Bit depth revisited

[Ozpeter]

In the 192kHz thread which most here are probably no longer reading, a side issue arose concerning bit depth.

In response to an earlier statement I posted -

If you compare a 16 bit recording vs a 24 bit recording, the additional 8 bits are added at the bottom, not somehow distributed across the full dynamic range. So the close-to-inaudible part of the dynamic range has a further 8 bits added to describe it. That's why when recording in 24 bits it isn't nearly so important to peak close to full scale - which is the chief point in using 24 bits to record.

This was then contested by a couple of forum members, to my surprise. I then posted a fuller description of the workings of bit depth, following which 'klaukholm' commented that it was an excellent explanation and that perhaps it should form the starting point for a new discussion of bit depth and audio. I'm not convinced that such a discussion is really required , and I should point out that it is a paraphrase of some explanations in a very old "Harmony Central" forum discussion, so I can claim little credit for it - but here it is, FWIW.

The first digit of a 16-bit word is called the Most Significant Bit (MSB). Thinking about a waveform as going up and down and up and down over a zero point, when the value is positive, the MSB is a "0", and when it's negative the MSB is a "1". So the "loudest" positive signal would be 0111 1111 1111 1111 (which is 32,767 expressed in base ten). A signal that's 6 dB down would be 0011 1111 1111 1111, or 16383. (32768, or 1000
0000 0000 0000, would actually be the lowest possible position below the zero crossing, which for all intents and purposes would be the same thing.)

Si we're not going from 1111 1111 1111 1111 down to 0000 0000 0000 0000, but from 0111 1111 1111 1111 (as close to 0 dB FS as we can get on the positive side of the zero crossing) down to 0000 0000 0000 0001 (one "step" above the zero crossing) to 0000 0000 0000 0000 (the zero crossing itself) to 1111 1111 1111 1111 (one step below the zero crossing) to 1000 0000 0000 0000 (the maximum amplitude on the negative side of the zero crossing). As we move to the right, each bit repesents a value closer and closer to the zero crossing, which is why as we increase our bit depth we can capture a wider dynamic range...we're capturing those signals closer to the zero crossing with more resolution. With 16 bits we can theoretically capture signals up to 96 dB below FS accurately. With 24 we can get closer and closer to the zero crossing and (theoretically) capture signals up to 144 dB down. Those extra bits are used when representing louder signals as well, but since they're representing such a quiet component of the signal they make no audible difference whatsoever.

Bear in mind that with digital audio we're always counting down from the top, not up. So putting it the most simple possible way, it's not unlike counting down from zero using two digits -

-01, -02, -03, -04, -05... -99

Now we could do the same with three digits -

-001, -002, -003, -004, -005... -099... -999

Now it's obvious that -01 = -001. And the two counting systems remain the same until the two digit system runs out of values at -99 (=-099). Then the three digit system carries on downwards from -100 down to -999. The fact that it does that doesn't alter the fact that the first 99 values are the same. It just goes a lot further down.

I'm assuming that few would dispute the above, but let's see.

However, when recording audio and you want to record at conservative levels to avoid digital overs, of course 24 bit recording has its place, and likewise in subsequent post-processing. But the theory set out above would indicate that a 24 bit original recording peaking to full scale will not generally sound any different from a 16 bit original recording under the same conditions, unless you are able to hear recorded components below around -96dB referenced to full scale. To determine that, generate a 24 bit -96dB sine wave. Mix it with the fortissimo end of a symphony peaking to full scale, and have the sine wave extend beyond the end, into what would otherwise be silence. Replay it at a tolerable level. Can you hear the sine wave (a) during the music or (b) afterwards? If you can, then you should be able to hear the benefit of the added dynamic range provided by a 24 bit recording. If you can't, then it would seem unlikely that you can hear the difference between a 24 bit recording and a 16 it recording, as that difference lies purely in the audio at levels 96dB below full scale.

[Karl Jackson]

I don't think your understanding of analog to digital conversion is accurate. Additional bits add accuracy to the conversion across the entire dynamic range, it doesn't add the ability to capture increasingly softer sounds, as you imply. My best analogy when I'm teaching this is climbing a mountain. The size of the mountain doesn't change, but the number and height of the steps leading to the top do. With 8 bit converters there are 2^n or 256 steps. 16 bit = 65536 steps. 24 bit = 16777216 steps. To put it another way, if the mountain is 1 Volt "tall", each 8 bit step is 3.91 mV high, whereas each 24 bit step is 59.5 nV high.

As a result, when recording a symphony orchestra the recordist can capture the actual loudness of the group with much greater accuracy, whether in loud passages or soft, with a 24 bit converter than with an 8 bit converter.

kj

Edit: I didn't realize this dead horse was so well beaten. Sorry for creating added noise on this off topic post.

