Roy,
> Here's a small example: we have a voltage of 9.37 volts.
But then how do you even have a voltage of 9.37V in the first place? How
did you measure it at that?
> First the quantization situation, we have a digital volt meter that
> measures to the nearest volt. The quantization error or noise is
> +/- .5 volts. No matter what we do we'll always read 9 volts and be
> off by .37 volts.
What about the noise in the signal? The noise may bias it up to 10V.
> Second we'll try the random noise situation: to be similar to the above
> let's say random noise of .5 volts is added to the 9.37 volts. So the
> voltage is at least bouncing around from 8.8 to 9.9.
How can you have 9.37V with .5V noise? You can't. I think there's some
level of basic understanding that's missing here...
> So any particular
> sample will have a value anywhere in that range. However if you make many
> samples and average them the result will converge on 9.37 because
> the errors
> due to random noise will cancel each other out.
Absolutely not. First off, you don't even know that the signal is really
9.37V, because it simply is not, it can't be with a noise of .5V. You can
only measure the signal to within +-.5V.
Film scanners take a STATIC measurement of the voltage from the CCD, and
even if you take more than one measurement, that does not mean you are going
to get more precise measurements...you can take a billion measurements, ALL
your measurements will be inaccurate to that noise, and "averaging" them
does NOT mean you then "average out noise" (I'm not talking about noise that
is greater than the noise floor, that's a different issue and getting rid of
that noise is a LOT easier, it's characterized). This noise could be caused
by some bias, and your "technique" has just failed.
If you can't ACCURATELY characterize the noise, you are not going to be able
to simply "remove it" by multiple sampling. When scanners do multiple
samples, they are removing noise that is way above the noise floor, that is
entirely different.
> I've been claiming a lot of stuff lately so I'm going to try to back it
> up with a real demonstration. Here's a step wedge file that I based on
> the 21step wedges that come with Piezography. The top part is the
> standard wedge with 21 gray steps from 0% K to 100% K in 5% steps.
> The bottom is a duplicate with lots of noise added.
You're merely spreading the tones out, so the AVERAGE tone ends up being the
same...it's just like using a coarse dither pattern.
> The PS command
> is Add Noise> 12.5% Gaussian if you want to try it yourself. The
> noise is a lot -- magnify to 400% on screen and see it, marquee a
> single step and check the histogram. What was 1 grayscale value now
> spans more than half the entire grayscale.
No matter, as it's the average that matters, and you didn't change the
average. I'm not quite clear what this test is supposed to "prove" or show.
You KNOW what the source of the noise is, and you KNOW that it's equally
dispersed throughout the patch...and this case is completely different than
taking voltage measurements, they do not relate.
Roy, how do you think dithering works?????
> Marquee and Histogram 2
> steps and there no obvious steps. However, print the file out on
> paper and the step wedge shows through loud and clear.
As well it should, I wouldn't expect it not to...given the fact that it's a
dither pattern it self, and the average density has not changed... Did you
expect it to somehow be different?
> Get out the
> densitometer and the gray tone measurements of each step match very
> well whether you measure the noise-less step or the noisy step. So
> the "signal" here is the 5% wedge, the "noise" is large enough to span
> many steps in the wedge, but its easy to resolve densities much
> closer than the noise level.
Nice try, but this is completely irrelevant to measuring voltages. Your
noise is completely characterized here, and is evenly dispersed. That is
NOT a given with CCD noise, so you can not make the same assumptions.
> Download this file, its a TIFF to insure there is no lossy compression.
>
>
> Austin, I hope you are willing to print this out with Piezo and
> measure some of the steps.
I did, and I don't see anything I wouldn't have expected. I think you think
you're showing something that you just aren't showing. Why do you believe
the average density on the patches SHOULD be different? Like I said, it's
simply like doing a coarse dither pattern. The densitometer has an area it
reads, and it averages the area that it reads...and if you add +- so much
percent noise over the area, the average is the same. I'm sure it's off by
something, but beyond the resolution of my desktop densitometer.
I believe you're simply comparing apples and oranges here, and your example,
though a nice try, just doesn't model the way film scanners and CCDs and
voltages typically work.
I know that there are many noise reduction techniques, but they do not apply
to film scanners.
Austin
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