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Topic: true 1-bit sampling (Read 8830 times) previous topic - next topic
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true 1-bit sampling

Hi

I reading through a book on digital audio at the momement dealing with 1-bit sampling. The authour states that one can move more of the noise (or distortion) floor of the quantisation error of a 1 bit sampling process to higher frequencies by increasing the sampling rate. A rate of 15 billion khz would give 100db of dynamic range. Now this sounds great and altough I can see in delta sigma processes that when the sampling error is fed into the next sample a greater amplitude sample would increase the amount of 1's or 0's if one is not using delta sigma but pure 1-bit sampling I cannot see how the bit stream would be different for two sine waves of the exact same frequency and phase yet of different amplitudes. So how would true 1-bit sampling differ between two such signals?

Thanks

true 1-bit sampling

Reply #1 –
Did the author say anything about "dithering"?

Cheers!
SG

true 1-bit sampling

Reply #2 –
Quote from: SebastianG on
Did the author say anything about "dithering"?

Cheers!
SG


There's plenty about dithering (a whole chapter in fact) and I know a 1-bit stream can't be dithered which is why there is a distortion floor and not a noise floor. I still don't see how two same frequency and phase waves of different amplitude would create a different bit stream (non delta sigma that is). There would be all ones above the zero crossing and all zero's below and they would have the exact same zero crossing. Would the difference be that the higher amplitude wave might round off to a 1 earlier after the zero crossing?

Thanks

true 1-bit sampling

Reply #3 –
Quote from: Zster on
Quote from: SebastianG on

Did the author say anything about "dithering"?

Cheers!
SG


There's plenty about dithering (a whole chapter in fact) and I know a 1-bit stream can't be dithered which is why there is a distortion floor and not a noise floor. I still don't see how two same frequency and phase waves of different amplitude would create a different bit stream (non delta sigma that is). There would be all ones above the zero crossing and all zero's below and they would have the exact same zero crossing. Would the difference be that the higher amplitude wave might round off to a 1 earlier after the zero crossing?

Thanks


If your 1 bit un-noise-shaped quantizer was properly dithered (simple uniform dither would do it in this case), you will not see what you assert coming out of the quantizer.

But why isn't delta-sigma "true one bit sampling", too? It's just PCM with noise shaping.
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J. D. (jj) Johnston
 

true 1-bit sampling

Reply #4 –
Quote from: Zster on
A rate of 15 billion khz...
Either you're pulling our leg or he is.

It's just not physically possible to generate a frequency at 15 THz (terahertz)... or PHz (Petahertz) if he's/you're using the long-scale billion. Somebody has put way too many zeroes in there! 

Cheers, Slipstreem. 

true 1-bit sampling

Reply #5 –
Quote from: Slipstreem on
Quote from: Zster on
A rate of 15 billion khz...
Either you're pulling our leg or he is.

It's just not physically possible to generate a frequency at 15 THz (terahertz)... or PHz (Petahertz) if he's/you're using the long-scale billion. Somebody has put way too many zeroes in there! 

Cheers, Slipstreem. 


He's talking theory not pratical. The book is "Digital Audio Explained" by Nika Aldrich and I'll quote the paragraph

"We can use a lower bit depth if we sample at a higher rate and then filter out all of the unneeded high frequency material. In fact we can record the entire audio range with only one bit if the sample rate is high enough" .....

"Recognizing this we can figure out what sample rate we would have to use if we wanted to capture the entire audible spectrum with 120db of dynamic range with only 1-bit. If we want to push the audible range noise floor from -6db to -120db FS and we subtract 3db for every doubling of the sample rate then we can calculate the sample rate we would need to use. It would take 38 doublings of the sample rate (above 40kS/s) to give us the recording that we want or about 1.1 10^13kS/s or 11,000,000,000,000KS/s." .... "The notion of sampling at rates like 11,000,000,000,000kS/s is unreasonable, but it represents the very valid point that a 1-bit recording can theoretically give us our required audio fidelity." He then goes on to show that by using a feedback of the error (delta sigma) one can shift the distortion floor out by more than 3db per doubling of the sample rate.

You see I can understand delta sigma because amplitude has a direct effect on the amount that is fed back into the next sample but with just one bit and no feedback I cant see it.

I think I see the point in dither as it would cause more 1's when the amplitude is higher but the author does not imply that you need to dither the signal before hand.

true 1-bit sampling

Reply #6 –
Quote from: Woodinville on
If your 1 bit un-noise-shaped quantizer was properly dithered (simple uniform dither would do it in this case), you will not see what you assert coming out of the quantizer.
But why isn't delta-sigma "true one bit sampling", too? It's just PCM with noise shaping.

I think that 1-bit signal cannot be properly dithered or noise-shaped. There's not enough quantization steps to accommodate the amplitude of dither or shaped dither w/o non-linear distortion.

true 1-bit sampling

Reply #7 –
Triangular Probability Density Function (TPDF) dither isn't possible in 1 bit because it requires a 2-LSB range (-1 to +1 or 0 to 2) for minimal sufficient dither.

However, rectangular dither (RPDF) is possible, which will decorrelate truncation distortion to prevent spurious frequencies. It normally is susceptible to noise modulation, which is why TPDF is preferred when possible.

Noise shaping of the dither is more problematic.
Dynamic – the artist formerly known as DickD

true 1-bit sampling

Reply #8 –
I still don't see how the samples of the waves can differ. I'm sure dither has nothing to do with it. I think I'll ask my maths genius friend for the answer.

true 1-bit sampling

Reply #9 –
> I'm sure dither has nothing to do with it.

That is why you fail. -- just kidding

It certainly has something to do with dithering. If you don't dither and simply round the sample values to +/-1 you always get something like a full scale square wave. Not only the sine's amplitude doesn't matter you'll also get lots of artificial harmonics. The purpose of dithering is to prevent this.

It has been mentioned that you can use a RPDF dither (IID with rectangular PDF, -1/2 LSB to +1/2 LSB).

Suppose the bits 0 and 1 correspond to levels -1 and +1. Using this kind of dithering digital silence turns into random noise (random string of 0 and 1 bits). The closer your signal's amplitude comes to +1 the more 1s you'll get and the closer your signal's amplitude comes to -1 the more 0s you'll get. Dithering turns the whole thing into a statistical process. You can model this kind of quantization process as additive noise. An RPDF dither makes the added noise's auto correlation to be a single pulse. It means, you added just white noise/clicks to your signal and no other artifical tonal parts. Hence, your sine's amplitude is preserved -- otherwise you would have "noise" with an auto-correlation other than a single pulse.

In this binary context, such an RPDF dither is equivalent to randomly choosing the quantization threshold between -1 and +1 for each sample. So, instead of comparing the values against a threshold of 0.0 you compare it against this random threshold and output a bit of 1 when the value is higher than this threshold and a 0 otherwise.

Cheers!
SG

true 1-bit sampling

Reply #10 –
That makes sense. So the final result is a form of PDM.

true 1-bit sampling

Reply #11 –
Quote from: Zster on
That makes sense. So the final result is a form of PDM.



No, it's just PCM. It looks a lot like PDM. But PDM can also be other things that do not obey the rules of lineary systems, and PCM with dithering does, even 1-bit systems.
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J. D. (jj) Johnston

true 1-bit sampling

Reply #12 –
Net stuttered
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J. D. (jj) Johnston