Quote:
Originally Posted by adamdea  
...
If you reduced by 24 db would you get 8 8 8 9 8 8 8 8 8 7 8 8 8. (I
said 24 db because you said it was a 6 db reduction in your example but
I confess I don't understand the 6 db equivalence. I just mean getting
quieter to the point where the peak becomes 9 and the trough 7) 

..and below that level it would just be 8s?
This is a 4-bit example. Each bit value is double the previous bit
(8-4-2-1 ... binary math).

Doubling in audio = approx 6dB.

with 4 bits you can only capture a total of 24dB of dynamic range... 

So if you reduced the level by 24dB you would have nothing! (silence)
Remember the smallest amount you can reduce by is 1 bit... so you just
keep taking one bit off both the top and bottom values until you are
left with all 8's. So in the 4-bit example, you can reduce a full scale
(15-0) wave 7 times/steps until both 15 and 0 become 8. (actually I've
just realised in all my examples it is 7 that is the zero-crossing
point, not 8 - doh! schoolboy error, I must learn to count).
unquote 

You can feel free to ignore the bit in brackets. If it is annoying just
move on.
[I understand the 6 dB = 1 bit equivalence. But in your example when
you reduced the peak and raised by trough by 1 from 14 to 13 you
described it as a 6db reduction. (I assumed from that that each
iteration represented 6db). I wonder whether the reduction by 1 should
have been a 3 dB reduction. I was simply reducing the peak etc by 4
more stages in the example I gave. 

There seem to be 7 reductions before your signal becomes 7s and 8s. I
cant see any way of reducing the a 15, 0 starting pair to get either 8
or 7- I wonder whether this is because the crossing point is the same
as the median value (7.5). If so I am now baffled as to how one
expresses silence in this 4 bit word.(pehaps you ignore 0, leaving 15
values and a happy median of 8).]

Now what was the point? Well I was tryign to work out whether we could
clarify the previous discussion by actually drilling down to the
numbers. I think that it is very difficult to discuss technical
(especially mathematical relationships  in everyday english. I have
noticed for example that on the news they often trip up when tryign to
discuss economic (or econometric) concepts like say the distinction
between the stock of public debt, the budget deficit, a change in the
budget deficit.
As to where this takes us 

1.1 You have pointed out that in information theory terms i have been
confusing two terms "precision" and "resolution". These seem to
correspond with 2 problems
a. the quantisation noise cuased by the limited choice of sample
values
b. the inabilty to produce a sound quieter than the lowest value
(apparently over or above the crossing value.)

1.2 It seems that an analog sytem has limited resolution but not
limited precision. If we return to your post  33, you described radio 3
as 13 bit. 
I assume that you meant that it had a snr of 78 dB. I have been making
the point at various stages (although apparently without using the
correct terminology) that whilst a digital recoding may have 16 bit
resolution and 16 bit precision, analogue systems can have 13 bit
resolution and unlimited precision. OR have I got this wrong and was
the 13 bit radio limited in both resoltuion and precision?   

1.3 I am assuming that if a sound gets quieter it not only gets closer
to the noise floor but the quantisation noise relative to signal
increases.  

II. My major concern was over the example of the 24 bit DAC which has
"only" 17 bit resolution. I think this means it can only resolve a
sound 102dB below peak. I am however confused as to what this means
about its precision. I understand that the DAC doesn't have to guess at
the values it is decoding, but isn't the output the same as if it were
reading a 17 bit recording with 7 random numbers on the end . Does this
not mean that not only does the noise floor increase but the
quantisation noise will increase?


-- 
adamdea
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