24 bit headroom

It's not an accounting gimmick.

Well-dithered 24 bit allows for about 150 dB of dynamic range, whereas 16 bit allows about 100 dB. If all your recordings are maxed out so the loudest parts hit zero dB on the meter, then a 24-bit recording is capable of recording sounds as quiet as 150dB lower than the loudest sound, which slightly exceeds the dynamic range of human hearing. It is going to contain 50dB more information than a 16-bit recording of the same thing.

16-bit recordings have a very adequate dynamic range of 100dB, which is not quite as wide as human hearing is capable of percieving, but it is plenty adequate to reproduce most meaningful program material in most playback scenarios. But that is only if you are using all 16 bits.

Let's say that you are recording a fairly dynamic clean-tone electric guitar part at 16 bits. And let's say you have PERFECTLY guessed the dynamic range and set your input gain so that the loudest peak of the loudest power chord comes in at -.01dB, just under digital clipping. Now let's say that the quieter parts of the single-note passages are coming in at -24dB (not unusual). No problem, since you were going to turn up those sections either with compression or distortion or by mixing it that way, right? Except, those 16bit fixed-point samples contain only 12bits of meaningful data in the quiet parts, and 4 bits of nothing (that 24 dB headroom). When you turn up those quiet parts, it is like blowing up a low-resolution photograph to a bigger size-- it looked fine when it was little (although you could hardly see it), but increasing the size reveals its grainyness and low resolution. and the decay and fade-out sections of those notes are even quieter, and have even lower resolution.

Anything that doesn't use all 16 bits of available dynamic range is not going to have a full 16 bit resolution. And when you turn up those lower resolution segments, then fade outs start to take on a "stepped" sound, transients and details start to sound grainy or digital, and your overall quality starts to sound less like "CD-quality" and more like mp3 or old video game quality, because you are essentially making 8- or 12-bit recordings, and then mixing them to a higher-resolution, 16-bit format.

And the above example was just a single-track electric guitar, which is an instrument with a relatively narrow dynamic range that is typically played without much in the way of performance dynamics. Imagine trying to record a symphony orchestra, or a dynamic piano piece, or a jazz combo, where the dynamic swings can vary a LOT, and are not always predictable. In order to avoid overs, you might need to set your levels so that your peaks are coming in at -6dB, and then find that your quietest sections are averaging -50 dB, which means that for the quiet passages (which might be most of the performance in a classical piece), you are actually only recording at 8-bit resolution.

This is one reason why, in the days of 16-bit recording, so many people were using compressors and limiters to bring up the signal before it was converted to digital. Because, if you didn't maximize the recording level, then you were going to lose all of that wonderful resolution and detail that digital offers.

Now we come to 24-bit. 24bit has MORE dynamic range than human hearing, which means that you could set up a mic in front of a symphony orchestra or whatever, set the record level so that the loudest peaks are coming in at -6, and capture more resolution than the human ear can hear. Even the sections that are at -50dB will still have 16 bits full of meaningful resolution, and that's at the extremely quiet parts. So if you want to turn those parts up, you can do so with no loss of resolution when it comes to the final 16-bit version, because you have enough "extra" information to provide additional detail and resolution when you bring up the level-- like blowing up a high-resolution photograph-- it looked great small, and it looks even better big.

If you are making a 24-bit mutitrack project that is eventually going to get mixed down to 16-bit digital, then you have WAY more dynamic range than you will ever be able to fit into your final mix. 8 tracks of 24-bit digital adds up to a total of 192 bits per sample (in the multitrack project) that all have to be mixed and dithered down to 16 bits, which means that 176 bits are going to be thrown away for every 16 bits we keep.

What does this all mean? that if you are making 16bit recordings, then it is critical to maximize your record levels as much as possible, which means that you will always need to run the risk of digital overs, or of unnaturally limiting the dynamic range of a performance that turns out louder than you were expecting it to be.

