i thought it is best to record around -18dbfs

[flatfinger]

southside;
you are in good company when you avoid calling it "resolution". I made a quick note of a section from a very long, most excellent ,thread over a t the gearsluts which featured Paul Frindell of sony Oxford pluggs design fame , . He said the following when trying to debunk the resolution concept/analogy......

You are NOT losing resolution - as I have said before you must lose that concept in order to go further in understanding how digital audio works. There is no such thing as resolution - ALL levels use ALL bits The only time it LOOKS as though they don't is on your editing screen.

All other ways of reducing gain in the analogue domain digitally (short of a motorised potentiomenter or banks of physical relays) will cause more degradation than simply reducing the level in the digital domain.

Reducing the level in the digital domain to even -50dBfs with modern DACs is better than using an analogue VCA.

I think that where I have the most difficulty in my still , admitadly rudimentary understanding of digital is the reconstruction from the encoded , discreet steps back into the continuous analoug.







Reggie , It certianly was'nt my intention to offend, I apologize as it would'nt be the first time I missed the point!!!!!!!

 

[Farview]

So the issue of the analog driver amp distorting before the converter reaches 0dbFS is NOT a gain staging problem?

No. It's a design problem.
BTW, are you thinking that a driver amp is a mic preamp? Katz is talking about the driver chip in a set of converters....12 years ago. This has nothing to do with gain staging, it has to do with a design flaw.

Back to mic preamps:
Since a preamp will react differently to high transients than it does to high RMS, the best compromise to match the levels is leaving headroom for the transients and keeping the RMS signal in the preamps sweet spot.


You could choose to redesign the entire system so that it works the way you want it to, but that isn't the way audio equipment is designed at the moment. If you are using off-the-shelf stuff, you will be dealing with these design realities.

Back to Bob Katz, what is his recommendation for the problem of driver amps distorting on some signals before 0dbFS is reached? Dont use the last few db's, or the last bit? As a temporary solution, yes.
His recommended solution? Install a driver amp with a higher undistorted output. More analog headroom.

Again, he is talking about the analog driver chip in an outdated set of converters.

back to preamps:
It's not headroom if you use it up with the RMS power of the signal. Again, coming up with a new standard is not going to help anyone that is using todays stuff.

I agree with him. I think it's called gain staging.

It's only gain staging when you move the sweet spot of everything in the chain up so that you can use all the bits. Of course, when you do that, what happens to the sounds that have a really high crest factor? Now, the RMS of things with huge transients are going to reside farther below the sweet spot of the preamp.

 

[Farview]

I think that where I have the most difficulty in my still , admitadly rudimentary understanding of digital is the reconstruction from the encoded , discreet steps back into the continuous analoug.

Just like that quote said that you have to give up the idea of resolution, you need to give up the idea of stairstepped audio. For two reasons.
1. It doesn't really work like that.
2. Anything 'lost' in the steps would have to be happening at a frequency above the nyquist frequency. We wouldn't be able to hear it anyway.

 

[snow lizard]

I'm sorry sir, I'm just not buying this dynamic resolution thing.

Ok.

Say we are dealing with audio at a set bitrate --16bit. The audio is dynamic but hits up to the full scale (0dbFS) and we want to turn it down 6db. So now we gotta fit 65,536 steps of volume changes into the 32,768 possible steps for the range between -12 and -6? Is this really what you are saying?

No.

Because it strikes me as rubbish. Hopefully I am misunderstanding what you are trying to say.

It's a popular belief that dynamic resolution in digital thins out as it goes down in level. This could be an explaination of why 16 bit converters calibrate analog line level to a higher digital level than 24 bit, where the problem is rendered moot.

This is really not an important discussion to get sidetracked on. The key is in how signed, fixed point quantization gets mapped onto the dBFS scale. People have tried to come up with the argument that higher levels results in more robust quantization as an excuse to print hotter levels. As levels drop, the quantization steps supposedly become greater. It's a confusing description, and a half-truth at best.

