Topic: 24-bit word length and setting levels (yes, again)

[Brian Skalinder]

From a post I made in another thread, succeeding in confusing myself about the benefits (or lack thereof) of running conservative levels when recording a 24-bit word length.  Any feedback from those with solid technical knowledge around my understanding as I describe it below?  I have little confidenced I'm understanding this correctly, and would appreciate any guidance people may offer.

Can you explain how added dynamic range allows you to run levels more conservative?  I'm not questioning the statement, I'm just trying to learn.

I'll try.    Although as I think through it again, it no longer makes sense.  At least not for 24-bit listening.  It may still make sense for 16-bit listening.  Here we go...


In an ideal world, we use all 24-bits of resolution, a full 144 dB of dynamic range.  Since 1 bit = 6 dBFS, the relationship between dBFS (our levels) and resolution (our bit-depth) looks like below, where S = signal and N = noise.  In this ideal case, if we set our levels to peak basically at 0 dBFS, we use all 24-bits of resolution, the full 144 dB of dynamic range.  The data is pure signal (no noise):

| -144 | -132 | -120 | -108 |  -96 |  -84 |  -72 |  -60 |  -48 |  -36 |  -24 |  -12 |   00 |  dBFS
--------------------------------------------------------------------------------------------
|   24 |   22 |   20 |   18 |   16 |   14 |   12 |   10 |   08 |   06 |   04 |   02 |   00 |  Bit-Depth
============================================================================================
|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|  Noise / Signal


However, none of our preamps and ADCs achieve full 24-bits of resolution - too much self-noise.  For the sake of discussion, let's assume that our preamp and ADC provides 108 dB of dynamic range, which equates to 18-bits of resolution.  Everything below -108 dBFS is noise, i.e. all the data in the least significant 6-bits is noise.  The new graph - again assuming our levels are effectively peaking at 0 dBFS - looks like the following, where S = signal and N = noise:

| -144 | -132 | -120 | -108 |  -96 |  -84 |  -72 |  -60 |  -48 |  -36 |  -24 |  -12 |   00 |  dBFS
--------------------------------------------------------------------------------------------
|   24 |   22 |   20 |   18 |   16 |   14 |   12 |   10 |   08 |   06 |   04 |   02 |   00 |  Bit-Depth
============================================================================================
|NNNNNN|NNNNNN|NNNNNN|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|  Noise / Signal


Now let's assume we change our levels to peak at -6 dBFS.  Since the noise-floor hasn't changed, we've now effectively reduced our resolution by 1-bit.  The new graph looks thus:

| -144 | -132 | -120 | -108 |  -96 |  -84 |  -72 |  -60 |  -48 |  -36 |  -24 |  -12 |   00 |  dBFS
--------------------------------------------------------------------------------------------
|   24 |   22 |   20 |   18 |   16 |   14 |   12 |   10 |   08 |   06 |   04 |   02 |   00 |  Bit-Depth
============================================================================================
|NNNNNN|NNNNNN|NNNNNN|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSSSSS|SSS...|  Noise / Signal


In this light, let's examine two qualifications of my previous statement (and one I've seen repeated often by others):

If our target listening resolution is 16-bit, we're in good shape - we're still capturing 17-bits of resolution, i.e. > 16-bits of resolution.  This assumes, though, that 17-bit resolution dithered down to 16-bit sounds the same as 18-bit resolution dithered down to 16-bit.  It also assumes any editing we perform does not benefit audibly from the extra 1-bit of resolution 18-bit provides v. 17-bit, prior to dithering down to 16-bit (or if there is an audible benefit, dithering down to 16-bit negates it).

If our target listening resolution is 24-bit (meaning an actual bit-depth of 18-bit), we're actually reducing our resolution from 18- to 17-bits.  Given that I hear a significant audible difference between 18- and 16-bit resolution, I'm inclined to say there's a significant audible difference between 17-bit and 16-bit (half of the difference between 18- and 16-bit).  I have not performed any specific testing of this suggestion, however.

Anyway...all this leads me to believe my previous conceptions were incorrect, and that we should run our levels as close to 0 dBFS as possible regardless of the final target resolution, but especially if our final target resolution is "24-bit". 

