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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 |
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....
Problem 2.5
From the geometry of Fig. 2.3, 7mm=35mm= z =500mm, or z = 100 mm. So the target size is 100 mm on the side. We have a total of 1024 elements per line, so the resolution of 1 line is 1024=100 = 10 elements/mm. For line pairs we divide by 2, giving an answer of 5 lp/mm....
Problem 2.11
Let p and q be as shown in Fig. P2.11. Then, (a) S1 and S2 are not 4-connected because q is not in the set N4 (p)u (b) S1 and S2 are 8-connected because q is in the set N8 (p)u (c) S1 and S2 are m-connected because (i) q is in ND (p), and (ii) the set N4 (p) \ N4 (q ) is empty....
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 3.5
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....
Problem 3.18
One of the easiest ways to look at repeated applications of a spatial ®lter is to use super-...
where the Gaussian lowpass ®lter is from Problemh4.13. Students whoistart directly 2 2 with the expression of the Gaussian highpass ®lter 1 ¡ e¡K D (u;v )=2D0 and attempt to raise it to the K th power will run into a dead end. The solution to this problem parallels the solution to Problem 4.13. Here, however, the ®lter will approach a notch ®lter that will take out F (0; 0) and thus will produce an image with zero average values (this implies negative pixels). So, there is a value of K after which the result of repeated highpass ®ltering will simply produce a constant image....
Problem 4.19
The equation corresponding to the mask in Fig. 4.27(f) is Eq. (3.7-4): As in Problem 4.15, where H (u; v ) = g(x; y ) = [f (x + 1; y ) + f (x ¡ 1; y ) + f (x; y + 1) + f (x; y ¡ 1)] ¡ 4f (x; y ): G(u; v) = H (u; v )F (u; v) h ej 2¼ u=M + e¡j 2¼ u=M + ej 2¼ v =N + e¡j 2¼ v=N ¡ 4 i...
Problem 5.21
The key to solving this problem is to recognize that the given function r 2 ¡ ¾ 2 ¡ r 2 =2 ¾ 2 h(r) = e ¾4 where r2 = x2 + y 2 , is the Laplacian (second derivative with respect to r) of the function h0 (r) = e¡r
2...
Problem 5.27
The basic idea behind this problem is to use the camera and representative coins to model the degradation process and then utilize the results in an inverse ®lter operation. The principal steps are as follows: 1. Select coins as close as possible in size and content as the lost coins. Select a background that approximates the texture and brightness of the photos of the lost coins. 2. Set up the museum photographic camera in a geometry as close as possible to give images that resemble the images of the lost coins (this includes paying attention to illumination). Obtain a few test photos. To simplify experimentation, obtain a TV camera capable of giving images that resemble the test photos. This can be done by connecting the camera to an image processing system and generating digital images, which will be used in the experiment. 3. Obtain sets of images of each coin with different lens settings. The resulting images should approximate the aspect angle, size (in relation to the area occupied by the background), and blur of the photos of the lost coins....
Problem 6.2
Denote by c the given color, and let its coordinates be denoted by (x0 ; y0 ). The distance between c and c1 is h i1=2 d(c; c1 ) = (x0 ¡ x1 )2 + (y0 ¡ y1 )2 :...
Problem 6.15
The hue, saturation, and intensity images are shown in Fig. P6.15, from left to right....
Problem 6.25
(a) The boundary between red and green becomes thickened and yellow as a result of blurring between the red and green primaries (recall that yellow is the color between green and red in, for example, Fig. 6.14). The boundary between green and blue is similarly blurred into a cyan color. The result is shown in Fig. P6.25....
Problem 6.28
The sketch is an elongated ellipsoidal ®gure in which the length lined up with the R-axis is 8 times longer that the other two dimensions. In other words, the ®gure looks like a blimp aligned with the R-axis....
Problem 7.22
They are both multi-resolution representations that employ a single reduced-resolution...
Problem 8.6
The conversion factors are computed using the logarithmic relationship 1 loga x = logb x: logb a Thus, 1 Hartley = 3.3219 bits and 1 nat = 1.4427 bits....
Problem 8.18
(a) Using the procedure described in Section 8.4.3, the decoded line is [W 1001 W W W W W W 0000 0010 W W W W W W ] where W denotes four white pixels - i.e., 1111. (b) - (c) Establish the convention that sub-blocks are included in the code string from left to right. Then, using brackets to clarify the decomposition steps, we get 1 [ [W 1001 W W W W W W ] [0000 0010 W W W W W W ] ] 1 [ 1 [W 1001 W W ] [W W W W ] ] [ 1 [0000 0010 W W ] [W W W W ] ] ] 1 [ 1 [ 1 [ [W 1001] [W W ] ] [ 0 ] ] [ 1 [ 1 [ [0000 0010] [W W ] ] [ 0 ] ] ] 1 [ 1 [ 1 [ 1 [W ] [1001] ] [ 0 ] ] [ 0 ] ] [ 1 [ 1 [ 1 [0000] [0010] ] [ 0 ] ] [ 0 ] ] ] 1 [ 1 [ 1 [ 1 [ 0 ] [11001] ] [ 0 ] ] [ 0 ] ] [ 1 [ 1 [ 1 [10000] [10010] ] [ 0 ] ] [ 0 ] ] ] Thus, the encoded string is 111101100100111100001001000, which requires 27 bits. The ®rst encoding required 28 bits....
Problem 8.23
(a) - (b) Following the procedure outlined in Section 8.6.2, we obtain the results shown in Table P8.23....
Problem 8.25 Y (0) = [1 + 2®¯ + 2®± + 6®¯° ± + 2° ± ] X (0) + [¯ + 3¯ ° ± + ± ] X (1) + [®¯ + 4®¯ ° ± + ®± + ° ± ] X (2) + [¯ °± ] X (3) + [®¯° ± ] X (4) + [¯ + 3¯ ° ± + ± ] X (¡1) + [®¯ + 4®¯ ° ± + ®± + ° ± ] X (¡2) + [¯ °± ] X (¡3) Thus, we can form the lowpass analysis ®lter coef®cients shown in Table P8.25-1. Table P8.25-1 Coef®cient Index §4 §3 §2 §1 0 Expression ®¯ ° ± =K ¯° ± =K (®¯ + 4®¯° ± + ®± + ° ± ) =K (¯ + 3¯° ± + ± ) =K (1 + 2®¯ + 2®± + 6®¯ ° ± + 2° ± ) =K Value 0.026748757 -0.016864118 -0.07822326 0.26686411 0.60294901 + [®¯° ± ] X (¡4) :...
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