in this framework, the use of that Dirac delta function changes a finite-energy signal into an infinite-energy signal (because the energy of $\delta(t)$ is infinite). For that matter, it changes a finite-power signal into an infinite power signal, for the same reason.

So your observation that the energy of the signal goes from finite to infinite is exactly correct.

For the purposes of distinguishing an "energy signal" from a "power signal"* you have to consider just the energy (or power) of the sequence $x_k$, where you re-cast (1) into the definition $$ x(k) = \sum_{-\infty}^\infty x_k \delta(t - kT_s). \tag2$$

Where $x_k$ is defined as $x_k = x(kT_s)$. The distinction here is that (1) and (2) are a handy model for the sampling process that unifies the math, where the series of $x_k$ is what you might get out of an ADC in the real world.

Now, we can extract the $x_k$. If $x(t)$ is of finite energy then the set of $x_k$ will** also be finite. If $x(t)$ is of finite power, then so will $x_k$. This is intuitively satisfying, because we would expect that converting a finite-energy continuous-time process into a discrete-time process should get us a finite-energy signal -- and it does.

In this context, if you take the Fourier transform of $x_k$ it will be either the DFT or the DTFT; in both of these cases the frequency domain of the transform only exists on an interval that spans $2\pi$ radians -- usually we choose $\omega \in [0, 2\pi)$ or $\omega \in [-\pi, \pi)$, but the math works if we choose any interval of the correct length.

At this point the spectrum does not repeat, because the domain of the spectrum is limited. So Parseval's Theorem holds, and the energy of an infinite-extent $x_k$ can be calculated from the integral over a finite span of frequencies of its discrete-time Fourier transform, or the energy of a finite-extent $x_k$ can be calculated from the sum of of it's discrete Fourier transform.

* I do dislike those terms, and sincerely hope they're a passing fad. Alas, I'm not sure what to replace them with, so I'm probably doomed to dealing with it.

** At least I believe that's the case. If it's not, then it's true in all but pathological cases.

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    $\begingroup$ As a side comment: I'm being very careful in my use of the term "modeled as" to talk about the sampling process. What the tidy, unified math (using $\delta(t)$) says it is, and what it really is (time-varying circuits with analog to digital converters and DACs and whatnot) are very different things. You can spend a lot of time reconciling the mathematical view with the reality -- that's time well spent, because to make this stuff work in practice you have to have one foot in mathemagic land, and the other foot in the real world. Lean too far one way or another, and you're lost. $\endgroup$
    – TimWescott
    Oct 24 at 4:25
  • $\begingroup$ Thank you for the subtleties that you put into the explanation. I can very much appreciate it given that I indeed am struggling between the real world ADC and the math behind it. I think the line, "Now, if x(t) is of finite energy then xk will** ...." captures the answer to my question. Are you saying that for sampling in real world (ADC), we need to, "IN OUR MIND" (no math), switch (1) to (2). Then, the process of DTFT takes care of the dilemma of not satisfying Parseval? Also, for (2), do you mean to switch () to [] to signify that delta train (arrows) is now discrete diracComb (dots)? $\endgroup$
    – scc28adi
    Oct 24 at 12:33
  • $\begingroup$ I need to figure out how to edit my answer so we don't need that clarification, but -- yes to your first comment, and no to the second -- the LH side of (1) and (2) are the same, I've just substituted in $x_k$ (where $x_k = x(T_s\ k)$) on the left side. $\endgroup$
    – TimWescott
    Oct 24 at 19:52
  • $\begingroup$ I think I understand you. $x_k$ is the discrete sequence obtained from (2). So (2) by itself is still continuous-time in nature, which spectrum is infinite energy. Am I right? Thanks again. $\endgroup$
    – scc28adi
    Oct 24 at 20:06
  • $\begingroup$ Yes, that's right. $\endgroup$
    – TimWescott
    Oct 24 at 21:05

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