From Wikipedia, the free encyclopedia
Jump to: navigation, search


I'm confused, I think the original is correct, with 16 (2^(24-20)) extra samples, there is no averaging, just a simple sum of the 16 samples. This keeps the answer as an integer, whereas an average would need a real number. 20:02, 15 February 2007 (UTC)

The article contained a fundamental error in the example of increasing the resolution from 20 bits to 24 bits by averaging 16 samples: when you average n samples, the signal/noise ratio increases only as the square root of n, not proportionally to n. You need to average 256 samples to obtain extra 4 bits of resolution. - I corrected it. Leocat 14 Oct 2006


I am redirecting sigma-delta modulation and delta-sigma modulation to here. I hope this is correct enough. - Omegatron 19:12, Apr 15, 2005 (UTC)

I don't think so. There's more to sigma-delta than oversampling! -- 18:28, 15 September 2005 (UTC)
I had actually landed here looking for a definition of Sigma-Delta modulation, making this page, as it stands, actually of no value to me. What it needs (preferably from someone who understands why it was redirected so) is a short paragraph explaining the redirection.

Any takers?


"After being sampled at 800 Hz, the signal could be digitally filtered to have a bandwidth of 100 Hz and then further downsampled to closer to 200 Hz."

Sigma-delta modulation[edit]

This is bizarre. I could swear I came here looking for sigma-delta modulation and complained that it isn't mentioned in the article at all. Then I left a message on the talk page about it. I must be thinking of some other article? Anyway both of them now redirect to delta-sigma modulation, as they should. — Omegatron 21:41, 13 October 2005 (UTC)

Heh. Posted on the wrong talk page. — Omegatron 19:50, 19 October 2005 (UTC)


Given that "oversampling" is a term that crops up in non-technical contexts quite often, (eg. advertisements for hi-fi gear, magazines) it would be appreciated if some kind soul could offer a dumbed down lay explanation that can be comprhended by non- or semi-technical people.. Editdroid 17:15, 17 February 2006 (UTC)

Kind of hard to do that. In order to understand what oversampling is, you have to understand the regular sampling theorem. Can you think of any specific contexts in which it is used that are not clear? — Omegatron 21:48, 14 October 2006 (UTC)
Editdroid is right. The introduction is far too technical and needs a common sense explanation perhaps using an analogy. As Nobel Prize winner Richard Feynman said "If we can't explain it to a fifth grader - we don't understand it ourselves." — Preceding unsigned comment added by (talk) 16:10, 20 June 2011 (UTC)
I've made some improvements. Let me know if this helps. ~KvnG 01:37, 2 January 2014 (UTC)
A better response to the 2011 comment might have been to clarify and source the concept of "oversampling ratio" rather than removing it. It's not so unusual in sources. Dicklyon (talk) 03:32, 2 January 2014 (UTC)
I added a one-sentence description. I think this is an improvement over having a whole section with formulas and greek letters. ~KvnG 15:15, 4 January 2014 (UTC)

Backward compatability[edit]

Backward compatability doesn't seem to be working with my various HiFi set-ups. Not sure if this is to do with oversampling or a.n.other it is all rather infuriating not to mention expensive. I wonder what the longer-term solution will be. Znethru

Oversample vs Supersample[edit]

Why no mention of "supersampling"? wasn't there a separate page for that before? —Preceding unsigned comment added by Spot (talkcontribs) 00:15, 14 July 2008 (UTC)

Dynamic Range vs. SNR[edit]

The article contains an error regarding calculation of Signal-to-noise ratio (SNR). "Leocat" (see earlier) attempted to correct the error, but is only partially correct.

Adding 2 numbers with same maximum equals a doubling of the maximum number space. So log2(N)=bits (log2 is log base 2). So averaging 16 times increases the Dynamic Range by 4 bits, that is correct.

This is only true if the noise can be modeled as AWGN (additive-white Guassian noise--normally distributed noise that is independent from sample to sample--thermal noise from an amplifier approximates this condition). If it's just quantization noise (not AWGN), it will likely always quantize to the same value, and both the noise and the signal will add coherently and there will be no SNR improvement. --Bryan Davis

However, the SNR increases by sqrt(N), (not by N as in the article). Summing up noise increases its amplitude by sqrt(N), summing up a coherent signal increases its average by N. As a result, the SNR (or signal/noise) increases by sqrt(N). In the example, that means while with N=16 there is an increase in Dynamic range by 4 bits, and the content of "coherent signal" increases by N, but the noise changes by sqrt(N)=sqrt(16)=4 in the example (don't confuse that with bits), so the SNR changes by 4 (that's "4 times"). —Preceding unsigned comment added by Wogaut (talkcontribs) 14:08, 11 April 2009 (UTC)

