# Talk:Nyquist–Shannon sampling theorem

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## Aliasing sinusoids whose frequencies are half the sampling rate

In the Critical frequency section, I want to change this:

But for any θ such that |cos(θ)| < 1, x(t) and xA(t) have different amplitudes and different phase. Ambiguities such as that are the reason for the strict inequality of the sampling theorem's condition.

to this:

But for any θ such that |cos(θ)| < 1, x(t) and xA(t) have different amplitudes and different phase (any sinusoid will alias xA if its amplitude is that of xA multiplied by sec(2πθ) where θ ≠ (k+1/2)π for integer k). Ambiguities such as that are the reason for the strict inequality of the sampling theorem's condition.

This directly relates the amplitude of any aliasing sinusoid as a function of the phase θ, rather than indirectly with the definitions x(t) = cos(2πBt + θ), xA(t) = cos(θ)*(2πBt). (This allows you to make this sort of graph.) If it's not immediately obvious to you why the amplitude gets divided by cos(2πθ) rather than just cos(θ), then the definitions alone aren't sufficiently serving their justice. X-Fi6 (talk) 03:58, 19 November 2012 (UTC)

We already have that sort of graph in the article (starting with File:CriticalFrequencyAliasing.png in 2006). It's not clear how the more "direct" expression makes it more clear. It's also not clear why you put that 2pi in there; the angle theta is in radians, I would presume, and multiplying radians by 2pi can't make any sense. Dicklyon (talk) 05:02, 19 November 2012 (UTC)

My interpretation of the suggestion is that he would have the section changed to something like this:
To illustrate the necessity of fs > 2B, consider the sinusoids:
$x(t) = \cos(2 \pi B t)\,$
$x_A(t) = \frac{\cos(2 \pi B t + \theta )}{\cos(\theta )}\ = \ \cos(2 \pi B t) - \sin(2 \pi B t)\tan(\theta ), \quad 0 < \theta < \pi/2.$
With fs = 2B or equivalently T = 1/(2B), the samples are given by:
$x(nT) = \cos(\pi n)$
$x_A(nT) = \cos(\pi n) - \underbrace{\sin(\pi n)}_{0}\tan(\theta ) = x(nT).$
That sort of ambiguity is the reason for the strict inequality of the sampling theorem's condition.
--Bob K (talk) 18:09, 21 November 2012 (UTC)

## New WikiProject Signal Processing

Wikipedia:WikiProject Signal Processing is a stub of a new project. We could use help, especially from someone with experience setting up project templates, or patient enough to figure out how to cobble them from another project. Dicklyon (talk) 19:24, 7 March 2013 (UTC)

## requirements for the Shannon theorem should be included into the article

the Shannon theorem only applies to:

1. smooth functions
2. square integrable functions

for reference see Google Books. --Biggerj1 (talk) 21:21, 4 May 2013 (UTC)

But the band-limited condition already implies smooth. And I'm not convinced the square integrable condition is necessary, though it certainly makes it easier to prove. Dicklyon (talk) 22:14, 4 May 2013 (UTC)
Ostensibly, it might be easier to require abs-integrable (which is stricter) to guarantee that any of the Fourier integrals converge, but even if you stick with the less strict square-integrable, the F.T. of non-time-limited sinusoids don't really exist. But, with the magic of dirac impulse functions, we can pretend that the F.T. of sinusoids do exist.
Personally, on this, on Dirac delta and on Dirac comb, I think the rigor that you get in a typical undergrad EE text should be sufficient. That is where you can treat the Dirac impulse as a "function" that is a limit of nascent delta functions all with unit area. But we know the mathematicians are unhappy with that. But I don't think it helps most people trying to learn the concepts to be that anal about it which is why most undergrad EE courses don't object to the Dirac delta being the "unit impulse function". 71.169.184.238 (talk) 00:23, 5 May 2013 (UTC)
Isn't square integrable more strict than absolute integrable? Either way, a sinusoid would be out. But sinusoids work fine with sampling and exact reconstruction, as long as they're not right at the Nyquist frequency, I think. You don't need the existence of Fourier transforms for the theorem to be true, so it doesn't really matter whether or not you like Dirac deltas. At least, that's my view as an engineer; but as you suggest, mathematicians may not agree. Dicklyon (talk) 03:37, 5 May 2013 (UTC)
Oh, I see you're right. The sinc function is square integrable, but not absolute integrable and that's the rub. It means the sinc is not BIBO stable, as I had noted in the sinc filter article already. Dicklyon (talk) 03:41, 5 May 2013 (UTC)

The theorem obviously applies to sinusoids, unless you can tell me what "smooth" function other than cos(2π f t), f < B, is bandlimited (< B) and fits these samples: cos(2π f n/(2B)), all integer values of n. Seems to me that both square integrable and abs-integrable are overly restrictive... sufficient but not necessary, which is likely the reason that sinusoid-loving EEs ignore them.
--Bob K (talk) 11:57, 5 May 2013 (UTC)

