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Science Forum Index » Optics Forum » FTs and Point Spread Function (PSF)
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| W. Watson |
 Posted: Tue Mar 25, 2008 1:35 am |
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Guest
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My post on Fourier Transforms (FTs) and Optics took some interesting turns.
Let me try to pick up on the original intent. It appears a necessary part of
using FTs is to have a PSF for the physical optical system. I would guess
this means if one has a simple lens only, one needs a PSF for it. If it's a
mirror, then another type of PSF. If this is remotely correct, where does on
get them? I would think for an "ideal" lens one would be available. For an
actual lens, one would be derived in some fashion by experimentation. Is
there a PST for an "ideal" pinhole? Comments?
--
Wayne Watson (Nevada City, CA)
Web Page: <speckledwithStars.net> |
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| Marc Reinig |
| Posted: Tue Mar 25, 2008 9:59 am |
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Guest
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Actually, the PSF of a perfect simple lens is the FT of the aperture of the
lens. Normally, this aperture would be circular. However, some telescopes,
such as Keck I and II, build the 10m lenses out of hexagonal sections a few
feet across. It's really hard to cast and grind a solid 10m mirror ;=).
This makes the aperture non circular, because the hexagons protrude at the
edge and gives a slightly different PSF.
BTW, there is not a different type of PSF for a mirror or a lens, they are
both lenses and work on identical principles: changing the path of the light
and its time of flight. Mirrors have fewer chromatic effects than lenses/
Marco
________________________
Marc Reinig
UCO/Lick Observatory
Laboratory for Adaptive Optics
"W. Watson" <wolf_tracks@invalid.com> wrote in message
news:_11Gj.10431$qS5.2367@nlpi069.nbdc.sbc.com...
Quote: My post on Fourier Transforms (FTs) and Optics took some interesting
turns. Let me try to pick up on the original intent. It appears a
necessary part of using FTs is to have a PSF for the physical optical
system. I would guess this means if one has a simple lens only, one needs
a PSF for it. If it's a mirror, then another type of PSF. If this is
remotely correct, where does on get them? I would think for an "ideal"
lens one would be available. For an actual lens, one would be derived in
some fashion by experimentation. Is there a PST for an "ideal" pinhole?
Comments? |
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| W. Watson |
| Posted: Tue Mar 25, 2008 7:04 pm |
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Guest
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Thanks, but maybe I'm missing something about the PSF or blur function. It
seems as though somehow must account for some physical optical system. If
one is using a pinhole or some much different apparatus than a simple
bi-convex lens, wouldn't the resulting frequency domain solution look much
different that that provided by a FT of the aperture function. How does one
account for other lens or filters in the optical path in determining the
frequency domain?
Marc Reinig wrote:
Quote: Actually, the PSF of a perfect simple lens is the FT of the aperture of the
lens. Normally, this aperture would be circular. However, some telescopes,
such as Keck I and II, build the 10m lenses out of hexagonal sections a few
feet across. It's really hard to cast and grind a solid 10m mirror ;=).
This makes the aperture non circular, because the hexagons protrude at the
edge and gives a slightly different PSF.
BTW, there is not a different type of PSF for a mirror or a lens, they are
both lenses and work on identical principles: changing the path of the light
and its time of flight. Mirrors have fewer chromatic effects than lenses/
Marco
________________________
Marc Reinig
UCO/Lick Observatory
Laboratory for Adaptive Optics
"W. Watson" <wolf_tracks@invalid.com> wrote in message
news:_11Gj.10431$qS5.2367@nlpi069.nbdc.sbc.com...
My post on Fourier Transforms (FTs) and Optics took some interesting
turns. Let me try to pick up on the original intent. It appears a
necessary part of using FTs is to have a PSF for the physical optical
system. I would guess this means if one has a simple lens only, one needs
a PSF for it. If it's a mirror, then another type of PSF. If this is
remotely correct, where does on get them? I would think for an "ideal"
lens one would be available. For an actual lens, one would be derived in
some fashion by experimentation. Is there a PST for an "ideal" pinhole?
Comments?
--
Wayne Watson (Nevada City, CA)
Web Page: <speckledwithStars.net> |
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| W. Watson |
| Posted: Tue Mar 25, 2008 7:48 pm |
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Guest
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Let me correct below some things in this passage.
---------------
Thanks, but maybe I'm missing something about the PSF or blur function. It
seems as though somehow must account for some physical optical system. If
one is using a pinhole or some much different apparatus than a simple
bi-convex lens, wouldn't the resulting frequency domain solution look much
different that that provided by a FT of the aperture function. How does one
account for other lens or filters in the optical path in determining the
frequency domain?
---------------
Not aperture function but irradiance distribution. I'm looking at this in
section 11.3.1 Linear systems of Hecht:
dIi(Y,Z) - S(y,z; Y,Z)*Io(y,z)dydz.
Io is irradiance of the object, Ii of the image, and S is the PSF.
blur spot rather than blur function. The aperture function is premature to
this material.
I think I jumped ahead of the game there. Nevertheless, my conundrum is
still where does the physical side of the optical system work its way into FTs?
Marc Reinig wrote:
Quote: Actually, the PSF of a perfect simple lens is the FT of the aperture of the
lens. Normally, this aperture would be circular. However, some telescopes,
such as Keck I and II, build the 10m lenses out of hexagonal sections a few
feet across. It's really hard to cast and grind a solid 10m mirror ;=).