But if you are making 24-bit multitrack recordings, then there is no need to maximize the recording levels, because you have more resolution than you need. You could set your record levels so that your highest PEAKS were coming in at -45dB and STILL have better-than CD quality sound. So there is absolutely no technical reason to compress or limit the incoming levels, or to obsess about getting your incoming level right up to 0dB, or any of it. You can just set the levels so your peaks are at -6 or -12 and not sweat it, because you still have WAY more resolution than is ever going to end up on the finished product. No clips, no squashed sound, no quantization artifacts, no worries.

That is the headroom part of it. As for the noise floor, that's up to your skill and the quality of your gear. But you're still better off recording at 24-bit, even if it's mostly just getting higher-resolution noise.

Cheers.

Hey Yep,

I didn't have the time to read your response, let alone write one that thorough. Just wanted to say thanks for contributing!

Great, thanks yep, now I have to go out and buy new converters!

Excellent dissertation.

Now where does all this extra resolution go when we make a 16 bit CD? Is it lost?

R

any amount of noise can be amplified through subsequent processing, for example compression

Ok, coming in, I understood how a 24 bit ladder circuit could register a smaller non-zero voltage than a 16 bit circuit could, but there just didn't seem to be enough advantage there to create the "scads of extra headroom" that I keep hearing about. This ties in with what ohhey was talking about.

From Yep and Name1432's responses I think I have the missing piece of the puzzle, but I want to make sure it's nailed down before trying to move ahead.

From this:

"...24 bit allows for about 150 dB of dynamic range, whereas 16 bit allows about 100 dB...." (Yep)

and this:

"...that's 6dB out of 96dB; but in 24 bit audio it's only 6dB out of 144dB..." (Name1432)

...I'd say my original supposition that the same signal is divided into more (24 bit) or less (16 bit) discrete digital level values is only part of the truth. If I'm reading between the lines here you are also telling me that 24 bit can sample higher absolute voltages without clipping than 16 bit can. Put another way, you aren't slicing the same amplitude spectrum into more or less pieces, you are also skewing it to the right (positive X) in the case of 24 bit. Since I'm not sure I have this, and I am sure I'm not using standard audio terminology, one more way of saying it is that 0 db on the meters is not the same sound pressure level in 16 bit as it is in 24 bit. 0 db in 24 bit corresponds to 144 db and only corresponds to 100 db on in 16 bit recordings?

If this is true, then 24 bit slices have to be smaller than 16 bit slices, for the rest of Yep's and Ohhey's post to add up, but not as small as you might expect. In my case I assumed they were equal to ((2**16)/(2**24)) times the size of the 16 bit slices, but this isn't true is it? If zero voltage equals zero voltage on both scales, and 0db equals 144 in 24 bit, and 100 in 16 bit, then the slice size ratio is the above value times 144/100, yes?

No....well, the idea seems right but sound pressure is logarithmic, and I'm not up to that kind of math tonight but I think I'm getting it. If that's the case then the "scads of headroom" only partly comes from the finer resolution as described by Yep, but also a significant part of it is being able to record a hotter signal?

Yes?

No?

Maybe?

Once I get this down, there's a good chance I'll have a few more q's but this has to come before I can really assimilate the rest.

jeff... in trying to answer your recent question i've lost confidence in my original posts, i need to research before i mislead you. i'm another who's not up for logarithms at this moment but i need to learn this so i'll post what i find

the way i've been thinking, 24-bit does not allow for higher voltages.

There's some confusion here on digital domain vs. analog.

The output voltage is entirely dependant on the DAC. It could be setup for 1,000 volts output at 0dBFS if they wanted it to be, the minimum sample value would still be exactly 144dB below that, 150dB effectively with dithering.

The same holds true for the input voltage, it depends on the analog circuitry of the ADC.

Generally, manufacturers loosly adhere to standards like +4dBu or -10dBV but they certainly don't have to.

Remeber that dB's are a logarithmic ratio and hold no specific value by themself. They are used for relative measurement, that's it. In the case of the above standards, the ratios are defined based on 4dB above a standard "unloaded" output (equivalent to 1mW into a standard audio load, which I forget the impedance) which works out to 0.775V. The other standard is 10dB below 1V.

