EELE 5310: Digital Image Processing. Ch. 3. Eng. Ruba A. Salamah. iugaza.edu

Transcription

1 EELE 531: Digital Image Processing Ch. 3 Eng. Ruba A. Salamah iugaza.edu 1

2 Image Enhancement in the Spatial Domain 2

3 Lecture Reading 3.1 Background 3.2 Some Basic Gray Level Transformations Some Basic Gray Level Transformations Image Negatives Log Transformations Power-Law Transformations Piecewise-Linear Transformation Functions Contrast stretching Gray-level slicing Bit-plane slicing 3

4 Principle Objective of Enhancement Process an image so that the result will be more suitable than the original image for a specific application. The suitableness is up to each application. A method which is quite useful for enhancing an image may not necessarily be the best approach for enhancing another images 4

5 2 domains Spatial Domain : (image plane) Techniques are based on direct manipulation of pixels in an image Frequency Domain : Techniques are based on modifying the Fourier transform of an image There are some enhancement techniques based on various combinations of methods from these two categories. 5

6 Good images For human visual The visual evaluation of image quality is a highly subjective process. It is hard to standardize the definition of a good image. A certain amount of trial and error usually is required before a particular image enhancement approach is selected. 6

7 Spatial Domain Procedures that operate directly on pixels. g(x,y g( x,y)) = T[f(x,y T[f(x,y)] )] where f(x,y f( x,y)) is the input image g(x,y g( x,y)) is the processed image T is an operator on f defined over some neighborhood of (x,y x,y)) 7

8 Mask/Filter Neighborhood of a point (x,y) can be (x,y) defined by square/rectangular using a subimage area centered at (x,y) The center of the subimage is moved from pixel to pixel starting at the top left corner. 8

9 Mask Processing or Filtering Neighborhood is bigger than 1x1 pixel Use a function of the values of f in a predefined neighborhood of (x,y) to determine the value of g at (x,y) The value of the mask coefficients determine the nature of the process. Used in techniques like: Image Sharpening Image Smoothing 9

10 Point Processing Neighborhood = 1x1 pixel g depends on only the value of f at (x,y x,y)) T = gray level (or intensity or mapping) transformation function: s = T(r) Where r = gray level of f( f(x,y x,y)) s = gray level of g( g(x,y x,y)) Because enhancement at any point in an image depends only on the gray level at that point, techniques in this category often are referred to as point processing. 1

11 Contrast Stretching Produce higher contrast than the original by: darkening the levels below m in the original image Brightening the levels above m in the original image 11

12 Thresholding Produce a two-level (binary) image T(r) is called a thresholding function. 12

13 3 basic gray-level transformation functions Negative nth root Log nth power Linear function Logarithm function Identity Inverse Log Negative and identity transformations Log and inverse-log transformation Power-law function nth power and nth root transformations Input gray level, r 13

14 Image Negatives Negative nth root An image with gray level in the range [, LL-1] where L = 2n ; n = 1, 2 Negative transformation : Log nth power s = L 1 r Identity Inverse Log Input gray level, r Reversing the intensity levels of an image. Suitable for enhancing white or gray detail embedded in dark regions of an image, especially when the black area dominant in size. 14

15 Image Negatives Example: Original Negative 15

16 Log Transformations s = c log (1 (1+r) Negative nth root c is a constant and r Log curve maps a narrow range of low gray-level values in the input image into a wider range of output levels. The opposite is true for higher values. Used to expand the values of dark pixels in an image while compressing the higher-level values. Log nth power Identity Inverse Log Input gray level, r 16

17 Example of Logarithm Image 17

18 Inverse Logarithm Transformations Negative nth root Do opposite to the Log Transformations Log nth power Used to expand the values of high pixels Identity Inverse Log in an image while compressing Input gray level, r the darker-level values. 18

19 Power-Law Transformations s = cr Input gray level, r Plots of s = cr for various values of γ (c = 1 in all cases) c and γ are positive constants Power-law curves with fractional values of γ map a narrow range of dark input values into a wider range of output values. The opposite is true for higher values of input levels. c = γ = 1 Identity function 19

20 Example 1: Gamma correction Monitor Gamma correction = 2.5 Monitor Cathode ray tube (CRT) devices have an intensityto-voltage response that is a power function, with γ varying from 1.8 to 2.5 The picture will become darker. Gamma correction is done by preprocessing the image before inputting it to the monitor with s = cr1/ =1/2.5 =.4 2

