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# Class Note for ECE 582 at UA-Comp Visn Dig Image Proc

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This 15 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Alabama - Tuscaloosa taught by a professor in Fall. Since its upload, it has received 23 views.

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Date Created: 02/06/15

Image Spatial Enhancement Lecture 4 EE582 Computer Vision amp Digital Image Processing Image Enhancement Purposes Process an image so that the result will be more suitable for a specific application Applications Human Visual subjective no standard Machine Perception objective better recognition results Domains Spatial manipulation of pixel values in an image Spectral modifying of Fourier transform of an image Image Spatial Enhancement Pixel Processing Linear Piecewise transformation lookup table Nonlinear transform Logarithm and Powerlaw Gamma Correction Histogram Processing Histogram normalization Equalization Histogram Matching Specification Area Processing Linear Filters smoothing and sharpening Nonlinear Filters max min median Thresholding Binary map Output gray level 5 3 basic graylevel l transformation functions Lil I Linear function Negative and identity transformations Logarithm function Log and inverse log transformation Powerlaw function nth power and nth root transformations Negative nth root Log nth power i Iden ity Inverse Log n n m m i am L i Input gray level r Gamma Correction Cathode ray tube CRT DLP display devices have an intensitytovoltage Ow response that is a power I function with y varying I from 18 to 25 433g 7 25 The picture will become darker Gamma correction is m done by preprocessing QUgt the image before inputting it to the display K with s cer 1 125 04 Contrast Stretching increase the dynamic range of the gray levels in the image b a lowcontrast image result from poor illumination lack of dynamic range in the imaging sensor or even wrong setting of a lens aperture of image acquisition c result of contrast stretching r1151 rmimo and r2152 rmale1 d result of thresholding 8 bit planes Bitplane 7 Bitplane 6 Bit Bit Bit plane 5 plane 4 plane 3 Bit Bit Bit plane 2 plane 1 plane 0 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 hm O39Fm i Example a i Dark image components of histogra m are concentrated on me iow side ot the gay ii i scaie i Bright image components of histogra m are concentrated on the 39 hiw Side otthe gay 1 a V saie g Example miiNWv Lowrconhmt image histogram is narrow and centered toward the miggie of the gray scaie Highncomrast image histogram covers broad range otthe gray scaie and the dsmbu on of pixeis is not too tar tron untonm With veryfew yerucai iines being much higher than the others y g H stogram transformation xrr mmrausrs Trsaus es a m rssrnqbs vamdar marmade mesmqu m the mhervd svsl h EISYVSlfm usvsl App 39ed to Image Let 110 dmote the PDF of rmdom varrab e r I156 denote the PDF of random varab e s If Mr and m are known and rim satis es condition a men My can be obtained us39ng a formula dr d NS pm sTrjprwdw yields P55 a random variable 5 characterized by a uniform probability function 5 1 Discrete transformation function The probability of occurrence of gray level in an image is approximated by prrk n k where k 0 1 Ll n The discrete version of transformation sk TrkZprJ j0 k 2 J wherek O 1 Ll Histogram Equalization Histogram equalization Example 7 before after Histogram equalization The quality is not improved much because the original image already has a broaden grayslevel scale Example No of pixels 2 4 3 2 3 2 2 4 3 4 3 2 2 3 5 4 4x4 image Gray scale 09 6 FINNhill Gray level 0123456789 histogram Gra y 0 1 2 3 4 5 6 7 8 9 LeveJ N 39 f 0 0 6 5 4 1 0 0 0 0 pixels k 2m 0 0 6 11 15 16 16 16 16 16 j0 WI 6 11 15 16 16 16 16 16 F1320 0 o 16 16 16 16 16 16 16 16 sx9 0 0 3393 6391 8394 9 9 9 9 9 g Example No of pixels 3 8 6 3 6 3 3 8 6 8 6 3 3 6 9 8 Output image Gray scale 09 6 FINNhill 0123456789 Graylevel Histogram equalization g Specification Histogram Matching 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 10 Consider The conTinuous domain LeT pr denoTe conTinuous probabiliTy densiTy funcTion of rayeve of inpuT image r LeT pzz denoTe desired specified conTinuous probabiliTy densiTy funcTion of graylevel of oquuT image 2 LeT s be a random variable wiTh The properTy s T lPrWdW lg HisTogram equalizaTion 0 Where w is a dummy variable of inTegraTion NexT we define a random variable 2 wiTh The properTy z g gzJ pztdts a HisTogram equalizaTion Where T is a dummy variable of inTegraTion Thus S Tquot Gz Therefore 2 musT saTisfy The condiTion z G1s G391Tr Assume 6391 exisTs and saTisfies The condiTion a and b We can map an inpuT gray level r To oquuT gray level 2 11 g Procedure Conclusion 1 Obtain the transformation function Tr by calculating the histogram equalization of the input image s M lpwdw Obtain the transformation function Gz by calculating histogram equalization of the desired density function 60 1 pz 0dr s g Procedure Conclusion 3 4 Obtain the inversed transformation function 6391 z 61 s G1Tr1 Obtain the output image by applying the processed graylevel from the inversed transformation function to all the pixels in the input image 12 g Example Assume an image has a gray level probabili ry densi ry func rion prr39 as shown 2r2 0 r 1 0 elsewhere pr lpwdw 1 g Example 7 We would like To apply The his rogr39am specifica rion wi rh The desired probabili ry densi ry func rion pZz as shown PZzl 2z 0 S z 1 2 z 2 pz 0 elsewhere 1quot z Ipzwdw 1 l r z 0 0 1 2 13 g Step 1 Obtain the Transformation function Tr T s Tr ipwdw 397 j 2w2dw One to one mapping 2 r function Z w 2w 0 0 1 r r2 2r g Step 2 Obtain the transformation function 62 z 2 2 0 Gzj2wdw 22 14 g Step 3 Ob rain The inversed Transformation func rion 6391 GzTr zzz r22r z 2r r2 We can guarantee Tha r 0 s 2 31 when 0 s r39 31 g Discrete formulation sk TrkZprJ k n k 012L 1 0 2 139 J n k Gzk Z pzzi sk k 012L 1 i0 zk G71Trk G 1sk k 012 L 1 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