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  • Hey all, just changed over the backend after 15 years I figured time to give it a bit of an update, its probably gonna be a bit weird for most of you and i am sure there is a few bugs to work out but it should kinda work the same as before... hopefully :)

Red Compression and Matrix Tests.

Antoine, an 8bit codec can store the dynamic range of film, or at least what it looks like :-)

But that's because film has a tonal curve that compresses highlights.

When we're talking dynamic range here, we're talking linear dynamic range, and for that you need at least as many bits are there are stops.

Graeme
 
Graham, thank you - that was a great explanation of something that's always confused me too, with regards to latitude and bit depth.

Can I just reel this off back to you, as I understand it, and check I've got it right (which I probably haven't!)?

Using Antoine's example of 20 stops on a 12-bit linear codec:
From my (new) understanding, the situation is that you can (or rather could) encode the 20-stop dynamic range within 12-bit linear, but by doing so you would introduce quantization noise - so, encoding at < 1 bit per 6db SNR is theoretically possible put practically undesirable.

Is that correct??

Thanks as always...
 
Well, you can't fit a 20 stop dynamic range into less that 20 bits if the data is linear. If you put a transfer curve on it, that makes the data non-linear, it will retain the look (I hope) of the extra dynamic range, but the dynamic range of the resulting image is still noise limited to whatever the bit depth is, even if the data is essentially noiseless.

Graeme
 
Ohh-kay...

I think I'm getting my head around this finally - please bear with me!

So essentially what you're saying is that by encoding the theoretical 20-stop image into 12-bit digital (with the transfer curve to essentially compress the dynamic range), you end up with an image that captures the high/low light detail of 20 stops from the scene, but has only 12 stops of usable dynamic range in the final file (i.e. for grading and manipulation in post)?

Am I getting there?!!
 
I think that's pretty much it, Dominic. I mean we've all seen film on VHS, but that doesn't mean VHS has the dynamic range of film :-)

Evin, I think RED RGB is 10bit Log or 10bit video gamma (ie REC709), both of which do a similar thing, but with different curves - they compress the data in such a way that you get your equivalent low noise levels in the shadows, but with a slight loss of precision in the highlights, which is a pretty good trade. The difference is in the curves that do that.

Graeme
 
Graeme Nattress said:
I think that's pretty much it, Dominic. I mean we've all seen film on VHS, but that doesn't mean VHS has the dynamic range of film :-)
Click to expand...
Lol, yeah ok - good point!

Thanks again Graeme for taking the time to go through that with me, it's really appreciated...
 
You're gonna hate me - another question!...

What would happened if you tried to encode a > 12 stop latitude image in 12-bit linear? Would it simply clip the highlights above the 12-stop point, or do something else entirely?

Cheers! ;)


EDIT: I might have answered my own question here - presumably that would be the choice of the person(s) implementing the encoding software - to either lose the top end, bottom end, or a combination of both?
 
You'd comprmise your noise levels, or loose highlights, or both, but it should be your choice.

Graeme
 
I'm so, so close to having my head around this!!

Can you clarify what you mean by "compromise your noise levels"? I've seen you mention that the amount of low-light detail is determined by the noise levels in an image, so does compromising noise levels essentially equate to "losing the bottom end", as I put it, or are we talking something different?
 
You'd comprmise your noise levels, or loose highlights, or both, but it should be your choice.

Graeme
 
Ah !

OK, now i understand a lot better !

...linear... !!!!

thanks Graeme
 
Thanks Antoine - happy to help as usual!

Graeme
 
ChrisForbes said:
dynamic range is the amount of information you get on a sensor between absolute black and absolute white. It doesn't have as much to do with the lens as it does the capture medium.

Think of a sunset. In the frame you will have the brightest point (the sun) and the darkest point (the shadows). The amount of information you can get in between those two point is your "range".
Click to expand...

Chris, now I got it 50% when it comes to dynamic range.
What is then one stop? and why it is called "stop"
Is it from aperture?
One stop = (how many) dB of change?

And why 12 stops equals 12 bits (72dB), coincidence?
Is one stop of aperture lowering/increasing light by 6dB (two times light level)?

Andrew
 
1 stop = 1 bit = 6db, roughly speaking.
 
I see dynamic range as being the difference between the darkest and brightest objects that can be differentiated from pure white and black in the same frame. But I don't consider dynamic range as being data enough to account for the number of samples between those two extremes. The gaps between the two extremes are filled with resolution, which might be 16 bits of linear samples, or 8 bits of logarithmic samples, or whatever number of samples in whatever weighted curve.

So what confuses me is how resolution is being lumped as a side-effect of dynamic range. Wouldn't the resolution be dependent on the capture device, rather than depending on the capture device's dynamic range? Couldn't you have trillions of steps of precision in the sample device but only be able to measure those trillions of samples between two extremely similar brightnesses?
 
Noise limits your ability to measure the difference between steps. If the difference between steps is smaller than the amount of noise in the system, then you can no longer differentiate between those steps. Say we sampled the Mysterium sensor at 32bit, or something silly like that. We've have gadzillions of steps, but you'd still be only able to measure about 2500 (that's what 11.3 stops is).
 
Graeme Nattress said:
Noise limits your ability to measure the difference between steps. If the difference between steps is smaller than the amount of noise in the system, then you can no longer differentiate between those steps. Say we sampled the Mysterium sensor at 32bit, or something silly like that. We've have gadzillions of steps, but you'd still be only able to measure about 2500 (that's what 11.3 stops is).
Click to expand...

So the sensor has a 2500 step linear resolution, and the dynamic range of the chip goes up to a little over 2000 times brighter than the darkest object it can recognize. Is the noise behavior of the chip a constant, or is there less chip noise in the low-end or high-end?
 
Well, a sensor doesn't have any steps as it's analogue, but it's pointless to digitize with a step size smaller than the amount of noise, as then you're making steps that are too small, and are wasteful.

I'm not that up on the noise characteristics of sensors - sorry. But you'll always notice noise in the dark regions, just because you've got less photons hitting the sensor, and hence more noise - just like the human eye in the dark.

Graeme
 
Is there a lowest level of lumen that the chip can noiselessly detect?
 
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