The dynamic range of 16 bits
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 21st June 2003, 06:15 PM #221 Christer   diyAudio Member   Join Date: Sep 2002 Location: Sweden Now Fred almost agrees with me, that we get a squarewave. However, if I get you right you are alternating between the input values 0001 and FFFF so we have only two output values, ie. a square wave, but the amplitude is "2 bits" pp (0001 - FFFF = 0002), ie. 1 bit peak, not 1/2 bit. To get the squarewave with 1/2 bit amplitude it seems we must add an offset to the signal as SY suggested and as I used in my program that generated such a squarewave. Well, now, what about larger amplitudes than 1/2 bit? Do we still get a square wave? No, except for signals of frequency 1/2 fs. Rather we do get at least three distinct levels, as in the Stereophile diagram and in Steves diagram. Well, I may be wrong, but this is my argument. It is just a proof sketch, for the lack of time and a reasonable way to write formulas (well, I could scan). Mathematics ahead!! "Theorem": With 16 bits, we can represent sine waves with real-valued amplitude arbitrarily close to, but not including the value 20log(2^-16) dB (which is approximately -96 dB). Proof sketch: One error that I, at least, have made is to not really bother about the case where the source has a value not corresponding to any of the quantization levels. It seemed not necessary, but of course it is. Let's suppose our input signal is a sine wave of amplitude A and with angular frequency w, that is Asin(wt). We restrict A to be a real number s.t. 0<= A <= 32767. For every sample, the real-number value of the function must be converted to an integer in the interval -32767..+32767, which I assume is most appropriately done by rounding off to the nearest integer. (Note that we cannot use the value -32768 when representing symmetric signals with no offset). Now, let's look at input signals at and just above the amplitude 0.5, and write the amplitude on the form 0.5+x, where x >= 0. Define this signal as the function f(x), that is f(x) = (0.5+x)sin(wt) , where x >= 0 and let r(x) be f(x) rounded off to the nearest integer. Now, consider the case x = 0.5, that is f(x) = f(0.5) = sin(wt), that is r(0.5) will have the values -1, 0 and +1. In each cycle we get a sample sequence of the form 0,...,0,1,...,1,0,...,0,-1,...,-1,0,...,0 that is, a signal of the form shown in Stereophile, or in Steves diagrams. We have one positive pulse of amplitde 1 and one negative pulse of value one. This is a "two-bit" peak to peak signal corresponding to a sine at level 20log(2^-15) which is approximately -90 dB. (I will round off all dB figures, not the formulas, to enhance readability). Then consider the case x = 0, which means f(x) = f(0) = 0.5sin(wt). Since the max and min values are 0.5 and -0.5 and these get rounded off to 0, we get r(0) = 0, that is no signal at all!!! Since f(0.5) = 2f(0), f(0) must be a sine at level 20log(2^-16) which is approximately -96 dB. So, this proves that we cannot record down to -96 dB then?? No, not at all. We have forgotten that the input signal can take on any real-valued amplitude between 0.5 and 1, i.e. we can have x values between 0 and 0.5 (and, of course larger too). Now consider an arbitrary value of x, s.t. 0 < x < 0.5. Obviusly the function r(x) will be of the alternating-sequence form shown for f(0.5), but the positive and negative peaks will be narrower, just as Steve showed in his diagrams. i won't go into trivial calculus details here, but we note that as the value of x decreases, the width of the pulses decreases (the width can be expressed in intervals of angular frequency, in time intervals or number of samples, though the latter case, gives us a further quantization in the time domain). To sum up, for arbitrary x > 0, r(x) will be a signal with one positive peak of value one and one negative peak of value one, and the remaining intervals of the cycle have the value zero. The width of the pulses decreases as x approaches 0, and the pulses disappear at x = 0. What does this mean? Well, it seems to me to mean that we can represent the function f(x) for all real values of x down to but not including 0. We get the smallest representable sine wave as the limes of f(x) when x approaches 0, and this corresponds to the amplitude 20log(2^-16), since it is the amplitude of this limes sine wave, although a signal of exactly amplitude 20log(2^-16) cannot be repsented. So there is a difference between -96 dB and -96 dB The conclusion would be that we can record sine waves with amplitude down to, but not including -96dB (where 96 is a rounded off value itself, to simplify).
Pedja
diyAudio Member

Join Date: Jun 2002
Here we go again…

Quote:
 Originally posted by RobM There are 65536 quantization steps, so LSB is 1/65536, or -96.33dB. Any other answer is wrong.
Give me that file which will show peak amplitude -96.3dB (or any below -90.3dB). In any analyzer.

