Category: Digital Image Processing Objective Questions

250+ TOP MCQs on Mathematical Tools in Digital Image Processing & Answers

Digital Image Processing Multiple Choice Questions on ” Mathematical Tools in Digital Image Processing”.

1. How is array operation carried out involving one or more images?
A. array by array
B. pixel by pixel
C. column by column
D. row by row
Answer: B
Clarification: Any array operation is carried out on a pixel by pixel basis.

2. The property indicating that the output of a linear operation due to the sum of two inputs is same as performing the operation on the inputs individually and then summing the results is called ___________
A. additivity
B. heterogeneity
C. homogeneity
D. None of the Mentioned
Answer: A
Clarification: This property is called additivity .

3. The property indicating that the output of a linear operation to a constant times as input is the same as the output of operation due to original input multiplied by that constant is called _________
A. additivity
B. heterogeneity
C. homogeneity
D. None of the Mentioned
Answer: C
Clarification: This property is called homogeneity .

4. Enhancement of differences between images is based on the principle of ____________
A. Additivity
B. Homogeneity
C. Subtraction
D. None of the Mentioned
Answer: C
Clarification: A frequent application of image subtraction is in the enhancement of differences between images .

5. A commercial use of Image Subtraction is ___________
A. Mask mode radiography
B. MRI scan
C. CT scan
D. None of the Mentioned
Answer: A
Clarification: Mask mode radiography is an important medical imaging area based on Image Subtraction.

6. Region of Interest (ROI) operations is commonly called as ___________
A. Shading correction
B. Masking
C. Dilation
D. None of the Mentioned
Answer: B
Clarification: A common use of image multiplication is Masking, also called ROI operation.

7. If every element of a set A is also an element of a set B, then A is said to be a _________ of set B.
A. Disjoint set
B. Union
C. Subset
D. Complement set
Answer: C
Clarification: A is called the subset of B.

8. Consider two regions A and B composed of foreground pixels. The ________ of these two sets is the set of elements belonging to set A or set B or both.
A. OR
B. AND
C. NOT
D. XOR
Answer: A
Clarification: This is called an OR operation.

9. Imaging systems having physical artefacts embedded in the imaging sensors produce a set of points called __________
A. Tie Points
B. Control Points
C. Reseau Marks
D. None of the Mentioned
Answer: C
Clarification: These points are called “known” points or “Reseau marks”.

10. Image processing approaches operating directly on pixels of input image work directly in ____________
A. Transform domain
B. Spatial domain
C. Inverse transformation
D. None of the Mentioned
Answer: B
Clarification: Operations directly on pixels of input image work directly in Spatial Domain.

250+ TOP MCQs on Unsharp Masking, High-boost filtering and Emphasis Filtering & Answers

Digital Image Processing Multiple Choice Questions 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
B. Unsharp masking
C. Butterworth filtering
D. None of the mentioned
Answer: B
Clarification: 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
Clarification: 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
Answer: A
Clarification: 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.

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
Answer: C
Clarification: 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)
Answer: D
Clarification: 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
Answer: A
Clarification: 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
Answer: D
Clarification: 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
B. Mask that excludes neighbors in 4-adjacancy
C. Mask that excludes neighbors in 8-adjacancy
D. None of the mentioned
Answer: A
Clarification: 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
Answer: C
Clarification: 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 _______
A. Unsharp masking
B. High-boost filtering
C. High frequency emphasis
D. None of the mentioned
Answer: C
Clarification: 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
Answer: D
Clarification: 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
Answer: D
Clarification: 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
Answer: A
Clarification: 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__________
A. Unsharp masking
B. High-boost filtering
C. Emphasized filtering
D. None of the mentioned
Answer: C
Clarification: 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
Answer: A
Clarification: Because of High frequency emphasis the gray-level tonality due to low frequency components is not lost.

250+ TOP MCQs on Smoothing Spatial Filters & Answers

Digital Image Processing Multiple Choice Questions on “Smoothing Spatial Filters”.

1. Noise reduction is obtained by blurring the image using smoothing filter.
A. True
B. False

Answer: A
Clarification: Noise reduction is obtained by blurring the image using smoothing filter. Blurring is used in pre-processing steps, such as removal of small details from an image prior to object extraction and, bridging of small gaps in lines or curves.

