adam@ste.dyn.bae.co.uk (Adam Curtin) (09/27/90)
Hi I'm considering buying a colour digitiser (video frame grabber) as a cheap substitute for a colour scanner. However, all of the images I've seen have black lines across them, something to do with the video signal and scanning and such, which spoil the image. My wife remembers from her physics undergraduate days that an image overlayed with a grating could be optically processed to remove the grating - light from the image was focused though a lens to come to a point at the fourier plane. Then, somehow, the image the other side was a magically cleaned up version of the original. (You can tell that I don't understand it very well :-). She can't describe it well enough for me to code it (even if I could understand it), but she tells me that the technique is often used for cleaning up images from satellites. Can anyone illuminate me? Adam -- A. D. Curtin . Tel : +44 438 753430 British Aerospace (Dynamics) Ltd. . Email: adam@ste.dyn.bae.co.uk PB 230, PO Box 19, Six Hills Way, . <This disclaimer conforms to RFC 1188> Stevenage, SG1 2DA, UK. . "My other car is an FJ1200"
tomg@hpcvlx.cv.hp.com (Thomas J. Gilg) (09/29/90)
> My wife remembers from her physics undergraduate days that an image overlayed > with a grating could be optically processed to remove the grating - light from > the image was focused though a lens to come to a point at the fourier plane. > Then, somehow, the image the other side was a magically cleaned up version of > the original. (You can tell that I don't understand it very well :-). There are many image processing/cleanup methods based on frequency domain filters. The usual computational sequence is: 1. Convert the Spatial Domain Image (ie, your digitized image) to the Frequency Domain. The Fast Fourier Transform (FFT) is popular here. 2. Analyze the Frequency Domain representation of your image, and then apply some selected mask to it. 3. Convert the Frequency Domain result back into the Spatial Domain. Magically, you'll have your cleaned image iff you did the right things in step #2. BTW - the inverse FFT is popular here. While the above is a standard sequence, and is usually more effective/faster than pure Spatial Domain methods, it does take time. Its my understanding that "optical" implementations of the above described computational sequence are possible. One lens setup deals with step #1, a few optical filters modify the image for step #2, and another lens setup converts the image back for step #3. While this allows for real time image processing, I'm not sure if its too handy for pre-digitized images. Thomas Gilg tomg@cv.hp.com Great reference: "Digital Image Processing" Rafael C. Gonzalez/Paul Wintz Addison-Wesley Publishing Company
ph@miro.Berkeley.EDU (Paul Heckbert) (09/29/90)
In article <1990Sep27.085647.13944@ste.dyn.bae.co.uk>, adam@ste.dyn.bae.co.uk (Adam Curtin) writes: >My wife remembers from her physics undergraduate days that an image overlayed >with a grating could be optically processed to remove the grating... This sounds like "deconvolution". If you've got an image a(x,y) that's been convolved with a linear, space-invariant filter h(x,y) to arrive at another image b(x,y) then you can un-filter using the following approach. Convolution in the spatial domain corresponds to multiplication in the frequency domain: SPATIAL DOMAIN FREQUENCY DOMAIN b(x,y) = a(x,y) * h(x,y) B(wx,wy) = A(wx,wy) . H(wx,wy) a(x,y) = b(x,y) deconvolved with h(x,y) A(wx,wy) = B(wx,wy) / H(wx,wy) where '*' denotes convolution, '.' denotes multiplication (wx,wy) are x and y frequencies F(wx,wy) is the Fourier transform of f(x,y) So if you have the degraded image b, and the point spread function (impulse response) of the filter h, you take their Fourier transforms to compute B and H, divide the first by the second frequency-by-frequency, and then compute the inverse Fourier transform to compute a. This works assuming that you know the degradation filter h, that it's linear and shift-invariant, and that H is always nonzero, and that the noise is zero or small. There are more sophisticated methods that don't assume knowledge of h (called blind deconvolution) and methods that minimize the amplification of noise during deconvolution, but I don't know them well enough to describe them. As a starter book on image processing, try: Anil K. Jain, Fundamentals of Digital Image Processing Prentice Hall, Englewood Cliffs, NJ, 1989. I don't know how the optical method you mentioned would work, except that lenses effectively compute a Fourier transform. I don't know how you would do the division optically, however. Maybe other netters can answer that. Paul Heckbert, Computer Science Dept. 570 Evans Hall, UC Berkeley INTERNET: ph@miro.berkeley.edu Berkeley, CA 94720 UUCP: ucbvax!miro.berkeley.edu!ph
turk@media-lab.media.mit.edu (Matthew Turk) (09/30/90)
