Dip3

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Dip3

  1. 1.   Prof. Duong Anh Duc
  2. 2.  To process an image so that output is “visually better” than the input, for a specific application.  Enhancement is therefore, very much dependent on the particular problem/image at hand.  Enhancement can be done in either: o Spatial domain: operate on the original image g (m, n) = T [f (m, n)] o Frequency domain: operate on the DFT of the original image G (u, v) = T [F (u, v)] where F(u, v) = F [f(m, n)], and G(u, v) = F [g(m, n)] 2
  3. 3. 3 Image Enhancement Techniques Point Operations Mask Operations Transform Operations Coloring Operations • Image Negative • Contrast Stretching • Compression of dynamic range • Graylevel slicing • Image Subtraction • Image Averaging • Histogram operations • Smoothing operations • Median Filtering • Sharpening operations • Derivative operations • Histogram operations • Low pass Filtering • Hi pass Filtering • Band pass Filtering • Homomorphic Filtering • Histogram operations • False Coloring • Full color Processing
  4. 4. 4  Output pixel value g(m, n) at pixel (m, n) depends only on the input pixel value at f(m, n) at (m, n) (and not on the neighboring pixel values).  We normally write s = T(r), where s is the output pixel value and r is the input pixel value.  T is any increasing function that maps [0,1] into [0,1].
  5. 5. 5 T(r) = s = L – 1 – r, L: max grayvalue
  6. 6. 6
  7. 7.  Increase the dynamic range of grayvalues in the input image.  Suppose you are interested in stretching the input intensity values in the interval [r1, r2]:  Note that (r1 – r2) < (s1 – s2). The grayvalues in the range [r1, r2] is stretched into the range [s1, s2]. 7
  8. 8. 8  Special cases: o Thresholding or binarization  r1 = r2 , s1 = 0 and s2 = 1 o Useful when we are only interested in the shape of the objects and on their actual grayvalues.
  9. 9. 9
  10. 10.  Special cases (cont.): o Gamma correction: s1 = 0, s2 = 1 and 10 2 21 12 1 1 ,1 , ,0 rr rrr rr rr rr rT g
  11. 11. 11 Gamma correction
  12. 12.  When the dynamic range of the input grayvalues is large compared to that of the display, we need to “compress” the grayvalue range --- example: Fourier transform magnitude.  Typically we use a log scale. s = T(r) = c log(1+ r ) 12
  13. 13. 13 Saturn Image Mag. Spectrum Mag. Spectrum in log scale
  14. 14.  Graylevel Slicing: Highlight a specific range of grayvalues. 14
  15. 15.  Example: 15 Original Image Highlighted Image (no background) Highlighted Image (with background)
  16. 16.  Bitplane Slicing: Display the different bits as individual binary images. 16
  17. 17. 17
  18. 18.  In this case, the difference between two “similar” images is computed to highlight or enhance the differences between them: g(m, n) = f1(m, n) – f2(m, n)  It has applications in image segmentation and enhancement 18
  19. 19. 19 f1(m, n): Image before dye injection f2(m, n): Image after dye injection g(m, n): Image after dye injection, followed by subtraction
  20. 20.  Noise is any random (unpredictable) phenomenon that contaminates an image.  Noise is inherent in most practical systems: o Image acquisition o Image transmission o Image recording  Noise is typically modeled as an additive process: g (m,n) = f (m, n) + (m, n) 20 Noisy Image Noise-free Image Noise
  21. 21.  The noise h(m, n) at each pixel (m, n) is modeled as a random variable.  Usually, h(m, n) has zero-mean and the noise values at different pixels are uncorrelated.  Suppose we have M observations {gi (m, n)}, i =1, 2, …, M, we can (partially) mitigate the effect of noise by “averaging” 21 M i i nmg M nmg 1 , 1 ,