2. What is the output of a smoothing, linear spatial filter?
A. Median of pixels
B. Maximum of pixels
C. Minimum of pixels
D. Average of pixels

Answer: D
Clarification: The output or response of a smoothing, linear spatial filter is simply the average of the pixels contained in the neighbourhood of the filter mask.

3. Smoothing linear filter is also known as median filter.
A. True
B. False

Answer: B
Clarification: Since the smoothing spatial filter performs the average of the pixels, it is also called as averaging filter.

4. Which of the following in an image can be removed by using smoothing filter?
A. Smooth transitions of gray levels
B. Smooth transitions of brightness levels
C. Sharp transitions of gray levels
D. Sharp transitions of brightness levels

Answer: C
Clarification: Smoothing filter replaces the value of every pixel in an image by the average value of the gray levels. So, this helps in removing the sharp transitions in the gray levels between the pixels. This is done because, random noise typically consists of sharp transitions in gray levels.

5. Which of the following is the disadvantage of using smoothing filter?
A. Blur edges
B. Blur inner pixels
C. Remove sharp transitions
D. Sharp edges

Answer: A
Clarification: Edges, which almost always are desirable features of an image, also are characterized by sharp transitions in gray level. So, averaging filters have an undesirable side effect that they blur these edges.

6. Smoothing spatial filters doesn’t smooth the false contours.
A. True
B. False

Answer: B
Clarification: One of the application of smoothing spatial filters is that, they help in smoothing the false contours that result from using an insufficient number of gray levels.

7. The mask shown in the figure below belongs to which type of filter?

A. Sharpening spatial filter
B. Median filter
C. Sharpening frequency filter
D. Smoothing spatial filter

Answer: D
Clarification: This is a smoothing spatial filter. This mask yields a so called weighted average, which means that different pixels are multiplied with different coefficient values. This helps in giving much importance to the some pixels at the expense of others.

8. The mask shown in the figure below belongs to which type of filter?

A. Sharpening spatial filter
B. Median filter
C. Smoothing spatial filter
D. Sharpening frequency filter

Answer: C
Clarification: The mask shown in the figure represents a 3×3 smoothing filter. Use of this filter yields the standard average of the pixels under the mask.

9. Box filter is a type of smoothing filter.
A. True
B. False

Answer: A
Clarification: A spatial averaging filter or spatial smoothening filter in which all the coefficients are equal is also called as box filter.

10. If the size of the averaging filter used to smooth the original image to first image is 9, then what would be the size of the averaging filter used in smoothing the same original picture to second in second image?

A. 3
B. 5
C. 9
D. 15

Answer: D
Clarification: We know that, as the size of the filter used in smoothening the original image that is averaging filter increases then the blurring of the image. Since the second image is more blurred than the first image, the window size should be more than 9.

11. Which of the following comes under the application of image blurring?
A. Object detection
B. Gross representation
C. Object motion
D. Image segmentation

Answer: B
Clarification: An important application of spatial averaging is to blur an image for the purpose of getting a gross representation of interested objects, such that the intensity of the small objects blends with the background and large objects become easy to detect.c

12. Which of the following filters response is based on ranking of pixels?
A. Nonlinear smoothing filters
B. Linear smoothing filters
C. Sharpening filters
D. Geometric mean filter

Answer: A
Clarification: Order static filters are nonlinear smoothing spatial filters whose response is based on the ordering or ranking the pixels contained in the image area encompassed by the filter, and then replacing the value of the central pixel with the value determined by the ranking result.

13. Median filter belongs to which category of filters?
A. Linear spatial filter
B. Frequency domain filter
C. Order static filter
D. Sharpening filter

Answer: C
Clarification: The median filter belongs to order static filters, which, as the name implies, replaces the value of the pixel by the median of the gray levels that are present in the neighbourhood of the pixels.

14. Median filters are effective in the presence of impulse noise.
A. True
B. False

Answer: A
Clarification: Median filters are used to remove impulse noises, also called as salt-and-pepper noise because of its appearance as white and black dots in the image.

