On 1 Des, 03:16, Jerry Avins <j...@ieee.org> wrote:> On 11/30/2011 10:22 AM, Rune Allnor wrote: > > > > > > > On 30 Nov, 16:13, Vladimir Vassilevsky<nos...@nowhere.com> �wrote: > >> Rune Allnor wrote: > >>> On 30 Nov, 14:35, Ling Chen<erie.stev...@gmail.com> �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? > >>> This could also be viewed as 'deconvolution'. > > >>> Given the two images x and y, the blurring operation is expressed > >>> as y = h (*) x where h is the system response to be found, and (*) > >>> denotes convolution. > > >>> Extracting this system function h is 'system identification' > >>> while extracting an unknown x from is 'deconvolution'. > > >> No. > > >> An image is an intensity picture. That is, abs(x) not x. > > > Wrong. abs(x) would indicate that there is a -x. > > How do you define negative intensity? > > Consider the shadow of a straight edge in the light of a point source. > If negative intensities existed, its intensity would be as sin(x)/x.They don't.> 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. Absolute values only make sense if the quantity they are applied to can take on negative values. Rune
image out-of-focus blur identification
Started by ●November 30, 2011
Reply by ●November 30, 20112011-11-30
Reply by ●December 1, 20112011-12-01
On 11/30/11 9:56 PM, Rune Allnor wrote:> On 1 Des, 03:16, Jerry Avins<j...@ieee.org> wrote: >> On 11/30/2011 10:22 AM, Rune Allnor wrote: >>> On 30 Nov, 16:13, Vladimir Vassilevsky<nos...@nowhere.com> wrote:...>>>> An image is an intensity picture. That is, abs(x) not x. >> >>> Wrong. abs(x) would indicate that there is a -x. >>> How do you define negative intensity? >> >> Consider the shadow of a straight edge in the light of a point source. >> If negative intensities existed, its intensity would be as sin(x)/x. > > They don't. > >> 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. > > Absolute values only make sense if the quantity they > are applied to can take on negative values.and why does it matter? okay, so here's my question, is can the example posed by Jerry be compared to what is happening with what a lens does and what happens when an image is captured out of focus. my naive understanding of what happens in the 2-dimensional space domain is that the intensity (or whatever quantity that the film emulsion or photodetector pixel measures) at a specific x,y point has a portion spread to neighboring pixels in a deterministic manner. it is convolved with a 2-dim impulse response that depends on {delta_x^2 + delta_y^2} like a gaussian or something. if the image is in focus, that impulse response gets pretty close to an impulse. now, if my understanding is not too fucked up, why cannot that impulse response be determined if you know the input and output to this 2-dim L"T"I system? -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●December 1, 20112011-12-01
>On 11/30/2011 12:01 PM, Vladimir Vassilevsky wrote: > > ... > >> Doctor Rune = idiot. > >Bah! > >> The blurring happens in the wave domain. >> If the intensity images are all that available, then the problem is >> intractable. > >Not intractible. Point-spread functions are computes routinely. > >JerryHave you ever seen a seriously blurred image (other than a hologram) successfully deblurred to something close to what a sharply focused optical system and non-moving objects would have produced. It wasn't the image that was blurred. It was the original light field. All the phase information was lost when that shined onto a sensor only capable of responding to amplitude. Image deblurring can do really useful things. It can reveal key information, like a runaway car's numberplate. It can sharpen up a slightly blurred image, and has done so very effectively for things like astronomy. What it doesn't do today, and probably never can, is take an arbitrarily blurred image and recreate the image that a properly focused optical system would have produced. The more blurred things get, the more the deblurring only approximates the right solution, and the worse the result looks. Steve
Reply by ●December 1, 20112011-12-01
On 1 Des, 05:13, robert bristow-johnson <r...@audioimagination.com> wrote:> On 11/30/11 9:56 PM, Rune Allnor wrote: > > > > > > > On 1 Des, 03:16, Jerry Avins<j...@ieee.org> �wrote: > >> On 11/30/2011 10:22 AM, Rune Allnor wrote: > >>> On 30 Nov, 16:13, Vladimir Vassilevsky<nos...@nowhere.com> � �wrote: > ... > >>>> An image is an intensity picture. That is, abs(x) not x. > > >>> Wrong. abs(x) would indicate that there is a -x. > >>> How do you define negative intensity? > > >> Consider the shadow of a straight edge in the light of a point source. > >> If negative intensities existed, its intensity would be as sin(x)/x. > > > They don't. > > >> 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. > > > Absolute values only make sense if the quantity they > > are applied to can take on negative values. > > and why does it matter?It matters in the same way it matters if you look for irrational solutions but are limited to use integer arithmetic, roots of real-valued polynomials become complex-valued, and so on. There are restrictions to the problem statement that limit what kinds of solutions are either feasible or possible to find, and what properties of the solution are permissible.> why > cannot that impulse response be determined if you know the input and > output to this 2-dim L"T"I system?A PSF can be *estimated*, if one uses the correct tools and algorithms etc. The issue is what properties of that PSF are permissible. Rune
