Re: Fixing an unfocused image theoretically straightforward?!
- From: acl <achilleaslazarides@xxxxxxxxxxx>
- Date: Mon, 04 Jun 2007 14:49:53 -0700
On Jun 5, 1:25 am, markw...@xxxxxxxxx wrote:
On Apr 25, 7:21 pm, Pat <gro...@xxxxxxxxxxxxxxxxxxxxxx> wrote:
There are, therefore, more possible sharp images than there are possible
unsharp images
Actually, this should read: there are more ways for sharp images to
*look different* (to the unaided eye) than there are for unsharp
images.
Nice proof. I was thinking of a different way to disprove it.
That may run into some basic problems, as explained below.
Assume there is an small object near your lens and you focus on something far
away (for example shooting through a screen window). The "out of
focus" image disappears.
It doesn't actually disappear so much as smear out and blend into the
rest of the image. The information is still there, but has become
intertwined with the rest of the figure. In a way, you might think of
it as analogous to what happens to the mind after the body dies.
What you say is true if you take each pixel as recording a real; but
it's not, it can only record an integer in a finite range. So you may
lose information when you do the convolution (optically).
The problem with a *disproof* is that the process is one-to-one
(though not necessaarily onto). In fact, there is a close analogy to
the relation (classical physics <-> quantum physics).
But it's not one to one, precisely because the signal is quantised by
a converter. If we had infinite bit depth then it would indeed be one
to one; but information is lost, not due to the convolution itself,
but due to the fact that this optical convolution might (to take an
extreme example) reduce the signal due to a delta function source to
the point where it's swamped by noise during recording.
The convolution by a Gaussian or any other distribution is given by
the kernel, which is defined by the convolution of <exp(Kx)>. For a
Gaussian this takes the form <exp(Kx)> = exp(Ka + (Ks)^2/2), where a
is the mean and s the spread. The inverse comes straight out of that:
one whose generating function is <exp(Kx)> = exp(-Ka - (Ks)^2/2).
The connnection to quantum theory is that every quantum state has
associated with it a distribution called its Wigner distribution.
Though not a probability distribution (as it would be in classical
physics), it is still the Gaussian deconvolution of a probability
distribution (The spread being directly related to Planck's constant).
The drawback of deconvolution is that since it has to try and piece
back together what's been smeared out all over the place, the process
is highly non-local. Since a blot gets smeared out all over the
figure, the source that reproduces the blot has to be drawn from the
entire figure. (Quantum theory also possess a similar property of a
high degree of non-locality, which is probably one of its most
distinguishing features). A filter implementing the process would
therefore be prone to noise generation, since it magnifies the high
frequency components substantially.
That's easily seen by just considering the case of what would result
if the smear you deconvolve is, itself, a Gaussian of the same size as
that you're deconvolving with. Then you get a point.
For discrete convolution (and this is something I actually worked out
once), the using a Gaussian on the pixels results in a filter that is
likewise one-to-one. The inverse, however, is ornery! Each pixel takes
contributions from the entire grid in an almost magical way. Another
drawback of straight deconvolution therefore is seen to arise from
this: an implementation of the process can be time-consuming. It
amounts to literally looking for needles in a haystack.
Well it seems clear that since you are essentially transferring
amplitude from low frequency components to high, noise will be a big
problem (amplified). But if the convolution is 1-to-1, shouldn't the
inverse also be? (although a practical implementation is another
story, I guess).
.
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