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Nt1310 Unit 4 Test Report

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Figure (2): (a) Original Image of Copter, (b) resulting image after basic histogram equalization of Copter, (c) comparison of original histogram (dark blue) versus equalized histogram (light blue) [1]

3.2-Histogram Mapping
It is more generalized than histogram equalization that allow us to change data that allow us get the resulting histogram matches some curve they call mapping sometimes histogram matching. The most common implementation of histogram mapping depending on three steps: 1) equalizing the original image, 2) histogram equalize the desired output image, 3) and apply the inverse of the second transformation to the original equalized image. nA T1 = FA (nA ) = ∫ pA ( y)dy (1)

0 nB T2 = FB (nB ) = ∫ pB (x)dx …show more content…

3.3-Discrete orthogonal transform
It is generally used in enhancement of images that have various properties we have many transforms such as: Discrete Hartley, Fourier, Cosine transforms, etc... All of these transforms has similar properties. Every transform has equation for transformation. 3.4-Alpha Rooting
Is a method that deals with many of orthogonal combinations as Fourier, Hartley, Haar Wavlet, and Cosine .the mathematics for Alpha Rooting:

X ( p, s)

α −1
=

X ( p, s) …show more content…

The output resulted from this focus on the high frequency content in the image without changing anything in the image phase. This result with an image enhanced in contrast sometime this enhancement results with ugly artifacts.

3.5- Logarithmic Transform Domain
Transform Domain allow us or gives us the ability to show the frequency content of the image, however it is uninformative or compacted. In figure (3) this will be obvious .By working on the problem we discovered that the solution is to take the logarithm of the image. This is done by making two steps as follow: The first step is to generate a matrix to restore in it the phase of the transform image and it will be used to store phase of the transform coefficients. The second step is to take the logarithm of the modulus of the coefficients as shown by the equation

Xˆ (i, j) = ln( X (i, j) + λ) (1)
Where λ is some shifting coefficient, usually set to 1, shifting coefficient is used to transform from continuous to discontinuous this results with more visible version of histogram, to return coefficients to standard domain the signal is exponentiated and restore the

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