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In Reply to: SACD equilivalent sampling rates posted by Tommart on January 19, 2001 at 20:15:46:
You cannot really extrapolate sampling rate to resolution , resolution is like shades of grey , you can take a zillion snapshots at a resolution of only black or white and only GET a resolution of black and white , you can take 10 snapshots where every colour is represented and the 10 snapshots will be more "accurate" than the zillion at black and white.
To put it really simply , the sampling rate is how many snapshots and the bits are how much info those snaphots contain , the more snapshots , the better one can represent motion over time , or the more accurately one can do that. More bits means that every pic has more info - pointless to use more sampling than nessecary without the info - Thing is , tho its not intuitive , one doesnt need a zillion samples to represent an action , one only needs to sample at the freq of the action x2 (nyquist)
In realistic terms - without getting too technical , DSD achives (with noise shaping) about a 20 bit resolution in the audible spectrum , its gets real bad in terms of noise at higher freqs.
At any rate , components and hificannot do better than 21 bits resolution (the dynamic range of the signal - IE diff tween softest and loudest level the signal can attain) due to the actual thermal noise of the component.
However there is an actual resolution and a notional resolution , IE actual may be 21 bits , however one can process at a higher rate.
What does this mean? Well when one does computation on numbers , there is a remainder , now the amount of bits used determines how this remainder is used , if there are not enough "spaces" to store all the digits of the remaineder , some get chopped off and results arent accurate , these inaccuracys become cumulative.
So having more resolution allows more accurate representation of the numbers.
In terms of digitising a signal , the resolution works more or less the same , the value chosen of the sound sampled in terms of its loudness has to be captured and the bigger the "register" you have , the more accurate the capture , lets say you have a number like 0 or 1 , then a 1 bit system can only capture the 0 or the 1 , a value like 0.5 will either be 0 or 1 , this is not that good as there is a huge amount of error (This is VERY simplified - doesnt quite work like that in real life) leading to what is called quantisation error (the difference tween the captured value and the real value)Just as an aside , I thought I would explain dither as well in a more understandable format , lets say you take a picture of a blue sky , you see a few clumps of red in this sky , its VERY noticeable. Now break up those clumps and add a bit more red (Decorrelating and adding MORE noise) and spread them around the blue sky at random (more evenly) , one will not notice "Clumps" , the sky will change shade to a slightly more purple , but wont appear "clumpy"
This is a very complex subject and I have tried to explain it in sort of laymens terms
A good article to read is listed below
Hope this helps
Rodney Gold
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