Reply by ●December 1, 20112011-12-01
On Nov 30, 9:56�pm, Rune Allnor <all...@tele.ntnu.no> wrote:> On 1 Des, 03:16, Jerry Avins <j...@ieee.org> wrote: > > > > > > > On 11/30/2011 10:22 AM, Rune Allnor wrote: > > > > On 30 Nov, 16:13, Vladimir Vassilevsky<nos...@nowhere.com> �wrote: > > >> Rune Allnor wrote: > > >>> On 30 Nov, 14:35, Ling Chen<erie.stev...@gmail.com> �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? > > >>> This could also be viewed as 'deconvolution'. > > > >>> Given the two images x and y, the blurring operation is expressed > > >>> as y = h (*) x where h is the system response to be found, and (*) > > >>> denotes convolution. > > > >>> Extracting this system function h is 'system identification' > > >>> while extracting an unknown x from is 'deconvolution'. > > > >> No. > > > >> An image is an intensity picture. That is, abs(x) not x. > > > > Wrong. abs(x) would indicate that there is a -x. > > > How do you define negative intensity? > > > Consider the shadow of a straight edge in the light of a point source. > > If negative intensities existed, its intensity would be as sin(x)/x. > > They don't. > > > 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. > > Absolute values only make sense if the quantity they > are applied to can take on negative values. > > Rune- Hide quoted text - > > - Show quoted text -Actually, sensors, films ect, record the amplitude squared - i,e., the intensity which is the energy. For example a simple (plane) hologram made of uniform parallel beam of light (the reference beam is also uniform and collimated) ends up recording sin(alpha*x)^2 which is nonnegative. The alpha is determined by the wavelength and the angle of incidence between the object and reference beams. See Bragg's law for more details. Even though Bragg's law is normally used for x-ray diffaction from crystall lattices, it also applies to holograms. 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. 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. Clay
Reply by ●December 1, 20112011-12-01
On 12/1/2011 6:31 AM, steveu wrote:>> On 11/30/2011 12:01 PM, Vladimir Vassilevsky wrote: >> >> ... >> >>> Doctor Rune = idiot. >> >> Bah! >> >>> The blurring happens in the wave domain. >>> If the intensity images are all that available, then the problem is >>> intractable. >> >> Not intractible. Point-spread functions are computed routinely. >> >> Jerry > > Have you ever seen a seriously blurred image (other than a hologram) > successfully deblurred to something close to what a sharply focused optical > system and non-moving objects would have produced. > > It wasn't the image that was blurred. It was the original light field. All > the phase information was lost when that shined onto a sensor only capable > of responding to amplitude. > > Image deblurring can do really useful things. It can reveal key > information, like a runaway car's numberplate. It can sharpen up a slightly > blurred image, and has done so very effectively for things like astronomy. > What it doesn't do today, and probably never can, is take an arbitrarily > blurred image and recreate the image that a properly focused optical system > would have produced. The more blurred things get, the more the deblurring > only approximates the right solution, and the worse the result looks.A broken picture has much in common with a broken implement. Restoration is usually hard and frequently impossible, but improvement to the point of usability can often be done. People often find "I can't fix it like new, but I can make it better" comforting to hear. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 1, 20112011-12-01
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 absolute value of any real number, negative or positive, is a positive number. I am not prevented from taking the absolute value of a positive number (inherently positive or only incidentally positive).> Absolute values only make sense if the quantity they > are applied to can take on negative values.They may only be useful for such numbers, but what makes sense depends on who does the sensing. 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. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●December 1, 20112011-12-01
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? 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. Rune
Reply by ●December 1, 20112011-12-01
steveu <steveu@n_o_s_p_a_m.coppice.org> wrote: (snip)> Have you ever seen a seriously blurred image (other than a hologram) > successfully deblurred to something close to what a sharply focused optical > system and non-moving objects would have produced.> It wasn't the image that was blurred. It was the original light field. All > the phase information was lost when that shined onto a sensor only capable > of responding to amplitude.Remember the first pictures from the Hubble telescope? In that case, they knew the PSF very accurately, and, for images with enough S/N deconvolution worked very well. 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. -- glen
Reply by ●December 1, 20112011-12-01
On 1 Des, 20:46, glen herrmannsfeldt <g...@ugcs.caltech.edu> 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=8-2 Rune
