Topic: Edirol R09 - ADC not good enough for 24bit, right?

[BayTaynt3d]

I don't think this is how linear quantization works - it's 1 bit per 6 dB of dynamic range, and that's fixed. So by going to 24-bit you're not dividing the same dynamic range into more bits, you're adding more dynamic range by (theoretically, anyways) decreasing the noise floor of the A/D chip. At least, that's how I understand it...

Nah. There IS MORE RESOLUTION as far as I understand this. How about some math...

So, can we at least agree on this to start? 16-bit has 96 db of dynamic range and 24-bit has 144 db of dynamic range?

If so, then here's how I do the math, but maybe I'm missing something here?

16 bit = 2^16 = 65,536 unique #s possible / 96 db of range = .0015 db per unit possible

24 bit = 2^24 = 16,777,216 unique #s possible / 144 db of range = .0001 db per unit possible

So, whether you're using the whole range or not, the fact is that there is 15 times the resolution happening at 24- vs. 16-bits, which is why when you normalize (aka zoom in) it holds up much better (see SD example above), which is also one of the reasons why there's not nearly as much need to run hot (cause at 24-bit, I can run 10 db under you running at 16-bit, then normalize it, and STILL end up having more resolution than you had if you ran your peaks perfectly hot up to 0db at 16-bit). This example of course is ignoring the quality of the ADC, but that's a different subject.

Am I understanding this correctly? I'm not trying to represent myself as someone who really knows this stuff, but I'm trying awfully hard, and I think I get it...

[trajhip2000]

I'm gonna have to dig out my copy of Pohlman and get back to you...

Steve

[it-goes-to-eleven]

Am I understanding this correctly?

By my way of thinking, yes...

[SparkE!]

I don't think this is how linear quantization works - it's 1 bit per 6 dB of dynamic range, and that's fixed. So by going to 24-bit you're not dividing the same dynamic range into more bits, you're adding more dynamic range by (theoretically, anyways) decreasing the noise floor of the A/D chip. At least, that's how I understand it...

Nah. There IS MORE RESOLUTION as far as I understand this. How about some math...

So, can we at least agree on this to start? 16-bit has 96 db of dynamic range and 24-bit has 144 db of dynamic range?

If so, then here's how I do the math, but maybe I'm missing something here?

16 bit = 2^16 = 65,536 unique #s possible / 96 db of range = .0015 db per unit possible

24 bit = 2^24 = 16,777,216 unique #s possible / 144 db of range = .0001 db per unit possible

So, whether you're using the whole range or not, the fact is that there is 15 times the resolution happening at 24- vs. 16-bits, which is why when you normalize (aka zoom in) it holds up much better (see SD example above), which is also one of the reasons why there's not nearly as much need to run hot (cause at 24-bit, I can run 10 db under you running at 16-bit, then normalize it, and STILL end up having more resolution than you had if you ran your peaks perfectly hot up to 0db at 16-bit). This example of course is ignoring the quality of the ADC, but that's a different subject.

Am I understanding this correctly? I'm not trying to represent myself as someone who really knows this stuff, but I'm trying awfully hard, and I think I get it...

We've been through this exercise before*** and it's meaningless to divide the number of dBs of dynamic range by the number of levels.  They both represent the same thing.  One of the numbers is on a linear scale and one of the numbers is on a logarithmic scale.  Each additional bit used in a binary number adds exactly 20 log 2 dB of dynamic range and it doubles the number of possible levels that can be expressed.  That's all there is to the math.  Try this:

Let's start out with 2, raise it to the 16th power, take it's log and multiply by 20.

20 log (2^16) = 20 log 65536 = 96.329598612473982468396446311838
               ^                 ^         ^
             bits              levels     dBs

That's all there is to the math and dividing dBs by bits or dividing bits by levels or dividing levels by dBs doesn't mean anything.  They all are ways of representing the same concept and that is what kind of resolution you get.  The more bits, the way more possible levels you get.  The more bits, the more dBs of dynamic range you get.  (In fact, you get approximately 6.0205999132796239042747778944899 dBs of dynamic range per additional bit, so just multiply the number of bits by 6.0205999132796239042747778944899 to get the possible dynamic range.  We usually round that off to 6 db per bit because it's easier to do the arithmetic.)

*** Look here for where this same topic was discussed previously:

http://taperssection.com/index.php/topic,77804.msg1037087.html#msg1037087

Nobody is arguing that 24 bit converters have the same resolution as an ideal 18, 19 or 20 bit converter. (Obviously a 24 bit converter has more resolution than any converter with a smaller number of bits in the encoded values that it produces.)  What we are arguing is that most 24 bit converters have no more accuracy than an ideal 18, 19 or 20 bit converter.  The reason they have nor more accuracy is that the lower 4, 5, or 6 bits are indistinguishable from the results you'd get by flipping a coin and calling heads a 1 and tails a 0.  It's as if noise is being added to the signal you are encoding and the amplitude of the noise occupies the lower 24, 30 or 36 dB of the dynamic range.

