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A few days ago I was in a class about electronics on my study. The first lesson was an introduction into the main subject, bits, analog signals, conversion, etc. An example was asked by the teacher: What is the maximum feasible amount of bits to store audio information in?

Some of the answers that were thrown through the class included "64bit, 32bit, 16bit, 8bit (yeah I know..)...".

then the teacher said it's about 18,19 bits, then you are reaching the upper limit because distortions, noise, etc begin to play a big role for audio recordings.

I do know the typical DVD/Studio quality is 24bit audio.

However this led me thinking: What is the maximum physical/real/electronic bitsize in which a piece of audio can be stored? would 32bit audio be overkill/contain too much noise?

Any explanation/sources on this?

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There's no maximum sample size. Your sample can have 2 megabytes, if you desire so. –  Dzarda Sep 2 at 13:15
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I think the teacher's question is a good one for introduction. It implies several important topics which will certainly be treated in the course: Signal-to-noise ratio (SNR), logarithmic scales (dB vs. log2), quantization effects, practical considerations ("balanced design") - and the relevance of marketing numbers like "32bit audio", "9600dpi", "40 megapixel smart phone camera", or "392kbit/s MP3". –  Hanno Binder Sep 2 at 14:47

6 Answers 6

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What is the maximum physical/real/electronic bitsize in which a piece of audio can be stored?

As Dzarda comments, this is not a sensible question, and it is not clear what you mean by 'piece'. If you mean sample, you can store it in as many bits as you can store. Typical HDs contain 1 TB and more, so 8 Tera Bits would be within reach.

will 32bit audio be overkill/contain too much noise?

It is overkill in the same way that it makes no sense to protect your bike with a very heavy chain that is closed with a soft plastic padlock. You'd better spend less money on the chain and use it to buy a better padlock.

Let's for the sake of argument say that the signal/noise ratio from the analog parts of your audio system corresponds to 16 bits. If you play back digital sound stored as 18 bits that adds 25% of that level worth of noise: it increased the noise by 25%. (from 100 to 125, in arbitrary units). 20 bits will increase it by 6.25%. 32 bits by 0.0015%. That is: assuming you have a perfect translation from digital to analog.

The cost of storage increases linearly with bit size, the cost of a full-range-accurate D/A converter raises almost exponentially when you approach a certain number of bits (~22?). So using more bits than the equivalent quality in the analog parts costs more, but the gain in quality diminishes. So it is simply not economical to use more bits: if you want to spend more money to get better quality, you should spend it on the analog parts. (I am not a audiophile, but AFAIK the speaker is often the weakest link.)

This is a common theme in engineering: it is not about doing individual parts as good as possible, but about a balanced design.

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Your 3rd paragraph (bit calculations) is a mass of confusion and contradictory to itself. Why does moving form 16 - 18 bits increase the noise by 25% (calculations to support that) - when there is a well known relationship to decreased noise at higher bit depths, by your argument if I decreased the bit depth to 14 bits the noise would decrease by 25%? what would then stop me from decreasing the bits to 1 bit for an ideal conversion? - clearly that is nonsense. Another contradiction in that at higher bit depths (32 bits) the noise barely increases? Very confusing. –  placeholder Sep 2 at 15:52
    
If the noise of the analog parts is equivalent to the noise of a 16 bit digital channel (assumption), then such a combination has equal noise contribution from the analog and digital parts, so the digital part adds 100% noise to the analog noise. Now moving from a 16 to 18 bits reduces that noise contribution of the digital part from 100% to 25% (relative to the analog noise contribution): the quantization error of an 18 bits channel is 1/4 of that if a 16 bit channel. –  Wouter van Ooijen Sep 2 at 17:30
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Not even close to being right. Independent noise sources add as RSS (root sum of squares) so two identical magnitude sources will increase total noise as sqrt(2). For RMS measure of an ideal ADC the SNR = 6.02*N + 1.76 (dB) which adds in quadrature with the analog noise, so moving from 16 -18 bits decreases the noise by 12.04 dB but added in quadrature you only see a decrease in noise form that sqrt(2) to sqrt(1+1/16) = 1.03 X so the digital aspect only contributes 3%. –  placeholder Sep 2 at 17:39
    
You are right about the RSS addition, and that indeed changes the numbers somewhat, but does not change the principle. –  Wouter van Ooijen Sep 2 at 17:44

Technology could allow you to store (almost) infinitely big (samples/sec) and infinitely deep (bits) data, and in fact lots of things do store this sort of thing: there's plenty of cameras that can record faster & higher detail than human eyes can see, for example 500 frames per second. Likewise there's scientific instruments such as seismometers which are (simplistically) a lot like microphones but far more sensitive than the human ear, and the recorded data is probably stored in more detail than a human could directly interpret if it was played back at real-world levels. However, these various devices are almost always used to capture things so we can analyse them in some other way: a wave on a graph, a slow-motion video, etc.

Going back to audio recording & playback, again there are scientific & test instruments which can sample, record, reproduce & generate far better quality (as in resolution/depth/accuracy) signals than humans can process, but there's not much point in having them in a recording studio.

