Digital Image Processing

Assignment 1 Solution ٍ

Due Date (29-2-2012)

Name : ___________________________Group : ___________ Marks : ________/20

Q1) Choose the correct answer of the following() :

1 .What is digital image processing?

a) A 2D function f(x,y) that can be observed by human visual system.

b) Is a representation of a two dimensional image as a finite set of digital values, called picture elements or pixels.

c) Processing digital images by means of operations on digital computers .

d) All of the above

2 .The process that expands the range of intensity levels in an image:

a) Contrast Stretching

b) Thresholding

c) Intensity level slicing

d) Bit-Plane Slicing

3 .Image Negatives the intensity transformation function means:

a) Mapping a narrow range of a low intensity values in the input into a wider range of output levels.

b) Revising the intensity values of an image.

c) Digitizing the coordinate values of an Image.

d) None of the above

4 .The function that converts the image into a binary image is:

a) Contrast Stretching Function

b) Power-law Transformation Function

c) Thresholding Function

d) Bit-Plane Transformation Function

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Q2) Select (T-True) for Correct statement and (F-False) for incorrect answer and correct the wrong answer:

1. Sampling (Quantization) means digitizing the amplitude values while quantization (Sampling) means digitizing the coordinate values.

( F )

2. In digital images 1sample (3 samples) per point represents (Red, Green, and Blue) ( F )

3. The first stage in image processing is image acquisition. ( T )

4. Log- Transformation (Power-Law) function curves with fractional values of ϒ map a narrow range of dark input values into a wider range of output values. ( F )

Q3) Discuss the following pictures in means of:

a. The Transformation function applied to each pictureb. The result of the transformation function applied

a. The transformation function applies is image negative.

b. The result is converting zero's (black pixels) into one's (white pixels)

a. The function applied is the gray level reduction (Quantization).

b.The result is reducing the gray levels until reaching only 2 gray levels balck and white.

c. What is the name of the fourth image in picture 2?

It is called false contouring it is caused by insufficient gray levels

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Q4) Suppose that a 3-bit image of size 64*64 = (4096) has the following intensity distribution nk = (212, 700, 923, 181,850, 656, 329, 245).

1 -Draw the histogram for the given input after filling the following table:

Pr(rk) = nk/mnnkrk

0.05212R0

0.17700R1

0.23923R2

0.04181R3

0.20850R4

0.16656R5

0.08329R6

0.06245R7

0 1 2 3 4 5 6 70

50.0

1.0

51.0

2.0

52.0

3.0

nm/kn = )kr(rP2nmuloC3nmuloC

2 -Fined the equalized values (sk) after applying the histogram equalization function on the previous values.

0 S0 = 7*0.05 = 0.35-2 1.54 =7) *0.05+0.17 =(S1-

3S2 = 7 * (0.05+0.17+0.23) = 3.15- 3S3 = 7*(0.05+0.17+0.23+0.04) = 3.43-

57)*0.05+0.17+0.23+0.04+0.20 = (4.83 = S4- 67)*0.05+0.17+0.23+0.04+0.20+0.16 = (5.95 = S5–

77)*0.05+0.17+0.23+0.04+0.20+0.16+0.08 = (6.51 = S6–

3

7S7 = 7*(0.05+0.17+0.23+0.04+0.20+0.16+0.08+0.05) = 6.86-

3 -Draw the equalized histogram (the output histogram after equalization).

0 2 3 5 6 70

50.0

1.0

51.0

2.0

52.0

3.0

)ks(rP3nmuloC1nmuloC

S0 = 0 = r0 = 0.05

S1 = 2 = r1 = 0.17

S2 = 3 = r2 = 0.23

S3= 3 = r3 = 0.04

S4= 5 = r4 =0.20

S5= 6 = r5 =0.16

S6= 7 = r6 = 0.08

0.06S7=7 = r7=

4