15. What is the maximum area of the cluster that can be eliminated by using an n×n median filter?
A. n2
B. n2/2
C. 2*n2
D. n

Answer: B
Clarification: Isolated clusters of pixels that are light or dark with respect to their neighbours, and whose area is less than n2/2, i.e., half the area of the filter, can be eliminated by using an n×n median filter.

250+ TOP MCQs on Homomorphic filtering & Answers

Digital Image Processing Questions and Answers for Campus interviews on “Homomorphic filtering-2”.

1. Which of the following fact is true for a image?
A. An image is the addition of illumination and reflectance component
B. An image is the subtraction of illumination component from reflectance component
C. An image is the subtraction of reflectance component from illumination component
D. An image is the multiplication of illumination and reflectance component

Answer: D
Clarification: An image is expressed as the multiplication of illumination and reflectance component.

2. If an image is expressed as the multiplication of illumination and reflectance component i.e. f(x, y)= i(x, y) * r(x, y), then Validate the statement “We can directly use the equation f(x, y)= i(x, y) * r(x, y) to operate separately on the frequency component of illumination and reflectance” .
A. True
B. False

Answer: B
Clarification: For an image is expressed as the multiplication of illumination and reflectance component i.e. f(x, y)= i(x, y) * r(x, y), the equation can’t be used directly to operate separately on the frequency component of illumination and reflectance because the Fourier transform of the product of two function is not separable.

3. In Homomorphic filtering which of the following operations is used to convert input image to discrete Fourier transformed function?
A. Logarithmic operation
B. Exponential operation
C. Negative transformation
D. None of the mentioned

Answer: A
Clarification: For an image is expressed as the multiplication of illumination and reflectance component i.e. f(x, y) = i(x, y) * r(x, y), the equation can’t be used directly to operate separately on the frequency component of illumination and reflectance because the Fourier transform of the product of two function is not separable. So, the logarithmic operation is used.I{z(x,y)}=I{ln⁡(f(x,y)) }=I{ln⁡(i(x,y)) }+I{ln⁡(r(x,y))}.

4. A class of system that achieves the separation of illumination and reflectance component of an image is termed as __________
A. Base class system
B. Homomorphic system
C. Base separation system
D. All of the mentioned

Answer: B
Clarification: Homomorphic system is a class of system that achieves the separation of illumination and reflectance component of an image.

5. Which of the following image component is characterized by a slow spatial variation?
A. Illumination component
B. Reflectance component
C. All of the mentioned
D. None of the mentioned

Answer: A
Clarification: The illumination component of an image is characterized by a slow spatial variation.

6. Which of the following image component varies abruptly particularly at the junction of dissimilar objects?
A. Illumination component
B. Reflectance component
C. All of the mentioned
D. None of the mentioned

Answer: B
Clarification: The reflectance component of an image varies abruptly particularly at the junction of dissimilar objects.

7. The reflectance component of an image varies abruptly particularly at the junction of dissimilar objects. The characteristic lead to associate illumination with __________
A. The low frequency of Fourier transform of logarithm of the image
B. The high frequency of Fourier transform of logarithm of the image
C. All of the mentioned
D. None of the mentioned

Answer: B
Clarification: The reflectance component of an image varies abruptly, so, is associated with the high frequency of Fourier transform of logarithm of the image.

8. The illumination component of an image is characterized by a slow spatial variation. The characteristic lead to associate illumination with __________
A. The low frequency of Fourier transform of logarithm of the image
B. The high frequency of Fourier transform of logarithm of the image
C. All of the mentioned
D. None of the mentioned

Answer: A
Clarification: The illumination component of an image is characterized by a slow spatial variation, so, is associated with the low frequency of Fourier transform of logarithm of the image.

9. If the contribution made by illumination component of image is decreased and the contribution of reflectance component is amplified, what will be the net result?
A. Dynamic range compression
B. Contrast enhancement
C. All of the mentioned
D. None of the mentioned

Answer: C
Clarification: The illumination component of an image is characterized by a slow spatial variation and the reflectance component of an image varies abruptly particularly at the junction of dissimilar objects, so, if the contribution made by illumination component of image is decreased and the contribution of reflectance component is amplified then there is simultaneous dynamic range compression and contrast stretching.

