24bit vs 16bit, the myth exploded!
Nov 26, 2010 at 5:18 PM Post #556 of 7,175
Quote:
"So lets say
 
with 1 bit we would resolve 6 dB into 2 values (e.g. 0 dB and 6 dB, nothing in between)
with 2 bit we would resolve 12 dB into 4 values (e.g. 0,4,8,12 dB, nothing in between)"
 
But that's strange, because people usually say to use 24-bits playback even if your source is 16-bits, because the source will be padded to 24-bits, and you get more range to waste with the digital volume control. But in this example, you just can't pad the 1-bit source to 2-bit at all (not without rounding 6 to 4 or to 8). So is everyone wrong in using 24-bit playback then (for 16-bit sources)?

 
Forget the 1 bit stuff if you didn't read the DSD, PDM articles. And this example/question doesn't make sense to me.
 
And please reconsider reading the first post...
 
(and I'm not sure if you even read Wikipedia articles, sorry if that's harsh)
 
Nov 26, 2010 at 6:58 PM Post #557 of 7,175


Quote:
Forget the 1 bit stuff if you didn't read the DSD, PDM articles. And this example/question doesn't make sense to me.  
And please reconsider reading the first post...
 
(and I'm not sure if you even read Wikipedia articles, sorry if that's harsh)

 
I'm fairly sure he's talking about PCM in this example I quoted.
 
But let's just keep it simple then: should I set my sound card to 24-bits, even if my source is 16-bits, or not? Will it be able to pad (in a lossless way) the data?
 
Why am I asking? Because if the steps are different in different bit-depths, then you cannot convert from one to the other without having to round stuff.
 
EDIT 2:
 
Did some serious tests here with an HEX editor and some wave files, and found out that you can pad perfectly from 16 bits to 24 bits.
 
If you have, say, (big endian) FF 7F then pad it to 24-bits you get 00 FF 7F. The first 00 are the least significant part and account for the smaller steps. So for every step possible in 16-bits, you get exactly 255 new steps in between in 24-bits.
 
I still don't understand how does this increase so much the dynamic range though, since the lowest point ("silence") is 00 00 00 (the same as 16-bits) and the highest is now FF FF 7F for the crest *, and FF FF 7F is just barely higher than FF 7F.
 
* It's not FF FF FF because after 7P it wraps around and represents the trough.
 
 
This image is what I found out through testing (HEX viewer and looking at the wave):

 
Nov 26, 2010 at 10:45 PM Post #558 of 7,175
Suppose that you have a system where the values are 0, 1, 2, 3, this is effectively 2 bit. and that you was a system that does 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, In your mind, you'd have doubled precision of the quantization.
Let's say that the digital values 0, 1, 2 ,3 represent 0, 1, 2, 3 mV out of your microphone, you want to record 0.5 mV steps, you simply double the pre amp values for your mike, making the the output values go from 0 to 7 mV by 1 mV steps which means you have doubled the dynamic range necessary to record the data. Effectively, to diminish the quantization error, you simply increase the dynamic range. You now have to use a 3 bit encoding with 0, 1, 2, 3, 4 ,5 ,6 ,7 as values.
 
Anyway, most recording are 24 bit today, allowing the smallest step to be set very low, and once digitized, every track (one for the piano, one for the singer...) is fed to the DAW (digital audio workstation) which performs its internal calculation on 32 and even 64 bit. And finally when you output your mastered song, you decide the values of your loudest and softest sounds considering that a variation smallest than the softest sound wouldn't be heard. This, for a CD is 16bit, ie. 96 dB of variation between the loudest and the softest sound, ie 65535 values, when you go up on and High Def, you get 144 dB, which means that either your loudest sound gets louder, or that you softest sound get softer, increasing dynamic range is the same as decreasing quantization error.
 
Except that, your listening room has noise for, even in an anechoic room you, if your hearing is very good, you'd hear your heart beat, the flow of your blood... So the softest sound, the smallest variation doesn't have to be that small, there's a limit for that. Now consider your 16 bit CD, your room noise floor is 25 dB (that's a very quiet room), you set the volume so that the highest sound reaches 120 dB (That's really loud), your softest sound is played at 24 dB, ie. lost in the noise floor. If you play so that the loudest sound is 90 dB, the softest sound plays below 0 dB which is below the audibility threshold. Of course this does not take into account dithering, which brings the perceivable dynamic range up to to 120 dB with a 16 bit encoding, considering a loudest level of playback at 120 dB, the small detail is right at the audibility threshold

As an addendum, your question was mostly answered in the first two paragraphs, the 3rd was mostly about why 16 bit is usually enough for playback in household conditions. That's with PCM encoding, SACd is a different story.
 