But I'm not 100% certain my understanding above is correct.  Anyone with more detailed technical knowledge able to chime in?


FWIW, two posts with lots more info regarding levels, other reasons why one may or may not want to run them hot (mixing with other channels, finding the sweetspot in your analog gear), etc.:

http://taperssection.com/index.php/topic,58384.msg906769.html#msg906769
http://taperssection.com/index.php/topic,58384.msg898317.html#msg898317

[BC]

If our target listening resolution is 16-bit, we're in good shape -

If our target listening resolution is 24-bit (meaning an actual bit-depth of 18-bit), we're actually reducing our resolution from 18- to 17-bits.  Given that I hear a significant audible difference between 18- and 16-bit resolution, I'm inclined to say there's a significant audible difference between 17-bit and 16-bit (half of the difference between 18- and 16-bit). 

Anyway...all this leads me to believe my previous conceptions were incorrect, and that we should run our levels as close to 0 dBFS as possible regardless of the final target resolution, but especially if our final target resolution is "24-bit". 

The above seems reasonable to me, with 24 bit playback in mind, you don't want to be throwing away dynamic range.

I think it is important not to hit 0 though. I just saw in recent measurements of an AD converter in  Pro Audio Review showing that the THD+n increases like crazy once you get in the -1 to 0 db range. Maybe the audibility of this is not that great in the real-world since clipped sections are usually very brief, but just something to keep in mind.

[Brian Skalinder]

Something else I wanted to add:  there must be more to 24-bit v. 16-bit than increased dynamic range.  If 24-bit v. 16-bit is all, and only, about dynamic range, then why do my concert PA recordings sound better at 24-bit than 16-bit?  No way my concert recordings have a dynamic range anywhere near 96 dB, much less 108 dB.  So I must be missing something in my previous post, which really only deals with increased bit-depth providing broader dynamic range.

For example, does the increased bit-depth really only increase the dynamic range, or does it also increase the available vertical steps used to measure the Y-axis of the waveform independently of the dynamic range?  I gotta believe the latter, else it seems my limited dynamic range concert recordings would sound the same as 16-bit concert recordings, since the concert's dynamic range doesn't approach, much less exceed, the dynamic range of either bit depth.

Anyone?

[shaggy]

I also think that the dynamic range issue is only a part of the overall improvement when going to 24bit.  I have been listening to some 24bit masters I have made vs. UV22HR dithered of the 24bit version and I have noticed a deeper soundstage.  I think that it has something to do with some overall spectral enchancement, there is more information that can be recorded at any given frequency.  I lifted this from the tweakheadz.com site:

"Bit Depth refers to the number of bits you have to capture audio.  The easiest way to envision this is as a series of levels, that audio energy can be sliced at any given moment in time.  With 16 bit audio, there are 65,536 possible levels.  With every bit of greater resolution, the number of levels double.  By the time we get to 24 bit, we actually have 16,777,216 levels.  Remember we are talking about a slice of audio frozen in a single moment of time. "

I suggest that others that can record 24bit and listen to it via a 24bit bit capable system listen carefully to their masters and the dithered versions, especially if the recording was done really close to the source or onstage.  Again, my theory is that there is alot of subtle information that can be recorded at 24bit vs 16bit, this is brought out even more when there is actually detail that CAN be recorded.  I think alot of PA taping from the section does not stand to gain that much from 24bit....while recording in a studio or onstage will.  In fact, I have noticed that if the PA or the mix is bad, you are gonna hear more of that bad sound when at 24bit. 

--------

On another note regarding the UV22HR dithering, has anyone noticed that when you dither a 24bit source that was not recorded with any peaks higher than -6dB will get the same with the 16bit source (a dithered product with no peaks higher than -6dB)?  Seems like the dithering process is not doing what it supposed to be doing or am I missing something here?  I thought that these programs mazimizes the bit usage (thus the special patented algorithms).  If there something better to use as a plugin to Wavelab, I would like to know about it.  Are people are using WAVES IDR?

