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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 2.10
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 2.18
With reference to Eq. (2.6-1), let H denote the neighborhood sum operator, let S1 and S2 denote two different small subimage areas of the same size, and let S1 + S2 denote the corresponding pixel-by-pixel sum of the elements in S1 and S2 , as explained in Section 2.5.4. Note that the size of the neighborhood (i.e., number of pixels) is not changed by this pixel-by-pixel sum. The operator H computes the sum of pixel values is a given neighborhood. Then, H (aS1 + bS2 ) means: (1) multiplying the pixels in each of the subimage areas by the constants shown, (2) adding the pixel-by-pixel values from S1 and S2 (which produces a single subimage area), and (3) computing the sum of the values of all the pixels in that single subimage area. Let ap1 and bp2 denote two arbitrary (but...
Problem 3.4
(a) The number of pixels having different gray level values would decrease, thus causing the number of components in the histogram to decrease. Since the number of pixels would not change, this would cause the height some of the remaining histogram peaks to increase in general. Typically, less variability in gray level values will reduce contrast. (b) The most visible effect would be signi®cant darkening of the image. For example, dropping the highest bit would limit to 127 the brightest level in an 8-bit image. Since the number of pixels would remain constant, the height of some of the histogram peaks would increase. The general shape of the histogram would now be taller and narrower, with no histogram components being located past 127....
Problem 3.8
We are interested in just one example in order to satisfy the statement of the problem. Consider the probability density function shown in Fig. P3.8(a). A plot of the transformation T (r) in Eq. (3.3-4) using this particular density function is shown in Fig. P3.8(b). Because pr (r) is a probability density function we know from the discussion in Section 3.3.1 that the transformation T (r) satis®es conditions (a) and (b) stated in that section. However, we see from Fig. P3.8(b) that the inverse transformation from s back to r is not single valued, as there are an in®nite number of possible mappings from s = 1=2 back to r. It is important to note that the reason the inverse transformation function turned out not to be single valued is the gap in pr (r) in the interval [1=4; 3=4]....
Problem 3.12
The purpose of this simple problem is to make the student think of the meaning of histograms and arrive at the conclusion that histograms carry no information about spatial properties of images. Thus, the only time that the histogram of the images formed by the operations shown in the problem statement can be determined in terms of the original histograms is when one or both of the images is (are) constant. In (d) we have the additional requirement that none of the pixels of g(x; y ) can be 0. Assume for convenience that the histograms are not normalized, so that, for example, hf (rk ) is the number of pixels in f (x; y ) having gray level rk , assume that all the pixels in g(x; y ) have constant value c. The pixels of both images are assumed to be positive. Finally, let uk denote the gray levels of the pixels of the images formed by any of the arithmetic operations given in the problem statement. Under the preceding set of conditions, the histograms are determined as follows: (a) The histogram hsum (uk ) of the sum is obtained by letting uk = rk +c; and hsum (uk ) = hf (rk ) for all k. In other words, the values (height) of the components of hsum are the same as the components of hf , but their locations on the gray axis are shifted right by an amount c. (b) Similarly, the histogram hdiff (uk ) of the difference has the same components as hf but their locations are moved left by an amount c as a result of the subtraction operation. (c) Following the same reasoning, the values (heights) of the components of histogram hprod (uk ) of the product are the same as hf , but their locations are at uk = c £ rk . Note that while the spacing between components of the resulting histograms in (a) and (b) was not affected, the spacing between components of hprod (uk ) will be spread out by an amount c....
Problem 3.16
With reference to Section 3.4.2, when i = 1 (no averaging), we have
2 g(1) = g1 and ¾g(1) = ¾ 2 : ´...
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....
Inserting this identity in the preceding integral yields Z1 (2¼)2 z2 ¾ 2 2 2 242 1 2 h(z ) = e¡ e¡ 2¾2 [w ¡j 4¼¾ wz¡(2¼ ) ¾ z ] dw ¡1 Z1 (2¼)2 z2 ¾ 2 22 1 2 = e¡ e¡ 2¾2 [w ¡ j 2¼ ¾ z] dw:
¡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...
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