21 Example 2: MRI a b (a) The picture is predominately dark (b) Result after power-law transformation with γ =.6 (c) transformation with γ =.4 c d (best result) (d) transformation with γ =.3 (washed out look) under than.3 will be reduced to unacceptable level. 21

22 Example 3 a b c d (a) image has a washed-out appearance, it needs a compression of gray levels needs γ > 1 (b) result after power-law transformation with γ = 3. (suitable) (c) transformation with γ = 4. (suitable) (d) transformation with γ = 5. (high contrast, the image has areas that are too dark, some detail is lost) 22

23 Piecewise-Linear Transformation Functions Advantage: The form of piecewise functions can be arbitrarily complex Disadvantage: Their specification requires considerably more user input 23

24 Contrast Stretching increase the dynamic range of the gray levels in the image (b) a low-contrast image (c) result of contrast stretching: (r1,s1)=(rmin,) and (r2,s2) = (rmax,l-1) (d) result of thresholding 24

25 Gray-level slicing Highlighting a specific range of gray levels in an image Display a high value of all gray levels in the range of interest and a low value for all other gray levels (a) transformation highlights range [A,B] of gray level and reduces all others to a constant level (result in binary image) (b) transformation highlights range [A,B] but preserves all other levels 25

26 Bit-plane Slicing One 8-bit byte Highlighting the contribution made to total image appearance Bit-plane 7 (most significant) by specific bits Suppose each pixel is represented by 8 bits Higher-order bits contain the majority of the visually Bit-plane (least significant) significant data Useful for analyzing the relative importance played by each bit of the image 26

27 Example An 8-bit fractal image 27

28 8 bit planes Bit-plane 7 Bit-plane 6 BitBitBitplane 5 plane 4 plane 3 BitBitBitplane 2 plane 1 plane 28

29 Next lecture Reading 3.3 Histogram processing Histogram Equalization Histogram Specification 3.4 Enhancement Using Arithmetic/Logic Operations Image Subtraction Image Averaging 29

30 Histogram Processing Histogram of a digital image with gray levels in the range [,L-1] is a discrete function h(rk) = nk Where rk : the kth gray level nk : the number of pixels in the image having gray level rk h(rk) : histogram of a digital image with gray levels rk 3

31 Example No. of pixels Gray level 4x4 image Gray scale = [,9] histogram 31

32 Normalized Histogram dividing each of histogram at gray level rk by the total number of pixels in the image, n p(rrk) = nk / n p( For k =,1,,L-1 p(rrk) gives an estimate of the probability of occurrence p( of gray level rk The sum of all components of a normalized histogram is equal to 1 32

33 Histogram Processing Basic for numerous spatial domain processing techniques Used effectively for image enhancement Information inherent in histograms also is useful in image compression and segmentation 33

34 Example Dark image Components of histogram are concentrated on the low side of the gray scale. Bright image Components of histogram are concentrated on the high side of the gray scale. 34

35 Example Low-contrast image histogram is narrow and centered toward the middle of the gray scale High-contrast image histogram covers broad range of the gray scale and the distribution of pixels is not too far from uniform, with very few vertical lines being much higher than the others 35

36 Histogram Equalization As the low-contrast image s histogram is narrow and centered toward the middle of the gray scale, if we distribute the histogram to a wider range the quality of the image will be improved. We can do it by adjusting the probability density function of the original histogram of the image so that the probability spread equally 36

37 Histogram transformation s = T(r) s sk= T(rk) T(r) rk 1 r Where r 1 T(r) satisfies (a). T(r) is singlevalued and monotonically increasingly in the interval r 1 (b). T(r) 1 for r 1 ٣٧

38 2 Conditions of T(r) Single-valued (one-to-one relationship) guarantees that the inverse transformation will exist Monotonicity condition preserves the increasing order from black to white in the output image T(r) 1 for r 1 guarantees that the output gray levels will be in the same range as the input levels. The inverse transformation from s back to r is r = T -1(s) ; s 1 38

39 Probability Density Function The gray levels in an image may be viewed as random variables in the interval [,1] PDF is one of the fundamental descriptors of a random variable 39

40 Applied to Image Let pr(r) denote the PDF of random variable r ps (s) denote the PDF of random variable s If pr(r) and T(r) are known and T-1(s) satisfies condition (a) then ps(s) can be obtained using a formula : dr ps(s) = pr (r) ds The PDF of the transformed variable s is determined by the graylevel PDF of the input image and by the chosen transformation function 4

