# The whole question of lens sharpness...

Started Jun 12, 2013 | Discussions thread
Re: The whole question of lens sharpness...

To follow on my previous post:

Alphoid wrote:

olliess wrote:

Alphoid wrote:

In theory, given a perfect model of the lens, I can completely undo any blur caused by the lens.

I took this statement to mean just what it said, namely that you could undo any blur given a perfect knowledge about the blur function of the lens. To me, that means lens defects, diffraction, defocus, and anything else that modifies a point in the image.

(above quoted for context)

[...]

Now moving on to your modified claim:

Original claim. You just misunderstood it

I will freely admit, my misunderstanding could be ongoing.    So let me just stick to points of clarification here...

1) The operation of masking the image with a fixed frame (e.g., a rectangular windowing) is equivalent to convolution in the frequency domain. Since the Fourier transform of a rectangle has infinite support, some of the variance below the Nyquist limit must be spread beyond the Nyquist limit. Do you agree?

I am not sure. I don't believe I agree, but I think I may be misunderstanding what you're trying to say. There are several things which I'm finding ambiguous (e.g. What operation in the optical chain corresponds to 'masking the image with a fixed frame?'). Can you write out the above being slightly more verbose/specific?

The lens blurs the original image; let's assume for simplicity it's exactly a convolution with a fixed PSF. Since the frame you capture is finite, some information that should have been within the frame has now been blurred outside the frame and is lost. Also, sources that should have been completely outside the frame can contaminate the captured frame.

In the spatial domain, your captured image (I) is the convolution of the PSF (P) and the original image (O), windowed by a rectangular window (R):

I = R . (O * P).

2) in the spatial domain, some of the image gets spread beyond the edge of the frame

This is correct, but not significant. The PSF is small. This would only matter if the PSF was a substantial portion of the image.

The PSF may or may not be small in extent. In theory, even the PSF due only to diffraction has infinite support, although the magnitude is small outside of a small extent.

PSF due to the lens is small, at least as it relates to sharpness. Places where it is has large extent but small magnitude contribute to contrast (rather than sharpness).

Now you're getting into subjective arguments about what constitutes sharpness as opposed to contrast (or resolution). Objectively, the stated goal is to undo the blur.

The assumption that the PSF is "small" (here taken to mean "contained within a small radius") is also problematic. You might reasonably claim that deconvolution works in the limit that both the blur and the noise are small, but the point is, for any given blur radius there will be some level of noise which breaks the inversion, and for any level of noise, there is some size to the blur beyond which the original signal cannot be recovered.

As an aside, the PSF for the Hubble Space Telescope was not particularly small.

Furthermore, no matter how "small" (in physical units) your PSF is, it will become relatively "large" with increasing resolution.

G(H(image)+noise)=image+G(noise)

For a sharpening filter, G>1 at high frequencies, so noise increases. In practice, this doesn't matter much at low ISO

H and G are linear operators. H(image) + noise is not a linear operator. G is the inverse of H but not of H + noise, so G is not the inverse solution of the problem, right?

I think we're arguing terminology and what we consider to be 'the system.' You get exactly what I said -- your original image plus a transformed version of the noise. This undoes any blur caused by the lens. Do you disagree?

To sort of repeat the argument made in my previous post, it's meaningless to say you've "undone" the lens blur when you've also added an unspecified amount of noise which may or may not obscure the gains of your inversion.

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