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21st June 2003, 05:15 PM  #221 
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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 realvalued 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 realnumber 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 "twobit" 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 realvalued 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 alternatingsequence 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). 
21st June 2003, 05:33 PM  #222  
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Join Date: Jun 2002
Location: Belgrade

Here we go again…
Quote:
Pedja 

21st June 2003, 05:50 PM  #223 
diyAudio Moderator

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.
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21st June 2003, 05:56 PM  #224 
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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. 
21st June 2003, 06:02 PM  #225  
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Join Date: Sep 2002
Location: Sweden

Quote:
merely pointed out that it was not necessary to add a 1/2bit 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. 

21st June 2003, 06:09 PM  #226  
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Join Date: Sep 2002
Location: Sweden

Re: Re: One, two, three... three or two, what do we need?
Quote:
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. 

21st June 2003, 06:23 PM  #227 
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Join Date: Sep 2002
Location: Sacramento, CA

Re: Re: One, two, three... three or two, what do we need?

21st June 2003, 06:45 PM  #228 
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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.

21st June 2003, 06:50 PM  #229  
diyAudio Member
Join Date: Sep 2002
Location: Sweden

Re: two heads are better than one and two voltage levels are smaller than three
Quote:
many levels. 

21st June 2003, 07:02 PM  #230 
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Join Date: Aug 2002
Location: Belgium

HAVING A FIELDDAY?
Hi,
Cheers,
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