Re: Accurate edge detection?

On Jul 24, 3:43=A0am, Martin Brown <|||newspam...@[EMAIL PROTECTED]
>
wrote:
> aruzinsky wrote:
> > On Jul 23, 3:16 am, Martin Brown <|||newspam...@[EMAIL PROTECTED]
>
> > wrote:
> >> aruzinsky wrote:
> >>> I was suggesting that, but I am not knowledgeable of every possible
> >>> optical effect. =A0Name one optical effect that can't be represented
=
by
> >>> a linear transformation.
> >> The most common ones that break modern deconvolution codes are a
point
> >> spread function that is not position invariant.
> >> The is the imaging
> >> system has off axis abberations (most real diffraction limited optics
> >> suffer some degradation as you go off axis).
>
> > That is still a linear transformation, as I previously mentioned for
>
> You are using "linear" in a way that is non-standard in the image
> processing community. The conventional definition is that in a linear
> transform the measured image is the strict convolution of a position
> invarinant point spread function and an unknown target image.
>
> Edge effects on modern optical scopes produce some level of coma so that
> the stars images are not circularly symmetric off axis.
>
> =A0 =A0 =A0 =A0 =A0v
>
> =A0> =A0 . =A0 =A0 =A0 <
>
> =A0 =A0 =A0 =A0 ^
>
> Typically they develop gull wings as you go significantly off axis.
>
> > the case of the bad Hubble telescope lens. =A0And, that problem was
> > tractable. =A0In fact, many common deconvolution methods work directly
> > in the spatial domain (no FFTs allowed) can be modified for this
> > problem.
>
> The most powerful ones almost invariably work in the Fourier domain. You
> need a fairly compact psf to work in the spatial domain and still get
> acceptable performance. Diffraction limited cicular telescope optics
> ring from unresolved stars with a characteristic Airey rings pattern.
> Off axis it becomes more messy.
>
> Hubble researchers found the residual spatial variation of the psf a
> real problem in high dynamic range images.
>
> http://adsabs.harvard.edu/abs/1994SPIE.2198.1349H
>
> There is a solution but it involves a lot more computation
>
>
>
> >> And the real killer in the general photography situation is depth of
> >> field. You can model the psf for a given subject distance, but when
> >> different parts of the same image are all at different unknown
distanc=
es
> >> you have a pretty much intractable problem. Or one that is only
> >> resolvable with binocular pairs of images and insane amounts of
> >> computing power.
>
> > It might be killer intractable, but I think it is linear.
>
> Again you are using linear in a nonstandard way. It is linear in the
> sense that each pixel is the result of summation of the scene blurred
> with an appropriate psf for the object distance in every direction.
>
> But that is not what is normally meant when linear is used in the
> context of image processing. The meaning is usually taken to be:
>
> D =3D P * T
>
> Where D is measured data, * is convolution,
> P is a spatially invariant psf and T is the target image.
>
>
>
>
>
>
>
> >> If the OP wants specifically to find sharp edges with the utmost
> >> precision and doesn't care about the computational cost then one of
th=
e
> >> Bayesian methods using all available prior knowledge will probably
giv=
e
> >> the optimum result. This reference is a bit old but is a start:
>
> >>http://adsabs.harvard.edu/abs/1980ITAC...25...36M
>
> >> There seem to be few papers on this application which suggests to me
> >> that it doesn't work well enough to be worth the computational cost.
>
> >> Regards,
> >> Martin Brown
> >> ** Posted fromhttp://www.teranews.com**-Hide
quoted text -
>
> >> - Show quoted text -
>
> > Don't forget that "optimal result" is with respect to some statistical
> > cost function and some individual cases of estimated edge location can
> > be grossly inaccurate.
>
> The question that you ask of the data determines the answer that you
> get. If you ask the wrong question then all bets are off.
>
> Regards,
> Martin Brown
> ** Posted fromhttp://www.teranews.com**-
Hide quoted text -
>
> - Show quoted text -

No.  The definition used practically everywhere is:

f() is linear iff

 f(a + b) =3D f(a) + f(b)

and

f(=E1x) =3D =E1f(x)


Convolution with a spatially variant psf is linear because it
satisfies those properties.  More specifically, matrix multiplication
is linear

A(a1*x1 + a2*x2)  =3D a1*Ax1 + a2*Ax2

and a discrete spatially variant PSF can always be represented by
matrix A.

Here is a simple 1D example of a constant PSF:

A =3D
[
1/2 1/2 0 ... 0 0
0 1/2 1/2 ... 0 0
=2E..
0 0 0 ... 1/2 1/2
]

Here is a simple 1D example of a spatially variant PSF:

A =3D
[
1/2 1/2 0  0
0 1/3 1/3 1/3
]