2019 Quiz 2 - Solutions
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Subject
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Date
Dec 6, 2023
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–
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Solutions
First name:
Student number:
Course Code:
ECE4445 / MBP4445 / ECE9201 / ECE9021 / BME9509 / MBP9509
INTRODUCTION TO DIGITAL IMAGE PROCESSING: QUIZ 2
INSTRUCTIONS
•
Print your name and student number at the top of
each
page in the space provided.
***** Please indicate the course code in which you are registered. *****
•
Closed book. All calculators are allowed but be prepared to show that the memory is empty.
•
Write answers in the space indicated in each question.
1.
[3 marks]
Table I gives the histogram
H
A
(
D
A
) of a digital image
A
(
i
,
j
). Using the algorithm
presented in class, design a point operation that equalizes the image’s histogram to produce an
output image
B
(
i,
j
). Specify the point operation by creating and filling in Table II below that
relates gray levels in the input image,
D
A
, to those in the output,
D
B
.
Show all work in the space
below Tables I and II. State any formulae you used, including how constants in the formulae
were determined.
Table I
Table II
D
A
H
A
(
D
A
)
D
A
D
B
0 (black)
1,000
0 (black)
1
0
1
2
1,000
2
3 (white)
0
3 (white)
Solution:
D
A
H
A
𝑆 = ∑ 𝐻
𝐴
(𝑖)
𝐷
𝐴
𝑖=0
D
m
/A
0
x
S
D
B
0
1000
1000
1.5
2
1
0
1000
1.5
2
2
1000
2000
3.0
3
3
0
2000
3.0
3
where
D
m
= 3,
𝐴
0
= ∑ 𝐻
𝐴
(𝑖) = 1000 + 0 + 1000 + 0 = 2000
3
𝑖=0
This question is similar to that from the lecture on October 3 (slide
6). I had told the class that this would be tested with different numer-
ical values.
ECE4445 / MBP4445 / ECE9201 / ECE9021 / BME9509 / MBP9509
–
2019 - Page 2/4
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First name:
Student number:
Course Code:
2.
[7 marks]
Table III gives the histogram,
H
1
(
D
1
), of a 3-bit digital image (IMAGE 1) with gray
levels
D
1
= 0, 1, …, 7. Table IV gives the histogram,
H
2
(
D
2
), of a second 3-bit image (IMAGE
2). Design a point operation that will match the histogram of IMAGE 1 to that of IMAGE 2.
Specify the point operation by filling in Table V that provides a value of
D
2
that each value of
D
1
maps to.
In the space below Tables III, IV and V, show all work, and explicitly state what
you are doing. State rules you used when inverting any functions. Use the reverse side of this
page if you need more space.
Table III
Table IV
Table V
D
1
H
1
(
D
1
)
D
2
H
2
(
D
2
)
D
1
D
2
0 (black)
0
0 (black)
0
0
1
4016
1
2470
1
2
2954
2
2536
2
3
1630
3
1953
3
4
800
4
1337
4
5
368
5
858
5
6
162
6
529
6
7 (white)
70
7 (white)
317
7
Step 1) Equalize Image 1 to get Image 3:
D
m
= 7
A
0
= 10000
D
1
H
1
𝑆 = ∑ 𝐻
1
(𝑖)
𝐷
1
𝑖=0
D
m
/A
0
x
S
D
3
0
0
0
0
0
1
4016
4016
2.8
3
2
2954
6970
4.9
5
3
1630
8600
6.0
6
4
800
9400
6.6
7
5
368
9768
6.8
7
6
162
9930
6.95
7
7
70
10000
7
7
Step 2) Equalize Image 2 to get Image 3:
D
m
= 7
A
0
= 10000
D
2
H
2
𝑆 = ∑ 𝐻
2
(𝑖)
𝐷
2
𝑖=0
D
m
/
A
0
x
S
D
3
0
0
0
0
0
1
2470
2470
1.7
2
2
2536
5006
3.5
4
3
1953
6954
4.9
5
4
1337
8296
5.8
6
5
858
9154
6.4
6
6
529
9683
6.8
7
7
317
10000
7
7
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Student number:
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Step 3) Invert table in step 2
D
3
D
2
Special considerations
0
0
-
1
0
Selected value for
D
3
= 0 since this is closest lower value
2
1
-
3
1
Selected value for
D
3
= 2 since this is closest lower value
4
2
-
5
3
-
6
4
Selected lower of the matching values of
D
3
= 6
7
6
Selected lower of the matching values of
D
3
= 7
Note: The dash means no reason is given since no special consideration was required because a unique
(
D
2
,
D
3
) pair existed and inversion was possible without further considerations.
Step 4) Fill in Table V
D
1
D
2
0
0
1
1
2
3
3
4
4
6
5
6
6
6
7
6
This question is similar to that from the lectures on October 7 and
10 (slides 10-13). I had told the class that this would be tested
with different numerical values.
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3.
[2 marks]
An image has an isolated cluster of purely black pixels on a purely white uniform
background. The area of the cluster is 3 pixels.
What happens to the cluster when the image
is filtered with a median filter of size of 7
x
7? Explain why this happens.
Answer:
It is removed (i.e., replaced with purely white background pixels) because there
will be more purely white pixels inside the filter window than dark pixels, so the central
element of the window after sorting, which is the median value, will always be purely white.
4.
[3 marks]
In an industrial application, X-ray imaging is to be used to inspect the inside of
certain composite castings. The objective is to look for voids in the castings, which typically
appear as small blobs in the image. However, due to noise, inspection of the castings is
difficult, and the decision is made to use averaging of multiple images to reduce the noise and
thus improve contrast. After numerous experiments, it is found that an increase in the
amplitude
signal-to-noise ratio by a factor of 3 is sufficient to reduce the noise to acceptable
levels. If the imaging device can produce 30 images/sec, how long would the casting have to
be imaged to achieve the desired decrease in noise?
Answer:
Figure out number of images (
M
) that are needed using
√𝑀
=
?𝑁?
̅̅̅̅̅̅
?𝑁?
= 3
since we are told that
a factor of 3 increase is desired.
⸫
M
= 9
The time required
t
is then
t
=
9 images
30 images/sec
= 0.3 sec
For instructor use only.
GRADE:
out of 15.
TA marking:
Median filtering was covered on October 25 and 28.
This question simply involves applying median filter-
ing. I cannot make all questions identical to what I
have given before in lectures and practice problems,
but if you understood how to apply median filtering,
you should have been able to do this question.
Image averaging was covered on Oct 18. I did promise
the class that I would not ask you to regurgitate the
derivation as many of you did not recall probability the-
ory, and I stuck to that promise. I also described the
type of question I would ask to the class.
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