[BayTaynt3d]

I by no means know much about this, but here's two conepts that I believe are also involved. For starters, any editing (at all, including fades, normalization, equalization, level balancing b/w channels, compression, reverb, etc.) creates quantization errors (at any bit rate). When you are working at 24-bit, those errors are virtually inaudible if I understand this correctly. So, that is a HUGE advantage right there for anyone who touches their sources in any way (especially when you know you are going to dither to 16-bit as your last step, but even still if you leave things at 24-bit for the master).

Now, the second piece of this I believe has to do with digital gain. See, when we record way down at -12 or something in 24-bit, we are going to raise it up to 0 one way or another (at least I will, either by normalization or maybe even with a little compression too). So, when you lose a whole bunch of range in the noisefloor and then even more by leaving a lot of headroom, you are left with a chunk in the middle. Then, when we normalize it up to 0, we essentially have to scale everything up up a little, almost like zooming in on a picture or something like that. If your picture started at 2 megapixels, then when you zoom in, you get artifacting (the jaggies), which in audio parlance would be quantization errors. BUT, if you started with an 8 megapixel image, you could zoom in a little, and it'll still look great b/c we had the resolution to do it without creating too many jaggies. I believe this concept is somewhat at the heart of the dynamic range issue. That we can give up a whole bunch of headroom, but still have plenty of resolution to zoom into (add digital gain) to get back to 0 such that we have more than enough resolution when dithering to 16-bit. That said, I wonder if you are going to leave it as a 24-bit master if that holds water any more?

Did any of that make sense to anyone?

[Brian Skalinder]

Did any of that make sense to anyone?

I understand why greater resolution provides better results during editing, which I think you're addressing with both of your comments, I think.  But I'm trying to understand what happens at the time of capture, before any editing.  If my original post is correct, we don't actually have 24-bits of signal with which to work while editing, but rather only 18-bits (per the example).  The 18-bits of resolution will provide more accurate editing, but I'm trying to understand whether we do, in fact, actually have only 18-bits of resolution.

The more I think about it, the more I believe we do, in fact, only have 18-bits of resolution (if we continue to use my example as a basis for discussion), due to limitations in our preamps and ADCs.  This means we can only utilize 108 dB of dynamic range, per my previous post.

But I think the piece that I was missing before is the fact that 18-bit audio provides a huge increase in precision on the Y-axis of the waveform, independently of the dynamic range.  The math:

• 16-bit audio = 2¹⁶ = 65,536 increments with which we may describe the audio in the Y-axis
  
• 18-bit audio = 2¹⁸ = 262,144 increments with which we may describe the audio in the Y-axis
  
  

In other words, while the extra 2-bits of resolution at our actual 18-bit resolution provide only 12 dB more dynamic range than 16-bit, the precision of our vertical axis has quadrupled relative to 16-bit.  So even when we're not capturing a broad dynamic range, the precision with which we capture -any- dynamic range has still quadrupled.

[shaggy]

In other words, while the extra 2-bits of resolution at our actual 18-bit resolution provide only 12 dB more dynamic range than 16-bit, the precision of our vertical axis has quadrupled relative to 16-bit.  So even when we're not capturing a broad dynamic range, the precision with which we capture -any- dynamic range has still quadrupled.

The analogy like when they taught the rudiments of calculus to me in high school, as the limit approaches infinity, the line becomes more respresentative of the actual function.  I understand this more as a function of time (sampling frequency) but it can be applied to any given sample as well.  The more packets of information you have at any given point, not just in the aspect of signal to noise but over all detail (which to me is the more important aspect rather than dynamic range as BayTaynt3d pointed out in the image analogy, less 'jaggies').  And this detail, when we talk of a sample itself, will contain information in the whole listening spectrum, more so than just increasing the frequency sampling rate (which will have a more profound effect on frequency range rather than detail when we talk about rates above 44.1khz).  At some point, to the listener, this will be indistinguishable from the original waveform...to some it might be at 16 or 18bit, to others it might be 20 or 24bit.

[live2496]

Hi Brian,
It's interesting that you drew the figures in terms of how it looks from metering. I tend to think of these things in reverse order due to the ordering of bits.

Anyway...all this leads me to believe my previous conceptions were incorrect, and that we should run our levels as close to 0 dBFS as possible regardless of the final target resolution, but especially if our final target resolution is "24-bit".