  22. 22.  In this case, we can show that:  Therefore, as the number of observations increases (M ), the effect of noise tends to zero. 22 nm M nmg nmfnmgE ,Var 1 ,Var ,,
  23. 23. 23 Noise-free Image Noisy Image Noise Variance = 0.05
  24. 24. 24 M =2 M =5
  25. 25. 25 M =10 M =25
  26. 26. 26 M =50 M =100
  27. 27. 27
  28. 28.  The histogram of a digital image with grayvalues r0, r1, …, rL – 1 is the discrete function  The function p(rk) represents the fraction of the total number of pixels with grayvalue rk.  Histogram provides a global description of the appearance of the image. 28 imageinpixels#total ewith valupixels# where, n rn n n rp kkk k
  29. 29.  If we consider the grayvalues in the image as realizations of a random variable R, with some probability density, histogram provides an approximation to this probability density. In other words, Pr[R = rk] p(rk) 29
  30. 30.  The shape of a histogram provides useful information for contrast enhancement. 30
  31. 31.  The shape of a histogram provides useful information for contrast enhancement. 31
  32. 32. The original image displayed The stretched image 0 0.5 1 0 5000 10000 The histogram of the original image 0 0.5 1 0 5000 10000 The stretched histogram 32
  33. 33.  Idea: find a non-linear transformation g = T (f ) to be applied to each pixel of the input image f(x, y), such that a uniform distribution of gray levels in the entire range results for the output image g(x, y). 33
  34. 34.  Let us assume for the moment that the input image to be enhanced has continuous grayvalues, with r = 0 representing black and r = 1 representing white.  We need to design a grayvalue transformation s = T(r), based on the histogram of the input image, which will enhance the image.  As before, we assume that: o T(r) is a monotonically increasing function for 0 r 1 (preserves order from black to white) o T(r) maps [0,1] into [0,1] (preserves the range of allowed grayvalues). 34
  35. 35. 35
  36. 36.  Let us denote the inverse transformation by r = T – 1(s). We assume that the inverse transformation also satisfies the above two conditions.  We consider the grayvalues in the input image and output image as random variables in the interval [0, 1].  Let pin(r) and pout(s) denote the probability density of the grayvalues in the input and output images. 36
  37. 37.  If pin(r) and T(r) are known, and T– 1(s) satisfies condition 1, we can write (result from probability theory):  One way to enhance the image is to design a transformation T(.) such that the grayvalues in the output is uniformly distributed in [0, 1], i.e. pout(s)=1, 0 s 1 37 sTr inout ds dr rpsp 1
  38. 38.  In terms of histograms, the output image will have all grayvalues in “equal proportion.”  This technique is called histogram equalization.  Consider the transformation 38 r o rdwwprTs 10,in
  39. 39.  Note that this is the cumulative distribution function (CDF) of pin(r) and satisfies the previous two conditions.  From the previous equation and using the fundamental theorem of calculus, 39 rp dr ds in
  40. 40.  Therefore, the output histogram is given by  The output probability density function is uniform, regardless of the input. 40 10for,11 1 1 1 s rp rpsp sTr sTrin inout
  41. 41.  Thus, using a transformation function equal to the CDF of input grayvalues r, we can obtain an image with uniform grayvalues.  This usually results in an enhanced image, with an increase in the dynamic range of pixel values. 41
  42. 42. 42 Original image pout after histogram equalization
  43. 43. 43
  44. 44.  For images with discrete grayvalues, we have o L: Total number of graylevels o nk: Number of pixels with grayvalue rk o n: Total number of pixels in the image 44 10and,10for,in Lkr n n rp k k
