On Jul 25, 2:15=A0pm, illywhacker <illywac...@[EMAIL PROTECTED]
> wrote:
> On Jul 23, 11:12 pm, slus...@[EMAIL PROTECTED]
wrote:
>
>
>
>
>
> > On Jul 23, 12:01 pm, illywhacker <illywac...@[EMAIL PROTECTED]
> wrote:
>
> > > On Jul 17, 7:54 pm, slus...@[EMAIL PROTECTED]
wrote:
>
> > > > I originally posted the following on the sci.optics newsgroup
becau=
se
> > > > it's more of an optical query. =A0Nevertheless, the folks here may
=
be
> > > > more familiar with the available references and research that's
bee=
n
> > > > done. Please forgive my statement of some things that will no
doubt=
be
> > > > obvious to many here. =A0(BTW: I also have a description of the
> > > > experiments we've performed for anyone interested.)
>
> > > >
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D
> > > > In the image processing community, there are discussions of edge
> > > > detection techniques. =A0For a simple case, imagine an opaque
knife
> > > > edge
> > > > with a uniform back light behind it. Further imagine there is a
len=
s
> > > > that images the knife edge onto a pixel detector of a camera.
>
> > > > In general, the transition from dark to light at the detector is
so=
me
> > > > smoothly varying function, not a sharp jump. =A0Diffraction, of
cou=
rse,
> > > > limits the ultimate sharpness of the edge image -- diffraction at
t=
he
> > > > edge itself, and diffraction at the aperture of the lens.
> > > > Aberrations
> > > > of the lens will also contribute to this edge smoothing.
>
> > > > The digital image as presented *by* the camera may take only a
pixe=
l
> > > > or two to transition from dark to light, or it might take many
more=
..
> > > > Regardless, digitization and pixel size and other factors such as
M=
TF
> > > > of the electronics themselves serve to mask the true edge function
>
> > > > A very common starting point in the discussions and papers about
ed=
ge
> > > > detection techniques is the assumption that the point at which the
> > > > slope of the edge is maximum represents the "true" edge position.
> > > > From then on, the various edge detection algorithms usually
present
> > > > different methods of more accurately calculating this maximum
slope=
,
> > > > especially in the face of optical and electronic noise, etc.
>
> > > > Nevertheless, it seems to me that the assumption that the maximum
> > > > slope represents the true edge is at least unmotivated (no matter
h=
ow
> > > > "common sense" it feels) if not wrong. =A0I have in mind the
pictur=
es
> > > > of
> > > > edge diffraction as produced by using Cornu's spiral. =A0The
locati=
on
> > > > of
> > > > the true edge, relative to the average intensity of the light area
> > > > (smoothing out the diffraction oscillations), looks to be at a
> > > > position of increasing slope as you go from dark to light, but NOT
> > > > maximum. =A0Furthermore, the edge is less than the 50% point of
pea=
k
> > > > light intensity. =A0(Another assumption sometimes made in image
> > > > processing is that the 50% point of the dark to light transistion
> > > > represents the edge.)
>
> > > > Does anyone have references they can point to (or their own
pesuasi=
ve
> > > > arguments) that describe where the true edge location should be
> > > > relative to the edge image function? =A0We have done a couple of
be=
nch
> > > > tests to suggest under the experimental conditions that the best
ed=
ge
> > > > location is about 41% to 46% of the range of the dark to light
> > > > transition. =A0We haven't yet completed our analysis regarding how
=
this
> > > > compares to the peak slope.
>
> > > > This problem has to have been tackled successfully before, but so
f=
ar
> > > > I've not found any good sources that address the optical issue,
onl=
y
> > > > software techniques.
>
> > > > Thanks!
>
> > > > Spencer
>
> > > Actually for a straight knife edge, it is pretty easy to see that
for
> > > any translation-invariant linear operator that is
reflection-invarian=
t
> > > in the direction along the edge (which, as Aruzinsky points out,
> > > covers a lot of optical ground):
>
> > > 1) the slope is an extremum at the edge position (in the continuum);
> > > and
>
> > > 2) the intensity at the edge position is the average of the
intensity
> > > at the extremes (of course, this is not a local measurement).
>
> > > The basic equation is as follows. Let J be the observed image (in
the
> > > continuum); let the knife edge be at point x0; let I1 and I2 be the
> > > intensities for x much less than x0 and x much greater than x0; let
> > > the kernel of the linear operator be G(x, y) (remember it is
> > > translation invariant). Let
>
> > > H(x) =3D \int dy G(x, y)
>
> > > and
>
> > > L(x) =3D \int_{-\infty}^{x} dx' H(x') .
>
> > > Obviously L(-\infty) =3D 0. Suppose that L(\infty) =3D 1 (this is
jus=
t
> > > normalization). Then
>
> > > J(x) =3D I1 + (I2 - I1) L(x - x0) .
>
> > > Note that J does not depend on y, as I did not. Since L(x) =3D L(-x)
=
by
> > > hypothesis, we have that L(0) =3D 1/2 and that L''(0) =3D 0.
>
> > > illywhacker;- Hide quoted text -
>
> > > - Show quoted text -
>
> > Well, this may take some thinking. Although I've worked in machine
> > vision since 1985, my degree is in physics and my area of expertise is
> > optics and lighting, not image processing.
>
> What I just described is physics. My PhD was in physics. The simplest
> model of the diffractive effect of an edge (a knife edge) and the
> optical system is a linear operator. Any linear operator with the
> properties that I described (note that edge effects are not being
> taken into account) will generate the type of image I described.
>
> > probably true, but doesn't this still simply assert that the maximum
> > slope/gradient of the real physical edge image corresponds to the true
> > edge location?
>
> It is not an assertion; it is a proof, under certain assumptions.
> Anyway, the answer to your question is no, since the image of a knife
> edge may not be a monotonically changing image (there may be
> diffraction fringes, for example). But the edge will lie at an
> extremum. If the image is monotonic, then yes, the edge position is
> the point of maximum gradient under the above assumptions (this does
> not take into account the discretization involved in the image). It
> is also a/the point where the image intensity is the average of its
> values at +/- \infty.
>
> The optics you can find in any textbook: you do not need research
> articles. But the main point is that you are dealing not with an
> optics problem (or rather, the optics problem is relatively trivial),
> but with an inference problem. You want to infer the position of the
> edge from the image that you have. This is not simply a question of
> inverting an equation because information has been lost. So although
> the image may be determined given the edge position, the edge
> position might not be determined given the image without further
> knowledge, i.e. given only the image, there are multiple solutions,
> which is what others have been telling you. This applies in
> particular if the edge geometry is unknown, or if there is noise in
> the system, but in any case the image discretization pretty much
> ensures that high frequency detail (like precise positions) is lost
> and must be reconstructed.
>
> So your problem is one of image processing (which is simply the study
> of inference problems like these), except that you want to be more
> precise about the optics part. If you know all the relevant
> parameters, your life will be much easier, otherwise you will have to
> infer these too.
>
> illywhacker;- Hide quoted text -
>
> - Show quoted text -
You said: "If you know all the relevant
parameters, your life will be much easier, otherwise you will have to
infer these too."
Indeed! Thanks very much for your time and advice. The same to
everyone else who responded.
Spencer
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