24 bit vs. 16 bit recording

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Did you bother to follow the link?

I did. read the first paragraph about "sampling theory" and assumed it was a foreign language.

on my recording device:
24 bits=16 tracks
16 bits=32 tracks

16 bits is better

EDIT:upon further inspection, I believe it IS a language foreign to me. I suspect it may be some derivitive of binary
 
I did. read the first paragraph about "sampling theory" and assumed it was a foreign language.

on my recording device:
24 bits=16 tracks
16 bits=32 tracks

16 bits is better

EDIT:upon further inspection, I believe it IS a language foreign to me. I suspect it may be some derivitive of binary
:laughings: yeah 16 would seem better in that case.
 
Each bit always represents the same amount of dynamic range regardless of the noise floor. Adding bit depth just pushes your digital noise floor farther down. The 16 most significant bits in a 24 bit signal represent the exact same voltages and the 16 bits in a 16 bit signal.

Interesting/confusing... I'd always thought that having a higher bitrate increases the amount of "slices in the pie", thus increasing the accuracy of the A/D conversion? I've always wondered how this relates to the S/N ratio, given my (apparently) incorrect understanding of bit depth.

I am not disagreeing with you at all, as I have significantly less knowledge in this subject. I'm just curious & would like to learn.

*Edit: Is audio bit depth not analogous to color bit depth? More bits = more colors between white and black?
 
Interesting/confusing... I'd always thought that having a higher bitrate increases the amount of "slices in the pie", thus increasing the accuracy of the A/D conversion? I've always wondered how this relates to the S/N ratio, given my (apparently) incorrect understanding of bit depth.

I am not disagreeing with you at all, as I have significantly less knowledge in this subject. I'm just curious & would like to learn.

*Edit: Is audio bit depth not analogous to color bit depth? More bits = more colors between white and black?

Without getting too far into the nuts and bolts of how binary word lenghts operate in PCM audio, bit depth represents voltage values of analog audio. More bits is equal to more resolution, or more differences between quiet and loud. Once these voltage values are translated to digital PCM audio, there are "steps" between quiet and loud. These steps are represented by binary numbers.

Binary is ones and zeroes. A one bit word length gives you two choices: zero or one. On or off. It's like a switch. If you want to represent more range than that, or higher numerical values you need to increase the word length.

a "2 bit" word length operates like a more complex switch. As you add bits you can represent higher numbers. One bit can only represent 2 possible numerical values while two bits can represent 4 numerical values.

00
01
10
11

The range of possible numerical values is calculated by 2 (the difference between zero and one of a bit) to the power of how ever many bits in the word length.

Every time you add one bit to the word length the range of numerical values that can be represented doubles.

24 bit audio offers way more resolution in amplitue or dynamics or the difference between quiet and loud or whatever you want to call it than 16 bit.

Because the scale is kind of top heavy as it relates to straight PCM audio, some people got the idea that if you completely overdrive your signal and fry it to a black, charcoalesque crisp on the way in in order to get your signal as close as possible to 0 dBFS, and in much the same way as The Price Is Right (which has an excellent theme song) without going over, that you are somehow doing yourself a favour because you're using more of the available numbers.

People caught on to this a lot faster than how to address the issue of proper gain scheduling.

In 16 bit audio it shouldn't be an issue anyway.

In 24 bit audio you'll get all of the "amplitude resolution" or whatever, that you would get in 16 bit audio at 0 dBFS, at a level of around -48 dBFS. So it becomes even more of a non-issue, and intentionally borking your own gain structure in the quest for "winning" at some half baked, audiological nitwit, bean counter aesthetic PCM version of "Plinko" or something, becomes even more ridiculous.

Some people say that this means 24 bit gives you more headroom. Not exactly correct, but it's shorter than the stuff that I just wrote.
 