41 Transformation function A transformation function of particular importance r in DIP is : Thus: s = T ( r ) = pr ( w )dw ds dt ( r ) = dr dr r d = pr ( w )dw dr = pr ( r ) dr ps ( s ) = pr ( r ) ds Substitute and yield 1 = pr ( r ) pr ( r ) = 1 where s 1 41

42 Transformation function It is important to note that T(r) depends on pr(r), the resulting ps(s) always is uniform, independent of the form of pr(r). r s = T(r)= p r ( w ) dw yields a random variable s characterized by a uniform probability function Ps(s ) 1 s 42

43 Discrete Transformation Function The probability of occurrence of gray level in an image is approximated by nk pr (rk ) = n where k =, 1,..., L-1 The discrete version of transformation k sk = T ( rk ) = pr ( rj ) j= k nj j= n = where k =, 1,..., L-1 43

44 Histogram Equalization Implementation Thus, an output image is obtained by mapping each pixel with level rk in the input image into a corresponding pixel with level sk in the output image such that: k nj j = n sk = where k =, 1,..., L-1 44

45 Example1 rk p(rk) ٠ ٢ ٤ ٦ ٨ ١٠ ١٢ ١٤ 45

46 Example1 Solution: rk p(rk) p(rk) (L-1)* p(rk) Sk s ps

47 Continue ٠ ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩ ١٠ ١١ ١٢ ١٣ ١٤ ١٥

48 Example2 rk p(rk) ٠ ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩ ١٠ ١١ ١٢ ١٣ ١٤ ١٥ 48

49 Example2 Solution: rk p(rk) p(rk) (L-1)* p(rk) Sk s ps

50 Continue ٠ ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩ ١٠ ١١ ١٢ ١٣ ١٤ ١٥

51 Example3 rk p(rk)

52 Example3 Solution: rk p(rk) p(rk) (L-1)* p(rk) Sk s ps

53 Continue ٠ ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩ ١٠ ١١ ١٢ ١٣ ١٤ ١٥

54 Example4 ٥٤

55 Histogram Processing Transformation functions used to equalized the histogram ٥٥

56 Histogram Matching (Specification) Histogram equalization has a disadvantage which is that it can generate only one type of output image. With Histogram Specification, we can specify the shape of the histogram that we wish the output image to have. It doesn t have to be a uniform histogram. ٥٦

57 Consider the continuous domain Let pr(r) denote continuous probability density function of gray-level of input image, r Let pz(z) denote desired (specified) continuous probability density function of gray-level of output image, z r Let s be a random variable with the property z We need to find z such that: s = T( r ) = g ( z ) = pz ( t )dt = s p ( w )dw r Thus: s = T(r) = G(z). Therefore: z = G-1(s) = G-1[T(r)] Assume G-1 exists and satisfies the condition (a) and (b). We can map an input gray level r to output gray level z 57

58 Discrete Formulation k sk = T ( rk ) = pr ( rj ) j= k nj j= n = k k =,1,2,..., L 1 G( zk ) = pz ( zi ) = sk k =,1,2,..., L 1 i = zk = G 1 [T ( rk )] = G 1 [sk ] k =,1,2,..., L 1 ٥٨

59 Histogram Matching Implementation 1. Find histogram of input image, and find histogram equalization mapping: 2. sk = k j= p r (r j ) Specify the desired histogram, and find histogram equalization mapping: k vk = p z ( z j ) j = 3. For every value of sk use the stored values of G to find the corresponding min value zq such that: 59

60 Histogram Matching Implementation ٦٠

61 Histogram Matching Example Consider an 8-level image with the shown histogram Match it to the image with the histogram ٦١

62 Example Continue 1. Equalize the histogram of the input image using transform s =T(r) ٦٢

63 Example Continue 2. Equalize the desired histogram v = G(z). ٦٣

64 Example Continue 3. Set v = s to obtain the composite transform k rk Sk 1 vk zk ٦٤

65 Example Image is dominated by large, dark areas, resulting in a histogram characterized by a large concentration of pixels in pixels in the dark end of the gray scale Image of Mars moon ٦٥

66 Image Equalization Result image after histogram equalization Transformation function for histogram equalization Histogram of the result image ٦٦

67 Solve the problem Since the problem with the transformation function of the histogram equalization was caused by a large concentration of pixels in the original image with levels near Histogram Equalization A reasonable approach is to modify the histogram of that image so that it does not have this property ٦٧

68 Histogram Specification Histogram Specification (1) the transformation function G(z) obtained from the specified histogram. (2) the inverse transformation G-1(s) ٦٨

