Talk:Nyquist–Shannon sampling theorem

Disagreeing with Kupfmuller filters

The following passage rings suspicious, it is obviously impossible to reconstruct arbitrary signals once they are sampled. If you want to keep this paragraph in, explain here why, or I will delete it.

The theory of reconstruction of the analog signal from samples requires a so-called Kupfmüller filter. Such a filter is a theoretical filter with certain properties, which can only be approximated in reality. The filter is supposed to be fed directly with the sample impulses and generates the original analog signal.

Loisel 07:09, 16 Sep 2004 (UTC)

I think there's some form of filtering or interpolation that can be used to *perfectly* reproduce a bandlimited sampled signal, instead of the usual approximate filtering of all harmonics. Related to the sinc function? I'm not sure, but assumed this is what they were talking about. - Omegatron 19:03, Sep 16, 2004 (UTC)

That would be the Nyquist-Shannon interpolation formula, Omegatron - Omegatron

The theory of discrete Fourier transforms tells you that if U is the discrete Fourier transform of u, then the discrete inverse Fourier transform of U is u again, without any error whatsoever. If you have a bandlimited signal f, sample it to u, take its DFT U, then take an IDFT of U, you'll recover u perfectly. If u has sufficiently many samples for the bandwidth of f, then f is given by the unique interpolating trigonometric polynomial for u. The coefficients of the interpolating polynomial are given by U. As you can see, there's no filter. Applying a sinc filter of the correct frequency to a bandlimited signal won't change it at all, since it's already bandlimited. So this Kupfmuler stuff is nonsense. Loisel 07:05, 20 Sep 2004 (UTC)

Oh I remember now. To convert the sampled signal back to analog you put it through a zero-order hold or the like, which creates the original signal in the baseband with harmonics every multiple of the sampling frequency, and you filter the harmonics. The sinc "filter" i was thinking of is actually what you convolute the signal by. sinc convolution in time domain = rectangular ideal filter in frequency domain. right? and no real filter is ideal, so technically you are only attenuating the higher frequencies, not destroying them completely. So I thought maybe this kupmuller stuff had to do with the sinc convolution, but nope. That's just an ideal filter. - Omegatron 13:29, Sep 20, 2004 (UTC)

Actually, since I will forget about this, I am deleting now. Restore it if you can justify it.

Loisel 07:11, 16 Sep 2004 (UTC)

The following disputed psychobabble also deleted:

In practice, the sampling data is usually not available any more. It has been quantized ("squeezed" or broken into discrete values) and digitized (converted into symbols, such as numbers). Digital to analog converter circuits are used to decode the digital information into analog, which is then filtered. These circuits are usually based on op-amps. Each quantized value is associated with a certain voltage level, and the op-amp circuits generate that voltage level when seeing the particular digital input signal.

Loisel 07:14, 16 Sep 2004 (UTC)

Psychobabble? I think you are using the wrong word. Anyway, what's wrong with a description of digital sampling? - Omegatron 19:03, Sep 16, 2004 (UTC)

All right, technobabble then. The text above is handwavy, fails to explain the terminology (like "quantized") and, since it's trying to explain why Kupfmuller filters don't have the "sampling data", irrelevant (by virtue of Kupfmuller filters being nonsense.)

If a signal is quantized to floating points, many purposes the quantization does not matter and the IDFT gives u from U to extremely high precision. If a signal is quantized for the purpose of compression, u can obviously not be recovered from V (my notation for quantized U) but this has nothing to do with the sinc filter or anything. If v is a quantized u, then the DFT V of v will yield v again to high precision using the IDFT. Digital to analog converters do not affect the quantization, and are no more affected by it than the software IDFT. Injecting op-amps adds nothing to the discussion. If you want to have a reference to DACs, just put "see also Digital to analog converters." One possible statement would be, "Measurement devices like Charge-coupled devices, antennas, geiger counters, etc... and output devices such as Cathode ray tubes, loudspeakers, Light-emitting diodes, the brakes of an Electronic Stability Program, etc... will add some error to the input or output. Such error can make high frequency information irrecoverable even when we have very many samples. This effect can sometimes (but not always) be regarded as the Nyquist theorem applied to the Fourier transform of the signal." Such a comment would have to be made in a non-technical context or include the caveats that "high frequency" may actually not be correct. In some instances, the low frequency data is the noise, and the high frequency is good. In other cases, the noise does not correspond to a neat range of requencies. See aliasing for details.

If you want to talk about hardware, make sure that what you're saying is true and has its place in an article with "theorem" in the title. Loisel 07:05, 20 Sep 2004 (UTC)

'just put "see also Digital to analog converters."'

Good enough for me. - Omegatron 13:29, Sep 20, 2004 (UTC)

Disagreeing with the theorem.

I just read this and the aliasing article, and I must say that I disagree with what's written in there.

Perhaps in engineering this is gospel and sacred, but many of the assertions and hypotheses are unstated, unmotivated and even perhaps incorrect. Why would signals with high frequencies be wrong? Why should we choose, of all the signals that alias to the same sampled signal, the one that has zeroes for the high frequencies? There are reasons for that, but they vary from case to case, and stating that using zero for the high frequency as a canon isn't scientific.

If anyone feels strongly about these articles, continue the discussion in here. If nobody pipes up, I'll change the articles around significantly. Loisel 04:41 Jan 24, 2003 (UTC)

Dou you mean we can generalize to: If S(f) = 0 outside a given interval of width 2W, then s(x) can be recovered. ? Here S(f) is defined for positive and negative f, so if the interval is 3W to 5W, S should also be zero for -5W to -3W. We can add this, especially if you can describe an application for this it is interesting. - Patrick 10:38 Jan 24, 2003 (UTC)

Not quite. Here is the short version. A signal s(x) is just another word for a function. There are many functions. If you sample functions, there are a lot fewer sampled functions. The map you use to go from a signal s(x) to a sampled version (s_k) will necessarely collapse several different signals into the same sampled version by the pigeonhole principle, this is called aliasing. Two signals s(x) and t(x) which collapse to the same sampled signals are called aliases of one another. In general it is preferable that the map L:s(x)->(s_k) produce sampled versions which correspond as closely to the original s(x) according to some metric, or perhaps according to the human brain. Experimentation has suggested that in many cases, the signal that most resembles the sampled version is the one which has the least high frequency components. However, this is simply a heuristic, and in some cases, it is patently false. For instance, if one is working on the real line (not with periodic function) with signals that are compactly supported, it is completely absurd to hope that the spectrum such a signal is also compactly supported. Let us assume that L is linear. Note that the space of sampled signals is of dimension n. Then we would like to choose, for each sampled signal (s_k), a pre-image L^{-1)(s_k)=s(x) which we believe gave rise to (s_k). With the assumption above in italics, then we can choose a linear space of dimension n as the pre-image L^{-1}(S)of the space of sampled signals S={(s_k)}. This is in fact the Nyquist-Shannon sampling theorem (and we could write it explicitly sort of the way it's written, although I suspect there's an off-by-one bug in the current statement.) With this presumption, the Nyquist-Shannon interpolation formula follows. In view of this, I'm not sure how to split the discussion between aliasing and this theorem, which is why my original comment was about both articles. Loisel 19:18 Jan 24, 2003 (UTC)