[okydoky]

Originally Posted by ozpeter

Eh? If you compare a 16 bit recording vs a 24 bit recording, the additional 8 bits are added at the bottom, not somehow distributed across the full dynamic range. So the close-to-inaudible part of the dynamic range has a further 8 bits added to describe it. That's why when recording in 24 bits it isn't nearly so important to peak close to full scale - which is the chief point in using 24 bits to record.

I really think this thread should be locked.

Originally Posted by plush

not right.

Originally Posted by oky****

hi,

well, i am not here to argue or teach, but 24 bit dynamic range is not a free pass to record at low, or unnecessarily reduced, levels, like some people may think. there is a lot of confusion about all that, apparently. most of it has to do with people trying to avoid clipping.

i'm not sure why you are referencing 16-bit recording. nobody was discussing that, so i am not sure what you are trying to get at there.

anyhow, 24 bits is not "the best thing ever", its just the best thing that current component will reasonably allow. you're not even getting 24 bits, as a practical matter.

another thing that is often overlooked is the fact that complex musical signals are made up lots of low level components. so when you see your piano peaking nicely at -3dBfs, it does not mean that all aspects of the sound are at that level. some harmonics are significantly lower in amplitude, and they are still very important to the timbre of the instrument.

here a quote re digital audio principles that you might find helpful. [emphasis added] [but don't be mad at me for just stating known facts].


"Analysis of the quantization error of low-amplitude signals reveals that the spectrum is a function of the input signal. The error is not noiselike (as with high-amplitude signals); it is correlated. At the system output, when the quantized sample values reconstruct the analog waveform, the in-band components of the error are contained in the output signal. Because quantization error is a function of the original signal, it cannot be described as noise; rather, it must be classified as distortion.

As noted, when quantization error is random from sample to sample, the rms quantization error E (sub)rms = Q(12) sup.1/2. This equation demonstrates that the magnitude of the error is independent of the amplitude of the input signal, but depends on the size of the quantization interval; the greater the number of intervals, the lower the distortion. However, the relevant number of intervals is not only the number of intervals in the quantizer, but also the number intervals used to quantize a particular level. A maximum peak-to-peak signal (as used in the preceding analysis) presents the best case scenario because all the quantization intervals are exercised. However, as signal level decreases, fewer and fewer levels are exercised as shown in Fig. 2.8. For example, given a 16-bit quantizer, a half-amplitude signal would be mapped into half of the intervals. Instead of 65,536 levels, it would see 32,768 intervals. In other words, it would be quantized with 15-bit resolution.

The problem increases as the signal level decreases. A very low-level signal, for example, might receive only single-bit quantization or might not be quantized at all. In other words, as the signal level decreases, the percentage of distortion increases. Although the distortion percentage might be extremely small with a high level, ) 0 dBFS, its percentage increases significantly at low-amplitude levels.

The error floor of a digital audio system differs from the noise floor of an analog system, because in a digital system the error is a function of the signal. the nature of quantization error varies with the amplitude and nature of the audio signal. For broadband, high amplitude input signals the quantization error is perceived similarly to white noise."


right.

Originally Posted by David Spearritt

ozpeter said: "If you compare a 16 bit recording vs a 24 bit recording, the additional 8 bits are added at the bottom, not somehow distributed across the full dynamic range."


Not sure this is correct Peter. With a 24 bit recording, the amplitude quantization is 256 times more abundant uniformly. As an example:

16 bit recording: Plus or minus 32767 amplitude steps possible.
24 bit recording: Plus or minus 8388607 amplitude steps possible.

If we send a sinewave waveform of 1 V P-P to each system set so that 1V peak is 0dB, the maximum amplitude words recorded will be 32767 and 8388607 respectively.

Now if we reduce the signal to 0.5 V P-P, the amplitude words will be 16384 and 4194304 respectively, so the top 0.5V is sliced up into 4194304/16384 or 256 more steps.

Originally Posted by oky****

hi,

correct. and the issue is that the ratio of the signal's amplitude to the amount of "steps" does worsen as you go lower in the bit depth. and of course quantization noise [to the extent that is of any moment]. there is effectively less accuracy down in the "basement". unavoidable with digital audio, it seems to me, dither notwithstanding.


right.

Originally Posted by David Rick

The quote below (apparently from Wikipedia) is a pretty good summary of the trouble with undithered digital audio.
Quote:
"Analysis of the quantization error of low-amplitude signals reveals that the spectrum is a function of the input signal. The error is not noiselike (as with high-amplitude signals); it is correlated. At the system output, when the quantized sample values reconstruct the analog waveform, the in-band components of the error are contained in the output signal. Because quantization error is a function of the original signal, it cannot be described as noise; rather, it must be classified as distortion.

