Digital Electronic MCQ Set 1
1. Which of the following is/are more commercially successful image enhancement method in mask mode radiography, an area under medical imaging?
a) Addition
b) Subtraction
c) Multiplication
d) Division
Answer
Answer: b [Reason:] In the given area of medical imaging, a mask of an X-ray image of a region of subject is captured using TV camera is subtracted from image of same region taken after injecting a contrast medium to the bloodstream. The subtraction result gives an enhanced detail of how a contrast medium propagates through the bloodstream.
This the best commercially successful method.
2. The subtraction operation results in areas that appear as dark shades of gray. Why?
a) Because the difference in such areas is little, that yields low value
b) Because the difference in such areas is high, that yields low value
c) Because the difference in such areas is high, that yields high value
d) None of the mentioned
Answer
Answer: a [Reason:] There remains a little change in some areas in the images to be subtracted that yields low value and so the result appears as dark shades of gray.
3. If the images are displayed using 8-bits, then, what is the range of the value of an image if the image is a result of subtraction operation?
a) 0 to 255
b) 0 to 511
c) -255 to 0
d) None of the mentioned
Answer
Answer: d [Reason:] The range of a result of a subtracted image is -255 m inimum to 255 max imum, if 8-bit channel is used to display the original images.
4. The subtracted image needs to be scaled, if 8-bit channel is used to display the subtracted images. So, the method of adding 255 to each pixel and then dividing by 2, has certain lim its. What is/are those lim its?
a) Very complex method
b) Very difficult to implement
c) The truncation inherent in division by 2 causes loss in accuracy
d) All of the mentioned
Answer
Answer: c [Reason:] The given method is quite simple and easy to implement, however it has the lim itation of accuracy loss because of truncation inherent in division by 2 and also that it doesn’t ensure the full range usage.
5. Which of the following is/are the fundamental factors that need tight control for difference based inspection work?
a) Proper registration
b) Controlled illum ination
c) Noise levels should be low enough so that the variation due to noise won’t affect the difference value much
d) All of the mentioned
Answer
Answer: d [Reason:] Proper Registration does special marking into the product in case two images are identical so as the difference won’t create any sense.
Controlled Illum ination is important because changes in illum ination can affect dramatically the difference image values.
Noise levels of a difference image must low enough so that the variation due to noise won’t affect the difference value much.
6. When can two random variables be uncorrelated?
a) Their covariance is 0
b) Their covariance is 1
c) Their covariance is -1
d) None of the mentioned
Answer
Answer: a [Reason:] The covariance of two random variables x i and x j given by: E [(x i – m i) (x j – mj)], E {.} is expected value of the argument and m is the mean. If this covariance turns out to 0, the variables are uncorrelated.
7. In Image Averaging enhancement method assumptions are made for a noisy image g(x, y). What is/are those?
a) The noise is correlated at every pair of coordinate (x, y)
b) The noise has average value 1 at every pair of coordinate (x, y)
c) All of the mentioned
d) None of the mentioned
Answer
Answer: d [Reason:] In Image Averaging enhancement method assumptions are made for a noisy image g(x, y) that at every coordinate (x, y) the noise has 0 average value and must be uncorrelated.
8. The standard deviation ‘σ’ at any point in image averaging: σḡ(x, y) = 1/√K σɳ(x, y), where ḡ(x, y) is the average image formed by averaging K different noisy images and ɳ(x, y) is the noise added to an original image f(x, y). What is the relation between K and the variability of the pixel values at each location (x, y)?
a) Increase in K, decreases the noise of pixel values
b) Increase in K, increases the noise of pixel values
c) Decrease in K, decreases the noise of pixel values
d) Decrease in K, increases the noise of pixel values
Answer
Answer: a [Reason:] As K increases, E {ḡ(x, y)} the expected value approaches f(x, y) the original image, i.e. decreasing the noise component.