Also, I should add that I don't intend to massacre the article, just to present the information so that the Fourier approach isn't presented as some sort of God-given truth, as well as offer a linear algebra explanation of the theorem. Loisel 05:51 Jan 25, 2003 (UTC)

Maybe, in the interest of conciseness, this article can stay as is, and the article under aliasing can fill in the blanks I discussed above. Loisel 08:34 Jan 27, 2003 (UTC)

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What does "the sampling frequency must be original perfectly from the sampled version." in the page meen - this change was made almost two months ago, and I have no idea what it means - can we revert to what the older edits said in this place: "the sampling frequency must be greater than twice the highest frequency of the input signal in order to be able to reconstruct the original perfectly from the sampled version."

Probably a typo where a line got deleted. --Damian Yerrick

I suggest that we leave the psychoacoustic stuff out of this particular article and limit the discussion strictly to the objective, mathematical properties of sampling that Nyquist originally studied. He was actually studying the reverse problem to the one usually mentioned in connection with his theorem, which was how fast one could signal in a bandlimited channel like a telegraph cable without intersymbol interference. His result was equivalent to its modern application in digital sampling. To that end, I would merge the discussion of oversampling into the main body and restate the theorem as it is now stated in the oversampled section, i.e., the sampling rate must be at least twice as high as the bandwidth of the signal being sampled to avoid any loss of information. --Phil Karn, KA9Q

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I liked the page very much but strongly suggest to remove the last paragraph (which I did) since the claim that the sampling problem is analogous to polynomial interpolation simply is wrong and misleading. -- Olivier Cappé, 11:23 Mar 8, 2004 (CET)

Also is known

It also is known as Whittaker-Nyquist-Kotelnikov-Shannon sampling theorem (example in Russia is's known as Nyquist-Kotelnikov theorem).

• Nyquist - 1928
• Kotelnikov - 1933
• Whittaker - 1935
• Gabor - 1946
• Shannon - 1948

Links:

• http://www.ifp.uiuc.edu/~minhdo/talks/Zurich_04/eth_intro2x2.pdf
• http://www.ics.ele.tue.nl/~fwillems/compri/total.pdf
• http://www.ieee.org/organizations/history_center/legacies/kotelnikov.html

--.:Ajvol:. 09:50, 24 Oct 2004 (UTC)

What is "W"?

The article states that

" F[s(x)] = S(f) = 0 for |f| ≥ W"

What is meant with the W? I think this should be explained in the article!

Thanks

W means bandwidth. So for example, an audio signal would have a one-sided bandwidth of 20 kHz; 20,000 cycles per second. If you took the frequency transform of an audio signal, it would be zero for everything higher frequency than 20 kHz (and for everything less than -20 kHz). Then, according to this section, you would need to measure that signal with a minimum time of 1/(2W) between each measurement to not lose any information. in this case 1/(2W) = 1/(2*20,000) = 1/40,000 = .000025 seconds. usually this is specified by just saying you have to measure it at least 40,000 times per second, or twice the bandwidth.

Is this clear enough? I will try to fit it into the article. - Omegatron 16:37, Nov 15, 2004 (UTC)

Hi Omegatron, thanks for your quick reply. It helped a lot! There only is one thing I haven't understood clearly: let's imagine a soundcard that samples audio input at 44 KHz (you see, the same with your explanation). What the soundcard does, is it takes a record of the voltage every .000025 seconds.

If you drew all those recorded values on a time-voltage diagram (time being the variable), you'd get a diagram full of dots. To reconstruct the original signal, you could draw lines between each dot and its next neighbor.

(*) That would be quite a close approximation, but you couldn't find out what happened between those .000025-second-snapshots. Maybe there was a high voltage burst (being a Dirac distribution for example) somewhere between the intervall 1.000025 s and 1.000030 s.

Maybe you understand the problem I see - but maybe I just made a mistake at some point in my thought.

Thank you for your help, --Abdull 18:57, 15 Nov 2004 (UTC)

Very correct. However, the dots with spaces between them actually DO have all the information in the original signal. As long as they are at most .000025 seconds apart. That's the whole essence of the theorem. You just have to get it back out in the correct way. I can't think of a simple explanation, but I will try. If you were to connect them as you described, with a straight line between (i think this is called a first-order hold), you would not reproduce the original signal. What you actually do in a real system is even cruder! You just make a horizontal line out from each dot like stairsteps (this is a zero-order hold, i think). The thing is, you are creating extra frequencies above 22 kHz when you do that, perfect multiples of the original frequencies. If you then filtered out those higher frequencies, you would smooth out the horizontal lines into exactly the original signal. As long as the original signal doesn't go above 22 kHz, you can reproduce it exactly with sampling at 44 kHz. It's hard to explain. (When you've figured it out, help me make the article easy for beginners to understand.) Here's some diagrams http://cnx.rice.edu/content/m10402/latest/ - Omegatron 19:23, Nov 15, 2004 (UTC)

In reply to (*): if the signal has no high frequencies (that is, if F, the Fourier transform of f is zero for any frequency w greater than W) then it is not possible to have a large spike between the samples. This is related to but not the same as the Nyquist sampling theorem. One manifestation of this truth is the Nyquist-Shannon interpolation formula.

If you don't have a very good understanding of the article as it is, please don't try to "make the article easy for beginners to understand." We'll just end up with something less good than what there is now.

Loisel 17:08, 23 Nov 2004 (UTC)

I meant "help us make the article easier to understand, by pointing out what needs to be added." It should be accessible to beginners while still being accurate. - Omegatron 20:33, Nov 23, 2004 (UTC)

I agree -- this is one of those things that I learned so long ago that I can't remember what it was like before I understood it. So, although it now seems obvious to me, it is difficult for me to explain to someone else.

Nevertheless, I'm going to try again, hoping something in what I say will inspire you to create a really good explanation. You, dear reader, whoever may be reading this.

OK, say that somewhere between the interval 1.000025 s and 1.000030 s, there's this huge spike that goes way up then comes back down again.

Every properly-designed sound card has an analog anti-aliasing filter to limit the bandwidth. If a signal is already band-limited, it goes right through that filter unchanged.

However, this spike is heavily "distorted" by that filter. If it's "small", it may be entirely clipped out by the filter. If it's "large", it will be rounded off and smeared over a much larger time period -- such a long time period that it effects at least one sample, probably more.

