On Jul 24, 3:04=A0pm, slus...@[EMAIL PROTECTED]
wrote:
> On Jul 24, 12:55=A0pm, aruzinsky <aruzin...@[EMAIL PROTECTED]
> wrote:
>
>
>
>
>
> > On Jul 23, 2:48=A0pm, slus...@[EMAIL PROTECTED]
wrote:
>
> > > On Jul 22, 11:59=A0am, aruzinsky <aruzin...@[EMAIL PROTECTED]
>
wro=
te:
>
> > > > On Jul 22, 8:04=A0am, slus...@[EMAIL PROTECTED]
wrote:
>
> > > > > Thank you for your interest and offer to review some sample
image=
s.
> > > > > Unfortunately, our concern is not on the software or algorithmic
=
side,
> > > > > but on the optical side. =A0As I mentioned in my original post:
>
> > > > > "Does anyone have references they can point to (or their own
> > > > > pesuasive
> > > > > arguments) that describe where the true edge location should be
> > > > > relative to the edge image function?"
>
> > > > I already did. =A0What is wrong with you?
>
> > > > > I also mentioned that I originally posted to sci.optics, then
add=
ed
> > > > > sci.image.processing because I thought the folks here might be
mo=
re
> > > > > aware of relavent references. =A0The (apparently) commonly held
b=
elief
> > > > > that the maximum slope of the edge image, or the midpoint
between=
the
> > > > > maximum and minimum grey level values of the edge image
correspon=
ds to
> > > > > the true edge location seems unmotivated or unsup****ted to me,
ev=
en
> > > > > though it "feels" like common sense. =A0
>
> > > > That is an algorithm, albeit a really stupid one that is not
common
> > > > sense, so don't tell me that you are not interest in algorithms.
> > > > Furthermore, that is not a commonly held belief, e.g., it is
common
> > > > knowledge that shock filters sharpen an edge around the inflection
> > > > point (the inflection point is preserved) and nobody, I know of,
> > > > pretends that shock filtering is deconvolution.
>
> > > > > We have also performed bench
> > > > > experiments that suggest that neither the max. slope nor the
grey
> > > > > level midpoint corresponds to the true edge, with errors being
in=
the
> > > > > range of 0.2 to 0.45 pixels in the experimental set up we used.
>
> > > > I hope my tax money didn't fund that busy work.
>
> > > > >=A0We
> > > > > are continuing our own investigation, but I have to believe that
> > > > > someone has already done this at some time and published their
> > > > > results. =A0I'd love to find out that our own efforts at
inventin=
g the
> > > > > wheel are unecessary. :-)
>
> > > > I already gave you a reference to a relatively simple algorithm
tha=
t
> > > > might do better than 0.2 pixels. =A0 It would be foolish not to
try=
it.
>
> > > Hmm. I appear to have offended you. =A0If it was because I had not
ye=
t
> > > acknowledged your reply to my query, I apologize. =A0I was, in fact,
> > > digesting your comments as well as the two references you provided.
I
> > > was also reading what I found upon looking up "edge interpolation"
as
> > > you suggested. =A0In the meantime, Andrew offered a suggestion that
I
> > > was more easily and quickly able to address, which included my
> > > referring back to, and quoting my original post. You may have
> > > interpreted that as my ignoring your suggestions.
>
> > > Regarding your references, I believe I can see the point(s) you are
> > > making, but I also don't see the relavence to the *physics* of
> > > correlating the image of an edge to the location of the real edge.
We
> > > might be talking past each other here, much as the Americans and
Brit=
s
> > > are separated by a common language. :-)
>
> > > Let's ignore pixels for a moment, and pretend we can sample the
image
> > > of a back lit knife edge with infinite resolution, casting God's eye
> > > upon the screen on which the light eventualy falls. Are you familiar
> > > with Fresnel diffraction at an edge? When one looks at a plot of
> > > intensity of the light around the "shadow" zone, one sees there is
> > > light in the region that would otherwise be forbidden if there was
no
> > > diffraction. =A0One also sees that "past" the edge, the light
intensi=
ty
> > > climbs, reaches a peak, then oscillates in a decaying fa****on,
> > > eventually reaching a uniform intensity.
>
> > > If you look closely at this plot, we see that the point on the light
> > > intensity curve that in fact corresponds to the true location of the
> > > knife edge is of an increasing slope, but is neither the maximum
slop=
e
> > > (gradient), nor an inflection point, nor, generally speaking, is it
> > > midway between the minimum light level and the average high light
> > > level, much less the peak. =A0Thus, edge finding techniques that
depe=
nd
> > > in some fa****on on maximum slope or gradient (Sobel, Canny ,etc.) or
> > > 50% threshold level, would not find the point on this edge image
that
> > > corresponds to the position of the true physical edge. Of course
this
> > > is just a start. =A0The edge image is further complicated by the
many
> > > other phenomena that occur when using lenses to place the image onto
=
a
> > > pixel detector.
>
> > > My own readings found, in many cases, discussion of edge detection
> > > schemes that do in fact start with the presumption that the maximum
> > > slope or gradient of an edge image should in some way correspond to
> > > the true location of the edge, but with no explanation of why the
> > > author believes this to be so. =A0My guess is that this is something
> > > that is simply "understood" by those working in the field, and is
> > > passed along from one person to another without question. =A0It's
> > > obvious that you hold no such belief, but at the level of my
> > > investigation it *is* common.
>
> > > As for time spent investigating this, rest assured that this is
> > > strictly a capitalistic venture using private money, snide comments
> > > not withstanding.