The closer you get to 0 dBFS the farther away you get from the dither added by the converter. However, to quantize any audio circuit you likely only need 20 bits. But, I would run the level as hot as you can without any possibility of going over the top.

The more level you record, the more detail you will get in the lowermost bits.

[BC]

The more level you record, the more detail you will get in the lowermost bits.

this is how I feel. I like to run comfortably warm (-3dB digi-peaks)

same here.

[BWolf]

In other words, while the extra 2-bits of resolution at our actual 18-bit resolution provide only 12 dB more dynamic range than 16-bit, the precision of our vertical axis has quadrupled relative to 16-bit.  So even when we're not capturing a broad dynamic range, the precision with which we capture -any- dynamic range has still quadrupled.

This is the key.  Like Freelunch mentioned earlier, 24 bit (or 18 or anything greater then 16) is about increasing your dynamic range, but really the advantage is the number of unique combinations of 1s and 0s that can be formed in the number of bits you have to work with.  This is really the reason that the dynamic range increases.  1 bit = 6 dB is no coincidence.  It is because the resolution of the bits and what they represent.

I have some great books from electrical engineering classes that explain this and have some great pics explaining it, I wish I could find them.  But let me see if I can explain it a little more.  Lets say you start with a dynamic range of 108 dB at 16 bits.   So we have 65,536 possible combinations.

108 dB / 65,536 bits = 0.0016479 dB/bit

(Now I realize that 16 bits can represent only 96 dB, which is 0.0014648 dB/bit, almost exactly the same.  I challenge anyone in the world to be able to tell the different between these step differences by ear.  Humans have a hard time detecting a 1 dB change, let alone a 0.0002 dB change)

Now lets assume you switch to 18 bits, so now there are 262,144 different combinations of 1s and 0s.  Assuming we still have 108 dB of dynamic range, you get

108 dB / 262,144 bits = 0.0004119 dB/bit

Alright, the same amount of dynamic range, but each bit represents a smaller "chunk" of sound.  With only increasing 2 bits of resolution, we have decreased the amount that each bit represents by fourfold.  That means that the 18 bit recording would have to change by 4 before the 16 bit changes by one.

Even if we say that because we increased by 12 dBFS because of the 2 extra bits, we have

120 dB / 262,144 bits = 0.0004577 dB/bit

So even with increasing the dynamic range, we still represent 120 dB with 18 bits 4 times better then 108 dB with 16 bits.  Hope this all makes sense.  If anyone has any questions, i'd be happy to try and field them.

Also, a side note, is that none of this means anything if the sampling rate isn't high enough.  You can completely eliminate the advantages of 24 bit recording by using a low sample rate.  These obviously have a direct relation to each other. 

[thegreatgumbino]

Also, a side note, is that none of this means anything if the sampling rate isn't high enough.  You can completely eliminate the advantages of 24 bit recording by using a low sample rate.  These obviously have a direct relation to each other. 

Alright, too much vodka to digest the rest of your post, Brad, but I am intrigued by this statement.  So, what is a "high enough" sample rate to reap the benefits of 24 bit?

[BWolf]

Also, a side note, is that none of this means anything if the sampling rate isn't high enough.  You can completely eliminate the advantages of 24 bit recording by using a low sample rate.  These obviously have a direct relation to each other. 

Alright, too much vodka to digest the rest of your post, Brad, but I am intrigued by this statement.  So, what is a "high enough" sample rate to reap the benefits of 24 bit?

Now this may cause some debate, but according to digital theory, the sampling rate needs to be no more then twice the highest frequency to capture the analog signal in the discrete time realm.  So if we are trying to capture the first 20,000 Hz, then the sampling rate needs to be no more then 40,000 times a second. 

Without getting into a lot of ugly detail, the more dynamic range we are trying to capture, the higher the sampling rate need be.  Given that most of us are recording at 96k, which is way higher then need be, this is a non issue.  But there is also a strong argument that when recording in 24 bit, no more then 48k in sampling is needed (or slightly higher to prevent aliasing).  Now I record at 24/96 even though I have been taught (and learned through experimentation in lab) that 24/96 and 24/48 will provide almost the exact same signal in the discrete world.