One could paraphrase the Nyquist-Shannon_sampling_theorem like this:

If the signal is so smooth and rounded off that it goes right through the antialiasing filter unchanged (or it's the output from such a filter), then nothing important happens between samples -- the signal simply rises or falls in a very smooth way between samples.

The mathematical details specify *how* smooth it has to be.

I like the pigeonhole principle explanation that Loisel gives, and hope it makes its way into the main article.

I hate to confuse people by bringing in further complexities, but I want to point out that this is not the only situation where a few samples can give us all the information there is to know about an infinite number of points.

For example:

• if you know some function is a straight line, you only need to know 2 points on it to know everything there is to know about that function, which contains an infinite number of points.
• If you know some shape is a perfect square, you only need to know a couple of points on each side of the square to know everything there is to know about the perimeter of the square, which contains an infinite number of points. (You don't necessarily need to be given the exact corners -- you can figure them out from other points).
• if you know that the signal is some kind of perfectly periodic triangle wave or sawtooth wave, and you know roughly what its repetition rate is, then it only takes around 6 (? or so) samples (a couple of samples on the falling line, a couple of samples on the next rising line, and a couple of samples on the next falling line) to measure everything there is to know about the the sawtooth wave -- the slope of the rising segments, the slope of the falling segments, the maximum value, the minimum value, the repetition frequency, and the "phase". (You don't necessarily need to be given the exact corners -- you can figure them out from the other points).

--DavidCary 22:58, 10 November 2005 (UTC)

I like that line. Here's my version: If the signal has already been smoothed enough by the anti-aliasing filter, then nothing important happens between samples — the signal simply rises or falls in a smooth, predictable way between samples. The mathematical details specify *how* smooth it has to be for this to happen.

I also like the geometric analogy.

• If you know that a function is a straight line, you only need to know two points and you can predict every other point along that line.
• If you know that a shape is a perfect circle, you only need to know three points to fully define it.
• If you know that a function is a sine wave, you only need to know three(?) points to fully define it.
• Similarly, it can be shown that if you know the signal is made out of sine waves that change no faster than f_s/2, then you only need to know points every T seconds.

Something like that. — Omegatron 01:23, 11 November 2005 (UTC)

Sampling theory for beginners

As a "beginner", I think the article is fairly good and accessible. However, I am left with a couple of questions that I think should be clarified in some way or another.

Suppose the signal consists of a simple sine wave, e.g. sin(2*pi*f*t) with f = 5Hz. Suppose I sample that signal at a frequency f = 11Hz. When plotted on graph paper, the sampled signal appears to be the product of two sine functions: one with the original frequency, and one with a lower frequency at 0.5Hz. The article states that the original signal is fully recoverable from the collection of sampling points. How is that possible? Is it fair to say that if we want a fairly accurate description of a signal, just by linear or quadratic interpolation between sampling points, then we need to sample at, say, 8 or 10 times the highest frequency? --Ahnielsen 09:55, 11 Mar 2005 (UTC)

Hi, that would be possible if there were a Fourier series for the Dirac comb. Since that's an impossibility, You cannot use the sampling theorem on sine or cosine functions. What You can do is to approximate those infinite waves by sinc(2*pi*0.000Hz*t)*sin(2*pi*5Hz*t), which decays at infinity. For perfect reconstruction, You will have to sum from minus to plus infinity, or at least over a very large intervall around the actual argument. In general, band limited functions are defined as $f(x)=\int _{-W/2}^{W/2}g(s)e^{isx}ds$, where g is square-integrable on the interval [-W/2,W/2]. So the above corresponds to two small boxes, each of area 1/2, around +-5Hz.

If You sample by averaging and reconstruct by step functions or linear interpolation, those oversampling factors of 8,10 or even 20 are appropriate (to get error levels below 1%). Because only local operations are involved, You can as well apply this procedure to sine functions.--LutzL 12:13, 24 May 2005 (UTC)

You can't use the sampling theorem on sinusoids? This is the first time I've heard that... - Omegatron 13:48, May 24, 2005 (UTC)

I've never heard it before, either. --DavidCary 22:58, 10 November 2005 (UTC)

Yes, that fact is right. It rises scepticism in everyone I tell this, but the minimal condition for the cardinal series to converge is that the sum $\sum _{k\neq o}|f(kT)/k|$ converges, the more common, but weaker condition is that $\sum _{k\in \mathbb {Z} }|f(kT)|^{2}<\infty$, which both are not satisfied by any sinoid. See papers and books by Butzer, Higgins and Benedetto. A weaker argument is already present in the article: if one samples a sinoid with frequency F at the Nyquist-frequency 2F, then result is an oscillating series with amplitude depending on the phase. For any sinoid with frequency equal to a fraction of the Nyquist frequency, the sampled series will be periodic so the reconstruction formula will diverge since it contains the harmonic series. The proof which Shannon gave — and which was quite common before people thought that a Dirac comb were a simple thing — involves the identity of a function on an interval with its Fourier-series in the L²-sense. A Dirac-Delta-distribution is not in L².--LutzL 15:00, 24 May 2005 (UTC)

I don't know L²-type math. Can you give a more laymen's explanation? You're saying that if I have a sinusoid at f and sample at 3f (>2f to fit the theorem but an exact multiple of f), there will be more than one possible waveform that fits those sample points using the Nyquist-Shannon interpolation formula? - Omegatron 15:36, May 24, 2005 (UTC)

In most cases, piecewise continuous is a good proxy for square integrable. At least the fourier-transform has to be a function, not a distribution. And no, there won't be any cardinal series, linked here under Shannon-Nyquist interpolation formula that fits those samples since they don't satisfy the convergence criterion mentioned above. So there would be divergence everywhere exept at the sampling points. Any partial sum of the cardinal series will of course exist, and divergence is very slow with the harmonic series, so numerical experiments will not at first glance show this behavior.--LutzL 16:10, 24 May 2005 (UTC)

Sorry, I don't understand. Don't know enough about distributions. Maybe someone else does. - Omegatron 17:43, May 24, 2005 (UTC)

Heh, well, if you ask me you're needing a formal proof to show that it, indeed, will work out correctly. The proof could use some plots to make it visually more understandable.

The frequency content of your signal can only have a sum of frequencies, but not products of frequencies. So even if it looks like there's a 0.5 Hz sine modulated on a 5 Hz wave, it's just your mind reading into it. :)

Plot a 5 Hz wave through those points and you'll see that it will match up. Keep in mind that you're nearing the nyquist frequency of 5.5 Hz. If you sample at 10 Hz then if you sample at the peaks of a sin wave then you'll get alternating spikes. So as you approach the nyquist frequency, you're samples will look more and more like the alternating spikes plot.

