What does a sensor SNR rating in dB actually mean?

Asked 1/26/2019

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I’m confused by sensor signal-to-noise ratio specifications given in decibels. For example, if a sensor is rated at 20 dB SNR at a certain ISO, how should that be interpreted? Does it refer to voltage, power, luminance, or something else? In practical terms, what does a 20 dB SNR mean for image data?

Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0

Photography Stack Exchange contributor

7y ago

2 Answers

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The answer is "C", something entirely different.

The signal-to-noise ratio used in imaging is the ratio of the mean signal value divided by the standard deviation of the signal. This is used because photon counts, luminous intensity, etc., are always-positive values that are basically probability distributions.

Regarding dB (decibel) scale, imaging uses the field-quantity ratio (i.e., 20×log(ratio)) and not the power quantity ratio (10×log(ratio)).

So a 20 dB SNR means the signal-to-noise ratio is 10(20/20) = 10, which represents only log2(10) ≈ 3.3 bits of signal above noise floor. An 8-bit image with values equally distributed in the range 0 - 255 has an RMS value of about 147.4. Thus, the standard deviation among pixels values for a 20 dB sensor producing an 8-bit image is 147.4 / 10 = 14.74, or nearly 4 bits of noise. That is a very poor SNR for a sensor. A sensor with 20 dB sensitivity at a certain ISO sounds like either a very poor sensor, or a very high ISO, or a combination of both.

See also:

Originally by user11924. Source · Licensed CC BY-SA 4.0

user11924

7y ago

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AI Answer

Generated from our catalog & community — verify before relying on it.

In imaging, an SNR value in dB usually means the ratio of the mean signal level to the noise standard deviation, expressed on a logarithmic scale.

For image sensors, this is generally treated as a field ratio, so:

SNR(dB) = 20 × log10(signal / noise)

So a 20 dB SNR means:

signal / noise = 10^(20/20) = 10

In other words, the useful image signal is 10× larger than the noise level.

It is not usually interpreted as “1% of full swing” in the simple voltage/power sense you described. In imaging, the noise figure is more commonly tied to the statistical variation of the signal itself, often using standard deviation.

A helpful way to think about 20 dB is that it corresponds to about log2(10) ≈ 3.3 bits of signal above the noise floor.

So the correct interpretation is closest to: something different from your two options — specifically, a logarithmic ratio of signal level to noise level used in imaging measurements.

UniqueBot

AI

7y ago

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