 
There is one thing I don't understand though:
 
Wouldn't it make more sense to use the extra bits not for increased dynamic range, but for more gradual steps in the quantization?
 
Suppose we have 3 bits (possible values: 000, 001, ... 111), and they represent, in order, -30, -20, -10, 0, 10, 20, 30, 40.
If we add one more bit, instead of going like -70, -60, ... , 70, 80 (which seems to be the analogous case for 16 vs. 24 bits if I understand correctly), why not -30, -25, -20, ... , 30, 35, 40, adding more steps instead of bigger range?
 
If someone could clarify things with some practical example like this it would be great!
wink.gif

 
EDIT: Better numbers for my example.

 
Nov 27, 2010 at 6:56 AM Post #559 of 7,175
Quote:
I still don't understand how does this increase so much the dynamic range though, since the lowest point ("silence") is 00 00 00 (the same as 16-bits) and the highest is now FF FF 7F for the crest *, and FF FF 7F is just barely higher than FF 7F.


FF FF 7F = 2^23-1 = 8388607 is just barely higher than FF 7F = 2^15-1 = 32767 ???
The multiplier is 2^8 = 256, or +48 dB !!!
 
And please, I'm begging you, read the first post.
 
Nov 27, 2010 at 8:16 AM Post #560 of 7,175


Quote:
Suppose that you have a system where the values are 0, 1, 2, 3, this is effectively 2 bit. and that you was a system that does 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, In your mind, you'd have doubled precision of the quantization.
Let's say that the digital values 0, 1, 2 ,3 represent 0, 1, 2, 3 mV out of your microphone, you want to record 0.5 mV steps, you simply double the pre amp values for your mike, making the the output values go from 0 to 7 mV by 1 mV steps which means you have doubled the dynamic range necessary to record the data. Effectively, to diminish the quantization error, you simply increase the dynamic range. You now have to use a 3 bit encoding with 0, 1, 2, 3, 4 ,5 ,6 ,7 as values.
 

 
Doesn't that confirm what I stated earlier.
With 2 dB and 0,1,2,3 in 1mV steps and then with 3 dB and doubled preamp values you have still 1mV steps but you have now 8 of theses steps for the same stuff reaching the microphone as before.
What ever comes after like mastering, filtering .... Effectively you have more values for the same recorded analog sound, more "resolution".
The point of view of the microphone is still having 1mV values so no more resolution than before, but more dynamic range.
It is that mV and SP stuff I can't get congruent with each other yet.
 
I'm going now, buying some Webers or Dickreiter ...
 
 
Quote:
 
Except that, your listening room has noise for, even in an anechoic room you, if your hearing is very good, you'd hear your heart beat, the flow of your blood... So the softest sound, the smallest variation doesn't have to be that small, there's a limit for that. Now consider your 16 bit CD, your room noise floor is 25 dB (that's a very quiet room), you set the volume so that the highest sound reaches 120 dB (That's really loud), your softest sound is played at 24 dB, ie. lost in the noise floor. If you play so that the loudest sound is 90 dB, the softest sound plays below 0 dB which is below the audibility threshold. Of course this does not take into account dithering, which brings the perceivable dynamic range up to to 120 dB with a 16 bit encoding, considering a loudest level of playback at 120 dB, the small detail is right at the audibility threshold

 
You can't hear tones which dB value is below noise floor dB value (or, if you only have your breath and heartbeat you couldn't hear sound below that ?) ?
Wouldn't that mean you couldn't here e.g. a triangel if it played simultaneously with a trumpet, because the trumpet is much louder ?
I know that noise floor contains many frequencies. I'm just curious, not stating you were completely wrong or something.
 
And I'm with you regarding the rest so far.
 
Nov 27, 2010 at 10:15 AM Post #561 of 7,175


 
 
You can't hear tones which dB value is below noise floor dB value (or, if you only have your breath and heartbeat you couldn't hear sound below that ?) ?
Wouldn't that mean you couldn't here e.g. a triangel if it played simultaneously with a trumpet, because the trumpet is much louder ?
I know that noise floor contains many frequencies. I'm just curious, not stating you were completely wrong or something.
 
And I'm with you regarding the rest so far.


Masking requires that the two frequencies are in proximity, the further apart they are the less masking works. You can hear some music beneath the noise floor but I cannot remember offfhand how much, I think 20db below the noise floor is not possible, but I may be wrong.. 
 