69 Result image and its histogram Original image After applingthe histogram matching The output image s histogram ٦٩

70 Notes Histogram specification is a trial-and-error process There are no rules for specifying histograms, and one must resort to analysis on a case-by-case basis for any given enhancement task. ٧٠

71 Readings From the book the following sections: 3.4: Enhancement Using Arithmetic/Logic Operations 3.4.1: Image Subtraction :Image Averaging 71

72 Enhancement using Arithmetic/Logic Operations Arithmetic/Logic operations are performed on pixel by pixel basis between two or more images Except NOT operation which perform only on a single image. We need only be concerned with the ability to implement the AND, OR, and NOT which are functionally complete. Pixel values are processed as strings of binary numbers. NOT operation = negative transformation ٧٢

73 Example of AND Operation The AND and OR operations are used for masking; for selecting subimages in an image. ROI processing. original image AND image mask result of AND operation 73

74 Example of OR Operation original image OR image mask result of OR operation ٧٤

75 Image Subtraction The difference between two images f(x, y) and h(x, y): g(x,y g( x,y)) = f(x,y f(x,y)) h( h(x,y x,y)) is obtained by computing the difference between all pairs of corresponding pixels from f and h. The key usefulness of subtraction is the enhancement of differences between images. 75

76 Image Subtraction 76

77 Mask mode radiography Note: the background is dark because it doesn t change much in both images. the difference area is bright because it has a big change ٧٧

78 Note The values in a difference image can range from 255 to 255, so some sort of scaling is required. One method is to add 255 to every pixel and then divide by 2. Another method is the min_max normalization: Where mina and maxa are the min and max values of the diff. image, and new_maxa = 255, and new_mina=. thus 78

79 Image Averaging consider a noisy image g(x,y) formed by the addition of noise η(x,y) to an original image f(x,y) g(x,y) = f(x,y) + η(x,y) 79

80 Image Averaging if noise has zero mean and is uncorrelated then it can be shown that if image formed by averaging K different noisy images. Then: and Thus if K increases, the variability (noise) of the pixel at each location (x,y) decreases in g(x,y). This means that g(x,y) approaches f(x, y) as the number of noisy images used in the averaging process increases. 8

81 Example a b c d e f a) original image b) image corrupted by additive Gaussian noise with zero mean and a standard deviation of 64 gray levels. c). -f). results of averaging K = 8, 16, 64 and 128 noisy images ٨١

82 Example 82

83 Reading 3.5: Basics of Spatial Filtering 3.6:Smoothing Spatial Filters 3.6.1: Smoothing Linear Filters 3.6.2: Order-Statistics Filters 3.7: Sharpening Spatial Filters The Laplacian Operator Use of First Derivatives for Enhancement 3.8: Combining Spatial Enhancement Methods 83

84 Spatial Filtering Origin x Simple 3*3 Neighbourhood y a b c d e f g h i Original Image Pixels e * r s t u v w x y z Filter 3*3 Filter Image f (x, y) eprocessed = v*e + r*a + s*b + t*c + u*d + w*f + x*g + y*h + z*i The above is repeated for every pixel in the original image to generate the processed image

85 Spatial Filtering use filter (can also be called as mask/kernel/template or window) the values in a filter subimage are referred to as coefficients, rather than pixel. our focus will be on masks of odd sizes, e.g. 3x3, 5x5, 85

86 Spatial Filtering 86

87 Spatial Filtering Process simply move the filter mask from point to point in an image. at each point (x,y), the response of the filter at that point is calculated using a predefined relationship. Or: 87

88 Linear Filtering Linear Filtering of an image f of size MxN filter mask of size mxn is given by the expression where a = (m-1)/2 and b = (n-1)/2 To generate a complete filtered image this equation must be applied for x =, 1, 2,, M-1 and y =, 1, 2,, N-1 88

89 Note An important consideration in implementing neighborhood operations for spatial filtering is the issue of what happens when the center of the filter approaches the border of the image. There are several ways to handle this situation: limit the excursions of the center of the mask to be at a distance no less than (n-1)2 pixels from the border. filter all pixels only with the section of the mask that is fully contained in the image. Other approaches include padding the image by adding rows and columns of s (or other constant gray level), or padding by replicating rows or columns. 89

90 Smoothing Spatial Filters Image sensors and transmission channels may produce certain type of noise characterized by random and isolated pixels with out-of-range gray levels, which are either much lower or higher than the gray levels of the neighboring pixels. One way to get rid of this kind of noise is to use the average of a small local region in the image so that the out-of-range gray levels can be suppressed. 9

