Talk:Analog-to-digital converter

Regarding musical application

This part should talk about the difference between sample rate and resolution regarding music recording and playback. Sample rate will increase the bandwith of the recording and nowadays 44kHz is too low. Theorist might call Nyquist theorem to the discussion, saying the normal person can't hear above 15-17kHz and that 44kHz is enough. The problem is that sound waves are produced mechanically by a speaker and if you simply dont send the hi-frequencies, it will vibrate differently, changing the sound. BTW, DVD is 96kHz. BTW, there is a musical producer that has detected a 95kHz frequency. The guys knew something was bad with his equipment, called in the tech guys, they didnt believe he could hear that high (experts usually hear 20kHz-22kHz) and made a series of test to see if the guy was really feeling it, and HE WAS!

Regarding resolution, 16bits are insufficient for today's standarts. Not only regarding the low SNR, but also regarding the depth, the dynamics of the music. Simply put, between 16bits and 20 or 24 bits, YOU WILL NOTICE the difference, even if you are not an expert. Once again, DVD quality is 24bits.

--- My two cents: we should isolate this discussion in another article, about psychoacoustics. This article would list scientific references about sound perception. I do know one or two articles about high and low frequency perception. Signal quantization is something, hearing quantized signals is another thing.

Now, I don't know what you mean when you say that DVD has 24bits and 96kHz. DVDs are MPEG2 streams, and not raw signals like CDs!!...

AFAIK, sampling rates and bit depths larger than 44100/16 are only required when the signals are going to be digitally processed. But this means "in consumer applications"... The real limits of human perception is a current area of research, an open question. -- nwerneck, 01 Dec 2005 20:25:49 -0200

The music perception is a highly controversial topic and not related to the ADC itself. So the question if the current CD audio standard is sufficient should not be part of the ADC article. If at all this would be a point with the CD standard or digital music part, but even there the controversial part is better avoided. 2A04:1C40:34C9:0:2139:39FF:F647:CF90 (talk) 20:03, 17 June 2022 (UTC)

Resolution and number of levels

Josecampos did some changes regarding the number of levels of the digitalization. I agreed only with some of his modifications... There are 256 levels in a 8 bit digitalization, from 0 to 255. But the difference from level to level is actually (V+-V-)/(2^8 - 1). So I fixed the article making a distinction between the number of levels and the number of "intervals"... Comments? -- NIC1138 18:44, 1 May 2007 (UTC)

12.47.224.7 15:57, 26 September 2007 (UTC) Jim Bach 12.47.224.7 15:57, 26 September 2007 (UTC) From: James.C.Bach@Delphi.com Date: 26-SEP-2007

The difference from level to level is (V+ - V-) / 2^N . . . . NOT (V+ - V-) / 2^n -1 . . . for a 3-bit ADC that is (V+ - V-)/8 not V+ - V-)/7!

In a 3-bit ADC there might only be 7 transitions from code-to-code . . . but the input voltage range (i.e. V- to V+) has 8 regions in it, representing codes "0" thru "7" ("000" to "111" for bit-bangers :-) ). These regions can be equal-width (at 1/8th of Vsupply) like the MicroChip PIC controllers, or they can have a 1/2-width "0" code and a 1.5-width "7" code like just about everyone else's ADC on the planet. Look at the diagrams (and in some cases, equations) provided in the URLs cited below.

Think about it this way . . . let's say the width of your hand represents the voltage range your A/D covers (i.e. HandWidth = V+ - V-) . . . and each finger represents an output code ("0" thru "4", you pick which hand and whether the thumb is LSB or MSB :-) ) . . . you have (presumably) 5 fingers . . . how many cracks (i.e. voltage transitions) do you have between the fingers? 4, right? So, would you estimate that the width of each of your fingers is HandWidth/(N-1) (i.e. HandWidth/4) OR HandWidth/N (i.e. HandWidth/5)? Obviously it is HandWidth/5, which is HandWidth/N. And, if you used your left hand your pinky reasonably approximates the "0" code of a real-world ADC (i.e. 1/2-width) and your thumb reasonably approximates the "4" ("max") code of a real-world ADC (i.e. 1.5-width).

