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AreBee

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Theoretical question
« on: May 01, 2016, 07:03:40 am »

All,

If a Bayer-filtered RAW image is downsized, saved to TIFF then upsized for subsequent print, is it true that colour and detail in the final image will be superior and inferior respectively to an image that would be returned if the RAW image had simply been upsized?
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Bart_van_der_Wolf

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Re: Theoretical question
« Reply #1 on: May 01, 2016, 07:29:54 am »

All,

If a Bayer-filtered RAW image is downsized, saved to TIFF then upsized for subsequent print, is it true that colour and detail in the final image will be superior and inferior respectively to an image that would be returned if the RAW image had simply been upsized?

Hi Rob,

If we are talking about downsizing the pixel dimensions, instead of reversible compression, there will be losses. Upsampling that result will usually not restore the losses (unless there was little or no detail in the image).

The amount of loss depends on the amount of downsampling, and the algorithms that were used. It also makes very little visible difference whether the original was a Bayer CFA original, or full RGB original capture. The luminance resolution will only lose some 6-7 percent of the resolution to the Bayer CFA demosaicing process, compared to full RGB captures, and the sensitivity of our eyes to subtle color losses is much lower than luminosity losses.

Cheers,
Bart
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AreBee

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Re: Theoretical question
« Reply #2 on: May 01, 2016, 08:17:33 am »

Thanks Bart,

Quote
If we are talking about downsizing the pixel dimensions, instead of reversible compression...

Yes, I had pixel dimensions in mind.

Quote
...there will be losses. Upsampling that result will usually not restore the losses...

Understood (except for your use of "usually" - how can it ever?). However, taking into account the competence of current software at upsizing, and that a significant increase in resolution is required to make a meaningful visible difference in a print, my thinking had been that although the final image would be less detailed, colour would be more accurate, and that this may be a worthwhile trade.

But it sounds like you lose more than you gain, except perhaps in extreme cases.
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GrahamBy

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Re: Theoretical question
« Reply #3 on: May 01, 2016, 09:15:21 am »


Understood (except for your use of "usually" - how can it ever?)

Imagine you had a photo of a plain green wall with perfectly uniform illumination :)
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AreBee

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Re: Theoretical question
« Reply #4 on: May 01, 2016, 09:57:26 am »

Graham,

Quote
Imagine you had a photo of a plain green wall with perfectly uniform illumination

The losses would be concealed, not restored.
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Bart_van_der_Wolf

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Re: Theoretical question
« Reply #5 on: May 01, 2016, 10:51:53 am »

Thanks Bart,

Yes, I had pixel dimensions in mind.

Understood (except for your use of "usually" - how can it ever?).

A totally uniform area, or a linear gradient, can be upsized without penalty.

Quote
However, taking into account the competence of current software at upsizing, and that a significant increase in resolution is required to make a meaningful visible difference in a print, my thinking had been that although the final image would be less detailed, colour would be more accurate, and that this may be a worthwhile trade.

While color might gain a bit by weighted averaging to a smaller size, we also lose some color resolution. The question then is, what's worse. I tend to take whatever realistic (luminance and color) resolution we have, and use that as a basis for the required final output size.

Quote
But it sounds like you lose more than you gain, except perhaps in extreme cases.

That's my take as well.

Cheers,
Bart
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Jack Hogan

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Re: Theoretical question
« Reply #6 on: May 01, 2016, 01:04:25 pm »

All,

If a Bayer-filtered RAW image is downsized, saved to TIFF then upsized for subsequent print, is it true that colour and detail in the final image will be superior and inferior respectively to an image that would be returned if the RAW image had simply been upsized?

The simplest demosaicing algorithm is to consider a Bayer rggb raw quartet as a single pixel: then r is R, b is B and G is (g+g)/2.  Color information has not been comingled with its neighbors by the demosaicing process but the image is 'half' the usual size.

This approach is very fast and easy (try it, it's available as dcraw -h) but it has two major drawbacks: aliasing and loss of resolution.  Either is WAY more objectionable to the typical observer than any microscopic local advantage in color fidelity.  Don't forget that the human visual system creates its own full resolution grayscale image from which it tells sharpness.  It is much less sensitive to changes in the chromatic components however, hence the various virtually transparent color subsampling schemes for storage and transmission.

Jack
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GrahamBy

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Re: Theoretical question
« Reply #7 on: May 01, 2016, 02:02:12 pm »



The losses would be concealed, not restored.

No, there would be no losses. Pixel dimension is about spatial sampling frequency: with lower sampling frequency, you are obliged to give up on high frequency signal. When you reconstruct, you (usually) assume the dignal filtered out was never there, so it is lost. In the case of a uniform zero-frequency signal, it happens that it never was there, and so the reconstruction is perfect.
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AreBee

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Re: Theoretical question
« Reply #8 on: May 01, 2016, 03:32:53 pm »

Jack,

Quote
This approach is very fast and easy (try it, it's available as dcraw -h) but it has two major drawbacks: aliasing and loss of resolution.  Either is WAY more objectionable to the typical observer than any microscopic local advantage in color fidelity.

Thank you.


Graham,

Quote
No, there would be no losses. Pixel dimension is about spatial sampling frequency: with lower sampling frequency, you are obliged to give up on high frequency signal. When you reconstruct, you (usually) assume the dignal filtered out was never there, so it is lost. In the case of a uniform zero-frequency signal, it happens that it never was there, and so the reconstruction is perfect.

I stand corrected. Thank you for your explanation.
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