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Additional info for Digital image processing. Solutions Manual
17 (a) Express ®ltering as convolution to reduce all processes to the spatial domain. Then, the ®ltered image is given by g(x; y) = h(x; y) ¤ f(x; y) where h is the spatial ®lter (inverse Fourier transform of the frequency-domain ®lter) and f is the input image. Histogram processing this result yields g0(x; y) = T [g(x; y)] = T [h(x; y) ¤ f (x; y)] ; where T denotes the histogram equalization transformation. If we histogram-equalize ®rst, then g(x; y) = T [f (x; y)] and g0(x; y) = h(x; y) ¤ T [f (x; y)] : In general, T is a nonlinear function determined by the nature of the pixels in the image from which it is computed.
The inverse is done in the same way. 3 The inverse DFT of a constant A in the frequency domain is an impulse of strength A in the spatial domain. Convolving the impulse with the image copies (multiplies) the value of the impulse at each pixel location in the image. 4 An important aspect of this problem is to recognize that the quantity (u2 + v2 ) can be replaced by the distance squared, D2 (u; v). This reduces the problem to one variable, which is notationally easier to manage. Rather than carry an award capital letter throughout the development, we de®ne w2 , D2 (u; v) = (u2 + v 2 ).
Obviously there are a number if variations of the basic theme just described. For example, additional intelligence in the form of tests that are more sophisticated than pixel-bypixel threshold comparisons can be implemented. A technique often used in this regard is to subdivide the golden image into different regions and perform different (usually more than one) tests in each of the regions, based on expected region content. 15 (a) From Eq. 4-3), at any point (x; y), K K K 1 X 1 X 1 X g= gi = fi + ´: K i=1 K i=1 K i=1 i Then K K 1 X 1 X Effi g + Ef´i g: K i=1 K i=1 But all the fi are the same image, so Effi g = f .