Interesting/confusing... I'd always thought that having a higher bitrate increases the amount of "slices in the pie", thus increasing the accuracy of the A/D conversion?
Using the pie analogy, a 24 bit pie is bigger than a 16 bit pie. (one and a half times bigger) Yes, it's cut up into more pieces, but the pieces are the same size as the piece in the 16 bit pie, there are just 24 of them instead of 16. (one and a half times more pieces)
 
Using the pie analogy, a 24 bit pie is bigger than a 16 bit pie. (one and a half times bigger) Yes, it's cut up into more pieces, but the pieces are the same size as the piece in the 16 bit pie, there are just 24 of them instead of 16. (one and a half times more pieces)

That doesn't make sense to me, but this is why I'm glad you're here. So you're saying a 24-bit sample can be 'louder' than a 16 bit sample? Or deals with higher voltage? This isn't making sense to me :/

*edit: I think the crux of my confusion is that 0dbfs relates to the highest voltage a system can handle (yes/no?), and zero volts is some arbitrary dbfs. Or is that exactly the point?

*edit 2: Wait... now I'm starting to think my understanding is totally flawed. In my mind, the system went something like this:

1 - Analog signal being received by ADC

2 - Dependent on sampling rate, ADC sends a '1' each time the signal is higher than was recorded in the previous sample, and '0' if it was not. (*now that I'm thinking about it, this can't be completely right, can it? There's no "-1" bit to indicate that it's decreasing in voltage... shit)

3 through x - not sure



*edit 3: WAIT... I just re-read snow lizard's post. Let me see if I got this right: the larger the word length, the more accurately the change in voltage between samples will be recorded? And thus, you can record a quieter signal with a higher degree of accuracy at 24 bits vs 16 bits vs 8 bits?
 
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*edit 3: WAIT... I just re-read snow lizard's post. Let me see if I got this right: the larger the word length, the more accurately the change in voltage between samples will be recorded? And thus, you can record a quieter signal with a higher degree of accuracy at 24 bits vs 16 bits vs 8 bits?


Essentially, yes.

The amount of headroom the system has will depend on where line level is calibrated at in the converters.
 
Cool, thanks man. Follow-up question, if there's enough high-frequency (>22khz) content of sufficient amplitude in the audio being converted by an ADC running at 44.1khz, couldn't it theoretically still affect the output audio from the DAC? I'm thinking it must, as the ADC will still be recording aspects of the >22khz signal intermittently, but will be unable to reproduce them as a >22khz signal (and arguably, anything approaching 22khz would be converted/reproduced at significantly lower accuracy, wouldn't it?).
 
That doesn't make sense to me, but this is why I'm glad you're here. So you're saying a 24-bit sample can be 'louder' than a 16 bit sample? Or deals with higher voltage? This isn't making sense to me :/
The added bits are at the quiet end of the dynamic range.

*edit: I think the crux of my confusion is that 0dbfs relates to the highest voltage a system can handle (yes/no?), and zero volts is some arbitrary dbfs. Or is that exactly the point?
This depends on the calibration of the converters. Assuming that the 16 and 24 bit converters are calibrated to the same level, 24 bit lets you record quieter signals than 16 bit does.

*edit 2: Wait... now I'm starting to think my understanding is totally flawed. In my mind, the system went something like this:

1 - Analog signal being received by ADC

2 - Dependent on sampling rate, ADC sends a '1' each time the signal is higher than was recorded in the previous sample, and '0' if it was not. (*now that I'm thinking about it, this can't be completely right, can it? There's no "-1" bit to indicate that it's decreasing in voltage... shit)
Nope.

*edit 3: WAIT... I just re-read snow lizard's post. Let me see if I got this right: the larger the word length, the more accurately the change in voltage between samples will be recorded? And thus, you can record a quieter signal with a higher degree of accuracy at 24 bits vs 16 bits vs 8 bits?
Yes.

The headroom thing comes into play because you can record a signal much lower without having to worry about running into quantization noise. This means that you can calibrate the converters such that 0dbVU (line level) relates to a lower digital level, which gives you more digital headroom.