  45. 45.  The discrete version of the previous transformation based on CDF is given by: 45 10for, 0 in 0 Lkrp n n rTs k i i k i i kk
  46. 46.  Consider an 8-level 64 x 64 image with grayvalues (0, 1, …, 7).  The normalized grayvalues are (0, 1/7, 2/7, …, 1). The normalized histogram is given in the table: k rk nk p(rk)= nk / n 0 0 790 0.19 1 1/7 1023 0.25 2 2/7 850 0.21 3 3/7 656 0.16 4 4/7 329 0.08 5 5/7 245 0.06 6 6/7 122 0.03 7 1 81 0.02 46
  47. 47. 47
  48. 48.  Applying the previous transformation, we have (after rounding off to nearest graylevel):  Notice that there are only five distinct graylevels --- (1/7, 3/7, 5/7, 6/7, 1) in the output image. We will relabel them as (s0, s1, …, s4). 48 100.1 198.0 195.0 7 689.0 7 681.0 7 565.0 7 344.0 7 119.0 7in1in0in 7 0 in77 6in1in0in 6 0 in66 5in1in0in 5 0 in55 4in1in0in 4 0 in44 3in1in0in 3 0 in33 2in1in0in 2 0 in22 1in0in 1 0 in11 0in 0 0 in00 rprprprprTs rprprprprTs rprprprprTs rprprprprTs rprprprprTs rprprprprTs rprprprTs rprprTs i i i i i i i i i i i i i i i i     
  49. 49.  With this transformation, the output image will have histogram 49 k sk nk p(sk)=nk / n 0 1/7 790 0.19 1 3/7 1023 0.25 2 5/7 850 0.21 3 6/7 985 0.24 4 1 448 0.11
  50. 50. 50  Note that the histogram of output image is only approximately, and not exactly, uniform. This should not be surprising, since there is no result that claims uniformity in the discrete case.
  51. 51. 51 Original image Equalized image
  52. 52. 52 Histogram of original image Histogram of equalized image
  53. 53. 53 Original image Equalized image
  54. 54. 54 Histogram of original image Histogram of equalized image
  55. 55. 55 Original image Equalized image
  56. 56. 56 Histogram of original image Histogram of equalized image
  57. 57. 57 Original image Equalized image
  58. 58. 58 Histogram of original image Histogram of equalized image
  59. 59.  Histogram equalization may not always produce desirable results, particularly if the given histogram is very narrow. It can produce false edges and regions. It can also increase image “graininess” and “patchiness.” 59
  60. 60.  Histogram equalization yields an image whose pixels are (in theory) uniformly distributed among all graylevels.  Sometimes, this may not be desirable. Instead, we may want a transformation that yields an output image with a prespecified histogram. This technique is called histogram specification.  Again, we will assume, for the moment, continuous grayvalues. 60
  61. 61.  Suppose, the input image has probability density pin(r). We want to find a transformation z = H(r), such that the probability density of the new image obtained by this transformation is pout(z), which is not necessarily uniform.  First apply the transformation This gives an image with a uniform probability density. 61 *10,in r o rdwwprTs
  62. 62.  If the desired output image were available, then the following transformation would generate an image with uniform density:  From the grayvalues we can obtain the grayvalues z by using the inverse transformation, z = G– 1( ). 62 **10,out z o zdwwpzG
  63. 63.  If instead of using the grayvalues n obtained from (**), we use the grayvalues s obtained from (*) above (both are uniformly distributed!), then the point transformation z = H(r) = G– 1[T(r)] will generate an image with the specified density pout(z), from an input image with density pin(r)  For discrete graylevels, we have 63 10for, and10for, 0 out 0 LkzpzG Lk n n rTs k i ikk k i i kk
  64. 64.  If the transformation zk G(zk) is one-to-one, the inverse transformation sk G– 1(sk), can be easily determined, since we are dealing with a small set of discrete grayvalues.  In practice, this is not usually the case (i.e., zk G(zk) is not one-to- one) and we assign grayvalues to match the given histogram, as closely as possible. 64