As noted, when quantization error is random from sample to sample, the rms quantization error E (sub)rms = Q(12) sup.1/2. This equation demonstrates that the magnitude of the error is independent of the amplitude of the input signal, but depends on the size of the quantization interval; the greater the number of intervals, the lower the distortion. However, the relevant number of intervals is not only the number of intervals in the quantizer, but also the number intervals used to quantize a particular level. A maximum peak-to-peak signal (as used in the preceding analysis) presents the best case scenario because all the quantization intervals are exercised. However, as signal level decreases, fewer and fewer levels are exercised as shown in Fig. 2.8. For example, given a 16-bit quantizer, a half-amplitude signal would be mapped into half of the intervals. Instead of 65,536 levels, it would see 32,768 intervals. In other words, it would be quantized with 15-bit resolution.

The problem increases as the signal level decreases. A very low-level signal, for example, might receive only single-bit quantization or might not be quantized at all. In other words, as the signal level decreases, the percentage of distortion increases. Although the distortion percentage might be extremely small with a high level, ) 0 dBFS, its percentage increases significantly at low-amplitude levels.

The error floor of a digital audio system differs from the noise floor of an analog system, because in a digital system the error is a function of the signal. the nature of quantization error varies with the amplitude and nature of the audio signal. For broadband, high amplitude input signals the quantization error is perceived similarly to white noise."

Originally Posted by oky****

hi,

yes, but it does not improve accuracy. it merely removes quantization noise created when the sampler attempts to resolve low level signals.

Originally Posted by ozpeter

David, you've shocked me! If what you say is true, then truncating a 24 bit recording to 16 bits wouldn't work, would it?

Bits and resolution - the way I've seen it explained elsewhere is like this.

Imagine you have a 16 bit ruler and a 24 bit ruler.

If the 16-bit ruler measured 100 cm (or 1000 mm) the 24 bit one should measure 150 cm (or 1500 mm).

The theoretical maximum dynamic range of a 16-bit recording is 96 dB and that the theoretical maximum dynamic range of a 24-bit recording is 144 dB, so the 24-bit ruler should be 1.5 times the length of the 16-bit ruler.

But in neither ruler would the gradations be evenly spaced. Sound (as expressed in dB) is logarithmic. As we went from the top of the rulers down to the bottom, the gradations would get closer and closer together. They would line up perfectly until we got to the bottom of the shorter ruler, after which they would continue to get closer and closer together until they got to the bottom of the second ruler. In this example, the 16-bit ruler is divided into 65,536 parts and the 24-bit ruler is divided into 16,777,216 parts. But remember, those parts are not evenly spaced. Using our 100 and 150 cm rulers, the "top" 100 cm of the 150 cm ruler is divided into 65,536 parts, and the "bottom" 50 are divided into 16,711,680. All that extra resolution is used to describe the "bottom" 48 dB dynamic range.

You get those extra 48 dB (theoretically) below the 96 dB you've already
got, not above. It's very simple to test this. Take a 16-bit signal that peaks at 9 dBFS (your "fully used" 16-bit signal) and run it into a device that has a 24-bit digital input. Does it peak at -48dB? No, it peaks at 0. 0 dBFS is always 0dBFS, whether it's on an 8-bit signal or a 24-bit signal. You can't go any higher, just lower.

Originally Posted by oky****

hi,

guess what, your right, it [truncation] doesn't work.

you're way off on the rest of your post. the bits are "evenly spaced" [except in some more exotic purpose-built sampling devices not useful in audio, where only a small part of the signal's dynamic range is of interest]

you appear to be confusing dBfs with bit resolution, among other things.


right.

Originally Posted by ozpeter

in this forum, I shouldn't have to be dealing with very common misconceptions about how digital audio works, but here goes...but it's important not to let wildly incorrect statements about the way digital audio works lie unchallenged - otherwise people read and believe them - especially in a forum which considers itself to conduct its discussions at a professional standard.

hi,

you apparently don't want to understand, and your analysis is wildly incorrect.

in 24 bit audio, the louder components are also resolved with greater accuracy than they are in 16 bit audio.

right.

[Ozpeter]

My best analogy when I'm teaching this is climbing a mountain.

Unfortunately that's a very bad analogy. You are descending the mountain, not climbing it, for a start. The rest I'll address in the morning. It's very late here.

[RobAnderson]

I definitely don't have the background to join this debate, and I'll probably end up with a bloody nose, but here is my own thought:

most gear in the signal chain before the converter has an audible and measurable noise floor - even good mic's add as much as 18dB of noise, preamps add their own EIN, etc. By the time we reach the converter, we have a "significant" amount of electronic noise that would be greater than the very smallest signal we could generate.

Add to that the fact that most gear has a maximum of some sort: mic's have a maximum SPL and are also limited by their own sensitivity, ditto for preamps - most can't take more than 24dB in and then have a limitation on their output before clipping.

