Rune Allnor <allnor@tele.ntnu.no> wrote: (snip, I wrote)>> There might be newer books by now, but the Jansson does describe >> non-linear convolution pretty well. Unlike linear deconvolution, >> it includes the constraint that the signal not go negative. >> Also, for absorption spectra, that the signal not go over one.> Did a search on amazon.com. Seems it is a new edition > to be expected in a few months:> http://www.amazon.com/Deconvolution-Images-Spectra-Second-Engineering/dp/0486453251/ref=sr_1_2?ie=UTF8&qid=1322768916&sr=8-2This seems to be the Dover reprint of the second edition. Some years ago (1996), I remembered having seen the book and wanted a copy. I sent e-mail to Jansson, and he said to wait until the new (2nd) edition came out. For some years now, used copies have been available for prices like Amazon now shows, $350. It is nice that the Dover edition will be available. It seems to be the same, second edition. (The first one was just Deconvolution of Spectra, not images. He considers the problem of slit size in IR spectra, where larger slits get larger signal that, with deconvolution more than makes up for the wider slit. As well as I remember, that isn't so much of a problem for visible (optical) spectra.) Otherwise, find a good physics or engineering library. -- glen
image out-of-focus blur identification
Started by ●November 30, 2011
Reply by ●December 1, 20112011-12-01
Reply by ●December 1, 20112011-12-01
Clay <clay@claysturner.com> wrote: (snip on amplitude, intensity, and holography)> When the hologram's image is reconstructed by illuminating it with a > facsimile of the original reference beam (really needs to be the phase > conjugate), the hologram then produces two wavefronts of the original > image where the 2nd is the phase conjugate.There is a sub-field of non-linear optics called phase-conjugate optics. Among others, one can build a device called a phase conjugate mirror which, when given an optical signal will generate the phase conjugate signal. The first use for such that I heard about was to correct for dispersion on long optical fiber runs. Near the middle of the fiber run, one puts a phase-conjugate device. Any dispersion that has accumulated up to that point is exactly reversed, such that continuing through the fiber will restore the original signal. I believe that there is a wikipedia article about them, and there have also been Scientific American articles.> If the hologram is > recorded in a medium that is thick relative to the fringe spacing > (spacing between the Bragg planes), then you have a volume grating and > it will only produce one of the two images at a time. If you desire > the other image than the one you are seeing, just reverse the > reference light's direction through the hologram.It will also mostly ignore light of the wrong wavelength, generating what is called a white-light hologram. Last I remember (when I took an optics lab class), you make the hologram with red (HeNe) light, but, due to shrinkage of the emulsion on processing, the image is green. -- glen
Reply by ●December 1, 20112011-12-01
On 1 Des, 21:42, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:> Rune Allnor <all...@tele.ntnu.no> wrote: > > (snip, I wrote) > > >> There might be newer books by now, but the Jansson does describe > >> non-linear convolution pretty well. Unlike linear deconvolution, > >> it includes the constraint that the signal not go negative. > >> Also, for absorption spectra, that the signal not go over one. > > Did a search on amazon.com. Seems it is a new edition > > to be expected in a few months: > >http://www.amazon.com/Deconvolution-Images-Spectra-Second-Engineering... > > This seems to be the Dover reprint of the second edition.I don't mind. That a book makes it to Dover is usually a good sign.> Otherwise, find a good physics or engineering library.No such thing this side of the pond! I eventually got my own library (in the sense 'room for books') where I could put my library (in the sense 'collection of books') and I was a bit scared: 5 sections of bookshelves, each containing six shelves of ca 80 cm width, 70% of which is filled with science & engineering books - that's more than 15 shelf-meters of books. Mad scientist... Rune
Reply by ●December 1, 20112011-12-01
Rune Allnor <allnor@tele.ntnu.no> wrote: (snip, someone wrote)>> >http://www.amazon.com/Deconvolution-Images-Spectra-Second-Engineering...>> This seems to be the Dover reprint of the second edition.> I don't mind. That a book makes it to Dover is usually > a good sign.Yes, I agree. I have many Dover books, and am glad that I have them. They are reasonably priced such that I can buy them even if I am not sure I need it yet. Also, I will feel better about suggesting it to people if it doesn't cost $350 or $688.88. (The original price seems to have been $145, according to Amazon.) -- glen
Reply by ●December 1, 20112011-12-01
