Re: A "slanted edge" analysis program

Don wrote: 
On Sun, 02 Oct 2005 16:07:04 +0200, "Lorenzo J. Lucchini"
<ljlbox@xxxxxxxxxx> wrote:


Now, whether you get ESF peaks is not really what I was addressing but
the fact that the whole concept of sharpening is based on this
selective contrast. So whether this causes ESF peaks or not, the image
has been (in my view) "corrupted". It may look good, and all that, but
I just don't like the concept.

I see, but I'm not sure sharpening can be dismissed as an optical illusion. 

It can, because the image is not sharpened only the contrast at both
sides of the border between dark and light areas is enhanced locally.

To really sharpen the image one would need to "shorten" the transition
from dark to light i.e. eliminate or reduce the "fuzzy" part and
generally that's not what's being done.

One simple proof of that is halos. If the image were truly sharpened
(the fuzzy transition is shortened) you could never get haloes! In the
most extreme case of sharpening (complete elimination of gray
transition) you would simply get a clean break between black and
white. That's the sharpest case possible.

The fact that you get halos shows that so-called sharpening algorithms
do not really sharpen but only "fudge" or as I would say "corrupt".

But my point is that sharpening algorithms should not necessarily produce haloes. I don't have proof -- actually, proof is what I'm hoping to obtain if I can make my program work! --, but just note that my hypothesis is just that: halos need not necessarily occur.

By the way - not that it's particularly important, but I don't think the "sharpest case possible" is a clean break between black and white, as at least *one* gray pixel will be unavoidable, unless you manage to place all of your "borders" *exactly* at the point of transition between two pixels.

From all I've understood, scanners (expecially staggered array ones)
soften the original image, and sharpening, when done correctly, is simply the inverse operation. 

But that's my point, exactly, it does not really reverse the process
but only tries to and that just adds to the overall "corruption".

Make a distinction between unsharp masking and similar techniques, and processes based on knowledge of the system's point spread function, which is what I'm trying to work on.

Unsharp masking just assumes that every pixel is "spread out" in a certain way (well, you can set some parameters), and bases its reconstruction on that.

*That*, I think, is its shortcoming. But if you knew *exactly* the way every pixel is "spread out" (i.e., if you knew the point spread function), my understanding is that you *could* then really reverse the process, by inverting the convolution.

Read below before you feel an urge to say that it's impossible because the process is irreversible...

Actually, "softening" and "sharpening" are just two specific case, the general concept being: if your optical system *corrupts* the original target it's imaging, you can *undo* this corruption, as long as you know exactly the (convolution) function that represents the corruption. 

In theory but not in practice. And certainly not always. You can't
reverse a lossy process. You can "invent" pixels to compensate
("pretend to reverse") but you can never get the lossy part back.

Now, this time, yes, what we're talking about is a lossy process, and as such it cannot be completely reversed.

But before giving up, we should ask, *what is it* that makes it lossy?
Well, I'm still trying to understand how this all really works, but right now, my answer is: noise makes the process lossy. If you had an ideal scanner with no noise, then you could *exactly* reverse what the sensor+optics do.

In real life, we have noise, and that's why you can't just do a deconvolution and get a "perfect" result. The problem you'll have is that you also amplify noise, but you won't be otherwise corrupting the image.

Sure, amplifying noise is still something you might not want to do... but pretend you own my Epson staggered CCD scanner: you have a scanner that has twice less noise than equivalent linear CCD scanners, but a worse MTF in exchange. What do you do? You improve the MTF, at the expense of getting noise to the levels a linear CCD scanner would should.
And, in comparison with a linear CCD scanner, you've still gained anti-aliasing.

Kennedy would agree :-) So let's quote him, although I can't guarrantee the quote isn't a bit out of context, as I've only just looked it up quickly.

--- CUT ---

[Ed Hamrick]
> The one thing you've missed is that few (if any) flatbed scanners
> have optics that focus well enough to make aliasing a problem when
> scanning film.  In this case, staggered linear array CCD's don't
> help anything, and just reduce the resolution.

[Kennedy McEwen]
If this was the situation then any 'loss' in the staggered CCD spatial
response would be more than adequately recovered by simple boost
filtering due to the increased signal to noise of the larger pixels

--- CUT ---

[snip]

Yes, in theory. In practice, my red channel has a visibly worse MTF than the green channel, for one.

That's *very* interesting!!! I wonder why that is?

Well, for one, who's to say that my scanner's white light source is white?
If red is less well represented in the light source than the other primaries, there will be more noise in the red channel.

Though noise should directly affect the MTF, AFAIK. 

Ehm, note that I meant to say "shouldn't" here.

But there are other possibilities: being a flatbed, my scanner has a glass. Who says the glass "spreads" all wavelengths the same way? 

But none of that should affect the results because you're dealing with
*relative* change in brightness along the edge. Now, in *absolute*
terms there may be difference between channels but if, for example,
red receives less light than other channels the *relative* transition
should still be the same, only the red pixels will be a bit darker.

Not darker: the scanner calibration will set the light as the whitepoint, so channels will still have the same brightness.
I agree on a second thought that the red source should be *really* dimmer than the other, for this to produce noticeably more noise.

But I think the glass hypothesis still stands: if the glass blurs red more than it blurs the other colors, well, here you have a longer edge transition, and a worse MTF.

> [snip]

by LjL
ljlbox@xxxxxxxxxx
.

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