05-78-0 Solution Set #. For the sampling function shown, analyze to determine its characteristics, e.g., the associated Nyquist sampling frequency (if any), whether a function sampled with s [x; x] may be recovered from the samples, the corresponding interpolation function, etc. This function may be written as the difference of two comb functions with the same separation parameter equal to : +! X +! X s [x; x] [x n x] [x n x ] n + X n n h ³ x n i! X + n h³ x x i! n n h COMB x i COMB x h COMB x i x COMB h ³ x n i! n ³ x x! n
S [ξ; x] F {s [x; x]} COMB [ ξ] COMB [ ξ] exp [ πiξ] COMB [ ξ] ( exp [ πiξ]) +! X [ ξ k] ( cos [πξ]+isin [πξ]) ξ! ( cos [πξ]) + µ cos π! +! ( cos [πk]) + ³ k! ( ) + ³ k ( ) if k is odd 0 if k is even COMB ξ ξ ξ! i sin [πξ]! i sin π k! i sin [πk]! i 0 k This means that the spectrum of the input object function will be translated to be centered at halfinteger values of ξ: F{f [x] s [x; x]} F [ξ] S [ξ; x] F [ξ] COMB ξ F [ξ] k F ξ k In other words, the central frequency of each replica of the object spectrum will be placed at a halfinteger value of ξ. The Nyquist frequency remains the same, but the spectrum of the interpolation function must be offset: X +! (F [ξ] S [ξ; x]) H [ξ] F ξ k RECT ξ x ˆF [ξ] ((F [ξ] S [ξ; x]) H [ξ]) ξ +
. A square CCD sensor with linear dimension of 7mm and pixel count along each axis of 04 pixels is used in a camera with a lens of focal length f 35mmthat is focused on a flat scene at a distance of z 500 mm. The schematic of the imaging system is shown below. Assuming ray optics (no diffraction), determine the resolution of the camera in line pairs per mm (analogous to cycles per mm for nonsinusoidal signals). This problem is actually pretty easy if you know the basics of optical imaging. The image formed by a lens with focal length f of an object located at distance z will be located at a distance z from the lens that satisfies the equations: + z z f 500 mm 35 mm z z z f z f 500 mm 35 mm 3500 93 mm 37.634 mm The magnification of the image is the ratio of the distances: M T z 3500 93 mm z 500 mm 7 93 0.075 which means that the image is smaller than the object and upside down (which has no consequence). The fact that the image is smaller suggests that the action of the lens is a minification rather than a magnification. The linear dimension of the sensor with 04 pixels is 7mm, which means that the linear dimension of the pixel (assuming 00% fill factor) is: 7mm 04 6.84 μm So the projected size of the pixel on the object is: 7mm 04 7 90.8 μm 93 A line pair consists of two adjacent pixels (one each white and black), which has a period of 90.8 μm 8.64 μm 0.8 mm So the spatial resolution in line pairs per millimeter is: line pairs 0.8 mm line pairs 5.6 mm 3
3. A flat object with uniform reflectance with value ρ is illuminated by a light source with intensity distribution h i [x, y] 55 exp ³[x i x 0 ] +[y y 0 ] The resulting image is quantized to k bits of intensity resolution. Assume that the eye can detect an abrupt change of gray scale of eight shades of intensity over the dynamic range. Determine the value of k where false contouring is visible. The profile of the object along the x-axis is a Gaussian function: h i [x, 0] 55 exp ³[x i x 0 ] +[0 y 0 ] This is plotted for x 0 y 0 0: i[x,0] 50 00 50 00 50-3 - - 0 3 x So we want to ensure that i 8 56 k k 56 3 k 5 8 4
4. Sketch the -D image that would be obtained for the previous problem if k. If k, there are 4 gray values i 0,,, 3, and the result is obtained by evaluating the greatest integer of i [x, 0] 56 µ h ³ i 64 exp [x x 0 ] +[0 y 0 ] + i[x,0] 4 3-3 - - 3 x 5
5. High-definition TV generates images with a vertical resolution of 5 scan lines and a width to height aspect ratio of 6:9. The system transmits 8 bits of data for each of the three additive primary colors (red, green, blue). Determine the number of bits per second necessary to store an HDTV program assuming no compression. zzzz first, find the number of pixels in the image. The number of pixels in each row is: 5 6 9 000 The number of pixels per frame therefore is: 5 000, 50, 000 and the number of bits per frame is the number of pixels multiplied by the number of bits per pixel: 5 000 8 354 0 6 In US television, there are 30 frames per second, so the number of bits per second is: 54 0 6 30.6 0 9 bits per second.6 0 9 04 04 04 8 0.9 gigabytes per second The remaining missing piece of information is the length of the program in seconds; for a one-hour program, the number of bits is:.6 0 9 3600 5. 83 0 bits per hour 679 gigabytes per hour o an uncompressed program will fill up your hard disk fairly fast, hence one reason to find useful video compression routines. 6
6. Use the program of your choice (Matlab, MathCAD, Excel, etc.) to do the following () construct and graph the following -D functions, () quantize to one bit by thresholding, and (3) quantize to one bit by error diffusion. h (a) f [n] cos π n i, 56 n 55 8 (a) f [n] before and after independent quantization to one bit (two levels); the graph of the thresholded function is scaled to have the same extrema as f [n]; (b)f [n] before and after error-diffused quantization to one bit; (c) detail view of (b), showing oscillations between the two output levels in the vicinity of the most rapid change in amplitude 7
h (b) f [n] cos π n i, 56 n 55 6 (a) f [n] before and after independent quantization to one bit (two levels); (b) magnified detail view of (a); (c) f [n] before and after error-diffused quantization to one bit; (d) detail view of (c), showing single oscillation between the two output levels near the most rapid change in amplitude. 8