>
> > > It may be of no interest to you, but others might find something
> > > interesting. =A0I'm cutting and pasting here a brief description of
o=
ur
> > > experiments that I also orignally posted on sci.optics:
>
> > > Using a
> > > chrome-on-glass bar target that is measured as 12.503mm long
> > > +/-0.003mm, we grabbed images with a telecentric lens and red LED
> > > collimated back light. =A0We then used a micrometer stage to move
the
> > > target 12.503mm (again, with an error of +/-0.003mm) -- so that the
> > > trailing edge at the new postion corresponds to the leading edge of
> > > the original position. =A0The direction of motion was perpendicular
t=
o
> > > the lens optical axis to within 0.25 degrees. =A0We then used
various
> > > edge detection methods, including the 2nd derivative of the edge
with
> > > a fit to a parabola, as well as the simple 50% local thresholding,
> > > among others. =A0What we found consistently through different
positio=
ns
> > > in the field of view and different new set-ups is that the trailing
> > > edge appeared to fail to move far enough to coincide with the old
> > > leading edge. =A0The error range was ~0.2 to 0.45 pixels, or about
> > > 0.015mm to 0.034mm in real world units -- always considerably larger
> > > than target and motion error, and alway in the same direction of
> > > error.
>
> > > Another way to interpret this result is that an object consistently
> > > looks larger than it actually is. (If the bar target appears to be
> > > 12.523mm, and you move it 12.503mm, the trailing edge will fail to
> > > "catch" the initial position of the leading edge.)
>
> > > We've tried this with different lens f-numbers from F/45 to F/8 and
> > > did not see an obvious trend that followed the aperture, but on this
> > > front we've only made a small number of tests -- 4 at each f-number.
> > > The errors average ~0.3pixels, with the standard deviation ~0.05.
>
> > > I will also add that we have used two different cameras (AVT
Stingray
> > > 145b and Sony XC-ST70) out of suspicion that a non-linear
responsivit=
y
> > > could be the culprit. =A0While differing slightly in detail, we saw
> > > essentially the same result.
>
> > > Spencer- Hide quoted text -
>
> > > - Show quoted text -
>
> Thank you very much for taking the additional time!
> Alright, let me make sure of my toddler steps here:
>
>
>
> > Consider a simple 1D example:
>
> > "Knife edge" is a discrete step function x.
>
> > x =3D [ 0 0 0 1 1 1 1 1]T
>
> > A =3D
> > [
> > 1/2 1/2 =A00 =A0 =A00 =A0 0 =A0 =A00 =A0 0 =A0 0
> > 0 =A0 =A0 0 =A01/2 1/2 =A00 =A0 =A00 =A0 0 =A0 0
> > 0 =A0 =A0 0 =A0 0 =A0 =A00 =A01/2 1/2 =A00 =A0 0
> > 0 =A0 =A0 0 =A0 0 =A0 =A00 =A0 =A00 =A0 0 =A01/2 1/2
> > ]
>
> So generally speaking, A represents the PSF of the lens, Fresnel
> diffraction at the edge, lens aberrations, etc. =A0Essentially the whole
> world of what goes into making the image of the edge no longer a step
> function, right?
>
>
>
> > A represents convolution with [1/2 1/2] followed by a factor of two
> > decimation. =A0You can see A is undetermined and not invertible.
>
> Decimation? I can see I'll need to pull out my old linear algebra
> text. =A0Or is this something more specific to image processing?
>
>
>
>
>
>
>
> > Your observed values would be
>
> > y =3D Ax =3D [ 0 1/2 1 1 ]
>
> > In estimating x from y, given A, you have to do something that vaguely
> > resembles an inversion of A to estimate x' of x.
>
> > For example, you can softly constrain the discrete derivatives of x'
> > to zero.
>
> > y' =3D [ yT 0T ]T
>
> > A' =3D [AT =A0 w*BT]T
>
> > w is scalar weight
>
> > B =3D
> > [
> > 1 -1 0 =A0... 0 0
> > 0 =A01 -1 ... 0 0
> > ...
> > 0 0 0 ... =A01 -1
> > ]
>
> > Now, A' is overdetermined
>
> > solve
>
> > min || y' - A' x' ||
> > =A0x'
>
> > ||.|| is a norm or possibly a function similar to a norm but not
> > necessarily convex. =A0L2 norm is not recommended because it will
cause
> > oscillations. =A0L1 norm recommended. =A0All of this has a statistical
> > interpretation which I won't get into for simplicity.
>
> > Your A will have to incor****ate a different PSF for diffraction
> > thingy.- Hide quoted text -
>
> > - Show quoted text -
>
> Yep, I definitely need to brush up on my linear algebra. =A0But let me
> see if I've at least got the gist of this.
>
> A cannot be known completely, nor possibly very well, at least not at
> first. =A0It also cannot be inverted. =A0Therefore, in estimating x from
> y, one needs to adjust parameters (in the more colloquial sense)
> during a process that is vaguely like inversion of A, and use this to
> find a function that operates on x to produce the known image result.
> This is likely an iterative process, comparing the results ...
>
> read more =BB- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
You have the gist of it, but for my purposes, image enlargement, A is
gestimated and that is good enough. My 1D example used a box kernel.
It turns out that the 2D box kernel is sufficiently accurate to model
the Sigma SD14 and DP1 pixel sensors. I think there are square
adjacent microlenses that account for this behavior. As far as I
know, I am the only one to provide a large variety of kernels for
image enlargement (as provided in my commercial software, SAR Image
Processor). It makes a visual difference when you assume the wrong
kernel, but it is not catastrophic as in the case pure deconvolution
(without decimation).
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