数字图像处理第二版(冈萨雷斯)习题答案.pdf

Digital Image Processing Second Edition Problem SolutionsStudent Set Rafael C. Gonzalez Richard E. Woods Prentice Hall Upper Saddle River, NJ 07458 or ii Revision history 10 9 8 7 6 5 4 3 2 1 Copyright c 1992-2002 by Rafael C. Gonzalez and Richard E. Woods 1Preface This abbreviated manual contains detailed solutions to all problems marked with a star in Digital Image Processing, 2nd Edition. These solutions can also bedownloaded from the book web site (). 2Solutions (Students) Problem 2.1 The diameter, x, of the retinal image corresponding to the dot is obtained from similar triangles, as shown in Fig. P2.1. That is, (d=2) 0:2 = (x=2) 0:014 which gives x = 0:07d. From the discussion in Section 2.1.1, and taking some liberties ofinterpretation, wecanthinkofthefoveaasasquaresensorarrayhavingontheorderof 337,000 elements, which translates into an array of size 580580 elements. Assuming equal spacing between elements, this gives 580 elements and 579 spaces on a line 1.5 mm long. The size of each element and each space is then s = (1:5mm)=1;159 = 1:3106m. If the size (on the fovea) of the imaged dot is less than the size of a single resolution element, we assume that the dot will be invisible to the eye. In other words, the eye will not detect a dot if its diameter, d, is such that 0:07(d) 0. We can generate the following table: Table P7.3 JPyramid ElementsCompression Ratio 011 155=4 = 1:25 22121=16 = 1:3125 38585=86 = 1:328 . . . 14=3 = 1:33 All but the trivial case (J = 0) are expansions. The expansion factor is a function of and bounded by 4/3 or 1.33. Problem 7.4 (a) The QMF fi lters must satisfy Eqs. (7.1-9) and (7.1-10). From Table 7.1, G0(z) = H0(z) and H1(z) = H0(z), so H1(z) = H0(z). Thus, beginning with Eq. (7.1-9), H0(z)G0(z) +H1(z)G1(z)=0 H0(z)H0(z) H0(z)H0(z)=0 0=0: Problem 7.1039 Similarly, beginning with Eq. (7.1-10) and substituting for H1(z), G0(z), and G1(z) from rows 2, 3, and 4 of Table 7.1, we get H0(z)G0(z) +H1(z)G1(z)=2 H0(z)H0(z) +H0(z)H0(z)=2 H2 0(z) H 2 0(z) =2 which is the design equation for the H0 (z) prototype fi lter in row 1 of the table. PROBLEM7.7Reconstructionisperformedbyreversingthedecompositionprocess; that is, by replacing the downsamplers with upsamplers and the analysis fi lters by their synthesis fi lter counterparts, as shown in Fig. P7.7. Figure P7.7 Problem 7.10 (a) The basis is orthonormal and the coeffi cients are computed by the vector equivalent 40Chapter 7 Solutions (Students) of Eq. (7.2-5): 0= h 1 p2 1 p2 i 3 2 # = 5p2 2 1= h 1 p2 1 p2 i 3 2 # = p2 2 so, 5p2 2 0+ p2 2 1= 5p2 2 1 p2 1 p2 # + p2 2 1 p2 1 p2 # = 3 2 # : Problem 7.13 From Eq. (7.2-19) we fi nd that 3;3(x)=23=2(23x 3) =2p2(8x 3) and using the Haar wavelet function defi nition from Eq. (7.2-30), obtain the plot shown in Fig. P7.13. To express 3;3(x) as a function of scaling functions, we employ