数字图像处理(双语)期中考试试卷答案.doc

考试试卷（答案）
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课程编号：课程名称：数字图像处理（双语）考试形式：
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学院：信息工程学院专业：电子系各专业考试日期：

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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 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 – 400nm） B。 (0。01 – 10 nm）
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= ( ） B。 D4 = （ 6 )
C。 D8 = （ 4 ) 0 0 0 1 2
1 2 7 6 5（p）
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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)
255
255
255
Exponential of the form s = T（r) = 255， 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分） （2） What kind of transformation does the function of (a) approximatelycomplete for an input of a gray intensity image？ （5分)
(1）
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so。
so
(a) , 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
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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.
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。
(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
0
–1/8
0
–1/8
1/2
–1/8
0
–1/2
0
from (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)
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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）
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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 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）。
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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 2n. 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
— n n
_
F（_）
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.

五、频域图像增强题2 （Image enhancement in Freq。 Domain 2) (20分）
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In a group of images there are (1） some
bright and isolated dots we are not interested in, （2） not sufficient resolutions， （3） not sufficient contrast and (4） the average gray level was changed whereas the correct DC value should be V。
Please propose solutions or processing steps respectively for these objectives by using the methods introduced in Chapter 3 or Chapter 4 of the textbook in our bilingual course Digital Image Processing。
Answer：
（1) Performamedian filtering operation, because the median filter is good at eliminating the isolated noises。
（2) Follow (1） by high-frequency emphasis。
(3） Histogram-equalize this result or some piece—wise linear stretch filters.
（4） Compute the average gray level, K0。 Add the quantity (V −K0） to all pixels.
If the blank space is not enough， you can write your answer on the reverse of the page。