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 local enhancement using mean and variance, the key condition is:

The variance within a local neighborhood must be sufficiently high to enhance the local features effectively.

Detailed Explanation:

• Mean: The mean is used to capture the average intensity of a local region, which helps to adjust the intensity values of the pixels within that region.

• Variance: The variance measures the spread of intensity values in the local neighborhood. A higher variance indicates more contrast or texture, while a lower variance corresponds to smoother or uniform areas.

Condition for Enhancement:

• The enhancement is applied more effectively in regions with higher variance, as these areas contain more detail (edges, textures). In regions with low variance, the enhancement is minimal since these regions are smooth and lack significant detail.

• Thresholding the variance: Often, a threshold is set for the variance. If the variance exceeds a certain threshold, the enhancement algorithm will apply a stronger enhancement to increase the contrast. If the variance is below the threshold, the enhancement is either reduced or not applied to avoid amplifying noise.

Mathematical Representation:

• Let $\mu (x,y)$ represent the local mean of the neighborhood centered at pixel $(x,y)$.
• Let σ2(x,y)\sigma^2(x, y)σ2(x,y) represent the variance of the neighborhood.

For enhancement:

• The variance σ2(x,y)\sigma^2(x, y)σ2(x,y) should be above a certain threshold to trigger significant enhancement.

This condition ensures that the enhancement algorithm boosts areas with detailed textures or features while leaving smooth, uniform areas unaffected or only lightly enhanced.