Re your interpolation question. The ideal interpolator is the sinc function because the freq domain equivalent is a rectangular function (a perfect low-pass filter). So anything less perfect of a low-pass filter will require extra room between your max frequency and nyquist frequency. You really would not want to sample @ 11 Hz if you have a real signal of 5 Hz because you'll be hard pressed to find a low pass filter that goes from 1 to 0 between 5 Hz and 5.5 Hz.

Perhaps when you understand this, you can change to article to answer the questions/confusions you have now...or post 'em and someone can work them in. Cburnett 10:22, 11 Mar 2005 (UTC)

Many thanks for your response, Cburnett. It was not quite what I was looking for, though. Many people have participated in the discussion on the Nyquist-Shannon Sampling Theorem. Most of these people are, I suspect, from an eletrical engineering background. I am a civil engineer, and I am not concerned with electronic signals. I am concerned with the dynamic behaviour of structures. By far the most popular method of structural analysis is finite element analysis. In this method, a solid structure is described by a (high) number of elements that are connected at nodes. The response of the structure is entirely determined by the displacement of the nodes. The displacement at any other point of the structure is usually given by linear or quadratic interpolation between nodes (this is inherent in the mathematical formulation of the elements). When a structure is subjected to dynamic excitation, stress waves propagate through the structure at finite speeds. Stress is accompanied by deformation. As a rule of thumb there should be at least 8 or 10 elements per wavelenght in order to capture the dynamic response of the structure. I was wondering, given the Nyquist-Shannon Sampling Theorem, why 3 or even 2.1 elements is not enough. But I think I am answering my own question. This is because the finite element method uses something as crude as linear interpolation to recover, as it were, the "true signal" (i.e. the deformed shape) from sampling points (nodes). Any opinion on this matter would be appreciated. --Ahnielsen 13:17, 15 Mar 2005 (UTC)

Yes, it definitely depends on the interpolation method used.

Actually, I think you can use linear interpolation, as long as you filter out the harmonics afterward. For electrical engineering you use a "stairstep" zero order hold. It creates harmonics which you filter to get the original signal. [1] The first order hold (these don't have articles???) is linear interpolation, and has a different harmonic structure, but you can treat it the same. [2]

Oops. Nope, that's wrong. The first order hold is not linear interpolation, as you can see in the image above. One of my professors told me that, once, though. Hmmph. Maybe it's somehow equivalent?

So I am not sure whether you can use linear interpolation and get the same results. I've heard of people using quadratic interpolation for quick processing of audio algorithms, but I believe they had a clause saying you should really oversample 4x or so if you want it to be accurate. - Omegatron 14:44, Mar 15, 2005 (UTC)

Here's some more info: http://cnx.rice.edu/content/m10402/latest/ http://cnx.rice.edu/content/m10788/latest/ - Omegatron 15:05, Mar 15, 2005 (UTC)

I have zero-order hold and first-order hold on my user page to be created along with Pass-band, stop-band, transition band. Surprised myself WP doesn't have them. :/ Cburnett 17:10, 15 Mar 2005 (UTC)

Some do. *creates redirects*  :-) Always check under similar names... Should we just group all the holds into nth-order hold or something? - Omegatron 17:44, Mar 15, 2005 (UTC)

Ahnielsen, it's always helpful to tell what you're really asking about to get a better response. :)

The nth order holds correspond to fitting n degree polynomial to the last n values (assuming a causal signal which is usually the case). I've not done the math but I suspect an infinite order hold would be the sinc function (perfect interpolator). Since you are using a low order hold/interpolator you'll need to have a higher sampling rate. Really, what you are doing is oversampling to account for your imperfect interpolator.

In the end, the point of the sampling theorem is to contain all the energy of the signal in the frequencies under the nyquist frequency. Oversampling gives you more samples to manipulate and less error (think law of large numbers). Cburnett 17:04, 15 Mar 2005 (UTC)

The infinite-order hold is a Gaussian function according to http://cnx.rice.edu/content/m10788/latest/ Check under the "Related info" box for more on that site. - Omegatron 17:44, Mar 15, 2005 (UTC)

The http://cnx.rice.edu/content/m10788/latest/ page doesn't make sense to me. I also expected a sinc() function, not a Gaussian function.

There seem to be 2 different kinds of "first-order hold". Both agree on drawing straight lines (perhaps with some non-zero slope) between sampling instants.

At least one Wikipedian, and also the author of [3] , place that straight line between sampling instants 3 and 4 is on a straight line extrapolating a straight line from sample 2 and 3 (because it's "no fair" using point 4 -- that would be non-causal). That line intersects sample 3, but is in general nowhere near point 4. I presume these people prefer "causal" filters involving the n preceding points. (Perhaps this is the "predictive first order hold"?)

Other wikipedians, and also the author of [4] and [5] , place that straight line between sampling instants 3 and 4 so it intersects both sample 3 and sample 4, using linear interpolation. I presume these people prefer "symmetric" filters involving roughly n/2 preceding and n/2 following points.

I wonder which type of "first order hold" that Lozano used?

Disagreement with "Important Note"

I have two complaints regarding the paragraph labelled "important note" at the articles beginning. First, while I generally agree with what is said, I think the paragraph should mention that while it is a necessary condition for sampling to be twice the bandwidth (as opposed to maximum frequency for the case of bandpass signals), it is not a sufficient condition. For example, a real signal with Fourier support over [1,10] Hz requires a 10 Hz sampling frequency (not 9 Hz!!) to prevent aliasing. As it is, the paragraph is somewhat misleading on this fact.

My second complaint is on the existence of this paragraph, even in a clarified form. Correct me if I'm wrong, but I don't believe basic Nyquist/WKS sampling ever considers such bandpass signals. It is certainly a straightforward extension from the basic theory, but I do not believe this generalization was covered in any of these 4 original papers. Since this paragraph actually deals with an extended topic which is related but not part of the article topic I am going to remove the paragraph. I suggest a new article on generalized sampling should be created and linked to. If someone does find it necessary to replace the paragraph, please at least rewrite it so it is not misleading. 8/18/2005

Hi, in some sense Your first example is wrong. That is, if You consider complex signals which can have negative frequencies. Because then the 9Hz sampling frequency is sufficient. But Your concern is right if You consider real signals with a positive frequency support of [1,10] Hz, as this is acompagnied with a negative part of the frequency support of [-10,-1] Hz. In this case, the smallest period of a nonintersecting periodization of the support is 20 Hz, and this corresponds to the lowest sufficient sampling frequency.