This makes the aperture non circular, because the hexagons protrude at the
edge and gives a slightly different PSF.
BTW, there is not a different type of PSF for a mirror or a lens, they are
both lenses and work on identical principles: changing the path of the light
and its time of flight. Mirrors have fewer chromatic effects than lenses/
Marco
________________________
Marc Reinig
UCO/Lick Observatory
Laboratory for Adaptive Optics
"W. Watson" <wolf_tracks@invalid.com> wrote in message
news:_11Gj.10431$qS5.2367@nlpi069.nbdc.sbc.com...
My post on Fourier Transforms (FTs) and Optics took some interesting
turns. Let me try to pick up on the original intent. It appears a
necessary part of using FTs is to have a PSF for the physical optical
system. I would guess this means if one has a simple lens only, one needs
a PSF for it. If it's a mirror, then another type of PSF. If this is
remotely correct, where does on get them? I would think for an "ideal"
lens one would be available. For an actual lens, one would be derived in
some fashion by experimentation. Is there a PST for an "ideal" pinhole?
Comments?
--
Wayne Watson (Nevada City, CA)
Web Page: <speckledwithStars.net> |
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| Salmon Egg |
| Posted: Wed Mar 26, 2008 1:22 am |
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In article <tpgGj.8430$Rq1.6482@nlpi068.nbdc.sbc.com>,
"W. Watson" <wolf_tracks@invalid.com> wrote:
Quote: Thanks, but maybe I'm missing something about the PSF or blur function. It
seems as though somehow must account for some physical optical system. If
one is using a pinhole or some much different apparatus than a simple
bi-convex lens, wouldn't the resulting frequency domain solution look much
different that that provided by a FT of the aperture function. How does one
account for other lens or filters in the optical path in determining the
frequency domain?
I have some difficulty understanding your post.
A real lens will have aberrations that will add to any blur arising from
focal errors. Moreover, the PSF is a function of angle. Thus, a
photograph will have different PDF applied to different parts of the
image.
bill |
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| W. Watson |
| Posted: Wed Mar 26, 2008 8:53 am |
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No doubt because I'm having trouble understanding the application of the FT
to optics and likely burdened by some misconceptions. See my post at 5:48 pm
yesterday in which I dig a small bit deeper into irradiance, blur and PSF
via Hecht. I see I misinterpreted irradiance (even its true
meaning--intensity, power) and its relationship to FTs.
I have no training in optics, but am acquainted with it through (way) past
courses in physics and an interest in image processing as applied to
astronomical images. It looks like the application of the FT varies in
optics depending on the intent.
In my "reading", basically trying to read between the equations, of Hecht,
2nd ed., I just sped ahead further ahead into his chapter of Fourier Optics.
It looks like my preconceived idea of how the FT is applied to optics is not
like the methods of Laplace and others I learned years ago in EE. He
eventually turns to a very interesting example (cut-off) on the modulation
(contrast) transfer function (MFT) that was of help.
Basically, I'm trying to tie the physical side of optics to the theoretical
in the sense of my EE experience long ago. They probably don't fit. Well,
I'm just going to go back to reading, the pluck out whatever understanding I
can out of the matter.
For me, this is a wrap on this thread.
Salmon Egg wrote:
Quote: In article <tpgGj.8430$Rq1.6482@nlpi068.nbdc.sbc.com>,
"W. Watson" <wolf_tracks@invalid.com> wrote:
Thanks, but maybe I'm missing something about the PSF or blur function. It
seems as though somehow must account for some physical optical system. If
one is using a pinhole or some much different apparatus than a simple
bi-convex lens, wouldn't the resulting frequency domain solution look much
different that that provided by a FT of the aperture function. How does one
account for other lens or filters in the optical path in determining the
frequency domain?
I have some difficulty understanding your post.
A real lens will have aberrations that will add to any blur arising from
focal errors. Moreover, the PSF is a function of angle. Thus, a
photograph will have different PDF applied to different parts of the
image.
bill
--
Wayne Watson (Nevada City, CA)
Web Page: <speckledwithStars.net> |
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| xiaowei |
| Posted: Fri Mar 28, 2008 4:23 am |
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Quote: No doubt because I'm having trouble understanding the application of the FT
to optics and likely burdened by some misconceptions. See my post at 5:48 pm
yesterday in which I dig a small bit deeper into irradiance, blur and PSF
via Hecht. I see I misinterpreted irradiance (even its true
meaning--intensity, power) and its relationship to FTs.
Itīs not simply irradiance, itīs the full complex "pupil function"
over the pupil that needs to be
fourier transformed (in k-space resp. aperture or sin(angle) space) to
get the PSF.
In most practical cases itīs not the irradiance that causes the
blurring (as in the simplest cases
it will only slightly change the shape of the PSF), but an wave front
aberrations that distort the PSF
and reduces the maximum acchievable energy concentration in the focus
(Strehl value).
However, in cases wher the so called sine-condition is not fulfilled
good enough - this will also show
up as a different irradiance distribution over the pupil, especially
the 3d shape of the PSF will
deviate, so high aperture parabolical mirror will have a different PSF
from a high aperture lens
system that (almost) fulfills the sine condition.
Those differences clearly show up even stronger if you are interested
in a vectorial description of the
light field near the focus which includes polarization.
Hope this helps
XW |
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