If I were designing a converter, I would know of these limitations and take them into account. It would make sense that I would not waste my increased resolution (i.e. my last 8 bits) in the range where there would be naught but electronic noise - I would spread it across the usable dynamic range of a system.

As far as the numbers go, I would think you can have the numbers represent anything you want them to - I could have my 24-bit steps be just as large as my 16-bit steps and so represent a greater dynamic range (but who needs 144 dB of dynamic range?); or I could spread my 24-bit steps across the same dynamic range as my 16-bit steps and so have less quantization error. I would also imagine that it is possible to do something in between.

I wish a designer would chime in here. I am far too limited by own ignorance and would like to know how it works in the practical sense.

[Corran]

This is all really fascinating to me. I don't have the tech background to make any meaningful posts but I am interested to hear from those that do.

One thing I did notice last week while recording a vocal recital - the entire dynamic range of the performance was about 60db. That's well within the limits for even 16bit audio. But depending on really who is "correct" here 24bits would still be an improvement, or it might not be anything different than 16bits.

[klaukholm]

It would be interesting if someone with a solid design background would chime in and explain what the challenges and limitations are in regards to bit depth, impulse response and the like.

I would also gladly take any suggestions as to a solid, yet pedagogical textbook that explains this thouroughly.

[Karl Jackson]

OK, because pictures are sometimes more useful than explanations, I threw together very quick and dirty unipolar (not peak to peak, but zero to peak) transfer functions for two ideal, linear ADCs, 3-bit and 4-bit, considering only analog input voltage and output digital code, and neglecting all types of error and filtering. Increasing from 3-bit to 4-bit increases resolution at all input voltage levels, not just at lower input voltage levels.

kj

The basis for the transfer function came from here: http://focus.ti.com/lit/an/sbaa147a/sbaa147a.pdf. FS = Full Scale

[RobAnderson]

Hi Karl

Thanks for the picture.

This diagram that you have I would understand to be an example of maintaining the same dynamic range, but spreading more bits across it to reduce your quantization error.

This would not be the same if we were using our increased values to represent a larger range; i.e. 111 is the top of your 3 bit scale; this could be equal to 0111 on your 4-bit scale, and in this respect would use the addition of the most significant bit increase the scope of the entire scale.

This, I think, is the crux of the matter - what are we using these additional bits for - greater scope or less error?

[Karl Jackson]

Originally Posted by RobAnderson

Hi Karl
This, I think, is the crux of the matter - what are we using these additional bits for - greater scope or less error?

What do you mean by scope? Do you mean that more scope = greater range of analog input voltage, perhaps -10V to +10V rather than -5V to +5V? This would mean that you would need to change the output voltage of your analog mixer if you changed your ADC.

On the other hand if by added scope you mean that the overall range stays the same, but that the region from -1V to +1V is broken down into smaller increments than the rest of the range, then you're talking about a non-linear converter, which I would agree is pretty much useless, so it's a good thing that ADCs aren't designed that way.

kj

[BrianHanke]

I don't understand why there is any controversy about this. There should be only one way to do the math, otherwise none of the digital audio equipment we use would work!

As far as I understand it, in 16 and 24 bit recording the minimum and maximum possible values represent the same original SPL, although the binary values will be different. In 24 bit recording we simply have more available values to represent the original signal with greater precision.

In other words, in 16 bit recording no signal (zero SPL) is represented by 0000 0000 0000 0000. In 24 bit recording no signal is represented by 0000 0000 0000 0000 0000 0000. In 16 bit recording the maximum signal is represented by 1111 1111 1111 1111. In 24 bit the maximum signal is represented by 1111 1111 1111 1111 1111 1111. Both these numbers represent the exact same original maximum SPL, even though one is a larger number.

A 24 bit measurement system gives us 16,711,680 more available values than a 16 bit one. Therefore, 24 bit recording is approximately 256 times more precise than 16 bit at all voltages, from no signal to maximum signal.

So why this talk about greater dynamic range in 24 bit if the minimum and maximum values remain constant? Simple: in 16 bit the top two available values are 1111 1111 1111 1110 and 1111 1111 1111 1111. In 24 bit, there are approximately 256 more available values between those two steps! The same is true at the bottom end, at the quietest dynamic ranges.

[chris319]

I don't understand why there is any controversy about this.

Well, this wouldn't be a proper audio forum unless this dead horse were beaten nine ways beyond dead

Using the old 1-bit-equals-6-dB rule of thumb, 16 bits can represent 96 dB of dynamic range while 24 bits can represent 144 dB. Modern electronic circuits top out at about 120 dB of dynamic range, so beyond 20 - 21 bits the issue of dynamic range becomes moot and the issue of resolution takes center stage. Given a measuring stick which is 144 cm in length, and the maximum height of the thing to be measured is 120 cm, you'll be able to measure the thing more precisely if your measuring stick is calibrated in tenths of a millimeter than in centimeters.

