[filmscanners] RE: Newish Digital Tech

["Paul D. DeRocco" <pderocco@ix.netcom.com>]

> From: David J. Littleboy
>
> There's a problem with your logic here. Although the Foveon chip measures
> "all the red", each individual site measuring red measures _the
> same amount
> of red_ as any individual site measuring red in a Bayer sensor.
> So the noise
> level in each measurement is, in principle, identical. To make
> your argument
> fly, you have to show that in reusing the smaller number of measurements,
> Bayer increases the noise. I don't think it does...

Oh, but it does, because correlated signals add linearly (i.e., in the
voltage domain) while uncorrelated signals add root-mean-square (i.e., in
the power domain).

For instance, consider a simple Bayer decoding algorithm that computes each
missing green value as the average of the four adjacent real green values.
(Real algorithms are more complex, but no amount of complexity will escape
the fundamental limitation of information theory.) This means that noise
from each green sensor is duplicated and divided by four into the adjacent
pixel locations that don't have sensors. If at a particular point in time, a
green sensor is generating, say, one millivolt of noise on top of the signal
itself. This will result in 0.25mV of noise added to each of the four
adjacent pixel locatons, so the total amount of noise added to the image is
2mV, spread out over five pixels. In a Foveon chip, the four adjacent pixels
have their own sensors which generate their own noise, but it will be
uncorrelated, meaning that it may be positive or it may be negative. When
you add the noise of one pixel to one quarter of the noise from the adjacent
four pixels, instead of getting 2mV, you get 1.414mV on the average, because
of the way two Gaussian bell curves add together.

The result is theoretically 3db better S/N for the green, and 6db better S/N
for red and blue.

--

Ciao,               Paul D. DeRocco
Paul                mailto:pderocco@ix.netcom.com

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