Hi Todd,
> This paper appears to speak to many of the issues discussed in
> this thread:
>
> <>
I had a chance to look over that paper. The diagram you mention (I believe
you were referring to the sinusoidal wave +-5V signal...) that looked like
the Higgins one, though is similar, is being used to show an example of 2's
compliment arithmetic. Nothing really to do with dynamic range, as far as
our discussion that is.
It says in a few spots, basically, that more bits is more dynamic
range...and when talking about what dynamic range N bits can hold...that's
true. Obviously, in a scanner, more bits may not give the SCANNER more
dynamic range, but you certainly need a minimum number of bits to be able to
represent all the different values.
The definitions they use for dynamic range, as I've said in another post,
are the same as I've been using, just don't forget that subtraction in logs
is the same as division using non-log numbers. See my other post on that...
This paper has a lot of superfluous information that is particular to DSPs
and audio, and not as applicable to the scanning issues we've been talking
about. Most scanners don't use DSPs that I'm aware of (at least the
consumer ones, like the SS4k etc.), the Leaf happens to...BTW, one for each
channel.
The paper also says, quite clearly "Note that the "6-dB-Per-Bit-Rule" is an
approximation to calculating the actual dynamic range for a given word
width.", as well as "In theoretical terms, there is an increase in the
signal-to-quantization noise or dynamic range by approximately 6 dB for each
bit added to the word-length of an ADC, DAC or DSP."...which says that the
dynamic range is DIRECTLY related to the number of bits, and each bit gives
you 6dB more dynamic range for audio. Of course, as we've discussed, that
is purely talking about being able to represent all the numbers of a
particular dynamic range...it has nothing to do with whether the bits are
actually "good". If you do the calculations your self, it shows how they
arrived at that:
2 bits = 20log2**2 = 12dB
3 bits = 20log2**3 = 18dB
Funny how that works out...
So, I simply don't see anything that's in that paper that conflicts with
what I've said. Perhaps you can elaborate on, specifically, where you
believe the difference is. Don't get caught up in any of the DSP
stuff...just stick to what they write about bits and dynamic range for the
A/D, that may give you less confusion.
BTW, the decibel (dB - deci being 10 BTW, so 10 deci-bels is 1 bel...) is
defined as 10log10 (power ratio), but since power goes as the square of
voltage or current (for constant impedance), the log of voltage or current
ratios can be expressed in dB, with a factor of 20 instead of 10...and since
that entire paper is talking about audio, they use 20 as the dynamic range
multiplier.
Austin
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