数字图像处理(双语)期中考试试卷答案

C. (Optical illusion.) D. (Simultaneous contrast.) 2、The visible spectrum consists of electromagnetic spectrum nearly in the range of wavelength: A. (10 – 400nm) B. (0.01 – 10 nm)
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四、 频域图像增强题 1 (Image enhancement in Freq. Domain 1) (20 分) Given a continuous function f(t) = cos(2πnt), (1) Its period T = ? 得分 评阅人 (2) Its frequency F = ? (3) Its Fourier transform F (jΩ) = ? (4) The Nyquist rate fs = ? (5) If it is sampled with a rate higher than fs, what’s the sampling function? What do the sampled function and its Fourier transform like like? (6) If it is sampled with a rate lower than fs, answer the same problem as (5). (7) If it is sampled with the Nyquist rate at the instants t = 0, ∆T, 2∆T, …, answer the same problem as (5). 得分 评阅人 Answer: (1) The period T = 1/n. (2) The frequency F = n. (3) F (jΩ)=
(a) The period is such that 2πnt =2π, or t =1/n. (b) The frequency is 1 divided by the period, or n. The continuous Fourier transform of the given sine wave looks as in Fig. P4.4(a) (see Problem 4.3), and the transform of the sampled data (showing a few periods) has the general form illustrated in Fig. P4.4(b) (the dashed box is an ideal filter that would allow reconstruction if the sine function were sampled, with the sampling theorem being satisfied). (c) The Nyquist sampling rate is exactly twice the highest frequency, or 2 n. That is, (1/ΔT) = 2n, or ΔT = 1/2n. Taking samples at t = ±ΔT,±2ΔT, . . . would yield the sampled function sin (2πnΔT ) whose values are all 0s because ΔT = 54 CHAPTER 4. PROBLEM SOLUTIONS
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Figure P4.4 1/2n and n is an integer. In terms of Fig. P4.4(b), we see that when ΔT = 1/2n all the positive and negative impulses would coincide, thus canceling each other and giving a result of 0 for the sampled data. (d) When the sampling rate is less than the Nyquist rate, we can have a situation such as the one illustrated in Fig. P4.4(c), which is the sum of two sine waves in this case. For some values of sampling, the sum of the two sines combine to form a single sine wave and a plot of the samples would appear as in Fig. 4.8 of the book. Other values would result in functions whose samples can describe any shape obtainable by sampling the sum of two sines.
一、 基础知识填空题(1,2 为单项选择,每空 3 分,3,4 为多项选择,每空 2 分,共 20 分), 1 1、When you enter a dark room on a bright day, it takes some time to see well enough, this is the visual process or visual phenomenon of A. (Brightness adaptation.) B. (Brightness discrimination.)
C. (400 – 700 nm) D. (700 – 1500 nm) 3、For V= {1}, the subsets S1 and S2are A. (m-connected) B. (8-connected )
C.(4-connected) D.(None of these 3) 4、Two pixels pandqare at the locations shown in the figure, their Euclidean, city-block and chessboard distances are respectively: A. De= ( C. D8 = ( B. D4 = ( 6 ) 4 ) 0 0 0 1 2 1 2 7 6 5(p) 3 4 5 6 7
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(c) , can not be decided from the information given by the figure (c). We can specify with an arbitrary positive number, for
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二、空域图像增强题 1(Image enhancement in the spatial domain) (20 分)
r Exponential of the form s = T(r) = 255 e , 0 ≤ r ≤ 255, with α beinga positive constant, are useful for constructing smooth gray-level transformation functions. (1) Start with this basic function and construct transformation functions having the shapes shown in the above figures. (15 255 255 分 ) (2) What kind of transformation does 255 the function of (a) approximatelycomplete for an input of a gray intensity image? （5 分）
三、空域图像增强题 2(Image enhancement in the spatial domain) (20 分) The White bars in the test pattern shown are 7 pixels wide and 210 pixels high. The separator between bars is 17 pixels. What would this image look like after application of 得分 评阅人 1. A 3×3 median filter? 2. A 7×7 median filter? 3. A 9×9 median filter? 4. A 15×15 median filter? (Note: in your answer, quantitative analysis is expected. ) Answer: The separator between bars is 17 pixels wide > 15, so none of the 4 filters can remove any black pixels. We can treat all black pixels as background. 1. Applying a 3 × 3 median filter removes the 1 pixel at the top left, top right, bottom left and bottom right locations of each bar, as shown in (a), which is an amplified corner of the filtered image. (a) 2. Applying a 7×7 median filter remove 6 pixels located at the top left, the top right, the bottom left, the bottom right of each bar, as shown in (b)，which is an amplified corner of the median filtered image. 0111110 0001000 3×3 1111111 7×7 01111110 1111111 01111110 3. Applying a 9×9 median filter remove 10 pixels located at the top left, the top right, the bottom left, the bottom right of each bar, as shown in (c). which is an amplified corner of the median filtered image 0000000 9×9 0011100 0111110 4. This time all bars are disappeared.