G6 vs newest sensors: Real difference in high ISO noize

Started 9 months ago | Questions thread
Øyvin Eikeland
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Re: G6 vs newest sensors: Real difference in high ISO noize
In reply to Anders W, 9 months ago

Anders W wrote:

Øyvin Eikeland wrote:

E-M5 set at ISO 12800:

Measured ISO: 5953

Manufacturer ISO: 12800

Dynamic Range: 7.5Ev

The usable dynamic range in this case would be:

7.5Ev-log2(100/5953)-log2(100/12800)=6.39Ev

right?

Not quite.

7.5 - log2(100/5953) - log2(100/12800) = 7.5 - log2(100) + log2(5953) - log2(100) + log2(12800) = 7.5 - 6.64 + 12.54 - 6.64 + 13.64 = 20.4

Perhaps you meant

7.5 - log2(100/5953) + log2(100/12800) = 7.5 - 6.64 + 12.54 + 6.64 - 13.64 = 6.4

But I'd just go straight away to

7.5 - log2(12800/5953) = 7.5 - 13.64 + 12.54 = 6.4

Sorry. Missed a parenthesis while retyping what I typed into scilab:

-- hide signature --

>7.5-(log2(100/5953)-log2(100/12800))

ans =

6.395545

Anyway, you confirmed the resulting figure. Thank you for that.

Another example:

E-M5 set at ISO 12800:

Measured ISO: 5953

Manufacturer ISO: 12800

SNR 18%: 20.9dB

The usable SNR in this case would be:

20.9-20*log10(12800/5953)=14.25dB

right?

Not quite. You can't insert the ISOs in the SNR formula in the manner you do. Provided that the sensor response is linear and fixed pattern noise can be ignored (which might be somewhat questionable assumptions in at least some cases, though perhaps not so much at the ISOs at issue here), the SNR increases by about 3 dB (about 3.01 to be somewhat more precise) when we increase exposure by one EV. So we have

20.9 - log2(12800/5953)*3.01 = 20.9 - 13.64*3.01 + 12.54*3.01 = 17.59

Ok,

I was guessing that this was a questinable result as it seemed a bit pessimistic. I am still not fully understanding this so please bear with me. The logic behind my original thought was that since olympus only utilise part of the range of the sensor, the numerator (signal) in the signal-to-noise ratio fraction needs to be reduced. The denominator (noise) would be the same. The amount of reduction of the numerator should be equal to the difference in gain I figured:

-->20*log10(12800/5953)

ans =

6.6494817

Can you tell me why the reduction should be only 3.01dB? (a really odd number?)

BR,

ØE

ps! I am an electrical engineer. I do not mind if you get a bit technical

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