Decibels aren't linear, so it kind of makes sense that you wouldn't need as many steps in the lower dB ranges. A reduction in level of 2 dB at 1/40th the average level doesn't exactly equate to the same thing as 2 dB from the average level itself. Logically, how can a uniform quantization step scale to the logarithmic dynamic range in decibels, and would it actually make sense for it to be top heavy on the decibel scale? OR, is quantization done logarithmically at all? That's where the confusion comes from.

Assuming that line level is -18 dB, you'd need a drop in voltage of x to reduce from -18dB to -21 dB. You'd need a drop in voltage of y to go from -80 dB to -83dB. The difference in dB's is 3 in either case. Does x = y?

IME, ANY quantization scheme should not be an excuse to overlook sensible gain staging. At the same time, people do not advocate printing RMS levels of -30 dB in 16 bit, while saying that you can actually get away with it in 24 bit even if it's less than ideal. People regularly claim that a level of -48 dBFS at 24 bit yields the maximum resolution of 16 bit. My previous posts were an attempt to scale the phenomenon, if it is infact correct.

The big problem is where people say "the best level is the maximum peak before clipping". No it isn't, for every reason already discussed in this thread.

My understanding is that there are as many amplitude steps (in a given bitrate, say 16) between 0 and -6 as there are between -90 and -96.

That would be very handy if line level were calibrated to -90 dBFS.

Only problem is the LSB sounds nasty/unnatural since it is simply on or off. For the digital device to be able to interpret any amplitude range between -96 and "digital black," more bits are needed because the steps will have to be much smaller. And changing this bitdepth doesn't screw up your amplitude by packing your 0 to -12 range now into the 0 to -6 range or something, it just extends that LSB deeper towards "digital black" by giving more possible amplitude values below what you already had.
Convince me otherwise.

The LSB is a problem area of the sample because it represents the zero crossing of the waveform.

In 16 bit audio:

2^16 = 65,536

This is the maximum number of quantization levels in 16 bit audio. With all the bits turned off, you're exactly at the zero line and they get signed interger values. It gets mapped as:

[-32,768 : +32,767]

as the negative and positive values that can be expressed in the waveform. One of those values (of the 65,536) doesn't count because it represents no signal. This creates difficulty in resolving the LSB. The result is noise at a very, very low level. Rather than adding more bits, it gets masked with dither in 16 bit. 24 bit moves the LSB to a dynamic range that your amp can not reproduce.

If you want to check out different quantization methods, there's fixed point, floating point, A-law and u-law. None of them actually change gain structure, so it's really not important.

Line level is important.

Because it strikes me as rubbish.

Used as an excuse to overdrive gain staging, it is.


sl

 

[SouthSIDE Glen]

southside;
you are in good company when you avoid calling it "resolution". I made a quick note of a section from a very long, most excellent ,thread over a t the gearsluts which featured Paul Frindell of sony Oxford pluggs design fame , . He said the following when trying to debunk the resolution concept/analogy......

Wow, I'm in an awkward position of somewhat critiqing a post that was meant to actually agee with me . And simultaneously questioning some quotes from a guy who sounds like he should be an expert in the field. BUT....

ALL levels use ALL bits

Technically true, but kind of misleading when taken out of context. An 8-bit (for example) word with the value 00010001 equals -24.375dBFS. But - as I understand it - so would the 5 bit word 10001, as long as each bit position represented the same thing. The leading three zeros in the 8-but word are not signifigant bits for that value at all, other than to act as placeholders IDing the value that the forth bit actually represents. The fact that it remains an 8-bit word and that therefore "all bits are used" is irrelevant in that case.

What's relevant is how many bits there are AFTER the first "1", i.e. how many bits are actually signifigant. And the longer that string of bits, the finer the precision of the translated value.

All other ways of reducing gain in the analogue domain digitally (short of a motorised potentiomenter or banks of physical relays) will cause more degradation than simply reducing the level in the digital domain.

Again, kind of bad if taken out of context. Definitely not true if one's problem is overdriving the ADC, which is one of the only reasons one would have for dropping the digital gain from the analog side.
---
Damn, this thread is as hard to nail down as a pile of warm jello.