> I don't know how the optical method you mentioned would work, except that > lenses effectively compute a Fourier transform. I don't know how you > would do the division optically, however. Maybe other netters can answer that. Yep, you can do multiplications in the Fourier domain easily with optics. And the division by H(wx,wy) is implemented as a multiplication by (1 / H(wx,wy)). But note that you can do this digitally as well. The impulse function of the filter whose frequency response is (1/H(wx,wy)) is itself a spatial filter h'(x,y), so once you find h'(x,y) you just convolute the image with it. In practice, though, this is kinda difficult, both because you probably don't know H(wx,wy) very accurately and because designing FIR filters (which is what we're talking about!) is not as straightforward as it sounds. Another good reference is Jae S. Lim's "Two-Dimensional Signal and Image Processing" (Prentice-Hall, 1990). But if I remember the original post properly, you're finding black lines which spoil your digitized images. I'd check your hardware, because these aren't normally part of the video signal! (Are you by any chance mixing 50 and 60 hz components?) Matthew Turk MIT Media Lab turk@media-lab.media.mit.edu 20 Ames St., E15-391 uunet!mit-amt!turk Cambridge, MA 02139 (617)253-0381
p560fgr@mpirbn.mpifr-bonn.mpg.de (Frank Grieger) (10/01/90)
In article <1990Sep27.085647.13944@ste.dyn.bae.co.uk> adam@ste.dyn.bae.co.uk (Adam Curtin) writes: >Hi > >I'm considering buying a colour digitiser (video frame grabber) as a cheap >substitute for a colour scanner. > >However, all of the images I've seen have black lines across them, something >to do with the video signal and scanning and such, which spoil the image. > >My wife remembers from her physics undergraduate days that an image overlayed >with a grating could be optically processed to remove the grating - light from >the image was focused though a lens to come to a point at the fourier plane. >Then, somehow, the image the other side was a magically cleaned up version of >the original. (You can tell that I don't understand it very well :-). >She can't describe it well enough for me to code it (even if I could understand >it), but she tells me that the technique is often used for cleaning up images >from satellites. > >Can anyone illuminate me? > >Adam >-- >A. D. Curtin . Tel : +44 438 753430 >British Aerospace (Dynamics) Ltd. . Email: adam@ste.dyn.bae.co.uk >PB 230, PO Box 19, Six Hills Way, . <This disclaimer conforms to RFC 1188> >Stevenage, SG1 2DA, UK. . "My other car is an FJ1200" Hallo Adam, the physical explanation of spatial filtering is very simple: Please consider an optical setup were a lense forms am image in its focalplane, which will be projected on a detektor by a second lense. We call this a telescope looking at an image at infinity. If there is no dirt on the lense or any other negative influence on the image formation, a paralell light beam will be focused to a lgiht spot in the focal plane. This spot is a diffraction pattern called an airy disc. If there are some dirt particles on the first lense, light will be scattered out of the light spot in the focal plane and will have negative influence on the image formation on the detector. If you now center a pinhole in the focal plane of the first lense which will cover up the light which was scattered by the dirt particles, the information about the dirt ist lost and the result is a clean image on the detector. This technique is for example used to produce clean laser light beams for holography. For more complicated setups and more general images you need more complicated masks in the focal plane of the lense. If you now would like to clean images in the computer, you have to know something about the mathematics of this phisical efect: In the experement we discribed above, we had an image in the infinity, a lense which projected the image into its focal plane where the image quality was improved by the positioning of a mask, and we had a secound lense which showed us the result on a detector. In terms of mathematics the lenses perform a Fourier-Transform on the original image. The Fourier-Transform is discribed by: H(f) = int { h(t) exp [ -i 2 PI f t ] } dt H(f) is the image in the Fourier domain, f is its coordinate vector, int denotes an integration, h(t) is the original image and t is the vector in the image plane. To clean images in the computer, you take the image, perform a Fourier Transform and multiply a binary mask to the result or better fit "bad" regions whith values in the surrounding. Finaly you perform an inverse Fourier-Transform to obtain the clean image. For your specific problem: The stripes in your original image, will be represented by peaks in the Fourier domain which you will have to remove. On the computer you should use the fast fourier transform algorithm, which might be in your math library or is discribed in the litterature. Litterature: ALSOP, L. E., and A. A. NOWROOZI, "Fast Fourier analysis," J. Geophys. Res. Vol. 71, PP. 5482-5483, November 15, 1966 GOODMAN, J. W., "Introduction to Fourier Optics," McGRAW-HILL BOOK COMPANY BRACEWELL, R., "The Fourier Transform and ist Applications," McGRAW-HILL BOOK COMPANY result. I hope this is the information you wanted. Frank