Checkout: http://www.freescale.com/files/microcontrollers/doc/app_note/AN2438.pdf?fsrch=1 http://www.embedded.com/columns/technicalinsights/60403334?_requestid=213222 http://www.maxim-ic.com/appnotes.cfm/appnote_number/1080/ http://zone.ni.com/devzone/cda/tut/p/id/3016

12.47.224.7 15:57, 26 September 2007 (UTC) Jim Bach 12.47.224.7 15:57, 26 September 2007 (UTC)

That formula is wrong...Q = (span)/(number_of_levels) ... —Preceding unsigned comment added by 87.16.20.206 (talk) 10:34, 3 May 2010 (UTC)

The formula is not wrong, the divisor needs to be the number of intervals, which is not 2^M. The argument above is that this does not take into account the common practice of fractional intervals at the extremes. This only makes even the slightest practical difference at very low number of bits - so lets consider the extreme example of one bit and a one volt full scale signal. With no fractional intervals there is only one interval and the resolution is 1.0V. It is stated above, however, that a common practice is 1.5 intervals after the top level and 0.5 intervals below the bottom level. This now makes 3.0 intervals altogether and a resolution of 0.33V. The formula with 2^M as the divisor which has been repeatedly inserted gives an answer of 0.5V which is incorrect for both cases. It only actually gives the correct answer in the special case when the top and bottom interval are both 0.5. This still does not make it a correct formula, it just happens to be spewing out the right answer by a coincidence. The article should either be left as it is, not dealing with fractional intervals, or else they should be properly explained and specifically included in the formula so that it always gives the right answer. SpinningSpark 13:26, 3 May 2010 (UTC)

Every literature says that Q = (span) / (number_of_levels). Mistake is because you consider that max voltage can read from ADC is Vref (and not Vref -1LSB). Then, for your example, when you have and ADC with a resolution of 1bit and a full range of 0..1V, you have 1LSB=1/2=0.5V -So the voltage that corresponds to D=1 is 0.5V (of course always +/- 1/2LSB in ideal case) not 1V Sorry for my english -- Frank Rossi —Preceding unsigned comment added by 87.16.20.206 (talk) 14:12, 3 May 2010 (UTC)

There is a lot of confusion in this area. Check out this book for instance. The formula is quoted as $\scriptstyle E_{\mathrm {fsd} }/2^{N}$ but then sneakily subtracts one when the calculation is actually carried out, effectively using $\scriptstyle E_{\mathrm {fsd} }/(2^{N}-1)$. SpinningSpark 15:03, 3 May 2010 (UTC)

Check this one, instead: http://www.analog.com/library/analogDialogue/archives/39-06/data_conversion_handbook.html (the bible of A/D D/A). That book you linked above is IMHO wrong (page 228): max input voltage not produce all one bits...but (vref-1LSB) voltage produce all ones...as you can read at analog.com link - That formula is anyway correct because the max number we can representate is always (2^number_ov_levels - 1) since we start from 0. In case of 1bit we have 2 levels so max number is 1 ( (2^1) -1 ) - But every step is always (span) / 2^number_of_level —Preceding unsigned comment added by 87.16.20.206 (talk) 15:39, 3 May 2010 (UTC)

Can you give me a chapter and page number? I don't want to read the whole book. If levels are equally spaced (including a whole level at the top and bottom) then the number of intervals will be 2^N+1. Everybody who uses 2^N is assuming half intervals at the end. Many of the references linked above in this post claim an interval of 1.5 at the top and 0.5 at the bottom, also resulting in 2^N+1. SpinningSpark 06:43, 4 May 2010 (UTC)

I tried to explain that formula is wrong. I'll make another time the same example. M=1 (A/D 1bit). VREF=1V so, according that formula, Q=1/( (2^1) - 1 ) and Number of intervals=1. We haven't one interval, but two intervals: [0...1/2LSB[ and [ (VREF/2) - 1/2LSB...VREF]. The first interval is coded with "0", the second interval is coded with "1". Infact, when I want to get volts from ADC result I have to do: Volts = VREF*ADC/(2^number_of_bits)...according you formula Volt would be VREF*ADC/1 !!!!! The proof that formula is wrong -- 87.18.147.138 (talk) 11:02, 4 May 2010 (UTC)Frank Rossi

The analog value represented by the all one codes is VREF -1LSB , so with 1bit of resolution and VREF=1V you have 1LSB = 1/2 = .5V and infact VREF - 1LSB= (1-.5) =.5V while,according your formula this "all-one-codes" value would be 1V and not .5V (source: http://www.analog.com/library/analogDialogue/archives/39-06/Chapter%202%20Sampled%20Data%20Systems%20F.pdf - page 2.4) --Dontronix (talk) 07:15, 5 May 2010 (UTC)