250+ TOP MCQs on Basic Intensity Transformation Functions & Answers

Digital Image Processing Interview Questions and Answers on “Basic Intensity Transformation Functions”.

1. Which of the following expression is used to denote spatial domain process?
A. g(x,y)=T[f(x,y)]
B. f(x+y)=T[g(x+y)]
C. g(xy)=T[f(xy)]
D. g(x-y)=T[f(x-y)]
Answer: A
Clarification: Spatial domain processes will be denoted by the expression g(x,y)=T[f(x,y)], where f(x,y) is the input image, g(x,y) is the processed image, and T is an operator on f, defined over some neighborhood of (x, y). In addition, T can operate on a set of input images, such as performing the pixel-by-pixel sum of K images for noise reduction.

2. Which of the following shows three basic types of functions used frequently for image enhancement?
A. Linear, logarithmic and inverse law
B. Power law, logarithmic and inverse law
C. Linear, logarithmic and power law
D. Linear, exponential and inverse law
Answer: B
Clarification: In introduction to gray-level transformations, which shows three basic types of functions used frequently for image enhancement: linear (negative and identity transformations), logarithmic (log and inverse-log transformations), and power-law (nth power and nth root transformations).The identity function is the trivial case in which output intensities are identical to input intensities. It is included in the graph only for completeness.

3. Which expression is obtained by performing the negative transformation on the negative of an image with gray levels in the range[0,L-1] ?
A. s=L+1-r
B. s=L+1+r
C. s=L-1-r
D. s=L-1+r
Answer: C
Clarification: The negative of an image with gray levels in the range[0,L-1] is obtained by using the negative transformation, which is given by the expression: s=L-1-r.

4. What is the general form of representation of log transformation?
A. s=clog10(1/r)
B. s=clog10(1+r)
C. s=clog10(1*r)
D. s=clog10(1-r)
Answer: B
Clarification: The general form of the log transformation: s=clog10(1+r), where c is a constant, and it is assumed that r ≥ 0.

5. What is the general form of representation of power transformation?
A. s=crγ
B. c=srγ
C. s=rc
D. s=rcγ
Answer: A
Clarification: Power-law transformations have the basic form: s=crγ where c and g are positive constants. Sometimes s=crγ is written as s=c.(r+ε)γ to account for an offset (that is, a measurable output when the input is zero).

6. What is the name of process used to correct the power-law response phenomena?
A. Beta correction
B. Alpha correction
C. Gamma correction
D. Pie correction
Answer: C
Clarification: A variety of devices used for image capture, printing, and display respond according to a power law. By convention, the exponent in the power-law equation is referred to as gamma .The process used to correct these power-law response phenomena is called gamma correction.

7. Which of the following transformation function requires much information to be specified at the time of input?
A. Log transformation
B. Power transformation
C. Piece-wise transformation
D. Linear transformation
Answer: C
Clarification: The practical implementation of some important transformations can be formulated only as piecewise functions. The principal disadvantage of piecewise functions is that their specification requires considerably more user input.

8. In contrast stretching, if r1=s1 and r2=s2 then which of the following is true?
A. The transformation is not a linear function that produces no changes in gray levels
B. The transformation is a linear function that produces no changes in gray levels
C. The transformation is a linear function that produces changes in gray levels
D. The transformation is not a linear function that produces changes in gray levels
Answer: B
Clarification: The locations of points (r1,s1) and (r2,s2) control the shape of the transformation function. If r1=s1 and r2=s2 then the transformation is a linear function that produces no changes in gray levels.

9. In contrast stretching, if r1=r2, s1=0 and s2=L-1 then which of the following is true?
A. The transformation becomes a thresholding function that creates an octal image
B. The transformation becomes a override function that creates an octal image
C. The transformation becomes a thresholding function that creates a binary image
D. The transformation becomes a thresholding function that do not create an octal image
Answer: C
Clarification: If r1=r2, s1=0 and s2=L-1,the transformation becomes a thresholding function that creates a binary image.