[BrianHanke]

Yeah, no kidding!!!

Originally Posted by chris319

Well, this wouldn't be a proper audio forum unless this dead horse were beaten nine ways beyond dead

Just to put the final nail in the coffin here, I encourage everyone to read this document from National Instruments: http://zone.ni.com/devzone/cda/tut/p/id/3016#toc4. Quoting the relevant section, starting with the fourth paragraph: "[...] a 3-bit ADC divides the range into 23 or eight divisions. A binary or digital code between 000 and 111 represents each division. The ADC translates each measurement of the analog signal to one of the digital divisions."

Looking at the graph we can clearly see that a 3 bit converter would divide the range (0 to 10 volts) into 8 equal sections. A 16 bit converter would divide the same range into 65,536 equal parts, thereby creating a much more accurate representation of the original analog signal.

[oshearer]

I have to preface this by saying that I'm still a student and therefore may be far out of my depth. To me, the idea of dynamic range and resolution are intrinsically linked as far as they relate to bit depth in digital audio. A signal at -96dB FS in 16 bit would be represented as 16 zeroes, but so would any incoming signal at a lower level than this. Therefore, since there is no more resolution available below this point, this is the functional "bottom" of the dynamic range of a 16 bit signal. On the other hand, -96dB FS in 24 bit would be represented as 0000 0000 0000 0000 1000 0000 (I think), and therefore there are another 7 bits of resolution below this level, which pushes the dynamic range of this signal that much lower. The only difference between the two is that 16-bit has no way to represent signals that fall between 0dB SPL and -96dB FS, while 24-bit does. If I'm way off with this, please don't hesitate to put me in my place.

Edit: just realized that this post is completely redundant, sorry.

[BrianHanke]

Using that logic wouldn't a hypothetical 128 bit ADC be able to reach all the way down to 768 dB below FS? That's some mighty quiet music (if that number even means anything)!

Originally Posted by oshearer

I have to preface this by saying that I'm still a student and therefore may be far out of my depth. To me, the idea of dynamic range and resolution are intrinsically linked as far as they relate to bit depth in digital audio. A signal at -96dB FS in 16 bit would be represented as 16 zeroes, but so would any incoming signal at a lower level than this. Therefore, since there is no more resolution available below this point, this is the functional "bottom" of the dynamic range of a 16 bit signal. On the other hand, -96dB FS in 24 bit would be represented as 0000 0000 0000 0000 1000 0000 (I think), and therefore there are another 7 bits of resolution below this level, which pushes the dynamic range of this signal that much lower. The only difference between the two is that 16-bit has no way to represent signals that fall between 0dB SPL and -96dB FS, while 24-bit does. If I'm way off with this, please don't hesitate to put me in my place.

Edit: just realized that this post is completely redundant, sorry.

[oshearer]

I don't think that shows a flaw in the logic. Sure a 128 bit ADC could reach that far down, but there would be no point in terms of dynamic range, only resolution. Even very good A/D converters only have a dynamic range of about 120 dB, therefore the difference between 20 bit and any greater bit depth is only realized in the practical realm as increased resolution.

Going to the theoretical side of things, the bottom limit of human hearing is .00002 pascals. Assuming we had equipment capable of recreating that signal without adding noise (impossible, I know), it would translate to approximately -150dB FS if no amplification were applied after the signal leaves the microphone (assuming a microphone with sensitivity=10mV/Pa and +18dBu=0dB FS). So even though the hypothetical 128 bit ADC would have resolution at that level, that number doesn't actually mean anything in relation to the physical limitations of human hearing.

[chris319]

16-bit has no way to represent signals that fall between 0dB SPL and -96dB FS, while 24-bit does. If I'm way off with this, please don't hesitate to put me in my place.

We're not talking dB SPL here. That's acoustical dB; we're talking about the digital representation of analog electrical signals.

If you're going to express digital levels in terms of dB below 0 dBFS, which is the proper way to do it, 16 bits can represent -96 dBFS to 0 dBFS; 24 bits can represent -144 dBFS to 0 dBFS. With modern audio circuitry capable of -120 dBFS (noise floor) to 0 dBFS, given 24 bits you've got more dB of range than you actually need -- like using a 12" ruler to measure a 10" thing (we all have a 10" thing, haven't we?). In addition, your 24 bit ruler is calibrated in (metaphorically speaking) 1/64ths of an inch as opposed to 1/16ths of an inch for your 16-bit ruler, which is why we say 24 bits have more resolution.

In digital audio you can't have more than 0 dBFS. Full scale is full scale and that's where signals get clipped.