On Dec 1, 3:57�pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:> Clay <c...@claysturner.com> wrote: > > (snip on amplitude, intensity, and holography) > > > When the hologram's image is reconstructed by illuminating it with a > > facsimile of the original reference beam (really needs to be the phase > > conjugate), the hologram then produces two wavefronts of the original > > image where the 2nd is the phase conjugate. > > There is a sub-field of non-linear optics called phase-conjugate > optics. �Among others, one can build a device called a phase > conjugate mirror which, when given an optical signal will generate > the phase conjugate signal. �The first use for such that I heard > about was to correct for dispersion on long optical fiber runs. > > Near the middle of the fiber run, one puts a phase-conjugate device. > Any dispersion that has accumulated up to that point is exactly > reversed, such that continuing through the fiber will restore > the original signal. �I believe that there is a wikipedia article > about them, and there have also been Scientific American articles. > > > If the hologram is > > recorded in a medium that is thick relative to the fringe spacing > > (spacing between the Bragg planes), then you have a volume grating and > > it will only produce one of the two images at a time. If you desire > > the other image than the one you are seeing, just reverse the > > reference light's direction through the hologram. > > It will also mostly ignore light of the wrong wavelength, > generating what is called a white-light hologram. > > Last I remember (when I took an optics lab class), you make the > hologram with red (HeNe) light, but, due to shrinkage of the > emulsion on processing, the image is green. > > -- glenThe white light hologram (the volume type and not the rainbow type) require the bragg object and reference beams to approach the film from opposite sides to create the standing wave pattern in the film. Thus you get a "reflection" hologram that is quite wavelength selective. If the volume hologram is made with beams both approaching from the same side, you don't get a frequency selective enough reflection to make the image very viewable in white light. However you can make an image plane hologram by making a master hologram of an object located say 6 inches from the plate. Then illuminate this hologram (master hologram) with a reversed reference beam and this hologram then projects a real image into space about 6 inches in front of the plate. You now place another plate into the center of the projected real image and also provide a collimated reference beam for this 2nd plate. This 2nd plate will become a white light hologram with a low coherence requirement since the image is centered about the plate so half of the image lies in front and the other half lies behind the plate. If the depth of field of the object is an inch or less, this makes for a reasonable white light hologram. The rainbow type is made by making the copy where the master hologram is only illuminated along a horizontal slit, thereby giving up one axis (vertical) of info keeping the left right info, These types may be recorded as plane gratings and hence foil holograms (like on a credit card) are rainbow types. Turn a rainbow hologram 90 degrees and you will notice the 3-D effect goes away. Clay
Reply by ●December 1, 20112011-12-01
On 12/1/2011 12:24 PM, Rune Allnor wrote:> On 1 Des, 18:07, Jerry Avins<j...@ieee.org> wrote: >> On 11/30/2011 9:56 PM, Rune Allnor wrote: >> >>> On 1 Des, 03:16, Jerry Avins<j...@ieee.org> wrote: >> >> ... >> >>>> Since there is no negative intensity, we actually get |sin(x)/x|. That >>>> makes the problem harder. >> >>> I still don't understand what you mean. I can't see >>> how a strictly positive quantity can be expressed >>> as and absolute value. > ... >> The intensity of an Airy disc is the absolute value of the "Mexican Hat" >> function. Intensities are always non-negative, but the functions we >> compute them from (or give rise to them in nature) may not be. > > But the images in question here are not computed, they > are measured. If the measurements are strictly positive, > why then the need for the absolute value?It may be superfluous, but that's not the same as it's being impermissible.> I like Clay's approach where image intensity has to do > with energy. In that case it's F(x,y)^2, not |F(x,y)|, > that determines the intensity image in the (x,y) plane.Yes. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 1, 20112011-12-01
Clay <clay@claysturner.com> wrote: (snip, I wrote)>> It will also mostly ignore light of the wrong wavelength, >> generating what is called a white-light hologram.>> Last I remember (when I took an optics lab class), you make the >> hologram with red (HeNe) light, but, due to shrinkage of the >> emulsion on processing, the image is green.> The white light hologram (the volume type and not the rainbow type) > require the bragg object and reference beams to approach the film from > opposite sides to create the standing wave pattern in the film. Thus > you get a "reflection" hologram that is quite wavelength selective. If > the volume hologram is made with beams both approaching from the same > side, you don't get a frequency selective enough reflection to make > the image very viewable in white light.I remember making them that way, but I forgot the math by now. You send the beam down through the plate, where it reflects off the object. We did them on 2in square plates. Some time ago, I saw some big (8in by 10in I believe) white light reflection holograms of microscopes. You could then look into the microscope and see an object through the microscope. (snip of more useful information about white-light holograms.) -- glen
