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
because
> > > it's more of an optical query. =A0Nevertheless, the folks here may
be
> > > more familiar with the available references and research that's been
> > > done. Please forgive my statement of some things that will no doubt
b=
e
> > > 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 lens
> > > that images the knife edge onto a pixel detector of a camera.
>
> > > In general, the transition from dark to light at the detector is
some
> > > smoothly varying function, not a sharp jump. =A0Diffraction, of
cours=
e,
> > > limits the ultimate sharpness of the edge image -- diffraction at
the
> > > 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 pixel
> > > or two to transition from dark to light, or it might take many more.
> > > Regardless, digitization and pixel size and other factors such as
MTF
> > > of the electronics themselves serve to mask the true edge function
>
> > > A very common starting point in the discussions and papers about
edge
> > > 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
how
> > > "common sense" it feels) if not wrong. =A0I have in mind the
pictures
> > > of
> > > edge diffraction as produced by using Cornu's spiral. =A0The
location
> > > 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 peak
> > > 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
pesuasive
> > > arguments) that describe where the true edge location should be
> > > relative to the edge image function? =A0We have done a couple of
benc=
h
> > > tests to suggest under the experimental conditions that the best
edge
> > > location is about 41% to 46% of the range of the dark to light
> > > transition. =A0We haven't yet completed our analysis regarding how
th=
is
> > > compares to the peak slope.
>
> > > This problem has to have been tackled successfully before, but so
far
> > > I've not found any good sources that address the optical issue, only
> > > 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-invariant
> > 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 just
> > 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!
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