Digital Image Processing Image Enhancement in the Spatial

ﺑﺴﻤﻪﺗﻌﺎﻟﻲ Digital Image Processing Image Enhancement in the Spatial Domain (Chapter 3) H. R. Pourreza

Objective The principal objective of enhancement is to process an images so that the result is more suitable than the original image for a SPECIFIC application Category of image enhancement • Spatial domain • Frequency domain H. R. Pourreza

Background g(x, y) = T [ f(x, y) ] H. R. Pourreza

Point Processing, Gray-Level Transformation Function s = T (r) Contrast stretching Thresholding H. R. Pourreza

Mask Processing Filtering g(x, y) = T [ f(x, y) ] T 0 1 -1 0 0 f(x, y) g(x, y) H. R. Pourreza 0 0

Basic Gray Level Transformation H. R. Pourreza

Image Negative s=L-1 -r H. R. Pourreza

Log Transformations s = c log(1+r) Pixel values dynamic range=[0 - 1. 5× 106] H. R. Pourreza

Power-Law Transformations s = cr s = r 2. 5 r = [1 10 20 30 40 210 220 230 240 255] s( = 2. 5) = [0 0 0 1 2 157 176 197 219 243 255] s( =. 4) = [28 70 92 108 122 236 240 245 249 253 255] H. R. Pourreza

Power-Law Transformations Automatic Selection H. R. Pourreza

Power-Law Transformations - Gamma Correction H. R. Pourreza

Power-Law Transformations c=1 = 0. 6 c=1 = 0. 4 c=1 = 0. 3 H. R. Pourreza

Power-Law Transformations c=1 =3 c=1 =4 c=1 =5 H. R. Pourreza

Piecewise-Linear Transformation Functions • Advantage • Arbitrarily complex • Disadvantage • More user input • Type of Transformations • • • Contrast stretching Gray-level slicing Bit-plane slicing H. R. Pourreza

Contrast Stretching Objective: Increase the dynamic range of the gray levels in the image Causes for poor image -Poor illumination -Wrong lens aperture -Wrong shutter speed -Lack of dynamic range in the imaging sensor H. R. Pourreza

Contrast Stretching Example 1 For image with intensity range [50 - 150] What should (r 1, s 1) and (r 2, s 2) be to increase the dynamic range of the image to [0 - 255]? s H. R. Pourreza r

Gray-Level Slicing Objective: Highlighting a specific range of gray levels in an image. H. R. Pourreza

Bit-Plane Slicing 255 138 30 65 12 201 180 111 85 MSB LSB MSB plane 1 1 0 0 0 1 1 0 0 LSB plane 1 0 1 1 1 1 255 0 1 138 1 1 30 1 0 65 0 0 12 1 0 201 1 1 183 1 1 111 1 0 85 H. R. Pourreza 1 0 1 0 1 1 0 1 0 0 0 1 1 0 0 1 0 1 1 0 0

Bit-Plane Slicing H. R. Pourreza

Bit-Plane Slicing H. R. Pourreza

Histogram Processing Histogram The histogram of a digital image with gray levels in the range [0, L-1] is a discrete function h ( r k) = n k, where rk is the kth gray level and nk is the number of pixels in the image having gray level rk. Normalized Histogram Dividing each value of the histogram by the total number of pixels in the image, denoted by n. p ( r k) = n k / n. Normalized histogram provide useful image statistics. H. R. Pourreza

Histogram Extraction using Matlab function Normalized_Hist = Img_Hist(img) [R, C]=size(img); Hist=zeros(256, 1); for r = 1: R for c=1: C Hist(img(r, c)+1, 1)=Hist(img(r, c)+1, 1)+1; end Normalized_Hist = Hist/(R*C); plot(Normalized_Hist) H. R. Pourreza

Histogram Processing H. R. Pourreza

Histogram Equalization Histogram equalization is used to enhance image contrast and gray-level detail by spreading the histogram of the original image. s = T ( r ) 0 r 1, where r and s are normalized pixel intensities Conditions for the transformation (a) T( r ) is single-valued and monotonically increasing in the interval 0 r 1 (b) 0 T ( r ) 1 for 0 r 1 1 H. R. Pourreza

Histogram Equalization Objective of histogram equalization Transform the histogram function of the original image pr(r) to a uniform histogram function. ps(s) = 1 0 s 1 pr(r) Equalizer Transformation ps(s) Discrete case Continuous case H. R. Pourreza

Input image Output image k rk pr(rk) sk 0 1 2 3 : : 128 129 130 : : 253 254 255 0 0. 004 0. 008 0. 011 : : 0. 505 0. 51 : : 0. 992 0. 996 1 0 0 : : 0. 004 0. 15 0. 05 : : 0. 005 0. 006 0. 004 0 0 : 0 0. 004 0. 154 0. 204 : : 0. 989 0. 994 1 /255 T (rk) H. R. Pourreza *255 0 0 : : 10 39 52 : : 252 253 255

