Explore topic-wise MCQs in Digital Image Processing.

This section includes 13 Mcqs, each offering curated multiple-choice questions to sharpen your Digital Image Processing knowledge and support exam preparation. Choose a topic below to get started.

1.

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

A. tTrue
B. tFalse
Answer» B. tFalse
2.

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__________

A. tUnsharp masking
B. tHigh-boost filtering
C. tEmphasized filtering
D. tNone of the mentioned
Answer» D. tNone of the mentioned
3.

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. tThe high frequency are emphasized
B. tThe low frequency are emphasized
C. tAll frequency are emphasized
D. tNone of the mentioned
Answer» B. tThe low frequency are emphasized
4.

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. ta = (A-1) and b = 0,A is some constant
B. ta = 0 and b = (A-1),A is some constant
C. ta = 1 and b = 1
D. ta = (A-1) and b =1,A is some constant
Answer» E.
5.

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. tH<sub>hfe</sub>(u, v) = 1 H<sub>hp</sub>(u, v)
B. tH<sub>hfe</sub>(u, v) = a H<sub>hp</sub>(u, v), a 0
C. tH<sub>hfe</sub>(u, v) = 1 b H<sub>hp</sub>(u, v), a 0 and b&gt;a
D. tH<sub>hfe</sub>(u, v) = a + b H<sub>hp</sub>(u, v), a 0 and b&gt;a
Answer» E.
6.

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 _______

A. tUnsharp masking
B. tHigh-boost filtering
C. tHigh frequency emphasis
D. tNone of the mentioned
Answer» D. tNone of the mentioned
7.

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

A. tMultiply the highpass filter by a constant
B. tAdd an offset to the highpass filter to prevent eliminating zero frequency term by filter
C. tAll of the mentioned combined and applied
D. tNone of the mentioned
Answer» D. tNone of the mentioned
8.

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. tH<sub>hb</sub>(u, v) = 1 H<sub>hp</sub>(u, v)
B. tH<sub>hb</sub>(u, v) = 1 + H<sub>hp</sub>(u, v)
C. tH<sub>hb</sub>(u, v) = (A-1) H<sub>hp</sub>(u, v), A is a constant
D. tH<sub>hb</sub>(u, v) = (A-1) + H<sub>hp</sub>(u, v), A is a constant
Answer» E.
9.

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. tComposite filter
B. tM-derived filter
C. tConstant k filter
D. tNone of the mentioned
Answer» B. tM-derived filter
10.

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. tH<sub>hp</sub>(u, v) = H<sub>lp</sub>(u, v)
B. tH<sub>hp</sub>(u, v) = 1 + H<sub>lp</sub>(u, v)
C. tH<sub>hp</sub>(u, v) = H<sub>lp</sub>(u, v)
D. tH<sub>hp</sub>(u, v) = 1 H<sub>lp</sub>(u, v)
Answer» E.
11.

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. tThe contribution of the image itself becomes more dominant
B. tThe contribution of the highpass filtered version of image becomes less dominant
C. tAll of the mentioned
D. tNone of the mentioned
Answer» D. tNone of the mentioned
12.

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. tHigh-boost filtering reduces to regular Highpass filtering
B. tHigh-boost filtering reduces to regular Lowpass filtering
C. tAll of the mentioned
D. tNone of the mentioned
Answer» B. tHigh-boost filtering reduces to regular Lowpass filtering
13.

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. tEmphasis filtering
B. tUnsharp masking
C. tButterworth filtering
D. tNone of the mentioned
Answer» C. tButterworth filtering