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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 |
Teaching Features of the Book
Undergraduate programs that offer digital image processing typically limit coverage to one semester. Graduate programs vary, and can include one or two semesters of the material. In the following discussion we give general guidelines for a one-semester senior course, a one-semester graduate course, and a full-year course of study covering two semesters. We assume a 15-week program per semester with three lectures per week. In order to provide -exibility for exams and review sessions, the guidelines discussed in the following sections are based on forty, 50-minute lectures per semester. The background assumed on the part of the student is senior-level preparation in mathematical analysis, matrix theory, probability, and computer programming. The suggested teaching guidelines are presented in terms of general objectives, and not as time schedules. There is so much variety in the way image processing material is taught that it makes little sense to attempt a breakdown of the material by class period. In particular, the organization of the present edition of the book is such that it makes it much easier than before to adopt signi®cantly different teaching strategies, depending on course objectives and student background. For example, it is possible with the new organization to offer a course that emphasizes spatial techniques and covers little or no transform material. This is not something we recommend, but it is an option that often is attractive in programs that place little emphasis on the signal processing aspects of the ®eld and prefer to focus more on the implementation of spatial techniques....
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 2.10
The width-to-height ratio is 16/9 and the resolution in the vertical direction is 1125 lines (or, what is the same thing, 1125 pixels in the vertical direction). It is given that the...
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 3.6
Let n be the total number of pixels and let nrj be the number of pixels in the input image...
Problem 3.14
Let g(x; y ) denote the golden image, and let f (x; y ) denote any input image acquired during routine operation of the system. Change detection via subtraction is based on computing the simple difference d(x; y ) = g(x; y) ¡ f (x; y ). The resulting image...
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, =...
e¡j 2¼ (ux=M + vy=N ) M ¡1 N ¡1 N M 1 XX f (x; y )e¡j 2¼(x[u¡ 2 ]=M +y[v ¡ 2 ]=N ) M N x=0 y=0...
Problem 4.15
The problem statement gives the form of the difference in the x-direction. A similar expression gives the difference in the y -direction. The ®ltered function in the spatial domain then is: g(x; y) = f (x; y ) ¡ f (x + 1; y ) + f (x; y ) ¡ f (x; y + 1):...
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 )...
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