10. In contrast stretching, if r1≤r2 and s1≤s2 then which of the following is true?
A. The transformation function is double valued and exponentially increasing
B. The transformation function is double valued and monotonically increasing
C. The transformation function is single valued and exponentially increasing
D. The transformation function is single valued and monotonically increasing
Answer: D
Clarification: The locations of points (r1,s1) and (r2,s2) control the shape of the transformation function. If r1≤r2 and s1≤s2 then the function is single valued and monotonically increasing.

11. In which type of slicing, highlighting a specific range of gray levels in an image often is desired?
A. Gray-level slicing
B. Bit-plane slicing
C. Contrast stretching
D. Byte-level slicing
Answer: A
Clarification: Highlighting a specific range of gray levels in an image often is desired in gray-level slicing. Applications include enhancing features such as masses of water in satellite imagery and enhancing flaws in X-ray images.

12. Which of the following depicts the main functionality of the Bit-plane slicing?
A. Highlighting a specific range of gray levels in an image
B. Highlighting the contribution made to total image appearance by specific bits
C. Highlighting the contribution made to total image appearance by specific byte
D. Highlighting the contribution made to total image appearance by specific pixels
Answer: B
Clarification: Instead of highlighting gray-level ranges, highlighting the contribution made to total image appearance by specific bits might be desired. Suppose , each pixel in an image is represented by 8 bits. Imagine that the image is composed of eight 1-bit planes, ranging from bit-plane 0 for the least significant bit to bit-plane 7 for the most significant bit. In terms of 8-bit bytes, plane 0 contains all the lowest order bits in the bytes comprising the pixels in the image and plane 7 contains all the high-order bits.

250+ TOP MCQs on Intensity Transformation Functions & Answers

Digital Image Processing Multiple Choice Questions on “Intensity Transformation Functions”.

1. How is negative of an image obtained with intensity levels [0,L-1] with “r” and “s” being pixel values?
A. s = L – 1 + r
B. s = L – 1 – r
C. s = L + 1 + r
D. s = L + 1 + r
Answer: B
Clarification: The negative is obtained using s = L – 1 + r.

2. The general form of log transformations is ____________________
A. s = c.log(1 + r)
B. s = c+log(1 + r)
C. s = c.log(1 – r)
D. s = c-log(1 – r)
Answer: A
Clarification: s = c.log(1 + r) is the log transformation.

3. Power-law transformations has the basic form of ________________ where c and ∆ are constants.
A. s = c + r∆
B. s = c – r∆
C. s = c * r∆
D. s = c / r.∆
Answer: C
Clarification: s = c * r∆ is called the Power-law transformation.

4. For what value of the output must the Power-law transformation account for offset?
A. No offset needed
B. All values
C. One
D. Zero
Answer: D
Clarification: When the output is Zero, an offset is necessary.

5. What is Gamma Correction?
A. A Power-law response phenomenon
B. Inverted Intensity curve
C. Light brightness variation
D. None of the Mentioned
Answer: A
Clarification: The exponent in Power-law is called gamma and the process used to correct the response of Power-law transformation is called Gamma Correction.

6. Which process expands the range of intensity levels in an image so that it spans the full intensity range of the display?
A. Shading correction
B. Contrast sketching
C. Gamma correction
D. None of the Mentioned
Answer: B
Clarification: Contrast sketching is the process used to expand intensity levels in an image.

7. Highlighting a specific range of intensities of an image is called _______________
A. Intensity Matching
B. Intensity Highlighting
C. Intensity Slicing
D. None of the Mentioned
Answer: C
Clarification: Highlighting a specific range of intensities of an image is called Intensity Slicing.

8. Highlighting the contribution made to total image by specific bits instead of highlighting intensity-level changes is called ____________________
A. Intensity Highlighting
B. Byte-Slicing
C. Bit-plane slicing
D. None of the Mentioned
Answer: C
Clarification: It is called Bit-plane slicing.

9. Which of the following involves reversing the intensity levels of an image?
A. Log Transformations
B. Piecewise Linear Transformations
C. Image Negatives
D. None of the Mentioned
Answer: C
Clarification: Image negatives use reversing intensity levels.

10. Piecewise Linear Transformation function involves which of the following?
A. Bit-plane slicing
B. Intensity level slicing
C. Contrast stretching
D. All of the Mentioned
Answer: D
Clarification: Piecewise Linear Transformation function involves all the mentioned functions.