[APML] CMY Filters and Signal-to-Noise

[Aplanatic@aol.com]

The use of carrots in my previous post has sent my browser into never-never 
land.  What a pain in the buttski, sorry.  Here's another attempt:

Chuck wrote:

> Using their conversion formulas for RGB and ignoring the weighting factor:
>  
>  R = (Y + M - C) = 2R
>  G = (Y + C - M) = 2G
>  B = (C + M - Y) = 2B
>  
>  The RGB channels have a factor of two which has to be same as stacking two
>  images. What I don't know is if doubling the orignal exposure is the same 
as
>  stacking two negatives, i.e., the signal to nise goes up by 1.4. With a 
> linear
>  CCD is probably is but wth non-linear film it could be more complicated.

Thanks for an interesting and thought provoking post, Chuck.  I have 
been puzzling over this for quite some time (since the original article).  
Here's my analysis.

As you did, let

C=G+B, etc.

Then the mean value of C, denoted mean(C), is

mean(C) = mean(G) + mean(B)

The squared-noise in the C-channel, denoted mean(C2) (for the average of 
C squared) is 

mean(C2) = mean(G2) + mean(B2) + 2*mean(GB)

where the 2*mean(GB) term is the cross-product (correlation) between the 
green and blue channels.  In a linear detector the cross-product should 
be zero since there is no physical mechanism that mixes independent 
photons of two different colors detected at two different times.  However, 
with film this is not necessarily so.  More on this in a moment.

Let's assume that mean(GB) equals zero, as it should for a CCD.

We then re-create the red channel by:

R = Y + M - C

The mean of this process is

mean(Y) + mean(M) - mean(C) = 2*mean(R)

The squared noise in this process (N^2) is

N^2 = mean(Y2) + mean(M2) + mean(C2)
which by the above is

N^2 = 2*( mean(R2) + mean(G2) + mean(B2) )

and the square of the signal-to-noise ratio (S/N)^2 is

(S/N)^2 = 4*mean(R)^2 / 2*( mean(R2) + mean(G2) + mean(B2) )

Thus, the signal-to-noise in the red channel is

S/N = sqrt[ 2*mean(R)^2 / ( mean(R2) + mean(G2) + mean(B2) ) ]

This is very interesting.  If each channel is exposed to the point that 
the noise backgrounds are equal then the three terms in the 
denominator are equal and one gets:

S/N = 0.82 S/N(native)

where S/N(native) is the native signal-to-noise ratio of just the 
R-channel alone.  This implies that there is actually a penalty 
(not an enhancement) in the signal-to-noise ratio for the CMY 
process.  Another way of seeing this is as follows:

1) You get twice the red signal from the CMY process.

2) The noise in each component of the CMY images are sqrt(2) bigger 
because you have twice the signal and background intensity.

3) Noise adds as root sum of squares (RSS), so the combining 
process (C+M-Y) adds sqrt(3) to the noise.

4) The signal is twice as big, but the noise grows by sqrt(6), so 
the S/N is degraded by a factor of sqrt(2/3).

Now, what if the backgrounds are not equal?  Well, that just means 
that some other (G or B) channel will take a bigger hit in S/N in favor 
of the red channel.

Finally, back to the correlation problem, <GB>.  In B&W film there IS 
a correlation between two photons detected at different times with 
different colors.  We know that a film grain requires multiple photons 
(of any color) to change states.  So there will be a cross-correlation 
term between photons of different colors.  How much is an interesting 
issue.  The cross term will increase the noise and further decrease 
the S/N of each combined channel, but probably not significantly.

Dave Rowe.

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