By Rafael C. Gonzalez, Richard E. Woods
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Extra info for Digital image processing. Solutions Manual
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.
A similar expression gives the difference in the y-direction. 16 47 From Eq. 6-2), G(u; v) = F (u; v) ¡ F (u; v)ej2¼u=M + F (u; v) ¡ F (u; v)ej2¼v=N = [1 ¡ ej2¼u=M ]F (u; v) + [1 ¡ ej2¼v=N ]F (u; v) = H(u; v)F (u; v); where H(u; v) is the ®lter function: i h H(u; v) = ¡2j sin(¼u=M )ej¼u=M + sin(¼v=N )ej¼v=N : (b) To see that this is a highpass ®lter, it helps to express the ®lter function in the form of our familiar centered functions: i h H(u; v) = ¡2j sin(¼[u ¡ M=2]=M )ej¼u=M + sin(¼[v ¡ N=2]=N )ej¼v=N : Consider one variable for convenience.
A brute-force implementation would require n2 ¡ 1 additions after each move. (b) The computational advantage is n2 ¡ 1 (n + 1)(n ¡ 1) A= = = n ¡ 1: n+1 (n + 1) The plot of A as a function of n is a simple linear function starting at A = 1 for n = 2. 17 29 position. Let f (x; y) and h(x; y) denote the image and the ®lter function, respectively. Assuming square images of size N £ N for convenience, we can express f (x; y) as the sum of at most N 2 images, each of which has only one nonzero pixel (initially, we assume that N can be in®nite).