As to the original paper, I've put in a link to a reprint some time ago into the german version, link section. So You can find, right below the fomulation of Theorem 1 on page 2 the statement "This is a fact which is common knowledge in the communication art", and in the second column "A similar result is true if the band W does not start at zero frequency but a higher value..." and mentions the possibilty of modulation. Of course, bandwith for "sampling=taking values" should be, as mentioned above, the smallest period for which a periodization of the frequency support is disjoint (disregarding boundaries). Because by then the Fourier transform of the signal coincides on its support with its periodization. Thus one can apply the identity of a sufficiently regular, periodic function to its Fourier series, which is at the heart of the sampling theorem, as You also will find in Shannons paper.--LutzL 10:55, 19 August 2005 (UTC)

Are you sure you got those right? If a signal is real, the negative and positive frequencies will be identical, so it would seem you only have to sample one side. If a signal is complex, you'd need to sample both positive and negative frequencies, since they can be different. - Omegatron 12:49, August 19, 2005 (UTC)

But You don't sample the frequencies, You sample the signal. You can have complex signals with frequency support only in [1,10] Hz. If You sample it with 9 Hz, You get 18 real numbers (symbols) per second. If You take the real part from such a signal, it has symmetric support, but You have to sample with 20 Hz, which coincides with 20 numbers/symbols per second. The negative frequencies of a real signal have the complex conjugate amplitude of the positive ones. Are You sure You understand what sampling does?--LutzL 14:18, 19 August 2005 (UTC)

Apparently not.

• If You sample it with 9 Hz, You get 18 real numbers (symbols) per second.

  You must mean a single complex number per sample? The article assumes a real-valued function. I'm certainly only used to real signals.

• Complex signals with independent spectra from [-10,-1] and [1,10] only need to be sampled at 9 Hz, but a real signal with a mirrored spectrum has to be sampled at 20? I don't get that. — Omegatron 01:59, 11 November 2005 (UTC)

Please read again my first example. It said "complex function" and "frequency support in [1,10]Hz". That means zero on [-10,-1]Hz. Then 9Hz sampling frequency is theoretically correct. And yes, the samples will be complex numbers, that is 9 complex numbers=18 real numbers per second (QAM has a related sample counting system). In reality signals are real, so one cannot have a spectrum restricted to [1,10]Hz. If there is a nonzero part of the spectrum in [1,10]Hz, there has to be a mirrored and complex conjugate part in [-10,-1]Hz. Thus they are not independend. Following the note in the undersampling section, there is no chance to choose a sampling frequency below 2*10Hz=20Hz (contrary to other situations, fct. with spectrum in [-10,-9]Hz and [9,10]Hz] can be sampled with 2*1Hz). The same holds for complex signals with independend spectra in [-10,-1] and [1,10]. They also have to be sampled at 20Hz, only that one gets 20 complex numbers=40 real numbers per second. This all under the assumption that by sampling one means taking values of the signal. With frequency multiplex methods it is possible to sample real functions with spectrum restricted to [-10,-1]Hz and [1,10]Hz with one polysample per second, the polysample consisting of 18 real numbers. Those techniques compute scalar products of the signal with model functions, technically realized by oversampling (A/D) followed by convolution="digital" filtering followed by downsampling.--LutzL 09:42, 11 November 2005 (UTC)

• It said "complex function" and "frequency support in [1,10]Hz". That means zero on [-10,-1]Hz.

  Oh. Oops.

• with spectrum in [-10,-9]Hz and [9,10]Hz] can be sampled with 2*1Hz

  That doesn't make sense. If [-10,-1] and [1,10] needs 20 Hz, then so would [-10,-9] and [9,10]. What's the difference between these examples?

• If you know that the signal is real, then you know that the spectrum is mirrored and complex conjugate, so it seems that the extra samples are redundant. The real signal from [-10,-1] and [1,10] has the same number of unique frequency components as the complex signal from [1,10]. Given only the positive frequency components, you can calculate all the negative ones. Maybe this isn't sampling theory, though.
• Speaking of which, which of these situations are covered in the original sampling theory? — Omegatron 15:21, 11 November 2005 (UTC)

• For the difference in the examples: See the undersampling paragraph. In the first, N=0 and only the Nyquist frequency and anything above is a sampling rate. For [9,10]Hz, N=9 and the lowest possible sampling frequency is 1Hz. This is covered in the "proof" section, although this is a real mess. And as I tried to clarify earlier, You don't have the spectrum, there is only a signal that can be sampled by measuring it. Only after that one can compute an approximate spectrum via FFT. But we are concerned with the sampling phase. To throw away the negative part means to perform a Hilbert transform, which is not very well behaved numerically. And You need the already sampled function.

• "Original sampling theory": I don't know what You understand by this term, but Shannon in the 1949 paper knew perfectly well that a general signal subspace, one that can serve as the model of a communication channel, needs only to be a function subspace that has an orthonormal basis generated by a finite number of functions per unit intervall, that is a basis of the type $g_{k}(t-nT)$, T the time unit, k=1,...,M the different types of basis functions (QAM has M=2, DVB-T has M=4096) and n varying over all integers, orthonormal meaning

$\langle g_{k}(t-nT),\,g_{l}(t-mT)\rangle =\delta _{k,l}\delta _{m,n}$, (using twice the Kronecker symbol).

Look it up, he explains the geometry of signal transmission directly after the citation, the link for the citation is a reprint of his 1949 paper. In the most simple cases, as in the basis band case, the only occuring basis function, here sinc with T=1 for its normalized version, is not only orthonormal to its nT-shifts, but also interpolating, so that the scalar products for a function in the signal subspace are actually values of this function. In other cases, sampling means orthogonal projection onto the signal space by means of computing the scalar products. However, in engineering books this is represented as analog filtering followed by "point-sampling". If You are interested in this topic, You should also read the paper of M. Unser (the guy that runs www.wavelet.org): Sampling:50 years after Shannon.--LutzL 16:03, 11 November 2005 (UTC)

Shannons original theorem

Cited from: Claude Elwood Shannon: Communication in the Presence of Noise 1949, probably circulated since 1941

Theorem 1: If a function f(t) contains no frequencies higher than W cps, it is completely determined by giving its ordinates at a series of points spaced 1/2W seconds apart.

This is a fact which is common knowledge in the communication art. The intuitive justification is that, if f(t) contains no frequencies higher than W, it cannot change to a substantially new value in a time less than one-half cycle of the highest frequency, that is, 1/2W. A mathematical proof showing that this is not only approximately, but exactly, true can be given as follows.

