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