Digital Image Processing Unsharp Masking Highboost Filtering Emphasis Filtering Online Exam Quiz
Digital Image Processing Unsharp Masking Highboost Filtering Emphasis Filtering GK Quiz. Question and Answers related to Digital Image Processing Unsharp Masking Highboost Filtering Emphasis Filtering. MCQ (Multiple Choice Questions with answers about Digital Image Processing Unsharp Masking Highboost Filtering Emphasis Filtering
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?
Options
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
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)?
Options
A : Composite filter
B : M-derived filter
C : Constant k filter
D : None of the mentioned
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”?
Options
A : Emphasis filtering
B : Unsharp masking
C : Butterworth filtering
D : None of the mentioned
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 __________
Options
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
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?
Options
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
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?
Options
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)
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?
Options
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
Which of the following is/ are a generalized form of unsharp masking?
Options
A : Lowpass filtering
B : High-boost filtering
C : Emphasis filtering
D : All of the mentioned
The frequency domain Laplacian is closer to which of the following mask?
Options
A : Mask that excludes the diagonal neighbors
B : Mask that excludes neighbors in 4-adjacancy
C : Mask that excludes neighbors in 8-adjacancy
D : None of the mentioned
To accentuate the contribution to enhancement made by high-frequency components, which of the following method(s) should be more appropriate to apply?
Options
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
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