In the 16 bit days, it was common for the converters to be calibrated so that line level was -10 or -12dbFS. That wasn't a lot of headroom to work with. Now, with 24 bit converters it is common for line level to be calibrated to -18 or -20 dbfs. So there is a lot more breathing room. -18dbfs calibration still gives you 30db more resolution (5 bits worth) than a full scale 16 bit file would.
 
Cool, thanks man. Follow-up question, if there's enough high-frequency (>22khz) content of sufficient amplitude in the audio being converted by an ADC running at 44.1khz, couldn't it theoretically still affect the output audio from the DAC? I'm thinking it must, as the ADC will still be recording aspects of the >22khz signal intermittently, but will be unable to reproduce them as a >22khz signal (and arguably, anything approaching 22khz would be converted/reproduced at significantly lower accuracy, wouldn't it?).
That is the reason why everything greater than 22.5kHz is filtered out. The converters never see anything that high.
 
Cool, thanks man. Follow-up question, if there's enough high-frequency (>22khz) content of sufficient amplitude in the audio being converted by an ADC running at 44.1khz, couldn't it theoretically still affect the output audio from the DAC? I'm thinking it must, as the ADC will still be recording aspects of the >22khz signal intermittently, but will be unable to reproduce them as a >22khz signal (and arguably, anything approaching 22khz would be converted/reproduced at significantly lower accuracy, wouldn't it?).

The anti-aliasing filter will limit the bandwidth available to the adc. At 44.1khz any signals > sample rate / 2 should not been seen by the adc.
 
Gotcha. Is this true regardless of sampling rate?
With 48k, everything over 24k is filtered out. With 96k, everything over 48k is filtered out. That's the only way digital sampling will work, it has to be a band limited signal that can not exceed half of nyquist.
 
With 48k, everything over 24k is filtered out. With 96k, everything over 48k is filtered out. That's the only way digital sampling will work, it has to be a band limited signal that can not exceed half of nyquist.

Like a speaker I saw advertised where it claimed the tweeter was capable of reproducing 96khz and was labeled as "DVD-Audio Compatible". :(
 
With 48k, everything over 24k is filtered out. With 96k, everything over 48k is filtered out. That's the only way digital sampling will work, it has to be a band limited signal that can not exceed half of nyquist.

That makes more sense. Is the filter analog? Also, does the accuracy of the converter drop off somewhat significantly towards the upper-end of its theoretical frequency response?

I really do appreciate y'all taking the time to explain this, by the way.
 
That is the reason why everything greater than 22.5kHz is filtered out. The converters never see anything that high.

22.05kHz. Modern oversampling converters allow the analog filters to be much shallower to reduce phase effects in the audible band. So the converters "see" much higher frequencies than that at the front end, and the final filtering is done in the digital domain in ways that can't be done in analog.
 
That makes more sense. Is the filter analog? Also, does the accuracy of the converter drop off somewhat significantly towards the upper-end of its theoretical frequency response?

I really do appreciate y'all taking the time to explain this, by the way.

I think some things have been oversimplified. Converters these days are generally the oversampling type. An 8x 44.1kHz oversampling converter runs at 352.8kHz. Its Nyquist frequency is 176.4kHz, which is as high as the analog filters have to go. That means they can be much shallower than filters that have to take out everything from 22.05kHz and up. Any further filtering can be done in the digital domain to get it down to 22.05kHz. That avoids some of the phase shift caused by steep analog filters.
 
I think some things have been oversimplified. Converters these days are generally the oversampling type. An 8x 44.1kHz oversampling converter runs at 352.8kHz. Its Nyquist frequency is 176.4kHz, which is as high as the analog filters have to go. That means they can be much shallower than filters that have to take out everything from 22.05kHz and up. Any further filtering can be done in the digital domain to get it down to 22.05kHz. That avoids some of the phase shift caused by steep analog filters.

Ahh, see that makes sense.
 
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