Nov 27, 2010 at 10:52 AM Post #562 of 7,175
also, pls recall that this entire "24bit vs 16bit, the myth exploded" thread contains numerous posts (from beginning to most-recent) presenting significant misinformation wrt fundamentals of sampling theory and digital signal processing -- neither of which are particularly simple topics...
 
for some reasonably-understandable texts providing technically-correct info -- and specifically to assist in understanding the relationships between bit depth, quantization error, resolution and dynamic range (which in this thread, sadly are oft-misrepresented) -- see the linked references in these posts:
 
from this thread:  #463
 
from the thread (short, easy reading) "Bit depth and sampling frequency explained":  #15#19 and #21
 
 
when i last checked, majority of the excellent and straightforward Pohlman text (Principles of Digital Audio) can still be accessed on the link to Google books; and the Analog Devices "Introduction to Sampled-Data Systems" presentation is available at the noted analog.com link
 
Nov 27, 2010 at 11:08 AM Post #563 of 7,175
Quote:
Quote:
I still don't understand how does this increase so much the dynamic range though, since the lowest point ("silence") is 00 00 00 (the same as 16-bits) and the highest is now FF FF 7F for the crest *, and FF FF 7F is just barely higher than FF 7F.


FF FF 7F = 2^23-1 = 8388607 is just barely higher than FF 7F = 2^15-1 = 32767 ???
The multiplier is 2^8 = 256, or +48 dB !!!
 
And please, I'm begging you, read the first post.

 
Dude, I've read the first post, I already told you I did. Seriously, you're just making yourself seem arrogant. No one else in this thread is behaving like this, chill out...
 
I was considering a different step size too.
If you pick up a 16-bits file with an FF 7F value, then pad it to 24-bits, you'll get 00 FF 7F. Now if you change it to FF FF 7F in an editor, all you'll see is a marginal, almost imperceptible difference in the displayed wave form. Because now, if you view it like I did, the possible steps are also 256 times smaller. You do not see the waveform shrink 256 times when you pad it to 24-bits, it's the steps that are reduced.
 
So I guess the problem here is that you can view it in two ways: the one above, considering the steps to be 256 times smaller, then the dynamic range is barely changed (this is the way I've been viewing it, because it feels more natural to me). Or if you consider steps of the same size, then dynamic range is indeed 256 times larger. This seems to be the correct approach, if I understand khaos974 explanation correctly.
 
Here is what you will see, if you pad a 2-bit source to 4-bit:

 
IMO, the most natural way to interpret it, is to think the possible steps have become smaller, and dynamic range has barely increased from 0011 to 1111 (it's a small difference here, because the steps are also 4 times smaller).
 
But it seems the correct way to interpret it is: the steps are the same size, but dynamic range quadrupled. When the file was padded, it was just "stretched" to space out in this new dynamic range.
 
Nov 27, 2010 at 11:31 AM Post #564 of 7,175
Quote:
Doesn't that confirm what I stated earlier.
With 2 dB and 0,1,2,3 in 1mV steps and then with 3 dB and doubled preamp values you have still 1mV steps but you have now 8 of theses steps for the same stuff reaching the microphone as before.
What ever comes after like mastering, filtering .... Effectively you have more values for the same recorded analog sound, more "resolution".
The point of view of the microphone is still having 1mV values so no more resolution than before, but more dynamic range.
It is that mV and SP stuff I can't get congruent with each other yet.
 
I'm going now, buying some Webers or Dickreiter ...

Yes, it does confirm what you said earlier, more or less, strictly speaking you don't get more resolution since the steps are still 1 mV, on the other hand you get up tu 7 mV, hence better dynamic range, same resolution. On the other hand, in practice, you've manage to gain more resolution for the recording with the use a a preamp. Hence, strictly speaking you gain only DR, loosely speaking you gain a bit of both. I don't understand you problem with mV and sound pressure, the sound pressure is converted via microphone to a voltage, a preamp is use to get the optimal voltage for the ADC to work with. Is that what you didn't understand ?
 
You can't hear tones which dB value is below noise floor dB value (or, if you only have your breath and heartbeat you couldn't hear sound below that ?) ?
Wouldn't that mean you couldn't here e.g. a triangel if it played simultaneously with a trumpet, because the trumpet is much louder ?
I know that noise floor contains many frequencies. I'm just curious, not stating you were completely wrong or something.
 