Let F(ω) be the spectrum of f(t). Then ${\begin{matrix}f(t)&=&{\frac {1}{2\pi }}\int _{-\infty }^{+\infty }F(\omega )e^{i\omega t}\,d\omega &\qquad &(2)\\&=&{\frac {1}{2\pi }}\int _{-2\pi W}^{+2\pi W}F(\omega )e^{i\omega t}\,d\omega &\qquad &(3)\end{matrix}}$ since F(ω) is assumed zero outside the band W. If we let $t={\frac {n}{2W}}\qquad (4)$ where n is any positive or negative integer, we obtain $f\left({\frac {n}{2W}}\right)={\frac {1}{2\pi }}\int _{-\infty }^{+\infty }F(\omega )e^{i\omega {\frac {n}{2W}}}\,d\omega .\qquad (5)$ On the left are the values of f(t) at the sampling points. The integral on the right will be recognized as essentially the nth coefficient in a Fourier-series expansion of the function F(ω), taking the interval -W to +W as a fundamental period. This means that the values of the samples f(n/2W) determine the Fourier coefficients in the series expansion of F(ω). Thus they determine F(ω), since F(ω) is zero for frequencies greater than W, and for lower frequencies F(ω) is determined if its Fourier coefficients are determined. But F(ω) determines the original function f(t) completely, since a function is determined if its spectrum is known. Therefore the original samples determine the function f(t) completely. There is one and only one function whose spectrum is limited to a band W, and which passes through given values at sampling points separated 1/2W seconds apart. The function can be simply reconstructed from the samples by using a pulse of the type ${\frac {\sin 2\pi Wt}{2\pi Wt}}.\qquad (6)$ This function is unity at t=0 and zero at t=n/2W, i.e., at all other sample points. Furthermore, its spectrum is constant in the band W and zero outside. At each sample point a pulse of this type is placed whose amplitude is adjusted to equal that of the sample. The sum of these pulses is the required function, since it satisfies the conditions on the spectrum and passes through the sampled values.

Mathematically, this process can be described as follows. Let xn be the nth sample. Then the function f(t) is represented by $f(t)=\sum _{n=-\infty }^{\infty }x_{n}{\frac {\sin \pi (2Wt-n)}{\pi (2Wt-n)}}.\qquad (7)$

A similar result is true if the band W does not start at zero frequency but at some higher value, and can be proved by a linear translation (corresponding physically to single-sideband modulation) of the zero-frequency case. In this case the elementary pulse is obtained from sinx/x by single-side-band modulation.

Remarks: "Band (of frequency) W" means support of the Fourier-transform F in [-2πW,2πW]. "cps" is "cycles per second".

Compare this to a paper with the same formula, but 30 years earlier (cited by Shannon): Edmund Taylor Whittaker: On the functions which are represented by the expansion of interpolation theory (1915)

Probably this could go into the article if it, despite being rather long, could count as scientific citation.--LutzL 08:59, 25 August 2005 (UTC)--minor corrections--LutzL 09:22, 26 September 2005 (UTC)

Frequency 'Support'

What the he** is a frequency support when its at home? Never heard of this one. can someone explain??--Light current 03:06, 2 October 2005 (UTC)

Don't look too hard... Support (mathematics). Cburnett 18:04, 17 October 2005 (UTC)

In fact, it is a little more interesting. A frequency support [A,B] means, that the fourier transform of the signal has its support (mathematics) in the intervall $[2\pi A,2\pi B]$. If an engineer speaks about it, she most likely is speaking about a real valued function with a fourier-transform that is zero outside the union of the intervalls $[-2\pi B,-2\pi A]\cup [2\pi A,2\pi B]$. Now try to find out what "bandwidth" means in each of these cases.--LutzL 13:40, 18 October 2005 (UTC)

Yes, it is 2*pi*B, fourier-transform is wrt. angular frequency.

Depends which transform you're using. f implies hertz; ω implies radians per second. This article uses f where you say it means radians per second?

Can we stick to only one of f or ω? Also can we stick to one of B or W? They are both used for the same thing here, right? I'm a little confused now. User:Omegatron/sig 20:05, 19 October 2005 (UTC)

Hi, I did believe that the consens on Wikipedia was to take the angular frequency fourier transform. The majority of the articles here and textbooks uses it. I too would prefer a transform wrt. normal frequency in Hz, because it is somewhat more "natural" and needs no constant factors. But by then all articles using the fourier transform, or at least all articles related to signal processing should be consistent in this.--LutzL 08:56, 20 October 2005 (UTC)

Well, we're allowed to use j instead of i for the imaginary number in electrical articles, so I don't know why we wouldn't be allowed to use f instead of ω in predominantly signal processing articles. I like consistency, too. But using the equations most commonly used by people who actually use them is good, too. As long as everything is tied together and explained I don't see why that would be a problem. Where was this consensus reached? User:Omegatron/sig 14:33, 20 October 2005 (UTC)

Well, by changing i to j, only the symbol has changed. By changing ω to f, not only the symbol, but the underlying function changes. This is a difference. By the way, from a mathematical point of view, both f and ω are simply symbols for variables, without further qualification. Sometimes f is even a function. I found this Fortran-like allocation of names always a little confusing. Also, one quickly runs out of "virgin" symbols.--LutzL 15:52, 20 October 2005 (UTC)

So we're using Ω now? I've never seen that used for frequency before except by mistake. User:Omegatron/sig 15:15, 24 October 2005 (UTC)

Hi, please change it back to anything that seems appropriate. I only changed it because some IP had it changed to H_s, which makes no sense to me. Ω is used in many recent papers in harmonic analysis dealing with sampling and interpolation formulas and generalisations of the Poisson summation formula. The space of bandlimited functions is then called a Paley-Wiener space $PW_{\Omega }$. One could suspect that Shannon used the capital letter W only because it was at the time complicated to realize greek letters with a typewriter and the small letters look identical. However, I never saw a manuskript of this time.--LutzL 20:38, 27 October 2005 (UTC)

The above quoted passage by Shannon says "frequencies higher than W cps", though. cps is the same as Hz. User:Omegatron/sig 05:53, 28 October 2005 (UTC)

Yes, and then he goes on to use the Fourier-integral wrt. the angular frequency, adjusting the support of the Fourier-transform to $[-2\pi W,2\pi W]$. But You are right in some way, since the index in PW is usually taken as angular frequency, so sampling frequency 1 corresponds in those papers to $PW_{\pi }$. Have I already said that for me it makes no difference, as long as transforms and sybols are used consistently? --LutzL 12:57, 28 October 2005 (UTC)

Yes, but it makes a difference to our readers. User:Omegatron/sig 13:57, 28 October 2005 (UTC)

Hi, I think I found a simple way for that. I also think $[f_{L},f_{H}]$ is easier to understand than $[\mathrm {A} ,\Omega ]$, which is an old joke (or pun?) dating back to biblical times meaning begin and end (of the greek alphabet).--LutzL 09:45, 11 November 2005 (UTC)

Definitely agree about $[f_{L},f_{H}]$, though remember that f implies cyclic frequency to most people; not angular. — Omegatron 14:37, 11 November 2005 (UTC)

Notation

What's with the := notation? I've only seen that in computer algebra systems. — Omegatron 19:33, 8 December 2005 (UTC)

Hi, that is also frequently used in math texts, meaning "left side is defined as right side". Obviously, in math texts one should avoid constructions as "n:=n+1" (of course, except in code examples), as math texts are considered as static.--LutzL 07:53, 9 December 2005 (UTC)

Edits by User metacomet

This article is listed as a mathematical theorem. So I hope one agrees that it should be mathematically correct and using consistent mathematical notations. To the points metacomet is unwilling to accept:

• The easiest to understand should be the variants of the Fourier transform. The articel this links to defines it using the usual mathematical convention as ${\hat {s}}(w)={\mathcal {F}}(s)(w):={\frac {1}{\sqrt {2\pi }}}\int _{\mathbb {R} }e^{iwt}s(t)\,dt$. So a "frequency component" of 1 Hz would show up at $w=2\pi /sec$. So there is a need to explain that the Fourier-transform as used in signal analysis is $S(f)={\mathcal {F_{N}}}(s)(f):=\int _{\mathbb {R} }e^{i(2\pi f)t}s(t)\,dt$.