91 Smoothing Spatial Filters Used for image blurring and for noise reduction Blurring is used in preprocessing steps, such as removal of small details from an image prior to object extraction bridging of small gaps in lines or curves noise reduction can be accomplished by blurring with a linear filter and also by a nonlinear filter 91

92 Two Types of Smoothing filters Smoothing Linear Filter Order-Statistics Filters 92

93 Smoothing Linear Filters These filters are called averaging filters or lowpass filters. Replacing the value of every pixel in an image by the average of the gray levels in the neighborhood will reduce the sharp transitions in gray levels. sharp transitions random noise in the image edges of objects in the image Thus, smoothing can reduce noises (desirable) and blur edges (undesirable) 93

94 3x3 Smoothing Linear Filters box filter weighted average the center is the most important and other pixels are inversely weighted as a function of their distance from the center of the mask 94

95 General form : smoothing mask Filter of size mxn (m and n odd) a g ( x, y) = b w(s, t ) f ( x + s, y + t ) s = a t = b a b w(s, t ) s = a t = b summation of all coefficient of the mask 95

96 Example 96

97 Example a). original image 5x5 pixel b). - f). results of smoothing with square averaging filter masks of size n = 3, 5, 9, 15 and 35, respectively. Note: big mask is used to eliminate small objects from an image. the size of the mask establishes the relative size of the objects that will be blended with the background. ٩٧

98 Example original image result after smoothing with 15x15 averaging mask result of thresholding we can see that the result after smoothing and thresholding, the remainders are the largest and brightest objects in the image. 98

99 Order-Statistics Filters (Nonlinear Filters) The response is based on ordering (ranking) the pixels contained in the image area encompassed by the filter example median filter : R = median{zk k = 1,2,,n x n} max filter : R = max{zk k = 1,2,,n x n} min filter : R = min{zk k = 1,2,,n x n} note: n x n is the size of the mask Median example: (1,2,2,2,15,2,2,25,1) rank to:( 1,15,2,2,2,2,2,25,1) Median=2 99

100 Median Filters Replaces the value of a pixel by the median of the gray levels in the neighborhood of that pixel (the original value of the pixel is included in the computation of the median) Quite popular because for certain types of random noise (impulse impulse noise salt and pepper noise) noise, they provide excellent noisenoise-reduction capabilities, capabilities with considering less blurring than linear smoothing filters of similar size. 1

101 Median Filters Forces the points with distinct gray levels to be more like their neighbors. Isolated clusters of pixels that are light or dark with respect to their neighbors, and whose area is less than n2/2 (one-half the filter area), are eliminated by an n x n median filter. Eliminated = forced to have the value equal the median intensity of the neighbors. Larger clusters are affected considerably less 11

102 Example : Median Filters 12

103 Sharpening Spatial Filters To highlight fine details in an image or to enhance detail that has been blurred, either in error or as a natural effect of a particular method of image acquisition. Blurring can be done in spatial domain by pixel averaging in a neighborhood, since averaging is analogous to integration, thus, we can guess that the sharpening must be accomplished by spatial differentiation. 13

104 Derivative operator The strength of the response of a derivative operator is proportional to the degree of discontinuity of the image at the point at which the operator is applied. Thus, image differentiation enhances edges and other discontinuities (noise) 14

105 First-order derivative a basic definition of the first-order derivative of a one-dimensional function f(x) is the difference f = f ( x + 1) f ( x) x 15

106 Second-order derivative similarly, we define the second-order derivative of a one-dimensional function f(x) is the difference f = f ( x + 1 ) + f ( x 1 ) 2 f ( x ) 2 x 2 16

107 Derivative operator 17

108 First and Second-order derivative of f(x,y) When we consider an image function of two variables, f(x,y), at which time we will dealing with partial derivatives along the two spatial axes. Gradient operator Laplacian operator (linear operator) f ( x, y) f ( x, y) f ( x, y) f = = + x y x y f ( x, y) f ( x, y) f = x y

109 Discrete Form of Laplacian from f = f ( x + 1, y) + f ( x 1, y) 2 f ( x, y) 2 x 2 f = f ( x, y + 1) + f ( x, y 1) 2 f ( x, y) 2 y 2 yield, f = [ f ( x + 1, y) + f ( x 1, y) 2 + f ( x, y + 1) + f ( x, y 1) 4 f ( x, y)] 19

