Histogram Specification MCQ [Free PDF] – Objective Question Answer for Histogram Specification Quiz

1. The technique of Enhancement that has a specified Histogram processed image as result, is called?

A. Histogram Linearization
B. Histogram Equalization
C. Histogram Matching
D. None of the mentioned

Answer: C

The histogram Specification method uses a specified Histogram, i.e. the shape of the histogram can be specified by self, to generate a processed image. And the same is also known as Histogram Matching.

 

2. In Histogram Matching r and z are the gray level of input and output image and p stands for PDF, then, what does pz(z) stand for?

A. Specific probability density function
B. Specified pixel distribution function
C. Specific pixel density function
D. Specified probability density function

Answer: D

In Histogram Matching, pr(r) is estimated from the input image while pz(z) is the Specified probability density function that the output image is supposed to have.

 

3. Inverse transformation plays an important role in which of the following Histogram processing Techniques?

A. Histogram Linearization
B. Histogram Equalization
C. Histogram Matching
D. None of the mentioned

Answer: C

In Histogram Matching or Specification, z = G-1[T(r)], r and z are the gray levels of input and output image and T & G are transformations.
In Histogram Linearization or Equalization s = T(r), r and s are the gray levels of input and output image and T is the only transformation.

 

4. In Histogram Matching or Specification, z = G-1[T(r)], r and z are the gray level of input and output image and T & G are transformations, to confirm the single value and monotonous of G-1 what of the following is/are required?

A. G must be strictly monotonic
B. G must be strictly decreasing
C. All of the mentioned
D. None of the mentioned

Answer: A

G being strictly monotonic confirms that the values of specified histogram Pz(zi) can’t be zero. That is G-1 is also single-valued and monotonic.

 

5. Which of the following histogram processing techniques is global?

A. Histogram Linearization
B. Histogram Specification
C. Histogram Matching
D. All of the mentioned

Answer: D

All of the mentioned methods modify the pixel value by transformations that are based on the gray level of the whole image.

h({});

 

6. What happens to the output image when the global Histogram equalization method is applied to the smooth and noisy area of an image?

A. The contrast increases a little bit with the considerable enhancement of noise
B. The result would have a fine noise texture
C. All of the mentioned
D. None of the mentioned

Answer: A

To an image’s smooth and noisy area, when the global histogram method is applied the contrast increases a little bit with the considerable enhancement of noise, while for the local method the result has a fine noise texture.

(A. Original image. (B. Result using global histogram equalization. (C. Result using local histogram equalization using 7*7 neighborhood about each pixel.

 

7. Let us suppose an image containing a quite small square under a large dark square with both having very close gray level values. If an image contains some of this such that the small squares can’t be visualized and some noise blurred enough to reduce its noise content as shown in fig. below, Which of the following method would be preferred for obtaining the small square clear enough?

A. Global histogram equalization
B. Local histogram equalization
C. All of the mentioned
D. None of the mentioned

Answer: B

For global histogram enhancement, the small squares have a very close gray value to larger squares and have a very small size to be influenced by the global histogram equalization method.
But, local histogram enhancement using a 7*7 neighborhood reveals the small square.

(A. Original image. (B. Result using global histogram equalization. (C. Result using local histogram equalization using 7*7 neighborhood about each pixel.

 

8. In terms of enhancement, what do mean and variance refer to?

A. Average contrast and average gray level respectively
B. Average gray level and average contrast respectively
C. Average gray level in both
D. Average contrast in both

Answer: B

In terms of enhancement, mean refers to the average gray level and variance to average contrast.
Given by, mean as: m = ∑_(i=0)^(L-1) ri p(ri ) and variance as: σ2(r) = ∑_(i=0)^(L-1) (ri –m)2 p(ri ).

Where, as is histogram component of the ith value of r, p(ri) is the probability occurrence of gray level ri and L is the max gray value allowed.

 

9. For a local enhancement using mean and variance, there is one condition: ms(x, y) ≤ k0 MG, where, MG is global mean, k0 a constant, and ms(x, y) a measure of gray value as light or dark at point (x, y). Then, which fact is true for k0?

A. It is a negative constant with values less than -1.0
B. It is a positive constant with values less than 1.0
C. It is an integer constant with values between -1.0 and 1.0
D. None of the mentioned

Answer: B

In the condition ms(x, y) ≤ k0 MG, k0 is a positive constant whose value is always less than 1.0.

 

10. For a local enhancement using mean and variance, there is one condition: σs(x, y) ≤ k2DG, where, MDG is global standard deviation, k2 a positive constant, and σs(x, y) a measure of contrast at point (x, y). Then, which fact is true for k2 if its value is less than 1.0?

A. Enhancement is being done in light areas
B. Enhancement is being done in dark areas
C. Enhancement is being done independently of the value of k0
D. None of the mentioned

Answer: B

In the condition σs(x, y) ≤ k2DG, k0 is a positive constant that helps in enhancing light areas if the value is greater than 1.0 and dark areas if the value is less than 1.0.

Scroll to Top