• The more controversial point is function notation. In mathematics a function is introduced as "$f:D\to V$ with f(x)=...". f(x) denotes the value at the point x, nothing else. Sometimes in school mathematics or engineering the function notation is abused. To write f(x) for clarity is perhaps tolerable, but to write $/mathcalF\{s(t)\}$ translates into "the Fourier transform of the constant s(t)", which does not exist (as function). The Fourier transform of the function s at frequency f has the notation used above.

--LutzL 18:47, 21 December 2005 (UTC)

By the way, using the notation ":=" is completely non-standard. The only place I have ever seen ":=" is as the assignment operator in the Pascal programming language. In that context, the notation ":=" means "is set equal to". You seem to be using ":=" to mean "is defined as". In my experience, it is acceptable to use the plain old "=" sign, or if you really want to be careful, then to use the "equivalent to" symbol, as in $A\equiv B$.— Preceding unsigned comment added by metacomet (talk • contribs)

Do your think it's possible that the reason the preceding comment was left unsigned is maybe just maybe that I forgot to sign it? So maybe instead of trying to make me look like a fool, you could just once cut me a break.... -- Metacomet 18:54, 21 December 2005 (UTC)

This is why I really think a translation page is necessary. The equiv sign is, for mathematically oriented people, only used in connection with remainder classes or to signigy that a function is identically constant to a given value.--LutzL 18:46, 21 December 2005 (UTC)

Response from Metacomet

This article is about signal processing and telecommunications theory, not mathematics. Furthermore, the disagreement that we are having is about notation, not about meaning. The notation that you are using is extremely confusing and difficult to understand. I am not interested in mathematical orthodoxy, I am interested in improving this article so that it is accessible by people other than just mathematicians. As I am sure you are aware, the purpose of Wikipedia is to provide information for a general audience – including people other than advanced theoretical mathematicians.

I don't know where you have gotton the strange notion that s(t) is a constant, but just plain s is a function. That makes no sense whatsoever. When I see just plain s, it indicates to me that you are talking about a complex number s which is the argument of a complex valued function. On the other hand, when I see s(t), it is absolutely clear to me, with no ambiguity whatsoever, that we are talking about a function s with an argument t. And in fact, by a very common convention, the argument t, especially in the context of this particular article, is very often meant to represent a real-valued, continuous domain that engineers and scientists like to call time.

So please don't lecture me about what is correct notation and what is incorrect notation. There is no such thing as correct or incorrect notation. Notation is to a large degree a subjective matter of taste and an objective matter of effectiveness. The test is whether a given notation is easy for the reader to understand clearly and unambiguously, not whether it meets the author's view of mathematical orthodoxy.

One more thing: throughout this entire article, the frequency domain is discussed using the symbol f in units of hertz. Nowhere does the article refer to angular frequency ω in radians per second. Yet all of a sudden, you want to introduce the Fourier transform in terms of angular frequency ω instead of frequency f because you think that angular frequency is more correct than plain old frequency. I have news for you: go out in the real world and talk to people who design real signal processing systems. They deal in the world of hertz, not radians per second. Oh, and they have no trouble converting from one to the other when they need to. It isn't all that difficult to multiply or divide by 2π. On the other hand, why not simply use the frequency form of the FT instead of the angular frequency form? It is just as real, and just as valid. And in many ways, it is far easier and more intuitive than the other.

-- Metacomet 16:31, 21 December 2005 (UTC)

Agree that the article needs more from a signal processing perspective. Mathematical rigor can be a good thing, but the article should be accessible to the people who actually use the theorem, and should use their conventions as well, at least at the beginning. The more concise definitions can be later in the article, perhaps.

The article is still confusing and cluttered.

Wikipedia:Make technical articles accessible applies to mathematics, as well. — Omegatron 16:57, 21 December 2005 (UTC)

Well, I hope You could just cool down a bit. Although sampling is something that is more frequently done in signal processing then in mathematics, the sampling theorem is a purely mathematical theorem telling something about some very ideal situation involving sharply bandlimited functions that are nowhere to be found in practice. That's why it has a mathematical category in the bottom. In practical sampling You can forget about the factor-2-rule of this theorem, since it is a theoretical, ideal limit for situations where You can wait for an indefinite amount of time and perform an indefinite number of operations. I even very much doubt that factors <3 are meaningful if near perfect reconstruction is demanded.

For the Fourier transform: please check where the link leads to and how the transform is defined there. Your assumptions and foreknowledge don't matter, since this is intended to be also read by people who don't have them. If You want consistency then make an article ((Fourier transforms in signal processing)) where this other definition with its slightly different properties is to be found and link to this one.

I always thought that even in FORTRAN they marked predefined meanings for variable names beginning with certain letters as old style. So bad news for You: in any scientific or technical article the symbol s can stand for anything, an integer, a real or complex number, a matrix or a sequence or even a function, just anything. That's why well written articles explain at least informally what every symbol stands for. But Your attitude explains why so few students in computer science, even with signal processing as specialty, don't even properly understand the Fourier transform or even polynomials and convolution. They just aren't able to read common math textbooks. So I expect that "${\mathcal {F}}:L^{2}(\mathbb {R} ,{\mathbb {C} })\to L^{2}(\mathbb {R} ,{\mathbb {C} })$ is a unitary linear function" is an incomprehensible statement for You, even though it is one of the fundamental properties that make the Fourier transorm useful in signal processing.