Reply by ●December 1, 20112011-12-01
On Nov 30, 5:35�am, Ling Chen <erie.stev...@gmail.com> wrote:> Hi there,Hi Ling If you are interested in a real answer, why don't you figure out which of the types of blur Clay has listed is the type you are asking about (or describe another)? Then you might get a real response. Dale B. Dalrymple
Reply by ●December 2, 20112011-12-02
On Dec 1, 5:18�pm, Jerry Avins <j...@ieee.org> wrote:> On 12/1/2011 12:24 PM, Rune Allnor wrote: > > > > > > > On 1 Des, 18:07, Jerry Avins<j...@ieee.org> �wrote: > >> On 11/30/2011 9:56 PM, Rune Allnor wrote: > > >>> On 1 Des, 03:16, Jerry Avins<j...@ieee.org> � �wrote: > > >> � � ... > > >>>> Since there is no negative intensity, we actually get |sin(x)/x|. That > >>>> makes the problem harder. > > >>> I still don't understand what you mean. I can't see > >>> how a strictly positive quantity can be expressed > >>> as and absolute value. > > ... > >> The intensity of an Airy disc is the absolute value of the "Mexican Hat" > >> function. Intensities are always non-negative, but the functions we > >> compute them from (or give rise to them in nature) may not be. > > > But the images in question here are not computed, they > > are measured. If the measurements are strictly positive, > > why then the need for the absolute value? > > It may be superfluous, but that's not the same as it's being impermissible. > > > I like Clay's approach where image intensity has to do > > with energy. In that case it's F(x,y)^2, not |F(x,y)|, > > that determines the intensity image in the (x,y) plane. > > Yes. > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������- Hide quoted text - > > - Show quoted text -Perhaps the impermissible comes from the requirement that |x| for sinusoids has a discontinuous 1st derivative whereas |x|^2 does not. Clay
Reply by ●December 2, 20112011-12-02
Ling Chen wrote:> > Hi there, > > Suppose an image is out-of-focus blurred. We have the original image > and its blurred version. What is the best approach to estimate the > blur function? That is, given y(m), x(m), how to we estimate the > point spread function (or impulse response function) h(m): > y(m) = h(m)*x(m) + n(m) > > where n(m) is some kind of background noise.If this is a strictly digital problem then you should be able to do this without n(m) term. Since you say you have the original and the blurred version, this sound like a digital problem with no involvement of any analog operations. If the image is blurred with a digital filter of the typical sort used to blur images (e.g. gaussian blur) then the frequency response of the blurring function can be decomposed into entirely positive cosines and the inverse function that will unblur the image can be precisely defined. The point is that type of blurring filter preserves all frequency and phase information that the digital image can support. Higher frequencies are attenuated but not completely eliminated. The only limitation to accuracy is the precision of the media used to store the inverse coefficients and the the blurred image. If you put the blurred version of the image back into 8 bit data containers that is going to seriously limit recovery of the original. If you store the blurred image and and inverse filter as double precision floats you should be able to recover the original so that the eye can't tell the difference. Another way to recover the original from the blurred image is to use an unsharp filter. If the original blurring filter is H[x,y] then adding an image that was created by applying 1-H[x,y] to the original will produce the original. Again all of this only applies if you are using digital processes with enough precision, but it can work to some extent for images where the blurring occurred outside the digital domain. However, if the image has not been digitally blurred, I can't see how you would have the unblurred version. -jim> > We are expecting to use some kind of minimized mean square error to > get a parametric and non-parametric estimation of the h(m). > > Please note, unlike general setup of image deblurring, where the > original image is not available. Our case is that both the original > image and its blurred version is available. We feel this problem is > somehow like a system identification problem, e.g., given the input > and output and unknown noise, identify the impulse response. > > Any advises, suggestions, pointers to right references would be highly > appreciated. > > Thanks.