Digital Electronic MCQ Set 2
1. Logic operations between two or more images are performed on pixel-by-pixel basis, except for one that is performed on a single image. Which one is that?
a) AND
b) OR
c) NOT
d) None of the mentioned
Answer
Answer: c [Reason:] Although all the Arithmetic/Logic operations between two or more images are performed on pixel-by-pixel basis, except for NOT that is performed on a single image.
2. Which of the following logical operator(s) is/are functionally complete?
a) AND
b) OR
c) NOT
d) All of the mentioned
Answer
Answer: d [Reason:] All the three logical operators given are functionally complete because all other logical operators can be implemented using these three.
3. While implementing logic operation on gray-scale images, the processing of pixel values is done as ____
a) String of integer numbers
b) String of floating numbers
c) String of binary numbers
d) None of the mentioned
Answer
Answer: c [Reason:] Logic operation on gray-scale images are done by processing of pixel values as string of binary numbers.
4. What is the equivalent for a black, 8-bit pixel to be processed under logic operation on gray scale image?
a) A string: 00000000
b) A string: 11111111
c) A string: 10000000
d) A string: 01111111
Answer
Answer: a [Reason:] Logic operation on gray-scale images are done by processing of pixel values as string of binary numbers, so, a black, 8-bit pixel is processed as a string of eight 0’s.
5. Which of the following operation(s) is/are equivalent to negative transformation?
a) AND
b) OR
c) NOT
d) All of the mentioned
Answer
Answer: c [Reason:] Applying NOT operator on a black, 8-bit pixel gives a white, 8-bit pixel, so, is equivalent to negative transformation.
6. Which of the following operations are used for masking?
a) AND, OR
b) AND, NOT
c) NOT, OR
d) All of the mentioned
Answer
Answer: c [Reason:] AND, OR operators are used for masking, while NOT works as negative transformation.
7. Two images having one pixel gray value 01010100 and 00000101 at the same location, are operated against AND operator. What would be the resultant pixel gray value at that location in the enhanced image?
a) 10100100
b) 11111011
c) 00000100
d) 01010101
Answer
Answer: c [Reason:] For AND operation results in 1 only for 1AND 1, else 0.
All the bits of the given gray value are operated similar resulting in 00000100.
8. Which of the following arithmetic operator is primarily used as a masking operator in enhancement?
a) Addition
b) Subtraction
c) Multiplication
d) Division
Answer
Answer: c [Reason:] Multiplication of one image by another is used as a gray-level mask.
Digital Electronic MCQ Set 3
1. The expression [∂2 f(x,y)/∂x2 +∂2 f(x,y)/∂y2] is considered as ___ where f(x, y) is an input image.
a) Laplacian of f(x, y)
b) Gradient of f(x, y)
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] The Laplacian for an image f(x, y) is defined as: ∇2 f=∂2 f/∂x2 + ∂2 f/∂y2 .
2. If the Laplacian in frequency domain is: 
where is the Fourier transform operator and F(u, v) is Fourier transformed function of f(x, y), then what is -(u2+ v2) is considered as?
a) Laplacian operation
b) Filtering operation
c) Shift operation
d) None of the mentioned
Answer
Answer: b [Reason:] The Laplacian in frequency domain is simply implemented by using filter:
H(u, v)= -(u2+ v2).
3. The Laplacian in frequency domain is simply implemented by using filter ____
a) H(u, v)= -(u2– v2)
b) H(u, v)= -(1)
c) H(u, v)= -(u2+ v2)
d) none of the mentioned
Answer
Answer: c [Reason:] Laplacian in frequency domain is: I[(∂2 f(x,y))/∂x2 +(∂2 f(x,y))/∂y2 ]= -(u2+v2)F(u,v), where ℑ is the Fourier transform operator and F(u, v) is Fourier transformed function of f(x, y) and -(u2+ v2) is the filter.