Histogram Equalization H. R. Pourreza

Histogram Equalization 255 H. R. Pourreza

rk pr(rk) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 0. 09 0. 06 0. 15 0. 3 0. 18 0. 05 0. 07 0 0 Normalized sk 0 0 0. 09 0. 15 0. 25 0. 4 0. 7 0. 88 0. 93 1 1 1 sk 0 0 1 2 4 6 11 13 14 15 15 15 Histogram Equalization 0. 4 0. 3 0. 2 0. 1 4 0. 25 8 0. 5 12 0. 75 15 1 Input image size 100 >>plot(sk) H. R. Pourreza

Histogram Matching Is histogram equalization a good approach to enhance the below image? H. R. Pourreza

Histogram Matching method generates a processed image that has a specified histogram. G (z) Desired Output image v pz(z) T (r) Input image s pr(r) z = G-1[T( s )] s v, since both will have approximately uniform pdf H. R. Pourreza s = G[T( r )]

Histogram Matching H. R. Pourreza

Histogram Matching H. R. Pourreza

Histogram Equalization H. R. Pourreza 0. 6 0. 12 0. 08 0. 0 : 0. 01 0. 005 0 0. 6 0. 72 0. 8 : 0. 95 1 153 184 204 : 230 242 255

Histogram Matching 1) Modify the histogram of the image to obtain pz(z). 2) Find the transformation function G (z) using the modified histogram in step 1. 3) Find the inverse G-1 (z) 4) Enhanced image is obtained by applying G-1 to the pixels of the histogram-equalized image shown in Fig. 3. 21. c H. R. Pourreza 1 G (z) 2 G-1 (z)

Local Histogram Enhancement Global Histogram 0. 98 0. 002 125 128 Local Histogram 250 255 0. 7 0. 3 178 255 H. R. Pourreza

Local Histogram Enhancement H. R. Pourreza

Histogram Statistics for Image Enhanc. Contrast manipulation using local statistics, such as the mean and variance, is useful for images where part of the image is acceptable, but other parts may contain hidden features of interest. H. R. Pourreza

Histogram Statistics for Image Enhanc. Let (x, y) be the coordinates of a pixel in an image, and let Sxy denote a neighborhood (sub-image) of specified size, centered at (x, y). The local mean and variance are the decision factors to whether apply local enhancement or not. H. R. Pourreza

Histogram Statistics for Image Enhanc. MG: Global mean DG: Global standard deviation E, k 0, k 1, k 2: Specified parameters Size of local area = 3 3 H. R. Pourreza E = 4, k 0 = 0. 4, k 1= 0. 02, k 2 = 0. 4

Histogram Statistics for Image Enhanc. E (x, y) Image 125 168 210 187 135 175 231 192 145 158 215 145 Mean 135 149 129 200 . . . . 174 164 187 H. R. Pourreza

Histogram Statistics for Image Enhanc. Original Image Enhanced Image H. R. Pourreza

Enhancement Using Arithmetic/Logic Op. Arithmetic/logic operations involving images are performed on a pixel-by-pixel basis between two or more images. Arithmetic Operations Addition, Subtraction, Multiplication, and Division Logic Operations AND, OR, NOT H. R. Pourreza

Enhancement Using AND and OR Logic Op. AND OR Logic operations are performed on the binary representation of the pixel intensities H. R. Pourreza

Enhancement Using Arithmetic Op. _ SUB Original 4 Upper-order Bit HE of Error H. R. Pourreza

Enhancement Using Arithmetic Op. _ SUB Mask Mode Radiography Problem: The pixel intensities in the difference image can range from -255 to 255. Solutions: 1) Add 255 to every pixel and then divide by 2. 2) Add the minimum value of the pixel intensity in the difference image to every pixel and then divide by 255/Max. Max is the maximum pixel value in the modified difference image. H. R. Pourreza

Enhancement Using Arithmetic Op. Averaging Original image H. R. Pourreza Noise with zero mean

Enhancement Using Arithmetic Op. Averaging f(x, y) g(x, y) K=8 K = 16 K = 128 K = 32 H. R. Pourreza Gaussian Noise mean = 0 variance = 64 - + x

Enhancement Using Arithmetic Op. Averaging K=8 Difference images between original image and images obtained from averaging. K = 16 K = 32 K = 128 H. R. Pourreza Notice the mean and variance of the difference images decrease as K increases.