Pedja

 21st June 2003, 06:50 PM #223 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland Christer, "offset" is a term that offputs some people. You can equally well call it "asymmetry" (as I've said, people don't normally call a trumpet's waveform "offset"), or "scalar gauge," or a bunch of other things. The imposition of a symmetry requirement about an arbitrarily selected bit would trivially be seen to require three values, i.e., a two bit argument. But that imposition of symmetry was not a qualification of any of the original claims that are being discussed. The claims went to dynamic range (resolution) only. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 21st June 2003, 06:56 PM #224 Christer   diyAudio Member   Join Date: Sep 2002 Location: Sweden Addendum to previous "theorem" I should, perhaps comment on one assumption I made in the "theorem" above. I assumed that the smallest signal we can represent is the smallest signal that does not get represented as a sequence of all zeroes. That is, it is the smallest signal that does not get completely lost in quantization error. One might argue that we should, perhaps, have some requirement for reconstruction also, but I think this is the right way to define it. As I have said previously, no sine wave can be represented in PCM and fully reconstructed, no matter what level it has, no matter what number of bits we use, and I am assuming a theoretical DAC and a theoretical LP filter. I need hardly say how much harder it would be in practice. So, if requiring reconstruction, beyong not being reconstructed as a DC level, we cannot really have any firm levels and say what the smallest signal is. We can only do so under some extra condition for the reconstructed signal (eg. less than a certain amoung of THD relative to the original signal), a condition that is not, to my knowledge, included in any PCM standard and thus would be an added requirement, not present in the original problem formulation. Adding for instance a THD limit as condition, would give us not one value for the dynamics, but an infinite number of values of the dynamics, one for each possible value of the max THD allowed.
Christer
diyAudio Member

Join Date: Sep 2002
Location: Sweden
Quote:
 Originally posted by SY Christer, "offset" is a term that offputs some people. You can equally well call it "asymmetry" (as I've said, people don't normally call a trumpet's waveform "offset", or "scalar gauge," or a bunch of other things. The imposition of a symmetry requirement about an arbitrarily selected bit would trivially be seen to require three values, i.e., a two bit argument. But that imposition of symmetry was not a qualification of any of the original claims that are being discussed. The claims went to dynamic range (resolution) only.
Yes, I agree. I didn't mean we should not allow "offsets", I
merely pointed out that it was not necessary to add a 1/2-bit
offset as I did in my program, after your suggestion. We can
get down to the -96dB limit even with signals symmetric
around zero, which my "theorem" says. Not sure what result
we get if taking away that constraint too, but I guess (time
for somebody else to the maths this time , that with offset
we can get down to the value 20log(2^-16) inclusive, while
with symmetry around zero we can only
come arbitrarily close to it. A distinction that is of no importance
whatsoever to others than us who are amused by theory. In
practice it becomes -96 dB in both cases.

Christer
diyAudio Member

Join Date: Sep 2002
Location: Sweden
Re: Re: One, two, three... three or two, what do we need?

Quote:
 Originally posted by RobM There are 65536 quantization steps, so LSB is 1/65536, or -96.33dB. Any other answer is wrong.
Pedja, assuming my maths is correct above, I would say it is
now clear that you will not get the correct answer from any
program. This would require the program to do the inverse
transformation from digital to analog domain (which is not
even possible for some data, the alternating unit sequence),
which would involve very heavy numerical computations
probably beyond what even a good PC today can do within
the time you are prepared to wait. I am sure the programs
just look at the actual digitalized samples and uses these
values as approximations.

Steve Eddy
diyAudio Member

Join Date: Sep 2002
Location: Sacramento, CA
Re: Re: One, two, three... three or two, what do we need?

Quote:
 Originally posted by RobM There are 65536 quantization steps, so LSB is 1/65536, or -96.33dB. Any other answer is wrong.
You look rather lonely there, Rob. Care for some company?

As they say, two heads are better than one.

se

Fred Dieckmann
diyAudio Retiree

Join Date: Oct 2002
Location: Spain or the pueblo of Los Angeles
two heads are better than one and two voltage levels are smaller than three

The the number of possible DAC output voltage levels for the smallest possible output signal.
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Christer
diyAudio Member

Join Date: Sep 2002
Location: Sweden
Re: two heads are better than one and two voltage levels are smaller than three

Quote:
 Originally posted by Fred Dieckmann The the number of possible DAC output voltage levels for the smallest possible output signal.
Hm, let's see, that's the "V sign", and V is roman 5, no that's too
many levels.

fdegrove
diyAudio Senior Member

Join Date: Aug 2002
Location: Belgium
HAVING A FIELDDAY?

Hi,

Cheers,
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Frank

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