ORIGINAL: nprime

Great, thanks yep, now I have to go out and buy new converters!

Excellent dissertation.

Now where does all this extra resolution go when we make a 16 bit CD? Is it lost?

R

Basically, yeah. The conversion process from 24bit to 16bit is pretty simple-- the computer just throws away the least important (quietest) 8bits. But your recording will still benefit noticably from doing all of your processing and mixing on 24bit signals, even if the above-mentioned "headroom benefit" is of no use to you at all.

Moreover, the use of good dither actually INCREASES the dynamic range of digital audio through a process that's more complicated than I want to describe right now. So a 16bit master can actually be made to capture and reproduce MORE than 16bits of information, and again, you benefit from working at 24bit for as long as you can. It's like taking a low-res photograph of a higher-resolution photograph-- you get better results than taking a low-res photo of another low-res photo.

Cheers.

Oh, and for those who are confused by dB (if you've made it this far)-- dB measurements are "logarithmic" NOT "arithmatic." What does this mean?

It means, with 150dB of dynamic range, the difference between the loudest sound and the quietest sound captured is about 30 TIMES more than with 100dB of dynamic range-- NOT 1.5 times more (as percieved by humans, anyway).

Cheers.

yep, small correction.

I believe that a doubling of power is a 3 dB increase, not a 6 dB increase as you stated.

Factoid: to make a signal sound twice as loud requires a ten times increase in power. Thats why a 200 watt amp is not twice as loud as a 100 watt amp.

R

ORIGINAL: nprime

yep, small correction.

I believe that a doubling of power is a 3 dB increase, not a 6 dB increase as you stated.

Factoid: to make a signal sound twice as loud requires a ten times increase in power. Thats why a 200 watt amp is not twice as loud as a 100 watt amp.

R

Technically, +6dB means a doubling of the signal voltage and -6dB means a halving of the signal voltage. You're right that signal voltage doesn't necessarily translate into "power," and my terminology was pretty vauge. I should have said "+6dB means doubling the signal strength" or something similar. 3dB is usually regarded as the smallest change in level that average people will percieve.

You're absolutely right about amplifier power-- to most people, an increase of 10dB SPL sounds "twice as loud," and typicially, that means ten times the electromotive power needed to get a 10dB increase in SPL. So you'd need a 1000W amp to get the same result as two 100W amps. But fear not, guitar players of the world-- adding speakers can add a lot more volume without increasing amplifier power.

This is why recording engineers will often refer to a 6dB increase as "twice the level" (because they're talking about signal strength), whereas people in sound reinforcement will almost always call 10dB "twice the power" (because they're talking about SPL and percieved volume).

Cheers.

Yeah, something like that. I actually don't know the value of different amplitudes in binary off the top of my head (believe it or not), but you got the right idea.

Here's a really crude but illustrative example (non-binary). Let's say you have three digits that you can use to assign any numerical value, but it HAS to be either zero or less (a negative value). Call this a "three bit non-binary system." So the highest possible value you could get is 000, and the lowest possible value would be -999 and the medium value would be -500 (or maybe 499 or 501, I forget, but whatever). So in this non-binary "3-bit" system, -999 is silence, and 000 is full volume. There are exactly one thousand possible values.

Now let's say you went to a "four-bit" system. Now you can have values ranging from -9999 (silence) to 0000 (full scale). There are TEN thousand possible values. Let's say you have a sample with amplitube value -5038, almost exactly half of the possible amplitube, and you want to convert it to the previous "3-bit" system. Just lose the last digit (the last bit) and you got -503. Still pretty much exactly in the middle of the scale, just a tiny fraction of a percent off from the higher-resolution value. That's how bit-rate conversion works. You could further reduce that to a "2-bit" system (one that goes from -99 to 00) by simply dropping off the 3 and you end up with -50, still halfway up the scale.

