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What is the precise definition of signal to noise ratio, as used by DxOmark? Please see here, after clicking Measurements, SNR 18% (it's not directly linkable).

I am looking for an explanation (in particular a precise definition) which is understandable by someone who has sufficient mathematical background, but no familiarity with this engineering field and its conventions.

To put it differently, given a greyscale image a_ij of a plain grey sheet (a_ij representing the intensity at pixel location (i,j)), how would one calculate the SNR? Also, does the calculation use the linear sensor data or the perceptual pixel values?

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SNR is the ratio of the log of power of the Signal to the log of power of Noise. So, it is Signal divided by Noise, but we don't take the amplitude of both but rather its powers. Power contained in a signal is amplitude squared. That is why, if you have the amplitude, you get a factor of 2 in the formula. Then there is an additional factor 10 relating to dB being the deci-Bell. 

SNR = 2 * 10 * log(signal/noise)

Here signal and noise are defined to be the mean and standard deviation. This assumes that we are taking a picture of a smooth, unicoloured surface. Thus we would expect the same lightness value across the image (or region of interest). This means our mean is the signal. Any deviation from that mean means that the camera measured the "wrong" value, i.e. added noise. Thus the standard deviation is taking to be the noise.

Finally, we get: 

SNR = 20 * log10(mean/std dev)
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  • \$\begingroup\$ This is a general definition of SNR, but not the whole story. It doesn't explain why a large sensor with more pixels will have a higher snr than a smaller sensor with the same pixel pitch but correspondingly fewer pixels. The standard deviation alone is independent of the number of pixels for a give per pixel noise level. Please see my comment on the main question. \$\endgroup\$
    – Szabolcs
    Nov 14, 2013 at 13:02
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    \$\begingroup\$ Sure. But if you want an answer to your second question ("why a large sensor..."), I suggest you ask it as a new question. Not a bad question but not the one you originally asked. \$\endgroup\$
    – Unapiedra
    Nov 14, 2013 at 13:36
  • \$\begingroup\$ That is not my question... It is clear to me why a larger sensor will show less noise when its image is viewed at the same size as the image from a smaller sensor. What is not clear is the precise meaning of the numbers DxO uses. \$\endgroup\$
    – Szabolcs
    Nov 14, 2013 at 15:43
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Talking about pixel level SNR.

SNR of down-sized image from Sensor A / SNR of down-sized image from Sensor B = sqrt(Ra / Rb) * (SNRa / SNRb) where Ra and Rb are the resolution (total pixel count) of Sensor A and Sensor B respectively; SNRa and SNRb are the pixel level SNR on Sensor A and Sensor B respectively.

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  • \$\begingroup\$ This is a derivation for Print SNR. I am not sure but I would guess 18% refer to the light intensity (light power per unit area) being applied for the measurement. \$\endgroup\$
    – photox
    Feb 28, 2016 at 4:27
  • \$\begingroup\$ 18% refers to the 18% of grey level. \$\endgroup\$
    – photox
    Feb 28, 2016 at 4:32
  • \$\begingroup\$ The definition at dxomark.com/About/In-depth-measurements/Measurements/Noise \$\endgroup\$
    – photox
    Feb 28, 2016 at 5:13

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