[oshearer]

Oh, I know they're different units, I was just using 0dB SPL as a clumsy way of saying "nothing." I would have been better off saying -∞dB FS. By the way, I like that ruler analogy.

[chris319]

I like the 10" thing analogy

[David Rick]

Originally Posted by Karl Jackson

OK, because pictures are sometimes more useful than explanations, I threw together very quick and dirty unipolar (not peak to peak, but zero to peak) transfer functions

Nice illustration, Karl. If it's not too much work, it would be good to post a bipolar version. Then everyone will understand why we can't just allocate bits above the noise floor. As I think you've mentioned, such a transfer function would generate horrible crossover distortion.

Kjetil: There are several popular books on digital audio. I've never looked at Ken Pohlman's book, but I have a copy of John Watkinson's The Art of Digital Audio, and it covers a lot of ground.

People who want to learn digital audio fundamentals without having to look at equations should consider Digital Audio Explained by Nika Aldrich.

There's a lot of DSP in modern audio equipment. Most working audio engineers have very limited DSP background, because the subject is inherently very mathematical. Dip more than your toe in the water, and you can easily drown in Z-transforms, analytic functions, and so forth. To do justice to the subject, one should probably study linear system theory first. But there's a recent textbook that turns this idea on its head, and uses DSP as a starting point.

DSP First: A Multimedia Primer

I can't say whether the authors (two of whom are very well known) actually succeed in this or not. Be warned that learning this subject is a "project", no matter how good the text. Also, you'll probably need to buy the student edition of MATLAB to work the problems and examples.

David L. Rick

[Ozpeter]

If I were designing a converter, I would know of these limitations and take them into account. It would make sense that I would not waste my increased resolution (i.e. my last 8 bits) in the range where there would be naught but electronic noise - I would spread it across the usable dynamic range of a system.

As far as the numbers go, I would think you can have the numbers represent anything you want them to - I could have my 24-bit steps be just as large as my 16-bit steps and so represent a greater dynamic range (but who needs 144 dB of dynamic range?); or I could spread my 24-bit steps across the same dynamic range as my 16-bit steps and so have less quantization error. I would also imagine that it is possible to do something in between.

It would indeed be handy if one could ignore the laws of mathematics but sadly you can't!

As far as I understand it, in 16 and 24 bit recording the minimum and maximum possible values represent the same original SPL, although the binary values will be different. In 24 bit recording we simply have more available values to represent the original signal with greater precision.

In other words, in 16 bit recording no signal (zero SPL) is represented by 0000 0000 0000 0000. In 24 bit recording no signal is represented by 0000 0000 0000 0000 0000 0000. In 16 bit recording the maximum signal is represented by 1111 1111 1111 1111. In 24 bit the maximum signal is represented by 1111 1111 1111 1111 1111 1111. Both these numbers represent the exact same original maximum SPL, even though one is a larger number.

A 24 bit measurement system gives us 16,711,680 more available values than a 16 bit one. Therefore, 24 bit recording is approximately 256 times more precise than 16 bit at all voltages, from no signal to maximum signal.

So why this talk about greater dynamic range in 24 bit if the minimum and maximum values remain constant? Simple: in 16 bit the top two available values are 1111 1111 1111 1110 and 1111 1111 1111 1111. In 24 bit, there are approximately 256 more available values between those two steps! The same is true at the bottom end, at the quietest dynamic ranges.

No, that's not how it works. I'll repeat the correct analysis -

The first digit of a 16-bit word is called the Most Significant Bit (MSB). Thinking about a waveform as going up and down and up and down over a zero point, when the value is positive, the MSB is a "0", and when it's negative the MSB is a "1". So the "loudest" positive signal would be 0111 1111 1111 1111 (which is 32,767 expressed in base ten). A signal that's 6 dB down would be 0011 1111 1111 1111, or 16383. (32768, or 1000
0000 0000 0000, would actually be the lowest possible position below the zero crossing, which for all intents and purposes would be the same thing.)

Si we're not going from 1111 1111 1111 1111 down to 0000 0000 0000 0000, but from 0111 1111 1111 1111 (as close to 0 dB FS as we can get on the positive side of the zero crossing) down to 0000 0000 0000 0001 (one "step" above the zero crossing) to 0000 0000 0000 0000 (the zero crossing itself) to 1111 1111 1111 1111 (one step below the zero crossing) to 1000 0000 0000 0000 (the maximum amplitude on the negative side of the zero crossing). As we move to the right, each bit repesents a value closer and closer to the zero crossing, which is why as we increase our bit depth we can capture a wider dynamic range...we're capturing those signals closer to the zero crossing with more resolution. With 16 bits we can theoretically capture signals up to 96 dB below FS accurately. With 24 we can get closer and closer to the zero crossing and (theoretically) capture signals up to 144 dB down. Those extra bits are used when representing louder signals as well, but since they're representing such a quiet component of the signal they make no audible difference whatsoever.