4. Assuming that the origin of F(u, v), Fourier transformed function of f(x, y) an input image, has been correlated by performing the operation f(x, y)(-1)x+y prior to taking the transform of the image. If F and f are of same size, then what does the given operation is/are supposed to do?
a) Resize the transform
b) Rotate the transform
c) Shifts the center transform
d) All of the mentioned
Answer
Answer: c [Reason:] The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0,0) is at point (M/2, N/2) for F and f of same size M*N.
5. Assuming that the origin of F(u, v), Fourier transformed function of f(x, y) an input image, has been correlated by performing the operation f(x, y)(-1)x+y prior to taking the transform of the image. If F and f are of same size M*N, where does the point (u, v) =(0,0) shifts?
a) (M -1, N -1)
b) (M/2, N/2)
c) (M+1, N+1)
d) (0, 0)
Answer
Answer: b [Reason:] The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0, 0) is at point (M/2, N/2) for F and f of same size M*N.
6. Assuming that the origin of F(u, v), Fourier transformed function of f(x, y) an input image, has been correlated by performing the operation f(x, y)(-1)x+y prior to taking the transform of the image. If F and f are of same size M*N, then which of the following is an expression for H(u, v), the filter used for implementing Laplacian in frequency domain?
a) H(u, v)= -(u2+ v2)
b) H(u, v)= -(u2– v2)
c) H(u, v)= -[(u – M/2)2+ (v – N/2)2].
d) H(u, v)= -[(u – M/2)2– (v – N/2)2].
Answer
Answer: c [Reason:] The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0, 0) is at point (M/2, N/2) and hence the filter is: H(u, v)= -[(u – M/2)2+ (v – N/2)2].
7. Computing the Fourier transform of the Laplacian result in spatial domain is equivalent to multiplying the F(u, v), Fourier transformed function of f(x, y) an input image, and H(u, v), the filter used for implementing Laplacian in frequency domain. This dual relationship is expressed as ___
a) Fourier transform pair notation
b) Laplacian
c) Gradient
d) None of the mentioned
Answer
Answer: a [Reason:] The Fourier transform of the Laplacian result in spatial domain is equivalent to multiplying the F(u, v) and H(u, v). This dual relationship is expressed as Fourier transform pair notation given by: ∇2 f(x,y)-[(u – M/2)2+ (v – N/2)2]F(u,v), for an image of size M *N.
8. Computing the Fourier transform of the Laplacian result in spatial domain is equivalent to multiplying the F(u, v), Fourier transformed function of f(x, y) an input image of size M*N, and H(u, v), the filter used for implementing Laplacian in frequency domain. This dual relationship is expressed as Fourier transform pair notation given by_______
a) ∇2 f(x,y)↔[(u –M/2)2+ (v –N/2)2]F(u,v)
b) ∇2 f(x,y)↔-[(u+M/2)2– (v+N/2)2]F(u,v)
c) ∇2 f(x,y)↔-[(u –M/2)2+ (v –N/2)2]F(u,v)
d) ∇2 f(x,y)↔[(u+M/2)2– (v+N/2)2]F(u,v)
Answer
Answer: c [Reason:] The Fourier transform of the Laplacian result in spatial domain is equivalent to multiplying the F(u, v) and H(u, v). This dual relationship is expressed as Fourier transform pair notation given by:∇2 f(x,y)↔-[(u – M/2)2+ (v – N/2)2]F(u,v), for an image of size M*N.
9. An enhanced image can be obtained as: g(x,y)=f(x,y)-∇2 f(x,y), where Laplacian is being subtracted from f(x, y) the input image. What does this conclude?
a) That the center spike would be negative
b) That the immediate neighbors of center spike would be positive.
c) All of the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] For the above given enhanced image the Laplacian subtraction suggest that the center coefficient of Laplacian mask is negative and so the center spike is negative with its immediate neighbors being positive.
