Smoothing Frequency Domain Filters MCQ [Free PDF] - Objective Question Answer for Smoothing Frequency Domain Filters Quiz

11. Using the feature of the reciprocal relationship of filter in the spatial domain and corresponding filter in frequency domain along with convolution, which of the following fact is true?

A. The narrower the frequency domain filter more severe is the ringing
B. The wider the frequency domain filter more severe is the ringing
C. The narrower the frequency domain filter less severe is the ringing
D. None of the mentioned

Answer: A

The characteristics feature of reciprocal relationship says that the narrower the frequency domain filter becomes it attenuates more low-frequency components and so increases blurring and more severe becomes the ringing.

12. State the statement true or false: “BLPF has sharp discontinuity and ILPF doesn’t, and so ILPF establishes a clear cutoff b/w passed and filtered frequencies”.

A. True
B. False

Answer: B

ILPF has sharp discontinuity and BLPF doesn’t, so BLPF establishes a clear cutoff b/w passed and filtered frequencies.

13. A Butterworth filter of what order has no ringing?

A. 1
B. 2
C. 3
D. 4

Answer: A

A Butterworth filter of order 1 has no ringing and ringing exists for order 2 although is imperceptible. A Butterworth filter of higher-order shows a significant factor of ringing.

14. 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”.

15. 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.

16. 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.

17. 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.

18. 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).

19. 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).

20. 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).

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