数字图像处理(双语)期中考试试卷答案.doc

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202020202000000100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。如有立即举手报告以便更换。 2、考试结束后，考生不得将试卷、答题纸和草稿纸带出考场。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. (Optical illusion.) D. (Simultaneous contrast.) 2、The visible spectrum consists of electromagnetic spectrum nearly in the range of wavelength: A. (10 400 nm) B. (0.01 10 nm)C. (400 700 nm) D. (700 1500 nm) 3、For V = 1, the subsets S1 and S2 are A. (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 respectively: A. De = ( 20 ) B. D4 = ( 6 )C. D8 = ( 4 ) 0 0 0 1 2 1 2 7 6 5(p)得分评阅人 3 4 5 6 7 (q) 4 5 6 7 3 第 6 页 共 6页二、空域图像增强题1(Image enhancement in the spatial domain) (20分)(a)(b)(c)255255255Exponential of the form s = T(r) = 255, 0 r 255, with 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)得分评阅人 (a) so . (b)so (c) , can not be decided from the information given by the figure (c). We can specify with an arbitrary positive number, for example, the same as in (a) or in (b).(2) The transform function of (a) is likely to complete the reverse transform.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 33 median filter?2. A 77 median filter?3. A 99 median filter?4. A 1515 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 33 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.2. Applying a 77 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)，whichis an amplified corner of the median filtered image.0111110 000100033 1111111 77 01111110 1111111 011111103. Applying a 99 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 isan amplified corner of the median filtered image 0000000 99 0011100 01111104. This time all bars are disappeared.(a)(b)(a)(c)(b)三、 频域图像增强题 (Image enhancement in the frequency domain)(20分) We know that a lowpass filter transfer function has the relation with that of the highpass filter (1) where is the corresponding highpass filter transfer function. Given a highpass filter mask 01/801/81/21/801/20from (1) and (2), how can you get the corresponding lowpass filter mask? And give the filter mask result.Apply the lowpass filter you have gotten to the image (a) in the below figure, which is the filtered image in (b),(c),(d),(e) and (f)? Why? Explain one by one why not others? (a) (b) (c) (d) (e) (f)得分评阅人 If the blank space is not enough, you can write your answer on the reverse of the page.四、 频域图像增强题1 (Image enhancement in Freq. Domain 1) (20分)得分评阅人 Given a continuous function f(t) = cos(2nt), (1) Its period T = ?(2) Its frequency F = ?(3) Its Fourier transform F (j) = ?(4) The Nyquist rate fs = ? (5) If it is sampled with a rate higher than fs, whats 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, 2T, , 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 (j)=(a) The period is such that 2nt =2, 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 illustratedin 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/T) = 2n, or T = 1/2n. Taking samples at t = T,2T, . . . wouldyield the sampled function sin (2nT ) whose values are all 0s because T =54 CHAPTER 4. PROBLEM SOLUTIONS- n n_F(_)Figure P4.41/2n and n is an integer. In terms of Fig. P4.4(b), we see that when T = 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.五、频域图像增强题2 (Image enhancement in Freq. Domain 2) (20分)得分评阅人 In a group of images there are (1) somebright and isolated dots we are not interested in, (2) not sufficient resolutions, (3) not sufficient contrast and (4) the average gray level