Basics of Spatial Filtering - Linear Spatial filtering are filtering operations performed on the pixel intensities of an image and not on the frequency components of the image. a = (m - 1) / 2 H. R. Pourreza b = (n - 1) / 2

Basics of Spatial Filtering Response, R, of an m n mask at any point (x, y) Special consideration is given when the center of the filter approach the boarder of the image. H. R. Pourreza

Nonlinear of Spatial Filtering Nonlinear spatial filters operate on neighborhoods, and the mechanics of sliding a mask past an image are the same as was just outlined. In general however, the filtering operation is based conditionally on the values of the pixel in the neighborhood under consideration, and they do not explicitly use coefficients in the sum-of products manner described previously. Example Computation for the median is a nonlinear operation. H. R. Pourreza

Smoothing Spatial Filtering - Linear Averaging (low-pass) Filters Smoothing filters are used - Noise reduction - Smoothing of false contours - Reduction of irrelevant detail Undesirable side effect of smoothing filters - Blur edges Weighted average filter reduces blurring in the smoothing process. Box filter H. R. Pourreza Weighted average

Smoothing Spatial Filtering _ Linear Averaging (low-pass) Filters n = filter size n=3 n=5 n=9 n = 35 n = 15 H. R. Pourreza

Smoothing Spatial Filtering Averaging & Threshold filter size n = 15 H. R. Pourreza Thrsh = 25% of highest intensity

Smoothing Spatial Filtering Order Statistic Filters Order-statistics filters are nonlinear spatial filters whose response is based on ordering (ranking) the pixels contained in the image area encompassed by the filter, and then replacing the value of the center pixel with the value determined by the ranking result. 3 3 Median filter [10 125 135 141 144 230 240] = 141 3 3 Max filter [10 125 135 141 144 230 240] = 240 3 3 Min filter [10 125 135 141 144 230 240] = 10 Median filter eliminates isolated clusters of pixels that are light or dark with respect to their neighbors, and whose area is less than n 2/2. H. R. Pourreza

Order Statistic Filters n=3 Average filter H. R. Pourreza n=3 Median filter

Sharpening Spatial Filters The principal objective of sharpening is to highlight fine detail in an image or to enhance detail that has been blurred. Image Blurred Image The derivatives of a digital function are defined in terms of differences. H. R. Pourreza

Sharpening Spatial Filters Requirements for digital derivative First derivative 1) Must be zero in flat segment 2) Must be nonzero along ramps. 3) Must be nonzero at the onset of a gray-level step or ramp Second derivative 1) Must be zero in flat segment 2) Must be zero along ramps. 3) Must be nonzero at the onset and end of a gray-level step or ramp H. R. Pourreza

Sharpening Spatial Filters H. R. Pourreza

Sharpening Spatial Filters Comparing the response between first- and second-ordered derivatives: 1) First-order derivative produce thicker edge 2) Second-order derivative have a stronger response to fine 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 first derivative for image enhancement. The principle use of first derivative is for edge extraction. H. R. Pourreza

Use of First Derivative for Edge Extraction Gradient First derivatives in image processing are implemented using the magnitude of the gradient. z 1 z 2 z 3 z 4 z 5 z 6 z 7 z 8 z 9 Roberts operator Gx = (z 9 -z 5) and Gy = (z 8 - z 6) Sobel operator Gx = (z 3+2 z 6 +z 9) - (z 1+2 z 4+z 7) and Gy = (z 7+2 z 8+z 9) - (z 1+2 z 2+z 3) H. R. Pourreza

Use of First Derivative for Edge Extraction Gradient Robert operator H. R. Pourreza Sobel operators

Use of First Derivative for Edge Extraction Gradient -1 -2 -1 -1 f(x, y) = [40, 140] …. 0 …. 0 -1 1 -1 0 1 -2 0 -1 0 2 0 1 1 0 200 400 400 1 H. R. Pourreza 0 2 0 1 0 0 400 400 …… 600 400 400 …. . . 400 0 0 0 400 0 … 0 … 0 0 100 200 0 0 0 200 0 0

Use of First Derivative for Edge Extraction Gradient H. R. Pourreza

2 nd Derivative _ Laplacian H. R. Pourreza

Use of 2 nd Derivative for Enhancement Laplacian Isotropic filter response is independent of the direction of the discontinuities in the image to which the filter is applied. H. R. Pourreza

Use of 2 nd Derivative for Enhancement Laplacian f(x, y) = [90, 100] 0 1 1 -4 1 1 0 0 0 … 0 10 0 … 10 -40 10 … 0 10 0 … 0 0 0 H. R. Pourreza 0 0 0 … … …

Use of 2 nd Derivative for Enhancement Laplacian 1 1 1 -8 1 1 1 If the center coefficient of the laplacian mask is negative H. R. Pourreza

Use of 2 nd Derivative for Enhancement Laplacian 0 g(x, y) = f(x, y) - 1 0 -4 1 = 1 0 H. R. Pourreza

Un-sharp Masking and High-boost Filtering High-boost filtering is used when the original image is blurred and dark. A>1 H. R. Pourreza

Un-sharp Masking and High-boost Filtering (b) H. R. Pourreza

Combining Spatial Enhancement Methods H. R. Pourreza

Combining Spatial Enhancement Methods H. R. Pourreza