This should be in the article! --Jimbo1qaz (talk) 16:47, 23 February 2015 (UTC)
"However, the SNR increases by sqrt(N), (not by N as in the article). " This is NOT TRUE! SNR is *always* used as a ratio of energy, not amplitude. The signal amplitude to noise amplitude is sqrt(N), but the signal energy to noise energy (SNR) is N. This is why SNR is computed in dB as 20*log10(S/N) not, 10*log10(S/N). Combining two signals coherently always doubles the SNR. Again, this only works with AWGN noise. --Bryan Davis


"For instance, to implement a 24-bit converter, it is sufficient to use a 20-bit converter that can run at 256 times the target sampling rate. Averaging a group of 256 consecutive 20-bit samples adds 4 bits to the resolution of the average, producing a single sample with 24-bit resolution."

I am no expert or engineer, but the abovementioned statements are unclear and seemingly contradictory. Does it mean running 256 x the target sampling rate (sentence 1), or simply running 256 20-bit samples then averaging it out? They are not the same. (talk) 02:30, 12 April 2010 (UTC)

I agree. --ANDROBETA (talk) 09:01, 17 October 2010 (UTC)
To help us improve this, please explain where specifically you see an internal contradiction. It is talking about running at 256 x target sample rate then averaging those in blocks of 256 samples to produce samples at the target rate. --Kvng (talk) 15:09, 17 October 2010 (UTC)
It could be either 256 20-bit A/D running in parallel at the sampling rate, or 256x the sampling rate with 1 A/D.

Source dispute[edit]

user:Cantaloupe2 has questioned a source I recently added to the article: Nauman Uppal (2004-08-30). "Upsampling vs. Oversampling for Digital Audio". Retrieved 2012-10-06. Without increasing the sample rate, we would need to design a very sharp filter that would have to cutoff at just past 20kHz and be 80-100dB down at 22kHz. Such a filter is not only very difficult and expensive to implement, but may sacrifice some of the audible spectrum in its rolloff. 

Audioholics runs a well-known forum but the ref is one of their published articles. ThIS material at Audioholics is published like a magazine with editorial controls and bylines. There is no claim that the magazine or the author of the cited article is an authority on the subject. Rather, it is a reasonable secondary reference and is not self published or otherwise obviously flawed. -—Kvng 18:42, 10 October 2012 (UTC)

Role of noise[edit]

With the exception of DC signals, noise is not required to gain resolution benefit of oversampling. -—Kvng 13:56, 5 February 2013 (UTC)

Not sure I agree with that, but what substantive change were you thinking about making? (talk) 05:47, 6 February 2013 (UTC)
I propose to remove the paragraph I have marked as dubious. The benefit described here is available for any changing signal, not just a noisy one. Trying to make this paragraph technically correct will take us out onto a tangent. Somewhere in the encyclopedia we have to have a discussion of sampling AC vs. DC signals. Successive approximation ADC is probably the best place to do that. -—Kvng 15:02, 10 February 2013 (UTC)
Well, it's not well written. It should link to Dither somehow. If the undithered signal is sufficiently broadbanded and intrinsically "noisy", no additional dither need be added for oversampling to add bits (one meaningful bit for each 4x oversampling factor). But if it isn't, and it need not be DC, but just something somehow synchronized to the sampling instances, then adding dither or noise would be necessary to gain bits or dynamic range (which is dB S/N + dB headroom). It would be nice to replace the dubious paragraph with something to this effect. I think the original creator of the dubious paragraph was trying to allude to that fact. (talk) 17:24, 11 February 2013 (UTC)
I agree that that's what they were trying to explain and it is true that dithering increases dynamic range. My point is that oversampling also improves dynamic range but it does so through a different mechanism which is not reliant on noise. -—Kvng 19:19, 13 February 2013 (UTC)
User:Thorney¿? has removed the {{dubious}} tag. I have just restored it. There is usable information between samples and oversampling can retrieve that. There is usable information between quantization levels and dither can retrieve that. ~KvnG 15:28, 29 May 2013 (UTC)
Sorry for jumping the gun regarding the tag. I agree that the paragraph should be removed and replaced with something clearer. What about putting something to the effect of what you just wrote User:Kvng - maintaining the important link to dither. Should this page point out that introducing dithering to increase resolution is only necessary when the signal doesn't span quantization levels? Thorney (talk) 00:10, 24 June 2013 (UTC)