Digital Image Processing. Image Enhancement in the Spatial Domain (Chapter 4)
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1 Digital Image Processing Image Enhancement in the Spatial Domain (Chapter 4)
2 Objective The principal objective o enhancement is to process an images so that the result is more suitable than the original image or a SPECIFIC application Category o image enhancement Spatial domain Frequency domain
3 Pixel neighborhood g(x,y) = T [ (x,y) ]
4 Point Processing, Gray-Level Transormation Function s = T (r) Contrast stretching Thresholding
5 Mask Processing Filtering g(x,y) = T [ (x,y) ] 1 T -1 (x,y) g(x,y)
6 Basic Gray Level Transormation Some Basic gray-level transormation or image Enhancement
7 Image Negative s = L r
8 Log Transormations s = c log(1+r) Pixel values dynamic range=[ ]
9 Power-Law Transormations s = cr s = r 2.5 r = [ ] s( = 2.5) = [ ] s( =.4) = [ ]
10 Power-Law Transormations - Gamma Correction
11 Power-Law Transormations c=1 =.6 c=1 =.4 c=1 =.3
12 Power-Law Transormations c=1 = 3 c=1 = 4 c=1 = 5
13 Piecewise-Linear Transormation Functions Advantage Arbitrarily complex Disadvantage More user input Type o Transormations Contrast stretching Gray-level slicing Bit-plane slicing
14 Contrast Stretching Objective: Increase the dynamic range o the gray levels Causes or poor image -Poor illumination -Lack o dynamic range in the imaging sensor -Wrong lens aperture
15 Contrast Stretching Example 1 For image with intensity range [5-15] What should (r 1,s 1 ) and (r 2,s 2 ) be to increase the dynamic range o the image to [ - 255]? s r
16 Gray-Level Slicing Objective: Highlighting a speciic range o gray levels in an image.
17 Bit-Plane Slicing MSB plane LSB plane LSB MSB
18 8-bit ractal image used or Bit-Plane Slicing
19 Bit-Plane Slicing
20 Histogram The histogram o a digital image with gray levels in the range [, L-1] is a discrete unction h(r k ) = n k, where r k is the kth gray level and n k is the number o pixels in the image having gray level r k. Histogram Processing Normalized Histogram Dividing each value o the histogram by the total number o pixels in the image, denoted by n. p(r k ) = n k /n. Normalized histogram provide useul image statistics.
21 Histogram Extraction using Matlab unction Normalized_Hist = Img_Hist(img) [R,C]=size(img); Hist=zeros(256,1); or r = 1:R or c=1:c Hist(img(r,c)+1,1)=Hist(img(r,c)+1,1)+1; end end Normalized_Hist = Hist/(R*C); plot(normalized_hist)
22 Histogram Processing
23 Histogram Equalization Histogram equalization is used to enhance image contrast and gray-level detail by spreading the histogram o the original image. s = T ( r ) r 1, where r and s are normalized pixel intensities Conditions or the transormation (a) T( r ) is single-valued and monotonically increasing in the interval r 1 (b) T ( r ) 1 or r 1 1
24 Histogram Equalization Objective o histogram equalization Transorm the histogram unction o the original image p r (r) to a uniorm histogram unction. p s (s) = 1 s 1 p r (r) Equalizer Transormation p s (s) s Continuous case r T ( r) pr ( w) dw s k T( rk ) Discrete case k j p r ( r j )
25 k : : : : Input image r k : :.5 p r (r k ) : : :.4 Output image T (r.15 k ).4 / * : : : : : : : s k : : : :
26 Histogram Equalization
27 Histogram Equalization 255
28 Histogram Matching Is histogram equalization a good approach to enhance the image?
29 Histogram Equalization p r (r k ) s k s : : :
30 Histogram Matching Histogram Matching method generates a processed image that has a speciied histogram. Desired Output image p z (z) z G (z) p z ( t) dt v T (r) s v, since both will have approximately uniorm pd Input image p r (r) r p r ( t) dt s z = G -1 [T( s )] s = G[T( r )]
31 Histogram Matching
32 Histogram Matching 1) Modiy the histogram o the image to obtain p z (z). 2) Find transormation unction G (z) using the modiied histogram in step 1. 3) Find the inverse G -1 (z) G (z) G -1 (z) 4) ApplyG -1 to the pixels o the histogram-equalized image.
