# Digital Image Processing Questions and Answers – Unsharp Masking, High-boost filtering and Emphasis Filtering

«
»

This set of Digital Image Processing Multiple Choice Questions & Answers (MCQs) focuses on “Unsharp Masking, High-boost filtering and Emphasis Filtering”.

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
c) Butterworth filtering
d) None of the mentioned

Explanation: 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

Explanation: 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: fhb = A f(x, y) – flp(x, y), where f(x, y) the input image, A is a constant and flp(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 to regular Highpass filtering
b) High-boost filtering reduces to regular Lowpass filtering
c) All of the mentioned
d) None of the mentioned

Explanation: High boost filtered image is modified as: fhb = (A-1) f(x, y) +f(x, y) – flp(x, y)
i.e. fhb = (A-1) f(x, y) + fhp(x, y). So, when A=1, High-boost filtering reduces to regular Highpass filtering.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. High boost filtered image is expressed as: fhb = A f(x, y) – flp(x, y), where f(x, y) the input image, A is a constant and flp(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 image becomes less dominant
c) All of the mentioned
d) None of the mentioned

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

5. If, Fhp(u, v)=F(u, v) – Flp(u, v) and Flp(u, v) = Hlp(u, v)F(u, v), where F(u, v) is the image in frequency domain with Fhp(u, v) its highpass filtered version, Flp(u, v) its lowpass filtered component and Hlp(u, v) the transfer function of a lowpass filter. Then, unsharp masking can be implemented directly in frequency domain by using a filter. Which of the following is the required filter?
a) Hhp(u, v) = Hlp(u, v)
b) Hhp(u, v) = 1 + Hlp(u, v)
c) Hhp(u, v) = – Hlp(u, v)
d) Hhp(u, v) = 1 – Hlp(u, v)

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

6. Unsharp masking can be implemented directly in frequency domain by using a filter: Hhp(u, v) = 1 – Hlp(u, v), where Hlp(u, v) the transfer function of a lowpass filter. What kind of filter is Hhp(u, v)?
a) Composite filter
b) M-derived filter
c) Constant k filter
d) None of the mentioned

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

7. If unsharp masking can be implemented directly in frequency domain by using a composite filter: Hhp(u, v) = 1 – Hlp(u, v), where Hlp(u, v) the transfer function of a lowpass filter. Then, the composite filter for High-boost filtering is __________
a) Hhb(u, v) = 1 – Hhp(u, v)
b) Hhb(u, v) = 1 + Hhp(u, v)
c) Hhb(u, v) = (A-1) – Hhp(u, v), A is a constant
d) Hhb(u, v) = (A-1) + Hhp(u, v), A is a constant

Explanation: For given composite filter of unsharp masking Hhp(u, v) = 1 – Hlp(u, v), the composite filter for High-boost filtering is Hhb(u, v) = (A-1) + Hhp(u, v).

8. The frequency domain Laplacian is closer to which of the following mask?
a) Mask that excludes the diagonal neighbors
d) None of the mentioned

Explanation: The frequency domain Laplacian is closer to 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

Explanation: 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 filter.

10. A process that accentuate 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 filter is known as _______
b) High-boost filtering
c) High frequency emphasis
d) None of the mentioned

Explanation: High frequency emphasis is the method that accentuate the contribution to enhancement made by high-frequency component. 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 filter.

11. Which of the following a transfer function of High frequency emphasis {Hhfe(u, v)} for Hhp(u, v) being the highpass filtered version of image?
a) Hhfe(u, v) = 1 – Hhp(u, v)
b) Hhfe(u, v) = a – Hhp(u, v), a≥0
c) Hhfe(u, v) = 1 – b Hhp(u, v), a≥0 and b>a
d) Hhfe(u, v) = a + b Hhp(u, v), a≥0 and b>a

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

12. The transfer function of High frequency emphasis is given as: Hhfe(u, v) = a + b Hhp(u, v), for Hhp(u, v) being the highpass filtered version of 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

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

13. The transfer function of High frequency emphasis is given as: Hhfe(u, v) = a + b Hhp(u, v), for Hhp(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 are emphasized
b) The low frequency are emphasized
c) All frequency are emphasized
d) None of the mentioned

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

14. The transfer function of High frequency emphasis is given as: Hhfe(u, v) = a + b Hhp(u, v), for Hhp(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__________
b) High-boost filtering
c) Emphasized filtering
d) None of the mentioned

Explanation: The transfer function of High frequency emphasis is given as: Hhfe(u, v) = a + b Hhp(u, v), for Hhp(u, v) being the highpass filtered version of image,a≥0 and b>a. When b increases past 1, the high frequency are emphasized and so the filtering process is better known as Emphasized filtering.

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

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

Sanfoundry Global Education & Learning Series – Digital Image Processing.

To practice all areas of Digital Image Processing, here is complete set of 1000+ Multiple Choice Questions and Answers. 