--LutzL 17:28, 21 December 2005 (UTC)

You are unable argue your case on the merits, so you revert to attacking me personally. Instead of presenting a solid case in favor of you ideas, or even showing the flaws in my arguments, you resort to demonstrating your superior intellect by putting down my intellect. That is not a very effective approach to convincing anyone that your ideas have any merit whatsoever. -- Metacomet 17:34, 21 December 2005 (UTC)

Please skim through the discussions above this one, I don't want to repeat myself on the same page. I told You in the post directly above this section that notation as s(t) is misleading, espacially if there is no explanation what s and t are, and that there are two different (by constants) versions of the Fourier transform in common use, so one has to state clearly which one one uses. I would agree that such distinctions are confusing in the introduction, but then one should not use the FT there at all and restrict the text to some diffuse but perhaps "intuitive" phrases like "frequency component". It was no insult to You but a general observation that very few people in signal processing know what a L^p-function space is. However, without this theory one can't understand in which sense the FT is invertible, which is a very important property in signal analysis. Please read everything above, omegatron explicitely stated that he doesn't know about those spaces.--LutzL 17:52, 21 December 2005 (UTC)

First, when I have some free time, I will read through all of the discussion above, as you suggested.

Second, I am well aware that there is more than one form of the Fourier transform. In fact, there are at least four different forms in common usage. I personally think that there is way too much time and energy expended arguing endlessly and mindlessly about which form is better, and which form we should use. Frankly, who cares? Everyone makes such a big f---ing deal about what really just amounts to simple scaling factors. Is it really all that difficult to undersand the difference between frequency in hertz and angular frequency in radians per second (and sampling frequency in radians per sample for that matter)? I have a confession to make: I use all three of these conventions all the time, and I readily switch back and forth amongst them without getting confused (most of the time). Wow. Again, who cares?

Third, I do know what L^p function space is, although I only just recently learned about it in a formal setting, and I must admit that I do not understand it at any great level of depth, nor do I really see why it is all that important or relevant. But that is beside the point. I would guess that most electrical engineers and signal processing practitioners do not have a deep background in function spaces. So if you start throwing around a lot of obscure notation without clearly defining what you mean (in English, not in math jargon), then you are absolutely going to confuse people. Worse yet, you will actually lose them right from the start, because they will immediately stop reading the article.

Fourth, I agree with Omegatron that this article is confusing and cluttered. I would like to make it less confusing and less cluttered. If you want to help achieve this objective, I would welcome the help. But please, let's not bury the general reader with a bunch of esoteric notation and confusing jargon in the very first three paragraphs!

-- Metacomet 18:08, 21 December 2005 (UTC)

I nowhere claimed that the "mathematical" version of the FT should be used. Quite to the contrary, even before You began editing the article, I've put in the conversion to the "real frequency" transform in the first occurence of the FT, again please read above for earlier discussions on this topic. But, I repeat myself, if You put direcly below the link to the "mathematical" FT without any further notice a formula using the "real frequency" FT, I call that confusing. Not that the state as it is now is better readable. As I said, leave out the FT off the introduction and use such equally obscure but better to imagine phrases as "frequency component".

If You want to note functions by s(t), fine, make a note at the top telling "In this article, mathematical notation as common in communication technology is used except in places marked "mathematical formulation"" and refer to a translation page, so that also a mathematician can understand what people in signal analysis are talking about. There it should also explained how You want to refer to a function value, say at 10. Is it s(10), which is confusing, since the letter t is missing from the function symbol, or is it s(t=10) or ...?--LutzL 18:34, 21 December 2005 (UTC)

This is a waste of time. I am done. Have a nice day. -- Metacomet 18:42, 21 December 2005 (UTC)

It seems to me that you (LutzL) do not really understand the purpose of Wikipedia. But I am really tired of arguing about it with you and several others. I have had enough. If you want to make Wikipedia into a Mathematics textbook, then go for it. I am not going to stop you. I will find a better way to use my time and energy. See 'ya. -- Metacomet 19:07, 21 December 2005 (UTC)

Sorry, I had the same impression about You. You seem to regard anything published here as Your private notepad, so that notations and necessary conversions don't need to be explained. If I may tell Ya, there are people in the outside world that don't have an education in what signal processing people believe to be mathematics (As there are people that don't know math, but they wouldn't be much interested in the sampling thm). So they have perhaps problems understanding ordinary mathematics notations, and then they are confronted with a seemingly different kind of math. Goodbye, farewell and a happy new year.--LutzL 19:21, 21 December 2005 (UTC)

Think about it this way. There are a few different types of people who will be reading this article:

1. Laymen who don't know anything about anything remotely related to sampling
2. Laymen who know a little about signal processing, but not much; probably audio-related
3. Engineers
4. Mathematicians

LutzL, what percentage of each do you think exist in our readership? — Omegatron 20:17, 21 December 2005 (UTC)

The sampling theorem for beginners

I have added a new section which contains almost no math but which hopefully describes the core of the sampling theorem in a sufficiently simple manner to be useful to someone not familiar with the concept. My suggestion is that this, or something like it, can be put somewhere in the beginning and that its content can be developed in more mathematical details later on in the article --KYN 01:20, 2 January 2006 (UTC)

Thank you for writing this section. I think you did an excellent job in explaining the concepts at an introductory level without oversimplifying things. It is also well written and presents the information in a logical manner. I hope you don't mind the edits that I made, but different people can bring additional ideas to the party, which often improves things. Again, well done. -- Metacomet 21:54, 2 January 2006 (UTC)

BTW, I agree with your assertion that a similar section at or near the beginning of many math and technical WP articles would make a major improvement in terms of accessbility and clarity. -- Metacomet 21:56, 2 January 2006 (UTC)

Historical background

I moved this section to the end of the article since it didn't make sense to me to have it between two technical sections. --KYN 21:31, 2 January 2006 (UTC)

Strict or non-strict inequality

There are now two versions of the condition for the Nyquist rate:

• It should be strictly larger than the largest frequency component of the sigal (intro)

• It should be larger or equal to the largest frequency component of the signal (section "Formal statement of the theorem")

My understanding of the theorem is that the second condition is not correct since it implies that the largest frequency component of the signal will in fact be aliased (as described in the "Critical frequency" section).

Is suggest that only the first version of the condition is used consistently in the article, and also that the "Critical frequency" section is removed since it only describes a special case of aliasing which is not really more interesting than any other. Also, "critical frequency" is here defined as synonomous with the Nyquist rate. Maybe move that to where the Nyquist rate is defined in the intro? --KYN 23:02, 2 January 2006 (UTC)

I agree with you on the issue of strict inequality. In practice, it really doesn't matter that much, because you will always tend to oversample at least a little bit to provide some margin of safety. Nevertheless, the article should state the theorem as a strict inequality, which is theoretically correct and avoids ambiguity. -- Metacomet 23:16, 2 January 2006 (UTC)

The section entitled "Critical frequency" is not all that significant to the article, but I am inclined not to remove it because I think it makes an interesting point, and helps justify why the sampling condition is a strict inequality. It may make sense to incorporate it within another, existing section. I am not sure it merits its own section. -- Metacomet 23:20, 2 January 2006 (UTC)