Hi Alan,
That makes sense, so the Drange would be what?
log(2^14.5) - log(2^1.5) = 3.9 which is plausible
For "number quantization levels", what specifically do you mean? the number of
bits?
"number quantization levels", or L, is the number of different binary
numbers you need and L<=2^b where b is the number of bits, in our case say
b=16 so L<=65536. Now, Dmax and Dmin are the max and min values
respectively of the sampled analog signal. If Dmin=0 and Dmax=65535 (i.e.
65536 numbers), then (Dmax-Dmin)/(L-1)=1. Now, for the particular case of
scanners, I do not know the max and min values, but I can definately tell
you that
log(2^14.5) - log(2^1.5) = (14.5-1.5)*log2=3.9, which may translate:
"The way the signal processing guys have said it to me is that the dynamic
range
coming out of an A/D converter is all the bits less the bottom 1.5 bits."
into (16-1.5)*log2=4.36 but again, I am not sure if this is the way to
calculate it....
Regards,
Dimitris