Bottom line: Resolution does not equal Accuracy and it is Accuracy that determines the S/N of the recordings we make and ultimately sets the achievable dynamic range in the recording.  Notice I'm talking about the dynamic range of the recording, not the dynamic range of the encoding scheme.  There's a difference and in the case of today's 24 bit A/D converters, its a big difference.

[Keyser Soze]

   This thread makes my head hurt 

[Gutbucket]

I think I get it. But here's a few questions to clear it up.

Question:  Is there a certain granularity of dynamic range within whatever total range is actually being encoded, be it 16, 18, 20 bits total or whatever that is a minimum describable level change?  Or, is there no step between those loudness levels and any infinitely small change of loudness value can be described, as long as that change in level lies within the noise floor and full scale?

Second part of the question:  If there is a certain minimum value for describing differences in loudness, is it the same for a 16 bit signal as a 20 bit one?  I'd think it would have to be if the only difference was that the total range was just that much larger for a 20bit signal.

Theoretical follow ups: If there is a certain loudness granularity and it doesn't vary with increasing bit depth, could that extra depth that is lost in the noise floor be used instead to increase the fineness of the individual loudness steps?  If they exist at all are they course enough to be audible?

[it-goes-to-eleven]

Second part of the question:  If there is a certain minimum value for describing differences in loudness, is it the same for a 16 bit signal as a 20 bit one?  I'd think it would have to be if the only difference was that the total range was just that much larger for a 20bit signal.

Good question... I think one wrinkle in the answer is that most A/Ds are 24 bit and must dither to give you 16 bit.  So I'd suppose those quantization errors might play a role in limiting the resolution of volume steps.


I think SparkE's explanation is very good and true..  24 bit provides way more digits than are significant given the accuracy of the system.  But you still need 24 bits to get the extra bits that DO matter, whether it is 18, 20 or whatever.

[poorlyconditioned]

Second part of the question:  If there is a certain minimum value for describing differences in loudness, is it the same for a 16 bit signal as a 20 bit one?  I'd think it would have to be if the only difference was that the total range was just that much larger for a 20bit signal.

Good question... I think one wrinkle in the answer is that most A/Ds are 24 bit and must dither to give you 16 bit.  So I'd suppose those quantization errors might play a role in limiting the resolution of volume steps.


I think SparkE's explanation is very good and true..  24 bit provides way more digits than are significant given the accuracy of the system.  But you still need 24 bits to get the extra bits that DO matter, whether it is 18, 20 or whatever.

Yes, you need 24 bits to get anything more than 16 (18, 20 or whatever).

BUT the ADC in there is only good to about 15 bits, so there are not extra bits to grab.  It is all noise below the 15th bit.

I suppose from a dithering point of view, this is cool.  You can record in 16 bits knowing that your last bit is essentially random.  But, you are never going to get any more than 16 bits.  This is not a headroom or preamp issue.  It is the ADC itself.  So, putting a preamp is front is not going to get you more bits.  Better sound, maybe, but not more bits.

  Richard

[Gutbucket]

Understood in this practical application.

I'm still curious if anyone knows what the actual quantization step size is between loudness levels. 
If it's one bit, then how can we store any loudness changes that are smaller than approx 6db?

In a similar vein, the actual quantization step size between frequencies is half the sampling rate is it not? Yet in that case, both the total range of values and the resolution within any particular sub range of the total frequency range increases with higher sampling rates.

I think that is the crux of the mental block.

I realize that those steps, both in the the vertical and horizontal axis of the waveform, are averaged out in the DAC when the signal is converted to analog.  But there must be a measurable value that is the minimum change in loudness value that can be stored.. Just as there is a minimum change in frequency that can be stored (or a minimum length event in the time domain).

[SparkE!]

Understood in this practical application.

I'm still curious if anyone knows what the actual quantization step size is between loudness levels. 
If it's one bit, then how can we store any loudness changes that are smaller than approx 6db?

In a similar vein, the actual quantization step size between frequencies is half the sampling rate is it not? Yet in that case, both the total range of values and the resolution within any particular sub range of the total frequency range increases with higher sampling rates.

I think that is the crux of the mental block.

I realize that those steps, both in the the vertical and horizontal axis of the waveform, are averaged out in the DAC when the signal is converted to analog.  But there must be a measurable value that is the minimum change in loudness value that can be stored.. Just as there is a minimum change in frequency that can be stored (or a minimum length event in the time domain).

If you're talking about how much increase in dB's you get by increasing the volume by one level, it depends on whether you are increasing the amplitude from 1 to 2 (6 dB increase) or whether you are increasing the amplitude from 65535 to 65536 (0.00013 db increase).

[Gutbucket]

If you're talking about how much increase in dB's you get by increasing the volume by one level, it depends on whether you are increasing the amplitude from 1 to 2 (6 dB increase) or whether you are increasing the amplitude from 65535 to 65536 (0.00013 db increase).

That's the number I was looking for.  Vanishingly small.