G.

 

[pipelineaudio]

I hope this becomes a sticky

 

[flatfinger]

1. It doesn't really work like that.

Dam!!!

 

[Tim Gillett]

True - BUT...

*Tape* had about the same operating level as the gear pounding on it. A signal that's hitting tape at 0dBVU (a "hot" tape level) is the same as a signal hitting digital at -18dBRMS (or wherever the converters are calibrated).

The signals some people want to record at would cook right through analog tape (figuratively speaking).

That, PLUS the fact that there's no inherent noise to be concerned with, it's a slam dunk.

A signal hitting a tape (hot or not so hot tape) at 0VU (depending on the nWbm you have the machine calibrated at) already had distortion in it. It only got more as you got higher. At 18 db above that initial admittedly small distortion it would indeed be pretty fruity. But that's tape! I've worked with it for 30 years.

Whereas with digital, the distortion at 0dbFS (peak indicated) will be no more than it was at -18dbFS.

Digital is not analog tape! Now there's a controversial point.

"you dont have to" is a very different argument from "you must not".

Yes my initial advice (track as high as you can but dont clip) was too stringent. It would do no harm at all but in practice, with good gear, one can be more laid back about getting the highest bit level. The dynamic limitations are still analog, not the 24 bits.

I feel what Glen said in #133 is balanced.

Tim

 

[Reggie]

No prob, flatfinger.

OK, I don't think I am having any luck making an intelligent argument, so I guess I will drop this for the moment and perhaps search out some solid facts (not that I don't trust the facts of a snow lizard, but hey, we are all just random dudes on the Interweb). Thanks for the brainfodder. But first:

Glen, are you sure the math works out like that? Here is an article on the linearity of the voltage value represented by each bit that I found interesting:
http://www.rogernichols.com/EQ/EQ_98-08.html
He is basically saying that with the actual bits, the 1100010100010111 or whatever, each bit in front of the LSB keeps doubling in value. This is how the 65,536 different values are of course derived. We're all there. BUT these values all represent linear voltage values. With 16-bit information, there are 65,536 steps between the negative clipping point and the positive clipping point. This means that each step is equal to 10 volts/ 32,767 steps, or approximately .0003518509476 volts per step. This would seem to refute the idea that the higher db levels have more steps representing the range than lower db ranges...although I guess this sort of explains why it is called "linear PCM" The term "linear" in Linear PCM means that each change in bit value represents the same change in the analog amplitude value.......the mind boggles...

 

[SouthSIDE Glen]

Glen, are you sure the math works out like that?

The math itself is the only thing that I am solidly sure of in this wacky world anymore . It's basic binary math, Reg. And Mr. Nichols basically confirms that, though he is explaining it in a slightly different way. It's actually basic stuff that anybody here should be ale to do pretty easily.

The only thing "unconventional" is the artificially (but not arbitrarily) chosen reference starting value scale where the MSB represents a particular 6dB range. The only way to reconcile the known dynamic ranges with that 6dB reference value, and have it actually all add up in a mathematically consistant way is by what I described in the response to you in post 138 (if someone else can come up with a different way of doing it that fits all the known constant decibel and range values and is mathematically consistant with the necessary binary math, I'm all ears.)

He is basically saying that with the actual bits, the 1100010100010111 or whatever, each bit in front of the LSB keeps doubling in value. This is how the 65,536 different values are of course derived.

Thsi is the same thing I described, except I described it using the MSB as the mental starting point rather than the LSB. The results are the same and the actual animal we are both describing is the same; I just described it from the direction that I thought might be more illuminating (at least that's the mental handle that finally fliped the light switch for me.)

BUT these values all represent linear voltage values. With 16-bit information, there are 65,536 steps between the negative clipping point and the positive clipping point. This means that each step is equal to 10 volts/ 32,767 steps, or approximately .0003518509476 volts per step.

No disagreement there, that's exactly true for a 16-bit word.