110 Result Laplacian mask 11

111 Laplacian mask implemented an extension of diagonal neighbors 111

112 Other implementations of Laplacian masks give the same result, but we have to keep in mind that when combining (add / subtract) a Laplacian-filtered image with another image. 112

113 Effect of Laplacian Operator As it is a derivative operator, It highlights gray-level discontinuities in an image Tends to produce images that have Grayish edge lines and other discontinuities, all superimposed on a dark, featureless background. ١١٣

114 Correct the effect of featureless background Easily by adding the original and Laplacian image. Be careful with the Laplacian filter used f ( x, y) 2 f ( x, y) g ( x, y) = 2 f ( x, y ) + f ( x, y) if the center coefficient of the Laplacian mask is negative if the center coefficient of the Laplacian mask is positive 114

115 115

116 Example a). image of the North pole of the moon b). Laplacian-filtered image with c). Laplacian image scaled for display purposes d). image enhanced by addition with original image ١١٦

117 Mask of Laplacian + addition To simply the computation, we can create a mask which do both operations, Laplacian Filter and Addition the original image. g ( x, y) = f ( x, y) [ f ( x + 1, y) + f ( x 1, y) + f ( x, y + 1) + f ( x, y 1) 4 f ( x, y)] = 5 f ( x, y) [ f ( x + 1, y) + f ( x 1, y) + f ( x, y + 1) + f ( x, y 1)]

118 Example ١١٨

119 Unsharp Masking and High-Boost Filtering fs ( x, y) = f ( x, y) f ( x, y) sharpened image = original image blurred image This operation is known as unsharp masking where fs(x, y) denotes the sharpened image obtained by unsharp masking, and f(x, y) is a blurred version of f(x, y). 119

120 High-boost filtering A slight further generalization of unsharp masking is called high-boost filtering. A high-boost filtered image, fhb (x,y) is defined at any point (x, y) as: where A 1. 12

121 High-boost filtering If we use Laplacian filter to create sharpen image fs(x,y) with addition of original image f ( x, y) f ( x, y) fs ( x, y) = 2 f ( x, y) + f ( x, y) 2 121

122 High-boost filtering yields if the center coefficient of the Laplacian mask is negative Af ( x, y) 2 f ( x, y) fhb ( x, y) = 2 Af ( x, y) + f ( x, y) if the center coefficient of the Laplacian mask is positive 122

123 High-boost Masks A 1 if A = 1, it becomes standard Laplacian sharpening 123

124 Example One of the principal applications of boost filtering is when the input image is darker than desired. By varying the boost coefficient, it generally is possible to obtain an overall increase in average gray level of the image, thus helping to brighten the final result. 124

125 Example 125

126 f Gx x f = = f G y y Gradient Operator first derivatives are implemented using the magnitude of the gradient. gradient f = mag ( f ) = [G + G ] 2 x f 2 f = + x y 2 2 y commonly approx. 2 f G x + G y the magnitude becomes nonlinear 126

127 Gradient Mask simplest approximation, 2x2 Gx = ( z8 z5 ) f = [G + G ] 2 x 2 y 1 and 2 z1 z2 z3 z4 z5 z6 z7 z8 z9 G y = ( z6 z5 ) = [( z8 z5 ) + ( z6 z5 ) ] f z8 z5 + z6 z5 127

128 Gradient Mask here z1 z2 z3 z4 z5 z6 z7 z8 z9 Roberts cross-gradient operators, 2x2 Gx = ( z9 z5 ) f = [G + G ] 2 x 2 y 1 and 2 G y = ( z8 z6 ) = [( z9 z5 ) + ( z8 z6 ) ] f z9 z5 + z8 z6 128

129 Gradient Mask Sobel operators, 3x3 z1 z2 z3 z4 z5 z6 z7 z8 z9 Gx = ( z7 + 2 z8 + z9 ) ( z1 + 2 z2 + z3 ) G y = ( z3 + 2 z6 + z9 ) ( z1 + 2 z4 + z7 ) f Gx + G y 129

130 Note the summation of coefficients in all masks equals, indicating that they would give a response of in an area of constant gray level. 13

131 Example 131

132 Example of Combining Spatial Enhancement Methods want to sharpen the original image and bring out more skeletal detail. problems: narrow dynamic range of gray level and high noise content makes the image difficult to enhance ١٣٢

133 Example of Combining Spatial Enhancement Methods solve : 1. Laplacian to highlight fine detail 2. gradient to enhance prominent edges 3. gray-level transformation to increase the dynamic range of gray levels 133

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135 135