CS 659图像处理

CS 659 Image Processing
Sample Midterm Exam (Closed Book, 90 minutes)
Covering Lectures 1~6. There are 5 questions. Each is 20 points.

1. (20 points)
   (a) Let a grayscale image be the size 100-by-200 with 256 gray levels. How many bits in
   memory are required to store this grayscale image without adding the overhead?
   (b) Let a color image be the size 100-by-200. How many bits in memory are required to store
   this color image without adding the overhead?
   (c) If we reduce the image size by a half in both row and column. What fraction in the memory
   size can we reduce?
   Answer:
   (a) 1002008=160,000 bits
   (b) 1002008*3=480,000 bits
   (c) or 0.25
2. (20 points) Let an image f of 4-by-4 and a mask g of 2-by-2 as follows. Calculate the
   convolution f * g and correlation f ° g. Note the origin is located at lower-left corner.
3. (20 points)
   (a) Let an image of 2-by-2 as follows. Perform image negative on this image. What is the
   resulting image?
4. 33
5. 255
   (b) Let an image of 2-by-2 as follows. Perform bit-plane slicing on this image. What are the eight
   images on each bit-plane, B0, B1, B2, B3, B4, B5, B6, and B7, from Least Significant Bit (LSB)
   B0 to Most Significant Bit (MSB) B7?
6. 33
7. 255
   Answer:
   (a) We use 255 to subtract each pixel.
8. 222
9. 0
   (b) We represent each pixel into 8-bit as:
   3=00000011,
   33=00100001
   170=10101010
   255=11111111
   Therefore,
10. (20 points)
    Let a binary image be f and a template be g as follows. Perform image matching using the
    equation: Note that is correlation and
    is the complement. What is the
    resulting image?
    1
11. (20 points) Explain the image equalization and image specification. Briefly describe both
    procedures to achieve their goals.
    Answer:
    In histogram equalization we are trying to maximize the image contrast by applying a gray level
    transform which tries to flatten the resulting histogram. It turns out that the gray level transform
    that we are seeking is simply a scaled version of the original image's cumulative histogram. That
    is, the graylevel transform T is given by T[i] = (G-1)c(i), where G is the number of gray levels
    and c(i) is the normalized cumulative histogram of the original image.
    r: the gray levels to be enhanced.
    Assume r continuous in [0,1]
    s=T(r)
    a) T(r) single-valued, monotonically increasing
    b) 0<=T(r)<=1
    When we want to specify a non-flat resulting histogram, we can use the following steps (called
    image specification):
12. Specify the desired histogram g(z)
13. Obtain the transform which would equalize the specified histogram, Tg, and its inverse
    Tg-1
14. Get the transform which would histogram equalize the original image, s=T[i]
15. Apply the inverse transform Tg-1
    on the equalized image, that is z=Tg-1
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