This would seem to refute the idea that the higher db levels have more steps representing the range than lower db ranges

Not at all. Again, look at the math itself; go to first principles: With 16-bit information there are 65,536 "steps". With 24 bits, there are 16,777,216 such steps, a difference by a factor of 256 just by adding 8 more bits.

Look at that from the standpoint of the decibel measurement. At 16 bits, one has 32,767 steps (only a single polarity is counted) to fill in a range of 96dB. That's less than 3/1000ths of a dB per "step"! That's only about 6,000 times less than what even the best golden ear can usually perceive. Moving that to 24 bits, which by the same recipe is 8,388,608 "steps" to fill in 144dB of dynamic range, and we now have less than 2/100,000ths of a dB per step. It's the difference between a ridiculously microscopic small number at 16 bits and a hyper-ridiculously electron-microscopic small number at 24 bits. At either alleged "resolution", it's so far below human hearing as to be completly unimportant and discardable as any kind of concern.

And finally, I think I mentioned it before (this thread is getting too damn long to keep track of ), but even if the difference in alleged "resolution" were important, taking a 24-bit value for which only the lower 16 bits are signifigant (i.e. the top 8 bits are zeroes) and digitally raising the volume to "use up all the bits" will do NOTHING to increase the resolution, because all it's doing is shifting the bit values over by 8 bits and then filling in the bottom 8 bits with zeroes. There is no increase in the precision of the binary value, and therefore no actual increase in "resolution" when boosting by digital gain only. And as an added dis-incentive to do so, one is actually just raising the volume of the noise floor (both recorded from analog and the digital bottom) by 8 bits, or 48dB. Only those with needles hanging from their arms really want a 48dB boost in noise volume.

G.

 

[snow lizard]

That's a great link, Reggie. If he's right about that stuff, it confirms what I'm talking about.

This means that each step is equal to 10 volts/ 32,767 steps, or approximately .0003518509476 volts per step. This would seem to refute the idea that the higher db levels have more steps representing the range than lower db ranges...

No. It doesn't refute.

If 10 volts is the entire range of the system, that would put a line level of 1 volt at -20 dBFS, which is good headroom!

If we were to take a 0 VU = 1 volt signal and take it to -6 dB VU, it would be at 0.5 volts. That's half the volts.

In digital, if we were to take a 0 dBFS signal and reduce it to -6 dBFS, it's like subtracting 1 bit. You end up with half the quantization slices. (We're all there!)

The idea that the bits scale to the volts makes perfect sense.

Now if we were to take the -6 dB VU = 0.5 volts signal and run it down to -12 VU, how many volts would we have?

(here's a link: )

http://www.answers.com/topic/decibel?cat=biz-fin


sl

 

[flatfinger]

Where is Nika Aldrich When you need him!

I've got to read his book again, but I'm afraid my head will explode if I do !!!!!!!!!!!!!!!!!!


Any how , I use the PSP vintage meters with 0VU set at -14dbfs and find my results to be satisfactory w/ no clipping. Thats complicated enough and Works for me !!

As you were.


Bet this about kills this thread: What was it about again??????????

 

[jrhager84]

This thread seriously makes my fuckin' head hurt....

 

[mrhotapples]

This thread makes me feel like I can make smarter decisions! Good thread. This should DEFINITELY be stickied.

 

[SouthSIDE Glen]

I hope this becomes a sticky

This thread seriously makes my fuckin' head hurt....

Good thread. This should DEFINITELY be stickied.

Well, it certainly is a sticky topic...or as they say across the pond, "a sticky wicket, eh wot?"

Seriously, though, I don't think I've seen anything stickied on this board since the days of Harvey's microphone dissertations. I don't think Dragon bothers with that feature any more, does he?

G.

 

[NYMorningstar]

This thread is full of as much bad info as it is good info. It should be deleted not stickied or whatever it's called. Save us all the trouble of reading it and maybe we can all get back to important stuff like mixing & mastering.

