Re: s/n ratio

[Caveman]

For an uniform exposure, you have the average film density Da, on the 
full frame. This is the "reference". An ideal film should have Da in 
each and every point. In the real case, you move the detector and get a 
Dmax and Dmin along a track. The peak noise would be the higher value 
between (Dmax-Da) and (Da-Dmin). Report that to Da and you have a ratio. 
Now the aperture of the detector will influence these values. IIRC 
there's a "standard" aperture used for measuring RMS.
To continue, you could under/over expose the film and do the same, and 
see how this is influencing the "noise" (grain).
Any expert in RMS here ?

cheers,
caveman
Bruce Rubenstein wrote:
Lets assume a detector with an electrical out put is being measured 
(viewed) with an oscilloscope. You will see an irregular wave form that 
corresponds to the p-p value of the noise (you can derive the RMS value 
from this). This is the same as looking at the noise floor for a 
circuit. The S/N ratio comes from comparing the average value of the 
noise floor to some other value, i.e., 0 dB. You need two different 
signals, with different average values that you are comparing to be able 
to use a ratio. Film grain, by itself is just a noise floor.

BR

[EMAIL PROTECTED] wrote:

Signal = ideal (image)
noise = difference between ideal and what's actually recorded
Now let's try an experiment. We expose the film to a uniform source of 
light. As an average on the whole frame, the "signal" is good, i.e. we 
got an average level of gray that corresponds to the exposure. Now 
let's take a smaller aperture - 1mm x 1mm. The recorded signal varies 
inside this area, due to the presence of grain. Basically that's how 
they measure RMS - they take a small aperture and measure the variance 
of signal when they move that aperture along a uniformly exposed area. 
What I was asking is if this could be expressed as a s/n ratio instead 
of the  classic "RMS" value.

cheers,
caveman