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Gonzalez, Woods. Solutions manual.. Digital image processing (2ed., PH, 2002)(209s).pdf |
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Size 1.6Mb Date Dec 9, 2006 |
One Semester Graduate Course (No Background in DIP)
The main difference between a senior and a ®rst-year graduate course in which neither group has formal background in image processing is mostly in the scope of material covered, in the sense that we simply go faster in a graduate course, and feel much freer in assigning independent reading. In addition to the material discussed in the previous section, we add the following material in a graduate course. Coverage of histogram matching (Section 3.3.2) is added. Sections 4.3, 4.4, and 4.5 are covered in full. Section 4.6 is touched upon brie-y regarding the fact that implementation of discrete Fourier transform techniques requires non-intuitive concepts such as function padding. The separability of the Fourier transform should be covered, and mention of the advantages of the FFT should be made. In Chapter 5 we add Sections 5.5 through 5.8. In Chapter 6 we add the HSI model (Section 6.3.2) , Section 6.4, and Section 6.6. A nice introduction to wavelets (Chapter 7) can be achieved by a combination of classroom discussions and independent reading. The minimum number of sections in that chapter are 7.1, 7.2, 7.3, and 7.5, with appropriate (but brief) mention of the existence of fast wavelet transforms. Finally, in Chapter 8 we add coverage of Sections 8.3, 8.4.2, 8.5.1 (through Example 8.16), Section 8.5.2 (through Example 8.20) and Section 8.5.3. If additional time is available, a natural topic to cover next is morphological image processing (Chapter 9). The material in this chapter begins a transition from methods whose inputs and outputs are images to methods in which the inputs are images, but the outputs are attributes about those images, in the sense de®ned in Section 1.1. We...
Two Semester Graduate Course (No Background in DIP)
A full-year graduate course consists of the material covered in the one semester undergraduate course, the material outlined in the previous section, and Sections 12.1, 12.2, 12.3.1, and 12.3.2....
which shows that a second pass of histogram equalization would yield the same result as the ®rst pass. We have assumed negligible round-off errors....
Problem 3.18
One of the easiest ways to look at repeated applications of a spatial ®lter is to use super-...
Problem 3.20
(a) Numerically sort the n2 values. The median is ³ = [(n2 + 1)=2]-th largest value. (b) Once the values have been sorted one time, we simply delete the values in the trailing edge of the neighborhood and insert the values in the leading edge in the appropriate locations in the sorted array....
¤ 1£ q Q + 2 (n2 ¡ q 2 )B n2 n where Q denotes the average value of object points. Let the maximum expected average value of object points be denoted by Qmax . Then we want the response of the mask at any point on the object under this maximum condition to be less than one-tenth Qmax , or ¤ q2 1£ 2 1 Q + (n ¡ q 2 )B < Qmax n2 max n2 10 from which we get the requirement · ¸1=2 10(Qmax ¡ B ) n>q (Qmax ¡ 10B ) for the minimum size of the averaging mask. Note that if the background gray-level is p 0, we the minimum mask size is n < 10q . If this was a fact speci®ed by the instructor, =...
Problem 4.4
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 ). Then we proceed as follows: 2 2 H (w) = e¡w =2¾ : The inverse Fourier transform is Z h(z ) = = = We now make use of the identity e¡
1...
Problem 4.14
(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. (4.6-2), i 1 h j 2¼v =N e + ej 2¼u=M + e¡j 2¼u=M + e¡j 2¼v =N F (u; v ) G(u; v ) = 4 = H (u; v)F (u; v ); 1 H (u; v) = [cos(2¼ u=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. As u ranges from 0 to M , the value of cos(2¼[u¡ M =2)=M ) starts at ¡1, peaks at 1 when u = M =2 (the center of the ®lter) and then decreases to ¡1 again when u = M . Thus, we see that the amplitude of the ®lter decreases as a function of distance from the origin of the centered ®lter, which is the characteristic of a lowpass ®lter. A similar argument is easily carried out when considering both variables simultaneously. where...
Problem 4.17
(a) Express ®ltering as convolution to reduce all processes to the spatial domain. Then, the ®ltered image is given by 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 )] 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. Thus, in general, T [h(x; y) ¤ f (x; y )] 6= h(x; y ) ¤ T [f (x; y )] and the order does matter. (b) As indicated in Section 4.4, highpass ®ltering severely diminishes the contrast of an image. Although high-frequency emphasis helps some, the improvement is usually not dramatic (see Fig. 4.30). Thus, if an image is histogram equalized ®rst, the gain in contrast improvement will essentially be lost in the ®ltering process. Therefore, the procedure in general is to ®lter ®rst and histogram-equalize the image after that. = T [h(x; y ) ¤ f (x; y )] ; g (x; y ) = h(x; y ) ¤ f (x; y )...
Problem 4.22
(a) Padding an image with zeros increases its size, but not its gray-level content. Thus, the average gray-level of the padded image is lower than that of the original image. This implies that F (0; 0) in the spectrum of the padded image is less than F (0; 0) in the original image (recall that F (0; 0) is the average value of the corresponding image). Thus, we can visualize F (0; 0) being lower in the spectrum on the right, with all values away from the origin being lower too, and covering a narrower range of values. Thatzs the reason the overall contrast is lower in the picture on the right. (b) Padding an image with 0zs introduces signi®cant discontinuities at the borders of the original images. This process introduces strong horizontal and vertical edges, where the image ends abruptly and then continues with 0 values. These sharp transitions correspond to the strength of the spectrum along the horizontal and vertical axes of the spectrum....
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