 

[SouthSIDE Glen]

If we were to take a 0 VU = 1 volt signal and take it to -6 dB VU, it would be at 0.5 volts. That's half the volts.

Or to use some real world values:

+4dBu = ~1.23 volts.
-10dBV (a.k.a. -7.8dBu) = ~0.32 volts.

Difference between the two values in absolute decibels = 11.8dB (very close to 12dB, or two 6dB "steps")

Starting at +4dBu, if a 6dB drop means halving the voltage, then we'd be at -2dBu = ~0.615 volts. A second 6dB drop would halve that value again, meaning that -8dBu = ~0.3075 volts.

Since we already know that -7.8dBu = ~0.32 volts, the calcualtion that -8dBu = ~-0.3075 volts looks to be about spot on target.

Bingo!

G.

 

[Reggie]

"Sticky" indeed...

BTW, there seems to be a typo in that roger nichols article. Each step in 16 bit should represent .00030518509476

OK, so this is blowing my mind a little bit. So far, looks like snow lizard is right about everything; except for maybe 0VU = 1V ; I thought 0VU = 1.228V ? .

Here is what I have learned (somewhat in approximations, and +/- for the values and voltages):
-18.4dbFS = value 4024 = 1.228V = 0VU
-20dbFS = value 3278 = 1.0004V
-22dbFS = value 2604 = .7947V
-26dbFS = value 1643 = .5014V

-12dbFS = value 8232 = 2.5123V
-6dbFS = value 16423 = 5.012V
-3dbFS = value 23198 = 7.08V
0dbFS = value 32767 = 10V

There are indeed a lot more possible "voltage steps" between -6 and 0dbFS, than between, say, -26 and -20. So I guess there are a lot finer choices in steps the closer to 0dbFS you get. Seems kinda goofy, but kinda necessary at the same time I suppose because your voltage has to double every 6db.

Here is another thought, say we have a waveform that bounces around -26 to -22 to -20, and we want to raise this waveform to full scale. The values would go from 1643, 2604, 3278 to 16423, 23198, 32767. Now there are many more inbetween steps that will not be used (because we were dealing with a smaller voltage range at the time), UNLESS perhaps you had recorded the signal to originally be in the -6 to 0 range. There would then be a much broader range of in between steps for the signal to pick from in the -6 to 0 range, and perhaps it gets represented more accurately?
The mind still boggles...

PS: I am not trying to suggest we should record as close to 0db as possible, just trying to get my mind around the inner workings.

 

[SouthSIDE Glen]

say we have a waveform that bounces around -26 to -22 to -20, and we want to raise this waveform to full scale. The values would go from 1643, 2604, 3278 to 16423, 23198, 32767. Now there are many more inbetween steps that will not be used (because we were dealing with a smaller voltage range at the time), UNLESS perhaps you had recorded the signal to originally be in the -6 to 0 range. There would then be a much broader range of in between steps for the signal to pick from in the -6 to 0 range, and perhaps it gets represented more accurately?

Not really. Remember that when we digitally increase the volume, all we're doing is moving bits over. The LSBs are just getting filled in by zeros, so the precision of the value is not really changing.

It's rather like in regular digital math. If you have a value of, say, 5, and you multiply that by 10 to try and get a 10x "closer look" at that value, you wind up with 50. No more real resolution as been added to the "5" by putting that 0 on the end.

G.

 

[Farview]

Here is another thought, say we have a waveform that bounces around -26 to -22 to -20, and we want to raise this waveform to full scale. The values would go from 1643, 2604, 3278 to 16423, 23198, 32767. Now there are many more inbetween steps that will not be used (because we were dealing with a smaller voltage range at the time), UNLESS perhaps you had recorded the signal to originally be in the -6 to 0 range. There would then be a much broader range of in between steps for the signal to pick from in the -6 to 0 range, and perhaps it gets represented more accurately?
The mind still boggles...

It won't get represented more accurately, but it doesn't matter. This all gets taken care of in the reconstruction process. Any of those 'lost' discrete steps would represent something happening at a higher than nyquist frequency.