Going back to Karl's mountain... for a start, we have to think in terms of steps descending the mountain. Let's imagine a 16 bit staircase and a 24 bit staircase. The 16 bit staircase takes us close to the bottom of the mountain, but there's a jump from the lowest step down to sea level. The 24 bit staircase is the same up to that point, but provides extra steps to avoid that jump at the bottom.

But note that in neither staircase are the steps evenly spaced. Sound as expressed in dB is logarithmic. As you go down the staircase from the top of the mountain, the height of each step gets smaller and smaller. (If it was a ruler, the graduations get closer and closer together). The two staircases line up perfectly until you get to the bottom of the 16 bit staircase, and then the 24 bit staircase continues towards sea level with a huge number of extra steps, this being necessary because the height of each step continues to reduce. So to get anywhere near the bottom of the mountain, you need an considerable number of extra diminishing steps.

We can examine this at a purely practical level too. Say somebody comes to you with a 16 bit recording, and a 24 bit recording, of the same material. Let's say it's a symphonic recording. You are asked to identify which is which. If the 24 bit recording had some magical quality that made it audibly superior at any point, you wouldn't have to go hunting for a very, very quiet passage and listen to it at an unnaturally high level to hear which was the version with the best reproduction of that low level information. But that's the only way you could do it, based on the Journal of the Audio Engineering Society (the study by Meyer and Moran, vol 55 issue 9, Sept 2007). Or is there some other attribute that you could describe that is evidenced by 24 bit audio vs 16?

Take another example - if you have a 24 bit recording and truncate it to 16 bits, if throughout the recording the audio is now represented by a 256th of its previous resolution, surely there would be a really dramatic degradation of the audio quality throughout? But that doesn't happen (again I refer back to Meyer and Moran's study). So how can we account for that? By simply understanding that only the component of the audio that is least audible has been changed - the audio at the very lowest levels.

And another one - what do you get when you truncate a 24 bit recording to 16 bits and invert it against the original (thus revealing the difference) - almost nothing. Certainly nothing that would support any assertion that above 96dB below full scale the 24 bit recording has a content which differs from the 16 bit version.

I could go on but hopefully I've clarified the incorrect assumptions in some preceding posts.

[oshearer]

Thanks for the book suggestions. I'm always trying to find out more about the inner workings of my equipment, both analog and digital, so these should be very helpful.

[David Spearritt]

Going back to Karl's mountain... for a start, we have to think in terms of steps descending the mountain. Let's imagine a 16 bit staircase and a 24 bit staircase. The 16 bit staircase takes us close to the bottom of the mountain, but there's a jump from the lowest step down to sea level. The 24 bit staircase is the same up to that point, but provides extra steps to avoid that jump at the bottom.

But note that in neither staircase are the steps evenly spaced. Sound as expressed in dB is logarithmic. As you go down the staircase from the top of the mountain, the height of each step gets smaller and smaller. (If it was a ruler, the graduations get closer and closer together). The two staircases line up perfectly until you get to the bottom of the 16 bit staircase, and then the 24 bit staircase continues towards sea level with a huge number of extra steps, this being necessary because the height of each step continues to reduce. So to get anywhere near the bottom of the mountain, you need an considerable number of extra diminishing steps.

Peter, I think talking SPL and logarithms is making things very difficult to understand.

Let's try this way. When we record a concert, we are recording voltage linearly. We are recording a voltage time waveform which has excursions plus and minus around zero volts. The voltage waveform is what you are looking at in your DAW when editing or mastering. It might help to not get sidetracked on SPL and logarithms and our hearing mechanism and such as this is all irrelevant.

Now that voltage time waveform is sampled at Fs, and it is amplitude sliced into even steps of voltage which is some fraction of the full scale voltage which will overload our A/D. The size or fraction of the voltage amplitude steps is full scale volts P-P divided by the resolution afforded by the bit depth.

Correct?

[chris319]

Now that voltage time waveform is sampled at Fs, and it is amplitude sliced into even steps of voltage which is some fraction of the full scale voltage which will overload our A/D. The size or fraction of the voltage amplitude steps is full scale volts P-P divided by the resolution afforded by the bit depth.

Correct?

Makes sense to me.

If it takes 2V P-P to light up all the bits of an A-to-D converter, 2V divided by 2^24 means each step represents 0.00000011921 volts.

[oshearer]

"Take another example - if you have a 24 bit recording and truncate it to 16 bits, if throughout the recording the audio is now represented by a 256th of its previous resolution, surely there would be a really dramatic degradation of the audio quality throughout? But that doesn't happen (again I refer back to Meyer and Moran's study). So how can we account for that? By simply understanding that only the component of the audio that is least audible has been changed - the audio at the very lowest levels."

