dBu to 0 dBFS
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Andy Peters
Well-known member
Ethan said:I understand the point you are making, but I've never liked the common audio analogy of "lowering the noise floor" with more bits, because even though the end result could be you have better SNR, that's not inherently the case just because we have more bits.
Except that it's not an analogy, it's actually what happens. As you increase the word length, the noise floor does indeed go lower.
-a
Aside from the decrease in quantization error noise with increased word length...or were you referring to something else?Andy Peters said:
For instance, assuming a fixed voltage level, without rounding error induced noise, how would the noise floor go lower?
[many apologies to Kambo for now really veering further :-X]
For every added bit you get an additional 6.02dB of dynamic range. You could call it a lowering of the noise floor or an increase in dynamic range.
Some of this has to do with the original sin of PCM digital audio labeling the top of the scale "0" instead of some kind of red line lower in the scale. There is no standard reference level so the difference between lowering the noise floor and increasing dynamic range is semantic and contextual.
Some of this has to do with the original sin of PCM digital audio labeling the top of the scale "0" instead of some kind of red line lower in the scale. There is no standard reference level so the difference between lowering the noise floor and increasing dynamic range is semantic and contextual.
M
mattiasNYC
Guest
Ethan said:
It is indeed about the quantization error noise, not the noise of the room or analog electronics. Many converters use single-bit delta-sigma converters to begin with anyway, and then map to PCM, so the analog signal is actually captured using far fewer than even 16-bits to begin with.
JRsaidalready said:
While the 6.02dB theoretical dynamic range per bit based on numbers of quanta may hold for modest digital word length, modern high performance convertors have analog, sounding, looking, and feeling, noise floors higher than the theoretical 6.02dB X number of bits related to the conversion technique. Don't hold your breath expecting a -144dB noise floor from a 24 bit convertor, not going to happen. The practical noise floor limits are complicated and have been discussed around this neighborhood before.
JR
ruffrecords
Well-known member
JohnRoberts said:
Exactly. Even if 144dB was possible, if we have +24dBu = 0dBFS then -144dB is -120dBu. A 150 ohm source has noise of -131dBu so you only need 10dB gain in the (perfect) mic pre for the source noise to exceed convertor noise.
Cheers
Ian
PCM and analog noise floors behave very differently. As always apples to apples comaprissons are almost impossible.
The only converter test I've ever needed is white noise or tape hiss. A good converter comes close. A bad converter falls down.
The only converter test I've ever needed is white noise or tape hiss. A good converter comes close. A bad converter falls down.
john12ax7
Well-known member
Gold said:
Interesting. I've never bought into the "you need X # of bits to equal analog" arguments, they just don't sound the same. But I've never considered using something like tape hiss to test a converters quality. I'm curious to try.
M
mattiasNYC
Guest
ruffrecords said:
Well the point wasn't to talk about how we needed to push the noise floor down to -120dBu, the point was that the move from 16 bits to 24 bits lowers the noise floor, and that in a 16-bit system quantization noise can actually be a problem. Applying the same calibration as above you'd look at -72dBu, which is quite the difference from -120dBu.
So again, the point was merely to illustrate that we increase dynamic range, not "increase resolution" the way we commonly think about it. In a sense, this is akin to what happens in the time domain where an increased sample rate doesn't really increase resolution (as many often think of it, intuitively) but instead increases the range of frequencies.
ruffrecords
Well-known member
mattiasNYC said:
OK, understood, but it begs the question what does define the resolution in a digital system?
Cheers
Ian
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mattiasNYC
Guest
I think you'll essentially have to look at the inner workings of PCM, since it isn't something we users "set".
Another way to look at it is the same way as with sample rate: Either we extend a ranger or we increase resolution. When it comes to sample rates, since frequency = cycles over time, increase sample rate = increased range for captured frequencies, not higher resolution of the frequencies captured. So in the "dynamic domain", more bits = increased range between maximum level and "noise floor".
Just like the technology takes care of any two samples being able to accurately describe any sine wave at a frequency less than half the sample rate, the technology takes care of us getting 6.x dB per bit. If you ask me to explain it I'll fail, because it's 4:30am and I'm still working and have forgotten any and all details.... If someone else explains it I'd appreciate it though, always good with a refresh.
Another way to look at it is the same way as with sample rate: Either we extend a ranger or we increase resolution. When it comes to sample rates, since frequency = cycles over time, increase sample rate = increased range for captured frequencies, not higher resolution of the frequencies captured. So in the "dynamic domain", more bits = increased range between maximum level and "noise floor".
Just like the technology takes care of any two samples being able to accurately describe any sine wave at a frequency less than half the sample rate, the technology takes care of us getting 6.x dB per bit. If you ask me to explain it I'll fail, because it's 4:30am and I'm still working and have forgotten any and all details.... If someone else explains it I'd appreciate it though, always good with a refresh.
Sure they do use less bits to start with, but at a much higher frequency, which mathematically compensates; however the sample length must be large enough for the math to be valid. It generally is.mattiasNYC said:
That's quite a strange assertion, since, by definition, resolution is the number of available discrete values used to describe the signal - also can be expressed as the exponent of 2 resulting in the same number (and identical to the number of bits).mattiasNYC said:
Anyone who thinks increasing SR increases resolution is mistaken, even if "many often think of it, intuitively". Anyone who's studied digital conversion shouldn't make this gross terminology error.
There is nothing like frequency resolution in digital conversion; it is continuous from zero to 1/2 SR. I believe you're referring to Fourier transform, which indeed discretizes frequency bins, but irrelevant here.mattiasNYC said:
No; more bits = increased range between maximum level and
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mattiasNYC
Guest
abbey road d enfer said:
No, I'm not talking about Fourier transforms. I was postulating an either/or scenario to illustrate that if we get higher frequencies captured with a higher sample rate we do indeed as you point out not get a higher resolution. That was my entire point.
abbey road d enfer said:
Yes, I'm talking about quantization noise. I thought that was obvious.
abbey road d enfer said:
It's not really a strange assertion.
It is, because it just contradicts the mathematical definition of resolution. The point in increasing the number of quantization steps is to increase the accuracy with which a value is assessed. It is not tied to any practical application.mattiasNYC said:
Dynamic range is a signal processing concept, that is indeed closely tied to resolution.
In fact, SNR, THD vs. level, DR and resolution are all intricately related in signal processing, but resolution can exist on its own in math.
It is true that the main operational benefit is rejecting the quantization noise low enough to make low-level distortion and noise not an issue any more.
