On Jul 23, 9:02=A0pm, "Science.Medical.Imaging List"
<pixel.to.l...@[EMAIL PROTECTED]
> wrote:
> On Jul 23, 2:12=A0pm, slus...@[EMAIL PROTECTED]
wrote:
>
>
>
>
>
> > On Jul 23, 12:01=A0pm, illywhacker <illywac...@[EMAIL PROTECTED]
> wrote:
>
> > > On Jul 17, 7:54=A0pm, 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. =A0No doubt what you say is
> > 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? =A0See my recent reply to Aruzinsky.
>
> > On the other hand, maybe I'm just being dense about the whole thing.
>
> > Regardless, thanks.
>
> > Spencer- Hide quoted text -
>
> > - Show quoted text -
>
> Spencer,
>
> I commend your repeated efforts to explain the doubts you have. This
> is a part of learning process, and I salute you!
> Please dont take offense by any objectionable posts. The purpose of
> this forum is to share and learn, not to make someone feel bad if they
> didnt know something and asked around.
> That is not the way to encourage learning. If everybody just blindly
> followed someone's inflated advice, there would be no innovation. Your
> honesty about the problem is commendable.
>
> Keep it up!- Hide quoted text -
>
> - Show quoted text -
Thanks very much! I appreciate it. I'm aware of the range of
personalities to be found on the web and use a reasonable set of
mental filters. Life is too short to get in a huff very often.
Your own attitude resonates with mine, and I find for the most part
that the golden rule works well. Thanks again.
Spencer
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