10. An enhanced image can be obtained as: g(x,y)=f(x,y)-∇2 f(x,y), where Laplacian is being subtracted from f(x, y) the input image of size M*Non which an operation f(x, y)(-1)x+yis applied.Unlike enhancing in spatial domain with one single mask, it is possible to perform the same in frequency domain using one filter. Which of the following is/are the required filter(s)?
a) H(u, v)= -[1 + u2+ v2].
b) H(u, v)= -[(u – M/2)2+ (v– N/2)2].
c) H(u, v)= [1 + (u – M/2)2+ (v – N/2)2].
d) All of the mentioned
Answer
Answer: c [Reason:] The filter H(u, v)= [1 + (u – M/2)2+ (v – N/2)2] is used to perform the same enhancement in frequency domain like in spatial domain.
11. Why is scaling of Laplacian filtered images necessary?
a) Because it contain high positive values
b) Because it contain high negative value
c) Because it contain both positive and negative values
d) None of the mentioned
Answer
Answer: c [Reason:] A Laplacian filtered image contain both positive and negative values of comparable magnitudes. So, scaling is necessary.
12. Which of the following fact is true for the masks that includes diagonal neighbors than the masks that doesn’t?
a) Mask that excludes diagonal neighbors has more sharpness than the masks that doesn’t
b) Mask that includes diagonal neighbors has more sharpness than the masks that doesn’t
c) Both masks have same sharpness result
d) None of the mentioned
Answer
Answer: b [Reason:] Including diagonal neighbor pixels enhances sharpness of the image. So, Mask that includes diagonal neighbors has more sharpness than the masks that doesn’t.
Digital Electronic MCQ Set 4
1. Which of the following fact(s) is/are true for the relationship between low frequency component of Fourier transform and the rate of change of gray levels?
a) Moving away from the origin of transform the low frequency corresponds to smooth gray level variation
b) Moving away from the origin of transform the low frequencies corresponds to abrupt change in gray level
c) All of the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] Moving away from the origin of transform the low frequency corresponds to the slowly varying components in an image. Moving further away from origin the higher frequencies corresponds to faster gray level changes.
2. Which of the following fact(s) is/are true for the relationship between high frequency component of Fourier transform and the rate of change of gray levels?
a) Moving away from the origin of transform the high frequency corresponds to smooth gray level variation
b) Moving away from the origin of transform the higher frequencies corresponds to abrupt change in gray level
c) All of the mentioned
d) None of the mentioned
Answer
Answer: b [Reason:] Moving away from the origin of transform the low frequency corresponds to the slowly varying components in an image. Moving further away from origin the higher frequencies corresponds to faster gray level changes.
3. What is the name of the filter that multiplies two functions F(u, v) and H(u, v), where F has complex components too since is Fourier transformed function of f(x, y), in an order that each component of H multiplies both real and complex part of corresponding component in F?
a) Unsharp mask filter
b) High-boost filter
c) Zero-phase-shift-filter
d) None of the mentioned
Answer
Answer: c [Reason:] Zero-phase-shift-filter multiplies two functions F(u, v) and H(u, v), where F has complex components too since is Fourier transformed function of f(x, y), in an order that each component of H multiplies both real and complex part of corresponding component in F.
4. To set the average value of an image zero, which of the following term would be set 0 in the frequency domain and the inverse transformation is done, where F(u, v) is Fourier transformed function of f(x, y)?
a) F(0, 0)
b) F(0, 1)
c) F(1, 0)
d) None of the mentioned
Answer
Answer: a [Reason:] For an image f(x, y), the Fourier transform at origin of an image, F(0, 0), is equal to the average value of the image.
5. What is the name of the filter that is used to turn the average value of a processed image zero?
a) Unsharp mask filter
b) Notch filter
c) Zero-phase-shift-filter
d) None of the mentioned
Answer
Answer: b [Reason:] Notch filter sets F (0, 0), to zero, hence setting up the average value of image zero. The filter is named so, because it is a constant function with a notch at origin and so is able to set F (0, 0) to zero leaving out other values.