[BWolf]

Now this may cause some debate, but according to digital theory, the sampling rate needs to be no more then twice the highest frequency to capture the analog signal in the discrete time realm.  So if we are trying to capture the first 20,000 Hz, then the sampling rate needs to be no more then 40,000 times a second. 

I think the "needs to be no more than twice" above should be "needs to be at least twice". I believe Nyquist specifies the minimum required sample rate, not the max.  But the "needs to be no more than 40,000 times a sec" is not correct.  My point here is not to be picky but to emphasize that 44.1K can *barely* represent a waveform at 20khz.

Right right.  Sorry if i wasn't clear.  THEORETICALLY, the sampling rate needs to be greater (not >=, but actually greater then) than twice the highest frequency sampled.  Nyquist does specify minimum, but also states that anything above that is overkill.  But with filters, and limits on electronic circuits, etc, oversampling is needed.  It was found that for a range ending at 20k, 44.1k samples per a second will represent sufficiently, thus becoming the standard.

Now for your experiment, its different.  Extereme accuracy is not what we are going for (accuracy, yes, but to that extreme, no way).  They are two very different applications.

[BJ]

In other words, while the extra 2-bits of resolution at our actual 18-bit resolution provide only 12 dB more dynamic range than 16-bit, the precision of our vertical axis has quadrupled relative to 16-bit.  So even when we're not capturing a broad dynamic range, the precision with which we capture -any- dynamic range has still quadrupled.

This is the key.  Like Freelunch mentioned earlier, 24 bit (or 18 or anything greater then 16) is about increasing your dynamic range, but really the advantage is the number of unique combinations of 1s and 0s that can be formed in the number of bits you have to work with.  This is really the reason that the dynamic range increases.  1 bit = 6 dB is no coincidence.  It is because the resolution of the bits and what they represent.

I have some great books from electrical engineering classes that explain this and have some great pics explaining it, I wish I could find them.  But let me see if I can explain it a little more.  Lets say you start with a dynamic range of 108 dB at 16 bits.   So we have 65,536 possible combinations.

108 dB / 65,536 bits = 0.0016479 dB/bit

(Now I realize that 16 bits can represent only 96 dB, which is 0.0014648 dB/bit, almost exactly the same.  I challenge anyone in the world to be able to tell the different between these step differences by ear.  Humans have a hard time detecting a 1 dB change, let alone a 0.0002 dB change)

Now lets assume you switch to 18 bits, so now there are 262,144 different combinations of 1s and 0s.  Assuming we still have 108 dB of dynamic range, you get

108 dB / 262,144 bits = 0.0004119 dB/bit

Alright, the same amount of dynamic range, but each bit represents a smaller "chunk" of sound.  With only increasing 2 bits of resolution, we have decreased the amount that each bit represents by fourfold.  That means that the 18 bit recording would have to change by 4 before the 16 bit changes by one.

Even if we say that because we increased by 12 dBFS because of the 2 extra bits, we have

120 dB / 262,144 bits = 0.0004577 dB/bit

So even with increasing the dynamic range, we still represent 120 dB with 18 bits 4 times better then 108 dB with 16 bits.  Hope this all makes sense.  If anyone has any questions, i'd be happy to try and field them.

Also, a side note, is that none of this means anything if the sampling rate isn't high enough.  You can completely eliminate the advantages of 24 bit recording by using a low sample rate.  These obviously have a direct relation to each other. 

so moving back to practical applications(sorry Brian, just a quick question!)...levels low(as teddy suggested) or levels high (as moke and gordon suggested)?  having only run 24bit with my new R-09, i read from the previously posted thread, and kept my levels low (around 8 on the r-09) which produced results ranging from -7 to -4 at which time i normalized in Audacity.  As I don't have any recordings to directly compare low to high levels with the R-09, should I run higher to get the -3 to -2 peaks? 

[SparkE!]

108 dB / 65,536 bits = 0.0016479 dB/bit

108 dB / 262,144 bits = 0.0004119 dB/bit

120 dB / 262,144 bits = 0.0004577 dB/bit
 

Dude...  There are 6.0205999132796239042747778944899... dB/bit, no matter how many bits you have or what dynamic range you have available.