Exam (elaborations) TEST BANK FOR Digital Image Processing 3ed Edition By Rafael C. Gonzalez and Richard E. Woods (Solution Manual)

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) 0.017 which gives x = 0.085d . From the discussion in Section 2.1.1, and taking some liberties of interpretation, we can think of the fovea as a square sensor array having on the order of 337,000 elements, which translates into an array of size 580 ×580 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] = 1.3×10−6 m. 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 0.085(d ) <1.3×10−6 m, or d < 15.3×10−6 m. Image of the dot x x/2 on the fovea Edge view of dot d d/2 0.2 m 0.017 m Figure P2.1 9 10 CHAPTER 2. PROBLEM SOLUTIONS Problem 2.2 Brightness adaptation. Problem 2.3 The solution is λ = c /v = 2.998×108(m/s)/60(1/s) = 4.997×106m=4997Km. Problem 2.4 (a) From the discussion on the electromagnetic spectrum in Section 2.2, the source of the illumination required to see an object must have wavelength the same size or smaller than the object. Because interest lies only on the boundary shape and not on other spectral characteristics of the specimens, a single illumination source in the far ultraviolet (wavelength of .001 microns or less) will be able to detect all objects. A far-ultraviolet camera sensor would be needed to image the specimens. (b) No answer is required because the answer to (a) is affirmative. Problem 2.5 From the geometry of Fig. 2.3, (7mm)/(35mm) = (z )/(500mm), or z = 100mm. 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 2.6 One possible solution is to equip a monochrome camera with a mechanical device that sequentially places a red, a green and a blue pass filter in front of the lens. The strongest camera response determines the color. If all three responses are approximately equal, the object is white. A faster system would utilize three different cameras, each equipped with an individual filter. The analysis then would be based on polling the response of each camera. This system would be a little more expensive, but it would be faster and more reliable. Note that both solutions assume that the field of view of the camera(s) is such that it is com11 Intensity Intensity 0 255 0 255 (x , y ) 0 0 G Equally spaced subdivisions (a) (b) Figure P2.7 pletely filled by a uniform color [i.e., the camera(s) is (are) focused on a part of the vehicle where only its color is seen. Otherwise further analysis would be required to isolate the region of uniform color, which is all that is of interest in solving this problem]. Problem 2.7 The image in question is given by f (x,y ) = i (x,y )r (x,y ) = 255e −[(x−x0)2+(y−y0)2] ×1.0 = 255e −[(x−x0)2+(y−y0)2] A cross section of the image is shown in Fig. P2.7(a). If the intensity is quantized using m bits, then we have the situation shown in Fig. P2.7(b), where G = (255 +1)/2m. Since an abrupt change of 8 intensity levels is assumed to be detectable by the eye, it follows that G = 8 = 256/2m, or m = 5. In other words, 32, or fewer, intensity levels will produce visible false contouring. 12 CHAPTER 2. PROBLEM SOLUTIONS Intensity 0 63 127 191 255 Image quantized into four levels 255 191 127 63 Figure P2.8 Problem 2.8 The use of two bits (m = 2) of intensity resolution produces four intensity levels in the range 0 to 255. One way to subdivide this range is to let all levels between 0 and 63 be coded as 63, all levels between 64 and 127 be coded as 127, and so on. The image resulting from this type of subdivision is shown in Fig. P2.8. Of course, there are other ways to subdivide the range [0,255] into four bands. Problem 2.9 (a) The total amount of data (including the start and stop bit) in an 8-bit, 1024× 1024 image, is (1024)2×[8+2] bits. The total time required to transmit this image over a 56K baud link is (1024)2 ×[8+2]/56000 = 187.25 sec or about 3.1 min. (b) At 3000K this time goes down to about 3.5 sec. 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