Unsharp Masking MCQ [Free PDF] - Objective Question Answer for Unsharp Masking Quiz

1. In frequency domain terminology, which of the following is defined as “obtaining a highpass filtered image by subtracting from the given image a lowpass filtered version of itself”?

A. Emphasis filtering
B. Unsharp masking
C. Butterworth filtering
D. None of the mentioned

Answer: B

In frequency domain terminology unsharp masking is defined as “obtaining a highpass filtered image by subtracting from the given image a lowpass filtered version of itself”.

2. Which of the following is/ are a generalized form of unsharp masking?

A. Lowpass filtering
B. High-boost filtering
C. Emphasis filtering
D. All of the mentioned

Answer: B

Unsharp masking is defined as “obtaining a highpass filtered image by subtracting from the given image a lowpass filtered version of itself” while high-boost filtering generalizes it by multiplying the input image by a constant, say A≥1.

3. High boost filtered image is expressed as f_hb = A f(x, y) – for(x, y), where f(x, y) is the input image, A is a constant, and f_lp(x, y) is the lowpass filtered version of f(x, y). Which of the following fact validates if A=1?

A. High-boost filtering reduces regular Highpass filtering
B. High-boost filtering reduces regular Lowpass filtering
C. All of the mentioned
D. None of the mentioned

Answer: A

High boost filtered image is modified as: f_hb = (A-1) f(x, y) +f(x, y) – f_lp(x, y)
i.e. f_hb = (A-1) f(x, y) + f_hp(x, y). So, when A=1, High-boost filtering reduces to regular Highpass filtering.

4. High boost filtered image is expressed as f_hb = A f(x, y) – f_lp(x, y), where f(x, y) is the input image, A is a constant and f_lp(x, y) is the lowpass filtered version of f(x, y). Which of the following fact(s) validates if A increases past 1?

A. The contribution of the image itself becomes more dominant
B. The contribution of the highpass filtered version of the image becomes less dominant
C. All of the mentioned
D. None of the mentioned

Answer: C

High boost filtered image is modified as: f_hb = (A-1) f(x, y) +f(x, y) – f_lp(x, y)
i.e. f_hb = (A-1) f(x, y) + f_hp(x, y). So, when A>1, the contribution of the image itself becomes more dominant over the highpass filtered version of the image.

5. If, F_hp(u, v)=F(u, v) – F_lp(u, v) and F_lp(u, v) = H_lp(u, v)F(u, v), where F(u, v) is the image in frequency domain with F_hp(u, v) its highpass filtered version, F_lp(u, v) its lowpass filtered component and H_lp(u, v) the transfer function of a lowpass filter. Then, unsharp masking can be implemented directly in the frequency domain by using a filter. Which of the following is the required filter?

A. H_hp(u, v) = H_lp(u, v)
B. H_hp(u, v) = 1 + H_lp(u, v)
C. H_hp(u, v) = – H_lp(u, v)
D. H_hp(u, v) = 1 – H_lp(u, v)

Answer: D

Unsharp masking can be implemented directly in the frequency domain by using a composite filter: H_hp(u, v) = 1 – H_lp(u, v).

6. Unsharp masking can be implemented directly in the frequency domain by using a filter: H_hp(u, v) = 1 – H_lp(u, v), where H_lp(u, v) the transfer function of a lowpass filter. What kind of filter is H_hp(u, v)?

A. Composite filter
B. M-derived filter
C. Constant k filter
D. None of the mentioned

Answer: A

Unsharp masking can be implemented directly in the frequency domain by using a composite filter: H_hp(u, v) = 1 – H_lp(u, v).

7. If unsharp masking can be implemented directly in the frequency domain by using a composite filter: H_hp(u, v) = 1 – H_lp(u, v), where H_lp(u, v) the transfer function of a lowpass filter. Then, the composite filter for High-boost filtering is __________

A. H_hb(u, v) = 1 – H_hp(u, v)
B. H_hb(u, v) = 1 + H_hp(u, v)
C. H_hb(u, v) = (A-1) – H_hp(u, v), A is a constant
D. H_hb(u, v) = (A-1) + H_hp(u, v), A is a constant

Answer: D

For given composite filter of unsharp masking H_hp(u, v) = 1 – H_lp(u, v), the composite filter for High-boost filtering is H_hb(u, v) = (A-1) + H_hp(u, v).