Oh, perhaps I should have asked for clarification... Are you saying "smooth and integrable", or, "smooth or integrable"?
--Bob K (talk) 14:04, 5 May 2013 (UTC)

You need both, square integrable and smooth. Both is provided by the condition band-limited. And no, you can't sample a sinusoid. That is, of course you can sample it, but the reconstruction formula will not converge. Since the reconstruction formula is essentially a representation in an orthonormal basis, it converges in the L2 sense exactly if the coefficient sequence is square summable, and the sample sequence of a sinusoid is not square summable because of its periodicity. But that was already discussed ad nauseatum years ago, so consult the archive pages to find the points of view of the participants.--LutzL (talk) 15:59, 5 May 2013 (UTC)

Since you just said that a sinusoid is not bandlimited (because it's not square integrable), I can see why it was discussed ad nauseum. But I'm not going to look, ... better things to do. And I see that it's the reconstruction part of the theorem that you think requires square integrability. (Thanks for that.) Maybe so, but I still think that this part also applies to sinusoids:
"If a function x(t) contains no frequencies ≥B hertz, it is completely determined by giving its ordinates at a series of points spaced 1/(2B) seconds apart."
--Bob K (talk) 17:29, 5 May 2013 (UTC)
The problem with finding interpolated values of a sinewave from samples via the Whittaker formula (dot product with sinc filter) is that the dot-product series has only conditional convergence; if you evaluate it by moving outward from the middle you'll approach the right answer, but other orders of evaluation may not do so. As for a sinewave not being "bandlimited", that's just because he took the space of bandlimited functions to be a subset of square-integrable functions. So, we engineers and mathematicians will continue to go our own ways. I'm with you (in most cases). Dicklyon (talk) 17:53, 5 May 2013 (UTC)
Reviewing archive 1, which I now recall better, I'd summarize LutzL's point as being that since the reconstuction formula is not absolutely convergent if the signals are not in L2, the proof of the theorem doesn't apply there and therefore the theorem doesn't hold there. Or maybe he didn't even say the latter clause. Either way, the fact the proof doesn't work outside of L2 does not mean that the theorem doesn't hold outside of L2; it means it might or might not hold, and should motivate us to find a better proof to cover a wider set. Not being a mathematician, I expect I won't see the subtle pitfalls of this approach, but consider proving it by specifying evaluating the sinc from the middle outward, in which case the series will approach a limit, I'm pretty sure. Or if that doesn't work, make a windowed version of the sinc and look at the limit as the window width goes to infinity. I'll be a monkey's uncle if that limit doesn't exist for sinewaves below the Nyquist frequency, for DC, and for WSS process, which are all outside of L2. Dicklyon (talk) 20:07, 5 May 2013 (UTC)
Returning to the start of this thread, which asserts that the Shannon theorem only applies to functions that are differentiable and square integrable. A more precise statement is that those are merely sufficient conditions (as far as we know). It has not been proven that those are necessary conditions. And indeed, we routinely sample sinusoids and linear combinations thereof all the time. That important point should stand high above the "ad nauseum" academic issues, not vice versa. -Bob K (talk) 23:01, 6 May 2013 (UTC)
True, but we also routinely sample signals that are not bandlimited; nothing prevents it. There's still an interesting mathematical question about the theorem; that is, what can be proved about when perfect reconstruction is possible. I think we're all comfortable noting that the conditions of this theorem are sufficient, but not necessary. Then the interesting question, mathematically, and maybe even to us engineers, is whether can we come up with a looser set of sufficient conditions, not including the signal being in L2. I think it's there, but I'm not enough of a mathematician to know whether my hand-waving outline of a proof could work rigorously. Dicklyon (talk) 23:58, 6 May 2013 (UTC)

To be mathematically clear, the theorem, or rather its careful proof (unfortunately not quite the one in the article), says that any band-limited square-integrable function:

1. is equal almost everywhere to a smooth (C-)function, and
2. can be sampled as specified, where the convergence is uniform, therefore a fortiori L2.

So smoothness is actually a consequence of the band-limited assumption. Mct mht (talk) 23:34, 6 May 2013 (UTC)

## Notation / Language

Since it is the job of mathematicians to discover mathematical concepts and communicate them via proofs, definitions, etc., (i.e., we construct the language of mathematics so others can use it), our experience with notation is far superior to that of engineers when it comes to communication of ideas - that is what we do.

As such, I think we should use the standard mathematical notation (f for function, k for frequency, etc) since that is consistent with the vast majority of mathematical texts as well as a large number of engineering texts on this subject. — Preceding unsigned comment added by 142.136.0.102 (talk) 16:05, 27 June 2013 (UTC)

Maybe you should point out the changes you want, and link a reference that does it that way, so we can see what you're proposing. Let's don't make this a mathematicians versus engineers dispute, as each group is pretty good at communicating to their own. Dicklyon (talk) 16:36, 27 June 2013 (UTC)

The tag says