I may be reading your argument incorrectly, but it seems as though you're stating that there is no increase in resolution at levels above -96dB FS. But if were to start with the highest possible value in a 16 bit system, 0111 1111 1111 1111, to decrease it by the smallest possible amount, we would change the least significant bit, getting 0111 1111 1111 1110. This leaves only one bit of resolution between these two values. If we were to replicate the same two values in a 24 bit system, we would have 0111 1111 1111 1111 0000 0000 and 0111 1111 1111 1110 0000 0000, which gives us a full 8 bits of resolution between the two values.

"A signal that's 6 dB down would be 0011 1111 1111 1111"

The 16 or 24 bit word does not actually represent a dB value. It is instead an approximation of the instantaneous voltage value of the incoming signal at the time when that particular sample was taken. As we get a longer word, we come closer to approximating that instantaneous value. A -6dB FS sine wave is not represented by a string of the above binary word, but a string of constantly changing binary words that approximate the voltage value at a given point in time.

Edit: Haha, I'm too slow a poster, my posts keep being made redundant as I'm writing them.

[oshearer]

Originally Posted by chris319

Makes sense to me.

If it takes 2V P-P to light up all the bits of an A-to-D converter, 2V divided by 2^24 means each step represents 0.00000011921 volts.

So then the least significant bit=0.00000011921, the bit one to the left = double that, and one more to the left = four times the LSB and so on, right? That would lead us to the MSB representing 1V either positive or negative?

Edit: Seems to make more sense that the MSB would toggle between +0.5V and -0.5V, then the next would go either 0.25V above or below the value set by the MSB, and so on down the line.

[David Rick]

Originally Posted by Ozpeter

Going back to Karl's mountain... for a start, we have to think in terms of steps descending the mountain. Let's imagine a 16 bit staircase and a 24 bit staircase. The 16 bit staircase takes us close to the bottom of the mountain, but there's a jump from the lowest step down to sea level.

Peter, this "jump" is the same size as all the other steps in the transfer function. You must believe this, or I will make you read some very dry data sheets!

The 24 bit staircase is the same up to that point, but provides extra steps to avoid that jump at the bottom.

There are extra steps between all the original step positions.

But note that in neither staircase are the steps evenly spaced. Sound as expressed in dB is logarithmic. As you go down the staircase from the top of the mountain, the height of each step gets smaller and smaller. (If it was a ruler, the graduations get closer and closer together). The two staircases line up perfectly until you get to the bottom of the 16 bit staircase, and then the 24 bit staircase continues towards sea level with a huge number of extra steps, this being necessary because the height of each step continues to reduce. So to get anywhere near the bottom of the mountain, you need an considerable number of extra diminishing steps.

I don't know where you got this idea, but it's wrong. The only place where one sees unequal step heights is in A-law and u-law codecs, used in telephony. But even those are likely built with fixed-step converter chips in modern implementations. The nonlinear coding is then done in software.

All modern audio converters have (ideal) transfer functions using fixed step size. Let's suppose that a 16-bit converter has a full scale input range of plus or minus one volt. All its steps are the size of the least significant bit:

LSB = 2^(-15).

For an ideal 24-bit converter with the same analog input range,

LSB = 2^(-23).

By the way, hearing is not really logarithmic. That's widely believed, but it is wrong. If you want to learn an approximate mathematical rule for loudness perception, it's actually a power law, not a logarithm. Here's how my psychoacoustics professor explained it to me nearly thirty years ago:

"Equal ratios in sound intensity yield equal ratios in perceived loudness, but those ratios are not the same.

We're all familiar with the rule of thumb that a 10 dB increase sounds about twice as loud. 10 dB is an intensity ratio of 10x, but the corresponding loudness ratio is 2x.

David L. Rick

[chris319]

So then the least significant bit=0.00000011921, the bit one to the left = double that, and one more to the left = four times the LSB, right?

Correct. All bits lit would be the sum of:

1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
65536
131072
262144
524288
1048576
2097152
4194304
8388608

This adds up to 16777215. From 0 to 16777215 is 16777216 values or 2^24.

[chris319]

But note that in neither staircase are the steps evenly spaced.

WRONG!

Sound as expressed in dB is logarithmic.

We're not talking about sound in dB, we're talking about converting linear voltages to digital bits.

[Ozpeter]

Originally Posted by chris319

WRONG!


We're not talking about sound in dB, we're talking about converting linear voltages to digital bits.

...which in audio translates, in practice, purely to considerations of the dynamic range of the system. I ask again, how do you hear the increased resolution which you are stating exists in a 24 bit system, as compared to a 16 bit system, above approximately 96dB below full scale? In what way is it useful or desirable? How do you account for the 16 bit vs 24 bit inversion phenomenon that I have described?