Thanks to everyone for replying...
Oh goody a test...
1. Why does each bit give us 6dB of dynamic range?dB = 20 x Log (V1/V2)
( i didnt pull that equation from memory, i knew of its existence)
2. What kind of effect occurs to the waveform when we have to round to the closest quantization step?I dont really know.
3. Are Signal to Noise Ratio and Dynamic Range the same? Why or why not?I dont think so, Dynamic range i think is theoretical, SNR is actual.
4. In a 16 bit system what is the SNR? If SNR and DR are different, what is the DR of a 16 bit system?Well, the dynamic range is 96db, i dont know what the SNR is.
5. If I have a 16 bit system that covers the range from 0V to 1V and I make it a 17 bit system, what happened to the size of the range that this converter covers?
well, it still overs 0 to 1V. Just with more steps.
I hope this will help.
Nika Me too
Look, i probably got everything above wrong...
And i guess, I might be asking this question, in a forum, where this stuff is probably, well, to put it badly, beneath you all, if thats the case i am sorry...I was just trying to get a clearer answer. I have been reading and reading and reading and reading.
This is how I understand it to work.
16 bit, gives a dynamic range of 96 db, for the reason I explained above.( i know its not EXACTLY 6db, close enough)
24 bit gives a dynamic range of 144db.
11111111 in binary, 8 bit word, equals in Decimal 256
this is where the bit depth/resoloution comes from, 8 bits gives 256 possible steps, 16 bit gives 65, 535 steps etc etc....
See the way I look at it, and its probably wrong, is that at the time of sample (say 44.1khz) the ADC, takes a snapshot, and determines the level of what it sees, to the resoloution of the bit depth created.
So, it looks and sees a signal , between 0 and -6db, and uses 16 bits to represent that level, in a 16 bit file, and 24 bits to represent that level in a 24 bit file.
In my mind there are more steps available, between 0 and -6db, in a 24 bit file, than there are in a 16 bit file, but my friend disagrees, to his mind, all the extra steps are below -96 db, and there is exactly the same amount of resoloution available to that 0 to -6db window, whether it is 16 bit or 24 bit.
Its almost like we see resoloution the exact opposite of each other. That I see more steps available at 0db than i do at -144db, where he sees the opposite.
I apologise , If i cant be clear enough.
But my basic question remains.
with a 24 bit file, is there more resoloution (steps available,possible values) to document a signal, between 0 and -6db, than there are in a 16 bit file
thanks for all your time and effort
Peter Knight