Now, in a really good multi-track studio you might want better quality than humans can discern as you are adding lots of things together, so the less error you introduce the better it'll come out in the final mix. Simplistically again; if you do all the hard sums using 4 decimal places your final answer may only need to be to 1 decimal place but might still come out better as you won't have lost as much in rounding errors.

In the final case (human consumption) there is only so much humans can discern so equipment is generally made to be good enough for that, because why would you do more work for no gain?

As an example: digital imaging has topped out at 8-bits-per-colour because the eye can't distinguish more than about 256 shades of grey / the total combination of 16.8 million colours & shades. We have 64-bit PCs and much better digital cameras these days, we could store 16 bits per colour, but people can't see 281,474,976,710,656 different colours and we'd waste a lot of effort capturing & storing that data.

Likewise, no-one will pay for a recording studio full of equipment that can hear, capture, record, and reproduce a fly farting at the back of the room over someone bashing a drumkit as no-one will ever hear it, even if it's there.

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There can actually be some major benefits going beyond 8 bits per color, and likewise 16 bits for audio, because both vision and hearing are quasi-logarithmic, but images and sounds need to be combined linearly. It isn't necessary to distinguish between something that's 99.5% full brightness and 100% full brightness, but if the brightest thing in a scene is 5% of full brightness, the difference between 0.2% of full brightness and 0.1% full brightness can be huge. –  supercat Sep 3 at 2:59
    
You're right - it is helpful to capture more data than you need, especially if you are going to process it (EG do a CSI: style "enhance!" on an image or recording to bring out details otherwise hidden or not using the full dynamic range available). You may capture a completely "black" picture where there are 100 levels of black, but the data is there to up the contrast and show the details. Of course, the higher the amount of data you capture, the more it all costs, and a lot of people never care enough about the benefit. –  John U Sep 3 at 8:52

Fun.. to play with some numbers. Let's assume 1 k ohm of source impedance. (You have to assume something.) So that's got ~4nV/rtHz of Johnson noise. For a 10kHz bandwidth, that's ~400nV of noise. OK and assume it's gained up to 5 Volts and stored. That's about 10^7 in dynamic range... 23 bits. (In real life there will be more noise...)

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You're assuming that the broadband noise floor represents some sort of absolute limit. It does not. Tests have shown that the human ear can pick out musical tones that are 10 to 20 dB or more below the broadband noise level. Psychoacoustics is a complicated subject. –  Dave Tweed Sep 2 at 14:10
    
@DaveTweed, Hey that's absolutely correct! (We've got an instrument with a sine wave buried in noise and I test it by listening.) OK, so add on a few more bits :^) I know very little about microphones..Do they have a noise limit that is not related to Johnson noise? Maybe Brownian motion of the element (coil.) –  George Herold Sep 2 at 14:20
    
Well, there's certainly the Brownian motion of the air (the concept only applies to fluids), which affects both the microphone and the eardrum. But again, that's broadband noise that does not represent any sort of absolute limit. –  Dave Tweed Sep 2 at 14:28
    
I agree with signal averaging we can keep drilling down into noise and finding signal. That makes the question open ended. SNR = 1 just seemed like a natural reference point. –  George Herold Sep 2 at 21:13

In the case of audio that is transmitted by telephone, the quantization levels for the A / D, are determined by the noise level modeling system. That is, we should not increase the levels of quantization, because noise within the converted values ​​is included. Moreover, the ear does not respond in a linear fashion, so that for optimizing the bandwidth of the signal transmitted by telephone, a nonlinear conversion is used, allowing to encode the audio in 8-bit and recover an intelligible signal.

Obviously, the quality of audio transmission for a phone, is not that is intended for a system of high fidelity audio.

In short, the theory states that there is no upper limit to the number of quantization levels of an audio signal, but in practice, the noise present in the system can put an upper limit. For more information see this link.

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32 bit audio stored in floating point format at is common in the professional industry. However, that is to reduce rounding errors during the digital processing where it gets heavily processed through all kinds of digital filters and transforms. On the recording or playing end of things I do not believe anyone can distinguish between 24 bits at 192kHz and deeper faster sampling. Probably not even bats.

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It appears that you need to understand some basic terms. There are AD<->DA converters of differing "bit" sizes, and different operating frequencies. The bit size, affects the accuracy of the "sample," while the frequency affects the sampling rate. Typically you have an audio signal which you want to digitize. So the first question you have to answer is, how accurate does the sample need to be? The higher the number of bits used, the higher the quality of the reproduced audio and the higher the cost of the converter. The higher the sampling rate, the higher the quality of the reproduced audio and the higher the cost of the converter. So, the first practical limit is imposed by the cost of the converter. There is another practical limit imposed by the "human sensibility." If our ears can not detect a difference between the original and the reproduced audio, then the number of bits and the sampling rate used to achieve that, will be "good enough." Based on this information, I believe your question should be:
What should be the maximum, practical, number of bits (of an AD <-> DA converter) be, to be able to reproduce an acceptable copy of an audio signal?
I believe that appropriate calculations would give approximately 18 bits with a sampling rate of 150K Hz.

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"Acceptable" is one of those specifications that mean absolutely nothing. –  Scott Seidman Sep 8 at 23:34

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