By Rafael C. Gonzalez, Richard E. Woods

ISBN-10: 0201180758

ISBN-13: 9780201180756

3rd iteration booklet that builds on hugely winning previous variants and the author's two decades of educational and commercial adventure in photo processing. displays new traits; rfile snapshot compression and knowledge compression criteria.

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**Example text**

It is easily veri®ed that the output image will have the required minimum and maximum values if we choose f ¡ f min a = max fmax ¡ fmin and f fmax ¡ f max fmin b = min fmax ¡ fmin where fmax and fmin are the maximum and minimum values of the input image. Note that the key assumption behind this method is that all images stay within the linear operating range of the camera, thus saturation and other nonlinearities are not an issue. Another implicit assumption is that moving objects comprise a relatively small area in the ®eld of view of the camera, otherwise these objects would overpower the scene and the values obtained from f0 (x; y) would not make a lot of sense.

27 Consider the following equation: f(x; y) ¡ r2 f (x; y) = f (x; y) ¡ [f (x + 1; y) + f (x ¡ 1; y) + f (x; y + 1) +f (x; y ¡ 1) ¡ 4f (x; y)] = 6f (x; y) ¡ [f (x + 1; y) + f(x ¡ 1; y) + f (x; y + 1) +f (x; y ¡ 1) + f (x; y)] = 5 f1:2f(x; y)¡ 1 [f (x + 1; y) + f(x ¡ 1; y) + f (x; y + 1) 5 +f (x; y ¡ 1) + f(x; y)]g £ ¤ = 5 1:2f (x; y) ¡ f (x; y) where f (x; y) denotes the average of f (x; y) in a prede®ned neighborhood that is centered at (x; y) and includes the center pixel and its four immediate neighbors.

9, taking the complex conjugate of an image mirrors it in the spatial domain. Thus, we would expect the result to be a mirror image (about both axes) of Fig. 41(e). 24 (a) and (b) See Figs. 24(a) and (b). (c) and (d) See Figs. 24(c) and (d). 25 Because M = 2n , we can write Eqs. 6. Next, we assume that the equations hold for n. Then, we are required to prove that they also are true for n + 1. From Eq. 6-45), m(n + 1) = 2m(n) + 2n : 52 Chapter 4 Problem Solutions Substituting m(n) from above, µ ¶ 1 M n + 2n 2 µ ¶ 1 n = 2 2 n + 2n 2 = 2n (n + 1) 1 ¡ n+1 ¢ = (n + 1): 2 2 Therefore, Eq.

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