Sorry, I wasn't clear enough, by "lost in the noise floor", I didn't mean you couldn't hear it, I should have use "at the same volume as the noise floor", it was only supposed to be a comparison of volume levels, my bad. On the other hand a propertly mastered and dithered CD can have 120 dB DR, which does indeed bring the softest sounds below the audibility threshold.
 
And I'm with you regarding the rest so far.

 
Nov 27, 2010 at 11:51 AM Post #565 of 7,175
What's your point exactly?

If your output is your computer, padding to 24 bit is useful if you need a digital volume control, otherwise outputing 16 bit is roughly equivalent, you DAC may do the padding automatically pre conversion anyway.
 
Quote:
...
IMO, the most natural way to interpret it, is to think the possible steps have become smaller, and dynamic range has barely increased from 0011 to 1111 (it's a small difference here, because the steps are also 4 times smaller).
 
But it seems the correct way to interpret it is: the steps are the same size, but dynamic range quadrupled. When the file was padded, it was just "stretched" to space out in this new dynamic range.

 
Nov 27, 2010 at 11:57 AM Post #566 of 7,175


Quote:
What's your point exactly?

If your output is your computer, padding to 24 bit is useful if you need a digital volume control...

This is my point.
tongue.gif

 
Nov 27, 2010 at 12:16 PM Post #567 of 7,175
Based on your sig, I suppose it's a question for foobar?
I believe foobar volume control is 32 bit precision, it then outputs at the selected bit depth. You may want to double check the info though.
 
Nov 27, 2010 at 12:31 PM Post #568 of 7,175


Quote:
Based on your sig, I suppose it's a question for foobar?
I believe foobar volume control is 32 bit precision, it then outputs at the selected bit depth. You may want to double check the info though.


Not really, I'm asking because I've been setting my Windows (Seven) mixing bit-depth to 24-bits (shared mode), so I can use the digital volume control without losing so much useful data.
 
It's also for curiosity's sake.
 
Nov 27, 2010 at 5:56 PM Post #569 of 7,175


Quote:
Quote:
Doesn't that confirm what I stated earlier.
With 2 dB and 0,1,2,3 in 1mV steps and then with 3 dB and doubled preamp values you have still 1mV steps but you have now 8 of theses steps for the same stuff reaching the microphone as before.
What ever comes after like mastering, filtering .... Effectively you have more values for the same recorded analog sound, more "resolution".
The point of view of the microphone is still having 1mV values so no more resolution than before, but more dynamic range.
It is that mV and SP stuff I can't get congruent with each other yet.
 
I'm going now, buying some Webers or Dickreiter ...

Yes, it does confirm what you said earlier, more or less, strictly speaking you don't get more resolution since the steps are still 1 mV, on the other hand you get up tu 7 mV, hence better dynamic range, same resolution. On the other hand, in practice, you've manage to gain more resolution for the recording with the use a a preamp. Hence, strictly speaking you gain only DR, loosely speaking you gain a bit of both. I don't understand you problem with mV and sound pressure, the sound pressure is converted via microphone to a voltage, a preamp is use to get the optimal voltage for the ADC to work with. Is that what you didn't understand ?
 

 
1. How is this fixed 6 dB/1 bit thing to mV to SP related, and is this a physical founded relation or could you do it differently?
2. When I'm playing a continous volume sweep e.g from 30 dB to 80 dB why is it that with 24 bit there is not more information about that volume sweep recorded as with 16 bit ? Is this just because of the mastering technics used or is this some kind of mathematical/physical law? I' m aware that also the dynamic range grows as makes your previous example clear (because the 7 mV is something more than before with the 3 mV max. which doubled is only 6 mV, so we have more range for a higher volume value as before, am I right?)
 
These are my main problems of understanding currently.
 
(The Dickreiter did not offer much information on that specific theme ... but I will read the links given in the post somewhat above)
 
EDIT: O.k. I'm out. I started to read the first 4 to 5 pages of the thread and the mentioned related threads and will continue by and by, also read the mentioned Posts and adjacent posts.
As it seems to me, the main problem in the discussion is, that some people like me argue theoretic and others regarding practical aspects.
So there is a "resolution" increase with more bits, but it is not used because it is not considered practical necessary due to other technics used to get it right, etc. etc.
Which technics are then considered better or even make audible difference seems to be more a matter of taste, belief or academical.
As you say in German: Die Augen essen mit.
 
Also I think no one here (of course including me) is able to fully comprehend the involved mathematics e.g. Nyquist et al.
 
I'm reading on ... :)
 
And I'm glad my Hifiman has a (no, two) PCM1704
wink.gif

 

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