Histogram Processing
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1 Histogram Processing The histogram of a digital image with gray levels in the range [0,L-] is a discrete function h ( r k ) = n k where r k n k = k th gray level = number of pixels in the image having gray level value r k
2 Normalization: h( r ) p( r ) = k = k n n k n n p( r k ) = total number of pixels in the image = probability of occurrence of gray level r k k p( ) = r k
3 p(r k ) =h(r k )/n r k = 255 k p( ) = r k
4 Uniformly distributed histogram yields HIGH CONTRAST IMAGE Hence, histogram processing requires the stretching of gray level uniformly over the entire gray level range. This is histogram equalization
5 Histogram p(r) r =
6 Histogram equalization p(s) 0 s
7 . Requires the distribution of histogram peaks uniformly over the entire gray level range 2. Equalize the height of the peaks Let us define the gray level values (r) to be Continuous and Normalized between [0,]
8 r = 0 Black r = White For any r, the transformation s = T( r) ; 0 r
9 Let us require that the transformation function T(r) should satisfy. T(r) is single valued and monotonically increasing in the interval 0 r 2. 0 T(r) for 0 r
10 s T(r) r Single valued and monotonically increasing
11 s T(r) Not single valued r Inverse transformation of the T(r) will not give the original image
12 Non-monotonic transformation function s T(r) r Transformation from black white Transformation from white black
13 Hence the requirements:. T(r) should be single valued which ensures the inverse transformation will exists The monotonicity condition preserves the increasing order from black to white in the output image 2. Condition (b) guarantees the output gray levels will be in the same range as that of input range
14 The gray levels (r or s) in an image random variable in the interval [0,] p r ( r) & p ( s) s are the PDFs of random variables r and s, respectively From the PDF theory of the random variable: ps ( s) ds = pr ( r) dr
15 To make the histogram uniform: ) ( = s p s ) ( ) ( ) ( ) ( r T d p s d p ds dr r p ds r r r r s r = = = = ω ω ω ω Put,
16 r s = T ( r) = p ( ω) dω r 0 Hence, the transformation function (T(r)) is equal to the CDF of random variable, r a)t(r) is singled valued and monotonically increasing function. Hence the first condition is satisfied. b) Values of T(r) or s lie in the range [0,]
17 For discrete values, 0,,..., ) ( = = L k n n r p k k r = = = k j j r k k r p r T s 0 ) ( ) (..., 0,,, ) ( 0 = = = = L k for n n r T s k j j k k
18 This will do the histogram equalization AUTOMATICALLY
19
20 Image averaging Noisy image can be enhanced by averaging it over a set of images A noisy image (g(x,y)) can be expressed as g( x, y) = f ( x, y) + η( x, y)
21 where f ( x, y) = Original image η ( x, y) = Noise If the noise η(x,y) is uncorrelated, then η( The expected value of the noise η(x,y) = 0 The expected covariance of the two random variables = 0
22 { g ( x, y)} i Represents a set of noisy image The average image is formed by, g( x, y) = K K i= g i ( x, y) K K = f + K i i= i= η i
23 Take expectation either side E K ( g) = E( f ) + E( η ) i i K i= i= K f i is a constant and same for all the images E { f } = f i i &
24 E{ η } = i 0 E { g( x, y)} = f ( x, y)
25 K=8 K=6 K=64 K=28
26
27 Scaling of the gray levels during mathematical operation + = unscaled scaled
28 - unscaled = scaled
29 Mask filtering in spatial domain Subimage Subimage filter, mask, kernel, template, or window
30 The subimages can be
31 Subimage is a small matrix whose elements are called COFFIECIENTS Preferably, the size of the mask should be odd If (m x n) size of the mask Then, m=2a+ and n=2b+ a, b are integers
32 Linear filter: = = + + = b b t a a s t y s x f t s w y x g ), ( ), ( ), ( The coefficients w(x,y) are defined by the user; depending on the nature of the filter required
33 eg., for an average filter of mask 3x3, each coefficient should be /9 W = /9
34 To perform the averaging, vary x =0,,2,.M- y =0,,2 N-
35 2 W = / Weighted average
36 = = = = + + = b b t a a s b b t a a s t s w t y s x f t s w y x g ), ( ), ( ), ( ), ( Hence the general expression for avg.
37 Mask reaches the border: (a) (b) Padding
38 Mask size --, 3 5, 9 5, 35
39 Averaging removes smaller details It leads to blurring
40
41 Non-linear filters: e.g., Median filter, variance filter Median filter is very powerful in removing random noise with considerably less blurring than linear avg. filter of similar size Median filters are effective for impulse noise.
42 Find median of the set of 5 numbers 5, 7, 4, 5, 3, 6, 8, 7, 2,, 0, 4, 4,, 3 a) Arrange them as 0,,, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 2, 4 Median = 5
43 Find median of the set of 4 numbers 87, 55, 67, 87, 98, 80, 45, 65, 75, 67, 87, 92, 87, 63 Arrange them 45, 55, 63, 65, 67, 67, 75, 80, 87, 87, 87, 87, 92, 98 Median = Avg(75,80) = 77.5
44
45 Noisy image Median filtered image (3x3 mask)
46 Comparison
47 a) Max filter The pixel value is replaced by the maximum calculated from the neighborhood. b) Min Filter The pixel value is replaced by the minimum calculated from the neighborhood. c) Mid point filter The pixel value is replaced by the 0.5*(min+max) calculated from the neighborhood.
48
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