|WikiProject Electronics||(Rated Start-class, Low-importance)|
Perhaps this article could be clarified... I'm confused as to which frequency is the actual "Nyquist frequency" as the article seems to somewhat contradict itself.
(a) The minimum frequency a signal should be sampled at to record it accurately.
(b) The maximum frequency a sampled signal can reproduce accurately.
i.e. in the CD example, a 44khz sample-rate can record frequencies up to 22khz. Is 22 or 44 the nyquist frequency?
— SimonEast 22:20, 22 Mar 2005 (UTC)
- The Nyquist frequency would be 22 kHz for CDs. Half the sampling rate. This would require a perfect interpolation filter if the signal has frequencies through 22 kHz. Since filters cannot be perfect then the signal really should be limited to less than 22 kHz. Cburnett 23:58, 22 Mar 2005 (UTC)
- This is actually incorrect. For a signal to be reconstructed perfectly, it has to be bandlimited. This means that if the frequency range of a signal is 0 to W, the bandwidth is W. To be considered "critically sampled," the signal must be sampled at more than twice the bandwidth, and this (2W) is known as the Nyquist frequency. Thus, for audio applications like CDs, (the standard audio upper limit is 20KHz), the Nyquist frequency would be 40KHz. The 44.1KHz gives the audio a slightly better range (around 22KHz, making the Nyquist frequency 44KHz, but we must sample above it, hence 44.1KHz). Basseq 01:00, 13 October 2005 (UTC)
- You've missed the whole point - which is an important point and very rarely made in discussions of the Nyquist theorem. The point is that if your Fs is 44.1kHz, you can't get to 22.05kHz and you can't even come very close to that, since the reconstruction process is impractical. So having the margin there is not to give you better range above 20k; it's to make the 20k bandwith feasible. Regarding the definition of 'Nyquist frequency', I've always understand that (or the 'Nyquist limit') to be 1/2 the sample rate, the only textbook I have at hand [Bowen&Brown] defines the 'Nyquist rate' to be 2 times the bandwidth. Both of these are consistent with wikipedia's current definitions of the two terms. I think this is one of those things (like the sign in the DFT exponent) which varies by author a little, but tends to be clear from the context. What is amazing is that most discussions of this issue clearly state that the sample rate must be greater than or equal to twice the highest frequency [including Bowen and Brown] when it is entirely obvious that a signal at exactly 1/2 the sample rate cannot be recovered; and in fact, as stated in this article, you need to have some margin, since in the real world there's much not difference between < and <=. (unsigned comment 16 August 2006 by 126.96.36.199)
- The definitions of Nyquist frequency and Nyquist rate are actually fairly consistent. You do see mistakes here and there, but most authors agree on what they are, and the wikipedia has it correct. One is a property of the system (Nyquist frequency, half the sample rate) and the other is a property of the signal (Nyquist rate, twice the bandwidth). Dicklyon 01:14, 17 August 2006 (UTC)
- Is the current wording less ambiguous? Cburnett 00:08, 23 Mar 2005 (UTC)
By the way, in the article the example for a transition band is 2000 Hz, which is rather large for today's lowpass filters. For example, the one used in the LAME MP3 encoder is pretty high-end, in that you can have a transition band of 400 Hz and still end up with a perfect cut-off lowpass 'slope'. In the article's example, 21600 Hz would then still be sampled 100% if you would resample from 48000 to 44100 Hz samplerate while encoding an MP3. 188.8.131.52 22:29, 17 May 2005 (UTC)
- You say "with a perfect cut-off", but that is impossible to achieve. For perfect, the transition band would have to be zero, which is (again) impossible to achieve. 2 kHz is an example to illustrate the point. In the end, the exact number is irrelevant to the context. Cburnett 01:39, May 23, 2005 (UTC)
- You don't need 'perfect cutoff', but you need it to be flat up to the transition band's low end, and zero above fs/2. So it's not a brickwall filter, but there's no aliasing except for stopband leakage. Yes, there are many audio processing systems which can significantly exceed 90% of the theoretical limit, since audio-rate processing is getting pretty cheap these days. Bear in mind that CDs were introduced in 1982 (according to wikipedia, anyway), and the design margin in the original example (90.7% of limit) is quite reasonable. (unsigned comment 16 August 2006 by 184.108.40.206)
- For reconstruction, the filter doesn't have to go to zero by fs/2, just by fs–B, which is a little higher; you get twice as much transition band as you thought. This would have been apparent in the mathematical proof of the Nyquist–Shannon sampling theorem before about 10 PM late night when it got reverted to Rbj's version. Dicklyon 01:14, 17 August 2006 (UTC)
- Yes, assuming (as is generally the assumption) that there's nothing at all in the signal you are trying to reconstruct above freq 'B'. In practical terms, that upper band may be empty, or may contain untrustworthy content, or maybe valid content that you don't mind attenuating somewhat, but you don't want images of it. There's another point here: If you are converting audio to a higher sample rate with a really large interpolator, you can, in principle, protect everything in the input up to, say, 0.495 fs, separating it carefully from the images at 0.505 fs and up, but you would be moving that marginal edge content (just below 0.495 fs) from a near-nyquist zone into a well-in-band area of your upsampled signal. This is not a good idea unless you know that any content present at that point is actually useful; it's possible that the input sampled signal has been generated by a system which has some aliasing there, so the signal you are protecting at, say 0.492 might really be an unrecoverable alias of a signal at 0.508. So the point is (a) it's a lot of extra work to get up there, and (b) it might be detrimental, when you look at the system as a whole. Conversely, when downsampling, you could filter everything above 0.505 of the output fs, while passing everything to 0.495 of the output fs, and this is fine, except when your downstream system doesn't handle the 0.495 signal well (i.e. a DAC might create an image at 0.505 Fs).In multi-rate systems, these issues need to be dealt with across an entire system, so that any imaging or aliasing generated in the near-nyquist zone of one block are rejected by the next block. This is far more practical than trying to make every block perfectly clean.
