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 values is greater than 1.0?

Question

For a local enhancement using mean and variance, there is one condition:

Answer 1

For a local enhancement using mean and variance, the key condition is that the variance within the local window should be high for enhancing the image features.

Condition for Local Enhancement:

• The variance within a local neighborhood (or window) of an image should be greater than a threshold value to apply enhancement effectively.
• Low variance regions often represent areas with little contrast or texture, such as uniform or smooth regions, which are not enhanced as much.
• The mean is used to determine the average intensity value of the local neighborhood, and it can be used to shift the intensity values for better contrast.

Mathematical Perspective:

For a given pixel $p(x,y)$, the local enhancement using mean and variance can be described as:

• Local Mean: μx,y=1N∑i,j∈W(x,y)I(i,j)\mu_{x,y} = \frac{1}{N} \sum_{i,j \in W(x,y)} I(i,j)μx,y​=N1​∑i,j∈W(x,y)​I(i,j), where $W(x,y)$ is the local window around pixel $(x,y)$ and $I(i,j)$ is the pixel intensity in the neighborhood.
• Local Variance: σx,y2=1N∑i,j∈W(x,y)(I(i,j)−μx,y)2\sigma^2_{x,y} = \frac{1}{N} \sum_{i,j \in W(x,y)} (I(i,j) - \mu_{x,y})^2σx,y2​=N1​∑i,j∈W(x,y)​(I(i,j)−μx,y​)2.

The enhancement occurs when the local variance is sufficiently high, indicating more texture or contrast in the region, allowing the image enhancement to accentuate local details effectively.

This condition helps to enhance detailed areas of the image while leaving uniform or smooth areas unaltered, ensuring the enhancement improves the image quality and features where necessary.