考试试卷(答案)试卷编号： ( )卷课程编号：课程名称：数字图像处理（双语）考试形式：适用班级：姓名：学号：班级：学院：信息工程学院专业：电子系各专业考试日期：一二三四五六七八九十总分累分人签名题分2222200000100得分考生注意事项：1、本试卷共5页，请查看试卷中是否有缺页或破损。

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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 ofA. (Brightness adaptation.)B. (Brightness discrimination.)C. (Optical illusion.)D. (Simultaneous contrast.)2、The visible spectrum consists of electromagnetic spectrumnearly in the range of wavelength:A. (10 – 400 nm)B. (0.01 – 10 nm)C. (400 –700 nm)D. (700 –1500nm)3、For V = {1}, the subsets S1 and S2areA. (m-connected)B. (8-connected )C. (4-connected)D. (None of these 3)4、Two pixels p and q are at the locations shown in the figure,their Euclidean, city-block and chessboard distances are(20分)Exponential of the form s = T(r) = 2552reα-, 0 ≤ r ≤ 255, with α being a 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分) (2) What kind of transformation does the function of (a) approximately complete for an input of a gray intensity image? （5分）(1)(a)so .(b)得分评阅人255255255((((20分)The White bars in the test pattern shown are 7 pixels wide and210 pixels high. The separator between bars is 17 pixels. What would this image look like after application of 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 pixelswide > 15, so none of the 4 filters can removeany black pixels. We can treat all blackpixels as background.1. Applying a 3×3 median filterremoves the 1 pixel at the top left, topright, bottom left and bottom right locations of each bar, as shown in (a), which is an amplified corner of the filtered image.2. Applying a 7×7 median filter remove 6 pixelslocated at the top left, the top right, the bottom left,the bottom right of each bar, as shown in (b)，whichis an amplified corner of the median filtered image.0111110 0001000 3×3 1111111 7×7 011111101111111 011111103. Applying a 9×9 median filter remove10 pixelslocated at the top left, the top right, thebottom left,the bottom right of each bar, as shown in (c). which isan amplified corner of the median filteredimage00000009×9 00111000111110得分 评阅人 (a) (b)(20分)Given a continuous function f(t ) = cos(2πnt ),(1) Its period T = ? (2) Its frequency F = ? (3) Its Fourier transform F (jΩ) = ?(4) The Nyquist rate f s = ?(5) If it is sampled with a rate higher than f s , what’sthe sampling function? What do the sampled function and itsFourier transform like like?(6) If it is sampled with a rate lower than f s, answerthe same problem as (5).(7) If it is sampled with the Nyquist rate at theinstants t = 0, ∆T, 2∆T, …, answer the same problem as (5). 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 transformof the given sine wave looks as in Fig. P4.4(a) (see Problem 4.3), and thetransform 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 reconstructionif the sine function were sampled, with the sampling theorem beingsatisfied).(c) The Nyquist sampling rate is exactly twice the highest frequency, or 2n . Thatis, (1/ΔT ) = 2n , or ΔT = 1/2n . Taking samples at t = ±ΔT ,±2ΔT , . . . wouldyield the sampled function sin (2πn ΔT ) whose values are all 0s because ΔT =54 CHAPTER 4. PROBLEM SOLUTIONS- n nF ()Figure P4.41/2n and n is an integer. In terms of Fig. P4.4(b), we see that when ΔT = 1/2nall the positive and negative impulses would coincide, thus canceling each otherand giving a result of 0 for the sampled data.(d) When the sampling rate is less than the Nyquist rate, we can have a situationsuch as the one illustrated in Fig. P4.4(c), which is the sum of two sine wavesin this case. For some values of sampling, the sum of the two sines combine toform a single sine wave and a plot of the samples would appear as in Fig. 4.8 ofthe book. Other values would result in functions whose samples can describeany shape obtainable by sampling the sum of two sines.得分 评阅人 得分 评阅人。