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Easily decompose the log argument! For any (ratio) R=S/N, then (please note that I highlighted the math formulas like codes, yet they are NOT codes):
A = Log( R) = Log(S/N) = Log(S) - Log(N)
leading simply to
-A = -Log( R) = Log(N) - Log (S) = Log(N/S) = Log(1/R)
Now, the definition of a logarithmic decibel scale for R (RdB) as,
RdB = 10A = 10 Log( R) = 10 Log (.. etc.
Or, converting between Amplitude (Am) and power of signal S (or noise N) using
AmS=S^2 and AmN=N^2 --> R=S/N = (AmS/AmN)^2
AmS (AmN) is the amplitude of signal (noise), again using basic math
RdB = 10 A = 10 Log ( R) = 10 Log ( (AmS / AmN)^2 ) = 2 * 10 * Log(AmS/AmN) = 20 Log (AmS/AmN)
RbD = 10 Log(S) - 10 Log (R) = SdB - NdB
SdB and NdB are the signal and noise, in the logarithmic decibel scale. Equivalent to:
RbB= 20 ( Log (AmS) - Log(AmN) ) = 2 AmSdB - 2 AmNdB
AmSdB and AmNdB are the signal and noise amplitudes, in a logarithmic, decibel scale.
In image processing it's more appropriate to define R as
M is the mean of the image pixels, and D is the standard deviation of the noise (calculated in few different ways, depending on the type of the imaging conditions usually). The rest is straightforward (parallel to the simple relations above).
You'll find that, taking ( -log( R) ) for an image reduces to only subtracting image noise from it's mean. If the noise was a fixed-type, the process means subtracting a constant from the entire image, leading to a homogenous intensity reduction.