G6 vs newest sensors: Real difference in high ISO noize

Started 6 months ago | Questions thread
Anders W
Forum ProPosts: 17,886Gear list
Like?
Re: G6 vs newest sensors: Real difference in high ISO noize
In reply to Øyvin Eikeland, 6 months ago

Øyvin Eikeland wrote:

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

Yes, I figured it must be a typo of one kind or another. But I still don't understand why you brought the 100s into the formula.

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.

No problem.

The logic behind my original thought was that since olympus only utilise part of the range of the sensor,

See Ulric's reply. It's not that Olympus doesn't utilize the entire range. It's rather that they leave more highlight headroom when calibrating things (metering relative to full-well capacity and sensor gain) than some other manufacturers. If you meter in the conventional way (which I have long since stopped doing), this will sometimes means that Olympus preserves highlights for you that would otherwise have been clipped and sometimes that you leave unused headroom that might otherwise have been exploited. Once you know how to expose for the highlights, this calibration issue does not matter any more.

the numerator (signal) in the signal-to-noise ratio fraction needs to be reduced.

If you do not expose up to the clipping point, this is correct.

The denominator (noise) would be the same.

No, that's where you go wrong. What you say holds for dynamic range, which is defined (in unlogged terms) as S/N where S is the signal and N is the noise floor. Various definitions of the noise floor exist (DxO uses the point where the unlogged SNR equals 1, i.e., 0 dB in logged terms) but no matter which of them we use, the denominator remains the same.

For SNR, however, it works differently. If we consider photon noise only, then the (unlogged) SNR is

S / sqrt(S)

In other words, the photon noise falls in absolute terms as you lower the signal but the signal falls more so that SNR falls nonlinearly (and nonproportionally) with reduced exposure.

BTW: I should have mentioned read noise as another of the factors we ignore, along with fixed pattern noise and the possibility of a non-linear sensor response, when we consider photon noise alone as I did above.

The amount of reduction of the numerator should be equal to the difference in gain I figured:

-- hide signature --

>20*log10(12800/5953)

ans =

6.6494817

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

It follows from what I said above that it takes a quadrupling rather than doubling of exposure to double the unlogged SNR. The difference in dB as we double the SNR is 20*log10(2) = 6.02 and 3.01 is half of that.

BR,

ØE

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

I figured something like that. I am nothing close to an engineer but I don't mind getting a bit technical either.

 Anders W's gear list:Anders W's gear list
Panasonic Lumix DMC-G1 Olympus OM-D E-M5 Olympus E-M1 Panasonic Lumix G Vario 14-45mm F3.5-5.6 ASPH OIS Panasonic Lumix G Vario 7-14mm F4 ASPH +21 more
Reply   Reply with quote   Complain
Post (hide subjects)Posted by
NopeNew
Keyboard shortcuts:
FForum PPrevious NNext WNext unread UUpvote SSubscribe RReply QQuote BBookmark post MMy threads
Color scheme? Blue / Yellow