Have a look at: http://www.national.com/appinfo/adc/files/ABCs_of_ADCs.pdf too. --Dontronix (talk) 07:56, 5 May 2010 (UTC)

The issue seems to be that there is without doubt 2^N-1 intervals between 2^N code values, but that is not the way resolution is defined. It is volts per step, not volts per interval which are not the same thing, because there are 2^N steps. I think a diagram is needed to make this clear in the article. I will see if I can produce something over the weekend. SpinningSpark 17:28, 6 May 2010 (UTC)

No, resolution is simply related to the number of steps; you talk about 8 or 16 or 24 bit resolution.- Wolfkeeper 19:12, 6 May 2010 (UTC)

Accuracy may be expressed in volts or decibels or whatever. There's a big difference between say, resolution and accuracy; the actual numbers associated with the input voltage are often not entirely linearly related.- Wolfkeeper 19:12, 6 May 2010 (UTC)

I appreciate the difference between accuracy and resolution and that resolution is normally expressed in bits. However, our article is currently stating that resolution can also be expressed in volts. Do you take issue with that? If so it would be an argument for removing the passage in its entirety, not just correcting the formula. SpinningSpark 20:24, 6 May 2010 (UTC)

I have had a try at editing this section into sense, including some new diagrams. Please let me know if I have got anything wrong. SpinningSpark 16:57, 16 May 2010 (UTC)

Regarding discussion on number of levels in an 8 bit A/D or D/A. The first coded level is zero. Therefore there are (256-1)or 255steps. The word steps is often used in A/D, D/A talk and is less confusing terminology than talking about intervals. — Preceding unsigned comment added by Alan Bate SEA (talk • contribs) 13:47, 22 February 2012 (UTC)

The use of 2^M-1 is incorrect. Even the source that is used for defining the resolution in bits uses 2^M . Another way to think of A/D conversion is the division of the full-scale range in half where a comparator is used to determine if the MSB is 1 or 0; then the next range is divided in half again, and again, until M times; thus the full-scale range is divided a total of 2^M times to give a final resolution size of the full-scale range divided by 2^M.75.174.53.165 (talk) 05:13, 24 January 2017 (UTC)

Yes, 0 is a useful and valid coding so a digital signal of M bits represents 2^M possible states. I have made minor corrections to the #Resolution section. ~Kvng (talk) 18:49, 3 February 2017 (UTC)

I just did some research into this, because I had the same issue myself when I saw the formulation in practice. Range/2^n is correct. The reason is that each bit doesn't represent a discrete analog value. It represents a small range of value and generally sets at the center of that range. For instance, if you're converting 3-bit to a range of 0-8 volts, then: 000 -> 0V-1V (0.5V) 001 -> 1V-2V (1.5V) 010 -> 2V-3V (2.5V) 011 -> 3V-4V (3.5V) 100 -> 4V-5V (4.5V) 101 -> 5V-6V (5.5V) 110 -> 6V-7V (6.5V) 111 -> 7V-8V (7.5V)

LordQwert (talk) 17:21, 13 March 2017 (UTC)

References

Aliasing misrepresented

The aliasing section says the following: "The frequency of the aliased signal is the difference between the signal frequency and the sampling rate. For example, a 2 kHz sine wave being sampled at 1.5 kHz would be reconstructed as a 500 Hz sine wave. This problem is called aliasing."

This is a gross misrepresenation of the topic. It mentions sampling, but omits reconstruction. It fails to take into account spectral folding and image bands, and egregiously it fails to mention that *all* frequencies on the frequency axis fold into the band 0 - fs/2.

Please, someone with the knowledge, expertise, and time, rewrite this section, and any other parts similarly afflicted! — Preceding unsigned comment added by 63.230.166.220 (talk) 20:50, 28 June 2012 (UTC)

I don't see a problem with this example. Reconstruction does not come into play here because the error occurs during sampling. ~Kvng (talk) 18:22, 31 January 2019 (UTC)

Nonlinearity

In Analog-to-digital converter § Non-linearity there is a claim that all ADCs have linearity issues. A 1-bit delta-sigma converter is designed to address this. I'm sure there are some residual non-linearities in supporting circuits but the core process is fundamentally linear. Should we soften this claim? In reviewing articles like this I've found a pattern of WP:NPOV issues from editors who can't seem to believe it is possible for digital systems to represent an analog signal accurately. ~Kvng (talk) 18:11, 31 January 2019 (UTC)