33 Local Histogram Enhancement Global Histogram.98.2 Local Histogram
34 Local Histogram Enhancement
35 Histogram Statistics or Image Enhanc. Contrast manipulation using local statistics, such as the mean and variance, is useul or images where part o the image is acceptable, but other parts may contain hidden eatures o interest.
36 Let (x,y) be the coordinates o a pixel in an image, and let S xy denote a neighborhood (sub-image) o speciied size, centered at (x,y). ). ( ] [ ) ( ), (, 2, 2 ), (,, xy xy xy xy xy S t s t s S t s S S t s t s t s S r p m r r p r m The local mean and variance are the decision actors to whether apply local enhancement or not. Histogram Statistics or Image Enhanc.
37 Histogram Statistics or Image Enhanc. M G : Global mean D G : Global standard deviation E, k, k 1, k 2 : Speciied parameters g( x, y) E. ( x, y) ( x, y) i m S xy k otherwise M G AND k 1 D G S xy k 2 D G E = 4, k =.4, k 1 =.2, k 2 =.4 Size o local area = 3 3
38 Histogram Statistics or Image Enhanc. (x,y) (x,y) E Image Mean
39 Histogram Statistics or Image Enhanc. Original Image Enhanced Image
40 Enhancement Using Arithmetic/Logic Op. Arithmetic/logic operations involving images are perormed on a pixel-by-pixel basis between two or more images. Arithmetic Operations Addition, Subtraction, Multiplication, and Division Logic Operations AND, OR, NOT
41 Enhancement Using AND and OR Logic Op. AND OR Logic operations are perormed on the binary representation o the pixel intensities
42 Enhancement Using Arithmetic Op. _ SUB Original 4 Upper-order Bit Error HE o Error
43 Enhancement Using Arithmetic Op. _ SUB Mask Mode Radiography Problem: The pixel intensities in the dierence image can range rom -255 to 255. Solutions: 1) Add 255 to every pixel and then divide by 2. 2) Add the minimum value o the pixel intensity in the dierence image to every pixel and then divide by 255/Max. Max is the maximum pixel value in the modiied dierence image.
44 2 ), ( 2 ), ( 1 1 ), ( ), ( ), ( 1 ), ( ), ( ), ( ), ( y x y x g k i i k y x y x g E y x g k y x g y x y x y x g Original image Noise with zero mean Enhancement Using Arithmetic Op. Averaging
45 Enhancement Using Arithmetic Op. Averaging (x,y) g(x,y) Gaussian Noise mean = variance = 64 K = 8 K = 16 K = 32 K = x
46 Enhancement Using Arithmetic Op. Averaging K = 8 Dierence images between original image and images obtained rom averaging. K = 16 K = 32 K = 128 Notice the mean and variance o the dierence images decrease as K increases.
47 Basics o Spatial Filtering - Linear Spatial iltering are iltering operations perormed on the pixel intensities o an image and not on the requency components o the image. a b g ( x, y) w( s, t) ( x s, y t) sa tb a = (m - 1) / 2 b = (n - 1) / 2
48 Basics o Spatial Filtering Response, R, o an m n mask at any point (x, y) mn R i1 w i z i Special consideration is given when the center o the ilter approach the boarder o the image.
49 Nonlinear o Spatial Filtering Nonlinear spatial ilters operate on neighborhoods, and the mechanics o sliding a mask past an image are the same as was just outlined. In general however, the iltering operation is based conditionally on the values o the pixel in the neighborhood under consideration, and they do not explicitly use coeicients in the sum-o products manner described previously. Example Computation or the median is a nonlinear operation.