6. Which of the following filter(s) attenuates high frequency while passing low frequencies of an image?
a) Unsharp mask filter
b) Lowpass filter
c) Zero-phase-shift filter
d) All of the mentioned
Answer
Answer: b [Reason:] A lowpass filter attenuates high frequency while passing low frequencies.
7. Which of the following filter(s) attenuates low frequency while passing high frequencies of an image?
a) Unsharp mask filter
b) Highpass filter
c) Zero-phase-shift filter
d) All of the mentioned
Answer
Answer: b [Reason:] A highpass filter attenuates low frequency while passing high frequencies.
8. Which of the following filter have a less sharp detail than the original image because of attenuation of high frequencies?
a) Highpass filter
b) Lowpass filter
c) Zero-phase-shift filter
d) None of the mentioned
Answer
Answer: b [Reason:] A lowpass filter attenuates high so the image has less sharp details.
9. The feature(s) of a highpass filtered image is/are _____
a) Have less gray-level variation in smooth areas
b) Emphasized transitional gray-level details
c) An overall sharper image
d) All of the mentioned
Answer
Answer: d [Reason:] A highpass filter attenuates low frequency so have less gray-level variation in smooth areas, and allows high frequencies so have emphasized transitional gray-level details, resulting in a sharper image.
10. A spatial domain filter of the corresponding filter in frequency domain can be obtained by applying which of the following operation(s) on filter in frequency domain?
a) Fourier transform
b) Inverse Fourier transform
c) None of the mentioned
d) All of the mentioned
Answer
Answer: b [Reason:] Filters in spatial domain and frequency domain has a Fourier transform pair relation. A spatial domain filter of the corresponding filter in frequency domain can be obtained by applying inverse Fourier transform on frequency domain filter.
11. A frequency domain filter of the corresponding filter in spatial domain can be obtained by applying which of the following operation(s) on filter in spatial domain?
a) Fourier transform
b) Inverse Fourier transform
c) None of the mentioned
d) All of the mentioned
Answer
Answer: a [Reason:] Filters in spatial domain and frequency domain has a Fourier transform pair relation. A frequency domain filter of the corresponding filter in spatial domain can be obtained by applying inverse Fourier transform on spatial domain filter.
12. Which of the following filtering is done in frequency domain in correspondence to lowpass filtering in spatial domain?
a) Gaussian filtering
b) Unsharp mask filtering
c) High-boost filtering
d) None of the mentioned
Answer
Answer: a [Reason:] A plot of Gaussian filter in frequency domain can be recognized similar to lowpass filter in spatial domain.
13. Using the feature of reciprocal relationship of filter in spatial domain and corresponding filter in frequency domain, which of the following fact is true?
a) The narrower the frequency domain filter results in increased blurring
b) The wider the frequency domain filter results in increased blurring
c) The narrower the frequency domain filter results in decreased blurring
d) None of the mentioned
Answer
Answer: a [Reason:] The characteristics feature of reciprocal relationship says that the narrower the frequency domain filter becomes it attenuates more low frequency component and so increases blurring.
Digital Electronic MCQ Set 5
1. Smoothing in frequency domain is achieved by attenuating which of the following component in the transform of a given image?
a) Attenuating a range of high-frequency components
b) Attenuating a range of low-frequency components
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] Since, edges and sharp transitions contribute significantly to high-frequency contents in the gray level of an image. So, smoothing is done by attenuating a range of high-frequency components.
2. Which of the following is/are considered as type(s) of lowpass filters?
a) Ideal
b) Butterworth
c) Gaussian
d) All of the mentioned
Answer
Answer: d [Reason:] Lowpass filters are considered of three types: Ideal, Butterworth, and Gaussian.
3. Which of the following lowpass filters is/are covers the range of very sharp filter function?
a) Ideal lowpass filters
b) Butterworth lowpass filter
c) Gaussian lowpass filter
d) All of the mentioned
Answer
Answer: a [Reason:] Ideal lowpass filter covers the range of very sharp filter functioning of lowpass filters.
