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日期：2019-02-21 09:12

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) 100*200*8=160,000 bits

(b) 100*200*8*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?

3 33

170 255

(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?

3 33

170 255

Answer:

(a) We use 255 to subtract each pixel.

252 222

85 0

(b) We represent each pixel into 8-bit as:

3=00000011,

33=00100001

170=10101010

255=11111111

Therefore,

4. (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

5. (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):

1. Specify the desired histogram g(z)

2. Obtain the transform which would equalize the specified histogram, Tg, and its inverse

Tg-1

3. Get the transform which would histogram equalize the original image, s=T[i]

4. Apply the inverse transform Tg-1

on the equalized image, that is z=Tg-1

[s]