  65. 65.  Consider the previous 8-graylevel 64 x 64 image histogram: 65 k rk nk p(rk)=nk/n 0 0 790 0.19 1 1/7 1023 0.25 2 2/7 850 0.21 3 3/7 656 0.16 4 4/7 329 0.08 5 5/7 245 0.06 6 6/7 122 0.03 7 1 81 0.02
  66. 66.  It is desired to transform this image into a new image, using a transformation z=H(r)= G– 1[T(r)], with histogram as specified below: 66 k zk pout(zk) 0 0 0.00 1 1/7 0.00 2 2/7 0.00 3 3/7 0.15 4 4/7 0.20 5 5/7 0.30 6 6/7 0.20 7 1 0.15
  67. 67.  The transformation T(r) was obtained earlier (reproduced below): 67 ri sk nk p(sk) r0 s0 = 1/7 790 0.19 r1 s1 = 3/7 1023 0.25 r2 s2 = 5/7 850 0.21 r3,r4 s3 = 6/7 985 0.24 r5, r6,r7 s4=1 448 0.11
  68. 68.  Next we compute the transformation G as before 100.1 7 685.0 7 565.0 7 235.0 7 115.0 000.0 000.0 000.0 7out1out0out 7 0 out77 6out1out0out 6 0 out66 5out1out0out 5 0 out55 4out1out0out 4 0 out44 3out1out0out 3 0 out33 2out1out0out 2 0 out22 1out0out 1 0 out11 0out 0 0 out00 zpzpzpzpzG zpzpzpzpzG zpzpzpzpzG zpzpzpzpzG zpzpzpzpzG zpzpzpzpzG zpzpzpzG zpzpzG i i i i i i i i i i i i i i i i      68
  69. 69.  Notice that G is not invertible. But we will do the best possible by setting o G– 1 (0) = ? (This does not matter since s 0) o G– 1 (1/7) = 3/7 o G– 1 (2/7) = 4/7 (This does not matter since s 2/7) o G– 1 (3/7) = 4/7 (This is not defined, but we use a close match) o G– 1 (4/7) = ? (This does not matter since s 4/7) o G– 1 (5/7) = 5/7 o G– 1 (6/7) = 6/7 o G– 1 (1) = 1 69
  70. 70.  Combining the two transformation T and G– 1, we get our required transformation H 70 r T(r)=s s G–1(s)=z r G–1 [T(r)]=H(r)=z r0 = 0 1/7 0 ? r0 = 0 z3= 3/7 r1 = 1/7 3/7 1/7 3/7 r1 = 1/7 z4= 4/7 r2 = 2/7 5/7 2/7 4/7 r2 = 2/7 z5= 5/7 r3 = 3/7 6/7 3/7 4/7 r3 = 3/7 z6= 6/7 r4 = 4/7 6/7 4/7 ? r4 = 4/7 z6= 6/7 r5 = 5/7 1 5/7 5/7 r5 = 5/7 z7= 1 r6 = 6/7 1 6/7 6/7 r6 = 6/7 z7= 1 r7 = 1 1 1 1 r7 = 1 z7= 1
  71. 71.  Applying the transformation H to the original image yields an image with histogram as below: 71 k zk nk nk/n (actual hist.) pout(zk) (spec. hist.) 0 0 0 0.00 0.00 1 1/7 0 0.00 0.00 2 2/7 0 0.00 0.00 3 3/7 790 0.19 0.15 4 4/7 1023 0.25 0.20 5 5/7 850 0.21 0.30 6 6/7 985 0.24 0.20 7 1 448 0.11 0.15
  72. 72. 72
  73. 73.  Again, the actual histogram of the output image does not exactly but only approximately matches with the specified histogram. This is because we are dealing with discrete histograms. 73
  74. 74. 74 Original image and its histogram
  75. 75. 75 Histogram equalized image Actual histogram of output
  76. 76. 76Histogram specified image, Actual Histogram, and Specified Histogram
  77. 77.  Used to enhance details over small portions of the image.  Define a square or rectangular neighborhood, whose center moves from pixel to pixel.  Compute local histogram based on the chosen neighborhood for each point and apply a histogram equalization or histogram specification transformation to the center pixel.  Non-overlapping neighborhoods can also be used to reduce computations. But this usually results in some artifacts (checkerboard like pattern). 77
  78. 78.  Another use of histogram information in image enhancement is the statistical moments associated with the histogram (recall that the histogram can be thought of as a probability density function).  For example, we can use the local mean and variance to determine the local brightness/contrast of a pixel. This information can then be used to determine what, if any transformation to apply to that pixel.  Note that local histogram based operations are non-uniform in the sense that a different transformation is applied to each pixel. 78

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