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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...
Problem 2.7
The image in question is given by f (x; y ) = i(x; y )r(x; y ) = 255e¡[(x¡x0 ) = 255e¡[(x¡x0 )
2...
Problem 2.15
(a) When V = f0; 1g, 4-path does not exist between p and q because it is impossible to get from p to q by traveling along points that are both 4-adjacent and also have values from V . Figure P2.15(a) shows this conditionu it is not possible to get to q . The shortest 8-path is shown in Fig. P2.15(b)u its length is 4. The length of the shortest m- path (shown dashed) is 5. Both of these shortest paths are unique in this case. (b) One...
Problem 2.20
The geometry of the chips is shown in Fig. P2.20(a). From Fig. P2.20(b) and the geometry in Fig. 2.3, we know that ¸ £ 80 ¢x = ¸¡z where ¢x is the side dimension of the image (assumed square since the viewing screen is square) impinging on the image plane, and the 80 mm refers to the size of the viewing screen, as described in the problem statement. The most inexpensive solution will result from using a camera of resolution 512 £ 512. Based on the information in Fig. P2.20(a), a CCD chip with this resolution will be of size (16¹) £ (512) = 8 mm on each side. Substituting ¢x = 8 mm in the above equation gives z = 9¸ as the relationship between the distance z and the focal length of the lens, where a minus sign was ignored because it is just a coordinate inversion. If a 25 mm lens is used, the front of the lens will have to be located at approximately 225 mm from the viewing screen so that the size of the...
Problem 3.5
All that histogram equalization does is remap histogram components on the intensity scale. To obtain a uniform (-at) histogram would require in general that pixel intensities be actually redistributed so that there are L groups of n=L pixels with the same intensity, where L is the number of allowed discrete intensity levels and n is the total number of pixels in the input image. The histogram equalization method has no provisions for this type of (arti®cial) redistribution process....
Figure P3.8. (b) From the discussion in Problem 3.8, it follows that if an image has missing gray levels the histogram equalization transformation function given above will be constant in the interval of the missing gray levels. Thus, in theory, the inverse mapping will not be single-valued in the discrete case either. In practice, assuming that we wanted to perform the inverse transformation, this is not important for the following reason: Assume that no gray-level values exist in the open interval (a; b), so that ra is the last gray level before the empty gray-level band begins and rb is the ®rst gray level right after the empty band ends. The corresponding mapped gray levels are sa and sb . The fact that no gray levels r exist in interval (a; b) means that no gray levels will exist between sa and sb either, and, therefore, there will be no levels s to map back to r in the bands where the multi-valued inverse function would present problems. Thus, in practice, the issue of the inverse not being single-valued is not an issue since it would not be needed. Note that mapping back from sa and sb presents no problems, since T (ra ) and T (rb ) (and thus their inverses) are different. A similar discussion applies if there are more than one band empty of gray levels....
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. Treating the constants in the last line of the above equation as proportionality factors, we may write...
Problem 4.6
(a) We note ®rst that (¡1)x+y = ej ¼(x+y) . Then, M ¡1 N ¡1 h i i 1 X Xh = f (x; y )ej ¼ (x+y) = f (x; y )ej ¼(x+y) e¡j 2¼ (ux=M + vy=N ) M N x=0 y=0 =
M ¡1 N ¡1 i yN 1 X Xh xM f (x; y )e¡j 2¼(¡ 2M ¡ 2N ) M N x=0 y=0...
Problem 4.26
Consider a single star modeled as an impulse ± (x ¡ x0 ; y ¡ y0 ). Then, from which f (x; y ) = K ± (x ¡ x0 ; y ¡ y0 )...
Problem 5.28
Using triangular regions means three tiepoints, so we can solve the following set of linear equations for six coef®cients: x0 y
0...
Problem 6.5
At the center point we have 1 1 1 1 1 R + B + G = (R + G + B ) + G = midgray + G 2 2 2 2 2 which looks to a viewer like pure green with a boot in intensity due to the additive gray component....
Problem 6.11
(a) The purest green is 00FF00, which corresponds to cell (7, 18). (b) The purest blue is 0000FF, which corresponds to cell (12, 13)....
Problem 6.17
(a) Because the infrared image which was used in place of the red component image has very high gray-level values. (b) The water appears as solid black (0) in the near infrared image [Fig. 6.27(d)]. Threshold the image with a threshold value slightly larger than 0. The result is shown in Fig. P6.17. It is clear that coloring all the black points in the desired shade of blue presents no dif®culties. (c) Note that the predominant color of natural terrain is in various shades of red. We already know how to take out the water from (b). Thus a method that actually removes the }background} of red and black would leave predominantly the other man-made structures, which appear mostly in a bluish light color. Removal of the red [and the black if you do not want to use the method as in (b)] can be done by using the technique discussed in Section 6.7.2....
Problem 6.19
The complement of a color is the color opposite it on the color circle of Fig. 6.32. The hue component is the angle from red in a counterclockwise direction normalized by 360 degrees. For a color on the top half of the circle (i.e., 0 · H · 0:5), the hue of the complementary color is H + 0:5. For a color on the bottom half of the circle (i.e., for...
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