In this case, the conversion doesn't do any rounding, it just drops off the last bit. So that conversion from -5038 (four bit) to -503 (three bit) probably SHOULD have been a conversion to -504, but we're pretty close nonetheless, and probably not gonna lose many album sales because of it. But, just in case you're worried, this is EXACTLY where dither comes in. Dither DOES cause the samples to "average" in-betwen values, and actually forces the system to achieve results that are technically less precise on a sample-by-sample basis, but that are MORE accurate overall.

Dither is a simple thing-- it's just low-level random noise mixed in with the signal to be converted, but it's effects are NOT simply to mask the "digitalness" of low-level digital sounds. It actually IMPROVES fidelity and dynamic range. But that's a topic for another thread.

Cheers.

Ok, I sketched it out on scrap paper and confidence levels are soaring. I think I have it. Gotta love this. Absorbing a fundamental principle doesn't happen everyday.

The sound of bank vault doors slamming open advances down the corridor towards me, echoing towards crescendo, untold treasure within is, oh yes I think it is MINE!

All I have to do now is the legwork necessary to SCOOP IT UP!

(dancing a touchdown jig around the computer chair...one lap....two laps...three laps...

...the older, wiser jeffers staring on in disapproving silence, picking his time to wag a stiff finger in my face..."down boy, DOWN!")

...sigh...

Ok, going all the way back to the original question, I have to conclude that there are TWO, significantly different definitions of the word "headroom".

In the analog world, the term refers to the span of signal strengths between the noise floor and a hard clip. In the digital world, headroom refers to the span of values between clip and audible stairstep effect? Yes?

If true, then under the analog definition of headroom, there should be no difference between 16 bit and 24 bit. Clip stays the same, noise stays the same, headroom stays the same, but in pure digital there is no analog noise so other effects become more important.

Which brings me to what Yep and Name1432 mentioned...a 6db induced noise..."digitizing audio at any bit depth introduces 6dB of digital noise"...is this deliberate, what you were referring to recently here Yep? Or is it an artifact of the DA conversion process, unwanted and accidental?

ORIGINAL: jeffers_mz
Which brings me to what Yep and Name1432 mentioned...a 6db induced noise..."digitizing audio at any bit depth introduces 6dB of digital noise"...is this deliberate, what you were referring to recently here Yep? Or is it an artifact of the DA conversion process, unwanted and accidental?

Jeff please put no weight on my earlier statement about 6db of noise. I was trying to recall something I read about ten years ago, and I might have gotten it totally wrong yesterday; sorry if I led you down the wrong path. I'm learning from Yep and others in this thread too at this point.

I do wonder if compressing-then-amplifying 24-bit signal could make the noise significantly audible; I'd assumed it could, but now I'm not sure

Um, no.

Noise-reduction strategies have been discussed at some length on these forums, and you should run a search and reply to one of the threads that has already been started if you want specific advice on noise-reduction. But for a whole bunch of reasons, recording at 24-bit for the purpose of specifically capturing extra noise that you later want to drop is not really likely to be a rewarding or useful strategy.

Remember, when you produce the finished CD, you are not only going to dither to 16bit, you are also going to mix all those 24-bit tracks together to a single track anyway. So you are ultimately going to lose a lot more than 8 bits per track by the time you get from a 12-track project down to a single stereo file.

And if your noise floor really is only in the bottom 8 bits of a 24-bit track (that is, it's -96dB or lower), then you already have a damn good noise floor and should have nothing to worry about.

I think you are in danger of outsmarting yourself here. 24 bit will give you more accurate, detailed, and sensitive recordings than 16bit, and it will allow you to process those recordings with less digital loss of fidelity. That's all. It will record your instrument, your ambience, and your noise more faithfully than 16bit will.

I don't know a whole lot about photography, but I think the analogies with digital photography are useful, because there doesn't seem to be as much "voodoo" to it in most people's minds. 24bit is like taking a higher-resolution photograph, and that's about it (or maybe like taking one with 24-bit color instead of one with a smaller number of colors). You get a more accurate picture of whatever's there. If the lens is dirty or the room is smoky or there's an ugly person in the field of view, it won't make that stuff prettier. It will only give you a better picture of the bad stuff.

Cheers.