8. The frequency domain Laplacian is closer to which of the following mask?

A. Mask that excludes the diagonal neighbors
B. Mask that excludes neighbors in 4-adjacency
C. Mask that excludes neighbors in 8-adjacency
D. None of the mentioned

Answer: A

The frequency-domain Laplacian is closer to the mask that excludes the diagonal neighbors.

9. To accentuate the contribution to enhancement made by high-frequency components, which of the following method(s) should be more appropriate to apply?

A. Multiply the highpass filter by a constant
B. Add an offset to the highpass filter to prevent eliminating zero frequency term by filter
C. All of the mentioned combined and applied
D. None of the mentioned

Answer: C

To accentuate the contribution to enhancement made by high-frequency components, we have to multiply the highpass filter by a constant and add an offset to the highpass filter to prevent eliminating zero frequency term by the filter.

10. A process that accentuates the contribution to enhancement made by high-frequency components, by multiplying the highpass filter by a constant and adding an offset to the highpass filter to prevent eliminating zero frequency term by the filter is known as _______

A. Unsharp masking
B. High-boost filtering
C. High-frequency emphasis
D. None of the mentioned

Answer: C

High-frequency emphasis is the method that accentuates the contribution to enhancement made by high-frequency components. In this, we multiply the highpass filter by a constant and add an offset to the highpass filter to prevent eliminating zero frequency term by the filter.

11. Which of the following is a transfer function of High-frequency emphasis {H_hfe(u, v)} for H_hp(u, v) being the highpass filtered version of an image?

A. H_hfe(u, v) = 1 – H_hp(u, v)
B. H_hfe(u, v) = a – H_hp(u, v), a≥0
C. H_hfe(u, v) = 1 – b H_hp(u, v), a≥0 and b>a
D. H_hfe(u, v) = a + b H_hp(u, v), a≥0 and b>a

Answer: D

The transfer function of High frequency emphasis is given as:H_hfe(u, v) = a + b H_hp(u, v), a≥0 and b>a.

12. The transfer function of High-frequency emphasis is given as H_hfe(u, v) = a + b H_hp(u, v), for H_hp(u, v) being the highpass filtered version of the image, a≥0 and b>a. for certain values of a and b it reduces to High-boost filtering. Which of the following is the required value?

A. a = (A-1) and b = 0,A is some constant
B. a = 0 and b = (A-1),A is some constant
C. a = 1 and b = 1
D. a = (A-1) and b =1,A is some constant

Answer: D

The transfer function of High frequency emphasis is given as: H_hfe(u, v) = a + b H_hp(u, v) and the transfer function for High-boost filtering is H_hb(u, v) = (A-1) + H_hp(u, v), A being some constant. So, for a = (A-1) and b =1, H_hfe(u, v) = H_hb(u, v).

13. The transfer function of High frequency emphasis is given as: H_hfe(u, v) = a + b H_hp(u, v), for H_hp(u, v) being the highpass filtered version of image,a≥0 and b>a. What happens when b increases past 1?

A. The high frequency is emphasized
B. The low frequency is emphasized
C. All frequencies are emphasized
D. None of the mentioned

Answer: A

The transfer function of High frequency emphasis is given as: H_hfe(u, v) = a + b H_hp(u, v), for H_hp(u, v) being the highpass filtered version of image,a≥0 and b>a. When b increases past 1, the high frequency is emphasized.

14. The transfer function of High frequency emphasis is given as: H_hfe(u, v) = a + b H_hp(u, v), for H_hp(u, v) being the highpass filtered version of image,a≥0 and b>a. When b increases past 1 the filtering process is specifically termed as__________

A. Unsharp masking
B. High-boost filtering
C. Emphasized filtering
D. None of the mentioned

Answer: C

15. Validate the statement “Because of High-frequency emphasis the gray-level tonality due to low-frequency components is not lost”.

A. True
B. False

Answer: A

Because of High-frequency emphasis, the gray-level tonality due to low-frequency components is not lost.

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