# Digital image processing - Solutions Manual by Gonzalez

By Gonzalez

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This result guarantees that the lowpass ®lter will act as a notch pass ®lter, leaving only the value of the transform at the origin. The image will not change past this value of K. 14 (a) The spatial average is 1 g(x; y) = [f (x; y + 1) + f (x + 1; y) + f (x ¡ 1; y) + f (x; y ¡ 1)] : 4 From Eq. 6-2), i 1 h j2¼v=N e + ej2¼u=M + e¡j2¼u=M + e¡j2¼v=N F (u; v) G(u; v) = 4 = H(u; v)F (u; v); where 1 H(u; v) = [cos(2¼u=M ) + cos(2¼v=N )] 2 is the ®lter transfer function in the frequency domain. (b) To see that this is a lowpass ®lter, it helps to express the preceding equation in the form of our familiar centered functions: 1 H(u; v) = [cos(2¼[u ¡ M=2)=M ) + cos(2¼[v ¡ N=2]=N )] : 2 Consider one variable for convenience.

In this case the four histograms just discussed would each have only one component. Their location would be affected as described (a) through (c). 13 Using 10 bits (with one bit being the sign bit) allows numbers in the range ¡511 to 511. The process of repeated subtractions can be expressed as K X dK (x; y) = a(x; y) ¡ b(x; y) k=1 = a(x; y) ¡ K £ b(x; y) where K is the largest value such that dK (x; y) does not exceed ¡511 at any coordinates (x; y), at which time the subtraction process stops.

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. , histogram equalization), he/she must discuss the same or similar types of assumptions.