50 Smoothing Spatial Filtering - Linear Averaging (low-pass) Filters Smoothing ilters are used - Noise reduction - Smoothing o alse contours - Reduction o irrelevant detail Undesirable side eect o smoothing ilters - Blur edges Weighted average ilter reduces blurring in the smoothing process. Box ilter Weighted average
51 Smoothing Spatial Filtering _ Linear Averaging (low-pass) Filters n = 3 n = ilter size n = 5 n = 9 n = 15 n = 35
52 Smoothing Spatial Filtering Averaging & Threshold ilter size n = 15 Thrsh = 25% o highest intensity
53 Smoothing Spatial Filtering Order Statistic Filters Order-statistics ilters are nonlinear spatial ilters whose response is based on ordering (ranking) the pixels contained in the image area encompassed by the ilter, and then replacing the value o the center pixel with the value determined by the ranking result. 3 3 Median ilter [ ] = Max ilter [ ] = Min ilter [ ] = 1 Median ilter eliminates isolated clusters o pixels that are light or dark with respect to their neighbors, and whose area is less than n 2 /2.
54 Order Statistic Filters n = 3 Average ilter n = 3 Median ilter
55 Sharpening Spatial Filters The principal objective o sharpening is to highlight ine detail in an image or to enhance detail that has been blurred. Image z z i Blurred Image The derivatives o a digital unction are deined in terms o dierences.
56 Sharpening Spatial Filters Requirements or digital derivative First derivative 1) Must be zero in lat segment 2) Must be nonzero along ramps. 3) Must be nonzero at the onset o a gray-level step or ramp Second derivative 1) Must be zero in lat segment 2) Must be zero along ramps. 3) Must be nonzero at the onset and end o a gray-level step or ramp ( x 1) ( x) x 2 x 2 ( x 1) ( x 1) 2 ( x)
57 Sharpening Spatial Filters
58 Sharpening Spatial Filters Comparing the response between irst- and second-ordered derivatives: 1) First-order derivative produce thicker edge 2) Second-order derivative have a stronger response to ine detail, such as thin lines and isolated points. 3) First-order derivatives generally have a stronger response to a gray-level step {2 4 15} 4) Second-order derivatives produce a double response at step changes in gray level. In general the second derivative is better than the irst derivative or image enhancement. The principle use o irst derivative is or edge extraction.
59 Use o First Derivative or Edge Extraction Gradient First derivatives in image processing are implemented using the magnitude o the gradient. x y mag( ) t z 1 z 2 z z z z 7 z 8 z 9 x y Roberts operator G x = (z 9 -z 5 ) and G y = (z 8 - z 6 ) Sobel operator G x = (z 7 +2z 8 +z 9 ) - (z 1 +2z 2 +z 3 ) and G y = (z 3 +2z 6 +z 9 ) - (z 1 +2z 4 +z 7 ) G x G y z 3
60 Use o First Derivative or Edge Extraction Gradient Robert operator Sobel operators
61 Use o First Derivative or Edge Extraction Gradient (x,y) = [4, 14]
62 Use o First Derivative or Edge Extraction Gradient
63 y x )], ( 4 1), ( 1), ( ) 1, ( ) 1, ( [ ), ( 2 1), ( 1), ( ), ( 2 ) 1, ( ) 1, ( y y y x y x y x y x y x y x y x y y x y x y x x 2 nd Derivative _ Laplacian
64 Use o 2 nd Derivative or Enhancement Laplacian Isotropic ilter response is independent o the direction o the discontinuities in the image to which the ilter is applied. 2 2 x 2 2 y 2
65 Use o 2 nd Derivative or Enhancement Laplacian (x,y) = [9, 1]
66 ), ( ), ( ), ( ), ( ), ( 2 2 y x y x y x y x y x g I the center coeicient o the laplacian mask is negative Use o 2 nd Derivative or Enhancement Laplacian
67 g(x,y) = (x,y) ), ( 2 y x = )], ( 4 1), ( 1), ( ) 1, ( ) 1, ( [ 2 y x y x y x y x y x Use o 2 nd Derivative or Enhancement Laplacian
68 Un-sharp Masking and High-boost Filtering High-boost iltering is used when the original image is blurred and dark. hb 2 A ( x, y) ( x, y) A > 1
69 (b) Un-sharp Masking and High-boost Filtering
70 Combining Spatial Enhancement Methods
71 Combining Spatial Enhancement Methods
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