4. Which of the following lowpass filters is/are covers the range of very smooth filter function?
a) Ideal lowpass filters
b) Butterworth lowpass filter
c) Gaussian lowpass filter
d) All of the mentioned
Answer
Answer: a [Reason:] Gaussian lowpass filter covers the range of very smooth filter functioning of lowpass filters.
5. Butterworth lowpass filter has a parameter, filter order, determining its functionality as very sharp or very smooth filter function or an intermediate filter function. If the parameter value is very high, the filter approaches to which of the following filter(s)?
a) Ideal lowpass filter
b) Gaussian lowpass filter
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] For high value of filter order Butterworth lowpass filter behaves as Ideal lowpass filter, while for lower order value it has a smoother form behaving like Gaussian lowpass filter.
6. Butterworth lowpass filter has a parameter, filter order, determining its functionality as very sharp or very smooth filter function or an intermediate filter function. If the parameter value is of lower order, the filter approaches to which of the following filter(s)?
a) Ideal lowpass filter
b) Gaussian lowpass filter
c) All of the mentioned
d) None of the mentioned
Answer
Answer: b [Reason:] For high value of filter order Butterworth lowpass filter behaves as Ideal lowpass filter, while for lower order value it has a smoother form behaving like Gaussian lowpass filter.
7. In a filter, all the frequencies inside a circle of radius D0 are not attenuated while all frequencies outside circle are completely attenuated. The D0 is the specified nonnegative distance from origin of the Fourier transform. Which of the following filter(s) characterizes the same?
a) Ideal filter
b) Butterworth filter
c) Gaussian filter
d) All of the mentioned
Answer
Answer: a [Reason:] In ideal filter all the frequencies inside a circle of radius D0 are not attenuated while all frequencies outside the circle are completely attenuated.
8. In an ideal lowpass filter case, what is the relation between the filter radius and the blurring effect caused because of the filter?
a) Filter size is directly proportional to blurring caused because of filter
b) Filter size is inversely proportional to blurring caused because of filter
c) There is no relation between filter size and blurring caused because of it
d) None of the mentioned
Answer
Answer: b [Reason:] Increase in filter size, removes less power from the image and so less severe blurring occurs.
9. The characteristics of the lowpass filter h(x, y) is/are___
a) Has a dominant component at origin
b) Has a concentric, circular components about the center component
c) All of the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] the lowpass filter has two different characteristics: one is a dominant component at origin and other one is a concentric, circular components about the center component.
10. What is the relation for the components of ideal lowpass filter and the image enhancement?
a) The concentric component is primarily responsible for blurring
b) The center component is primarily for the ringing characteristic of ideal filter
c) All of the mentioned
d) None of the mentioned
Answer
Answer: d [Reason:] The center component of ideal lowpass filter is primarily responsible for blurring while, concentric component is primarily for the ringing characteristic of ideal filter.
11. Using the feature of reciprocal relationship of filter in 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
Answer: a [Reason:] The characteristics feature of reciprocal relationship says that the narrower the frequency domain filter becomes it attenuates more low frequency component and so increases blurring and more severe becomes the ringing.
12. Which of the following defines the expression for BLPF H(u, v) of order n, where D(u, v) is the distance from point (u, v), D0 is the distance defining cutoff frequency?
a)
b)
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] BLPF is the Butterworth lowpass filter and is defined as:
.
13. Which of the following defines the expression for ILPF H(u, v) of order n, where D(u, v) is the distance from point (u, v), D0 is the distance defining cutoff frequency?
a)
b)
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] ILPF is the Ideal lowpass filter and is defined as:
.
14. 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
Answer: b [Reason:] ILPF has sharp discontinuity and BLPF doesn’t, so BLPF establishes a clear cutoff b/w passed and filtered frequencies.
15. A Butterworth filter of what order has no ringing?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: a [Reason:] A Butterworth filter of order 1 has no ringing and ringing exist for order 2 although is imperceptible. A Butterworth filter of higher order shows significant factor of ringing.