- A picture on this page would be worth a thousand words in terms of explaining the basic idea - just a sawtooth 1 -1 1 -1 function. Paul Matthews 11:07, 3 October 2006 (UTC)
- The reference to time in the article kind of limits the scope of Nyquist frequency to temporal signals. There are other signals for which the term is equally relevant, for example, digital photography, in which the frequency is spatial rather than temporal. Victor Engel (talk) 04:23, 30 January 2008 (UTC)
Merge with nyquist Rate?
In the nyquist article it's mentioned that the transition band for the antialius filter is 2000hz. This should not be stated without first defining the order (more specificaly the poles and zero) of the filter since the fiter characteristics define the transition band. It's very confusing without this information, at first glance I thought the band pass was 2000hz. --03:01, 9 October 2012 (UTC)Wpkelley (talk)
Given that the Nyquist rate and Nyquist frequency are such closely interrelated terms, I think that this article could be covered in a single section in the nyquist rate article with nothing lost. HatlessAtless (talk) 21:39, 10 June 2008 (UTC)
- I took that malformed tag out, since it was a redlink. And I didn't see a corresponding proposal at Nyquist rate. Maybe your caps key is broken. Dicklyon (talk) 23:03, 10 June 2008 (UTC)
Nyquist rate and epsilon
I was about to revert this edit but spent so long thinking about it that someone else beat me to it by the time I got round to pressing the button. So even though I didn't revert it (honest guv) here are my reasons for doing so. Firstly the edit is introducing a subtle discussion into the lede concerning Nyquist rate, but this article is about Nyquist frequency, it should not be in the lede, possibly it should not be in this article. It also introduced a parameter, epsilon, that is not defined. One cannot say that the Nyquist rate is defined as twice the max frequency and then go on to say it is something different. Possibly the system will not work at exactly the Nyquist rate (for reasons of imperfect low-pass filters? which is discussed in the body of the article) but that is a different matter. I was also unhappy that the quoted reference (presumably this book) does not get any hits for "Nyquist" according to a google search. SpinningSpark 00:13, 17 October 2009 (UTC)
- I reverted it; I think he doesn't understand the idea of a bound; I hadn't even noticed that the topic of the change wasn't the topic of this article. Dicklyon (talk) 02:02, 17 October 2009 (UTC)
This page seems to be very centred around those with an engineering background - there is a plethora of undefined terms such as 'foldback' or 'bandwidth'. Since I came to this page to hear some form of discussion of the optimal sampling rate for a DFT, this wasn't a great experience. Is there really no stronger bound on the sampling rate than 'at least twice the frequency of the signal'? What if I don't know the signal frequency a priori? 220.127.116.11 (talk) 15:44, 3 November 2009 (UTC)
- The guideline "at least twice the frequency of the signal" seems to be already plenty of guidance for you; if you don't even know the frequencies in your signals, this tells you that you need to find out. The article is unavoidably technical, as it's a technical topic. If the terms like bandwidth are not linked to appropriate articles for more info, that can be fixed. I added a small explanation of folding. Dicklyon (talk) 18:09, 5 November 2009 (UTC)
- I disagree that 'technical topics' are necessarily unclear to the layperson; a good writing style should make the broad structure of the meaning clear - and since my problem is that I'm exploring the state-space of an unstable linear magnetohydrodynamic problem; if I can /find/ the (complex) frequencies I've finished my task. Thanks for the added info. 18.104.22.168 (talk) 11:21, 9 November 2009 (UTC)
Nyquist frequency as cut-off frequency?
While the Nyquist frequency is generally the cut-off frequency in designs due to its importance, "cut-off frequency" is certainly not a synonym of Nyquist frequency. If nobody has an issue with this I will remove it. Oldmaneinstein (talk) 21:50, 16 April 2010 (UTC)
- Completely agree, that is either a misunderstanding or taken from an eccentric source. SpinningSpark 23:13, 16 April 2010 (UTC)
Proving the DFT is periodic is not helpful
The recently added proof of the DFT's periodicity won't help the typical person who comes here to learn what Nyquist frequency means. He might not even know what a Fourier transform is. Notice that the figure depicting aliases and folding is not even a spectrum of frequencies. It shows just one sinusoid at 0.6 fs and three other frequencies that would produce the same samples. Then it shows (by the red lines) an example of what would happen if that sinusoid was altered in frequency and amplitude between 0.5 fs and 1.0 fs.
Also, the DFT is a sampling of the DTFT (see Sampling_the_DTFT), whose periodicity is obvious by inspection of:
Nyquist rate vs. Nyquist frequency; apparent contradiction
I spotted the following, apparent contradiction at the beginning of the article: first it is stated:
"The Nyquist frequency, ... is ½ of the sampling rate of a *discrete signal* processing system"
Then in the discussion on the distinction between Nyquist rate and frequency in the next paragraph, it is stated that the Nyquist frequency is a property of a continuous-time system, and the Nyquist rate a property of a discrete-time system.
Since I am no engineer/mathematician, I can't judge whether this is actually a contradiction or whether I'm just missing something; but it is confusing. — Preceding unsigned comment added by 22.214.171.124 (talk) 11:58, 16 May 2013 (UTC)