Re: PSF function.. and convolution help needed

On Apr 15, 12:18 am, "Marc Reinig" <Ma...@xxxxxxxxxxxxxxxxx> wrote:
<odprt...@xxxxxxxxx> wrote in message

news:06c55b30-32f7-4b13-b928-02a63ccda6d1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Hi all,

I am having a hard time understanding the meaning, of the PSF and the
convolution term in optic  .
I would be very glad if anyone can help.

- what is the meaning of PSF in optical system
- does the PSF is a function that I can create  s only when I try to
create an image with  a lens, and a point source?
(is there a PSF for other source?)

The PSF (Point Spread Function) of an optical system is what the system does
to a point source object in the process of imaging it.  For a round pupil
with no aberrations in the system it will be a sync squared function  ( (
sin(x)^2 ) / ( x^2 ) whose extent is a function of the diameter of the pupil
and the focal length of the system.  This function has a central bright core
and a series of dark rings and decreasing intensity bright rings around it..
Any aberrations in the system will make the image of a point source
different and you will have a less than ideal PSF, i.e., the image of a
point in a non ideal system will be spread out compared to an ideal system..

If you are trying to image two point sources close together, each will
produce its own PSF in the image plane.  As the points sources get closer
and closer the PSFs will get closer and closer.  At some point, you will not
be able to tell if you have two points or an extended source that is larger
than a point.  Generally, when   the peak of on PSF is at the same position
as the first null of another PSF, that is considered the point beyond which
you cannot tell you have two points.

Generally, four your purpose, since it is a "point" spread function, it is
only used to describe the effect on point sources.

and how you calculate it?

See below for calculations.  However, for complex instruments, you would
usually measure it rather than calculate it.

what is the meaning of convolution in Optic ??

Light passing through a focusing lens also is passing through a pupil,  area
at the lens through which light is allowed to pass.  Think of the pupil as a
function that is 1 inside and 0 everywhere outside.  This assumes the pupil
has no aberrations in either intensity or phase, i.e., light passes through
uniformly in intensity inside the aperture and is slowed down uniformly
inside the aperture.

Now think of the light coming to the aperture.  The light on one side is
L0(x,y).  The aperture function is A(x,y).  After passing through the
aperture the light is L1 (x,y) = L0(x,y) A(x,y).  i.e., at each point x,y
the light after the aperture will either be L0(x,y) because it is multiplied
by 1; or 0 because it was multiplied by 0.

The light at the image plane turns out to be the Fourier transform of the
light at the pupil plane.  An important relationship in math is that the
Fourier transform of a product is the Fourier transform of the convolution
of the elements and vice versa.  So,

    F( X Y ) = F( X * Y )  and F( X * Y ) = F( X Y )  where * indicates the
convolution.

So, the amplitude of the light at the image plane is the Fourier transform
of the light at the pupil plane:

    Image(x,y) = F( L1(x,y) ) = F( L0(x,y) A(x,y) ) = F( L0(x,y) ) * F(
A(x,y) )

It is the convolution of the Fourier transform of the light at the pupil
plane with the Fourier transform of the Pupil function.

If the source light is a point source, the PSF would be the intensity in the
image plane.  Intensity if the square of the light amplitude, we perceive
intensity, not whether the E field of the light is + or -.

For an example of how this can be used, let's imagine a camera with a non
perfect lens and/or pupil aperture.  The image taken will be less than
ideal, but it will be the convolution of the Fourier transform of the bad
pupil/lens with the Fourier transform of the light at the pupil.  If you
could take the camera in isolation you could characterize it with a point
source to measure the bad PSF and develop a mathematical expression for it..

You know the image is the convolution of the bad PSF with the Fourier
transform of the light at the pupil.  You could then take the image and
deconvolve it (using the expression of the bad PSF) to get an expression for
the light at the pupil when the image was taken.  Then you could take that
expression and process it with the expression for a perfect PSF and get the
image as it would have been recorded with a perfect camera.

The theory is better than the practice and it is very computationally
intense.  However, it has been used to sharpen the original Hubble images
taken with the original improperly designed lens, it is used to enhance
images by law enforcement to correct for all manner of defects in
surveillance video, etc.  It is however a slow process, does not produce
enhancement back to having a perfect lens in the original camera, and not
the way you see it in CSI, NCIS, .... ;=)

Marco
________________________
Marc Reinig
UCO/Lick Observatory
Laboratory for Adaptive Optics

Thanks for a super informative answer!
(I am learning it :) )
.

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