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Histogram processing

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This slide show explains the concept of histogram processing.

This slide show explains the concept of histogram processing.

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  • 1. DIGITAL IMAGE PROCESSINGIMAGE ENHANCEMENT Histogram Processing by Paresh Kamble
  • 2. Histogram ProcessingHistogram: It is a plot of frequency of occurrence of an event. freq. of occurrence 0 1 2 3 event
  • 3. Histogram Processing Histogram of images provide a global description of their appearance. Enormous information is obtained. It is a spatial domain technique. Histogram of an image represents relative frequency of occurrence of various gray levels. Histogram can be plotted in two ways:
  • 4. Histogram Processing First Method: X-axis has gray levels & Y-axis has No. of pixels in each gray levels. nkGray Level No. of Pixels (nk) 40 0 40 30 1 20 2 10 3 15 20 4 10 5 3 6 2 10 0 1 2 3 4 5 6 Gray Levels
  • 5. Histogram Processing Second Method: X-axis has gray levels & Y-axis has probability of occurrence of gray levels. P(µk) = nk / n; where, µk – gray level nk – no. of pixels in kth gray level n – total number of pixels in an image Prob. Of Occurrence 1 0.8Gray Level No. of Pixels (nk) P(µk) 0.7 0 40 0.4 0.6 1 20 0.2 0.5 2 10 0.1 0.4 3 15 0.15 0.3 4 10 0.1 0.2 5 3 0.03 0.1 6 2 0.02 n = 100 1 0 1 2 3 4 5 6 µk
  • 6. Histogram Processing Advantage of 2nd method: Maximum value plotted will always be 1. White – 1, Black – 0. Great deal of information can be obtained just by looking at histogram. Types of Histograms: P(µk) P(µ k) 0 rk 0 rk P(µk) P(µ k) 0 rk 0 rk
  • 7. Histogram Processing Advantage of 2nd method: Maximum value plotted will always be 1. White – 1, Black – 0. Great deal of information can be obtained just by looking at histogram. Types of Histograms: P(µk) Dark Image P(µ k) Bright Image 0 rk 0 rk P(µk) Low Contrast Image P(µ k) Equalized Image 0 rk 0 rk
  • 8. Histogram Processing The last graph represent the best image. It is a high contrast image. Our aim would be to transform the first 3 histograms into the 4th type. In other words we try to increase the dynamic range of the image.
  • 9. Histogram Stretching1) Linear stretching: Here, we don’t alter the basic shape. We use basic equation of a straight line having a slope. (Smax - Smin)/(rmax - rmin)Where, Smax – max gray level of output imageSmin – min gray level of output imagermax – max gray level of input imagermin – min gray level of input image 0 min max 0 min max
  • 10. Histogram Processing S = T(r) = ((Smax - Smin)/(rmax - rmin))(r - rmin) + SminSmaxSmin rmin rmax
  • 11. Histogram ProcessingEx. 1) Perform Histogram Stretching so that the new image has a dynamic range of 0 to 7 [0, 7]. Gray Levels 0 1 2 3 4 5 6 7 No. of Pixels 0 0 50 60 50 20 10 0
  • 12. Histogram ProcessingEx. 1) Perform Histogram Stretching so that the new image has a dynamic range of 0 to 7 [0, 7]. Gray Levels 0 1 2 3 4 5 6 7 No. of Pixels 0 0 50 60 50 20 10 0Soln:- rmin = 2; rmax = 6; smin = 0; smax = 7; slope = ((smax – smin)/(rmax - rmin)) = ((7 - 0)/(6 - 2)) = 7 / 4 = 1.75.S = (7 / 4)(r - 2) + 0; r (7 / 4)(r - 2) = SS = (7 / 4)(r - 2) 2 0 =0 3 7/4 = 1.75 = 2 4 7/2 = 3.5 = 4 5 21/4 = 5.25 = 5 6 7 =7
  • 13. Histogram Processing Ex. 1) Perform Histogram Stretching so that the new image has a dynamic range of 0 to 7 [0, 7]. Gray Levels 0 1 2 3 4 5 6 7 No. of Pixels 50 0 60 0 50 20 0 1060 6050 5040 4030 3020 2010 10 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
  • 14. Histogram ProcessingEx. 2) Perform Histogram Stretching so that the new image has a dynamic range of 0 to 7 . Gray Levels 0 1 2 3 4 5 6 7 No. of Pixels 100 90 85 70 0 0 0 0
  • 15. Histogram ProcessingEx. 2) Perform Histogram Stretching so that the new image has a dynamic range of 0 to 7 . Gray Levels 0 1 2 3 4 5 6 7 No. of Pixels 100 0 90 0 0 85 0 70Dynamic range increases BUT No. of pixels at each gray level remains constant.
  • 16. Histogram Equalization2) Histogram Equalization:o Linear stretching is a good technique but not perfect since the shape remains the same.o Most of the time we need a flat histogram.o It can’t be achieved by Histogram Stretching.o Thus, new technique of Histogram Equalization came into use.o Perfect image is one where all gray levels have equal number of pixels.o Here, our objective is not only to spread the dynamic range but also to have equal pixels at all gray levels.
  • 17. Histogram Equalization nk Normal Histogram rnk Equalized Histogram nk Linearly Stretched Histogram S S
  • 18. Histogram Equalizationo We have to search for a transform that converts any random histogram into flat histogram.o S = T(r)o We have to find ‘T’ which produces equal values in each gray levels.o The Transform should satisfy following 2 conditions:(i) T(r) must be single value & monotonically increasing in the interval, 0 ≤ r ≤ 1.(ii) 0 ≤ T(r) ≤ 1 for 0 ≤ r ≤ 1 0 ≤ S ≤ 1 for 0 ≤ r ≤ 1Here, range of r is [0, 1] (Normalized range) instead of [0, 255].
  • 19. Histogram EqualizationThese two conditions mean:1) If T(r) is not single value: S T(r) S T(r) S2 S1 S1 r1 r2 r r1 r2 r fig. (a) fig. (b)
  • 20. Histogram Equalization In fig.(a) r1 & r2 are mapped as single gray level i.e. S1. Two different gray levels occupy same in modified image. Big Drawback! Hence, transformation has to be single value. This preserves order from black to white. A gray level transformation with single value and monotonically increasing is shown in fig.(b).2) If condition (ii) is not satisfied then mapping will not be consistent with the allowed range of pixel value. S will go beyond the valid range.
  • 21. Histogram Equalization Since, the Transformation is single value & monotonically increasing, the inverse Transformation exists. r = T-1(S) ; 0 ≤ S ≤ 1 Gray levels for continuous variables can be characterized by their probability density Pr(r) & Ps(S). From Probability theory, we know that, If Pr(r) & Ps(S) are known & if T-1(S) satisfies condition (i) then the probability density of the transferred gray level is Ps(S) = [Pr(r). dr/ ds] r = T-1(S) ----(a)Prob. Density (Transformed Image) = Prob. Density (Original Image) x inv. Slope of transformation
  • 22. Histogram Equalization Now lets find a transform which would give us a flat histogram. Cumulative Density Function (CDF) is preferred. CDF is obtained by simply adding up all the PDF’s. S = T(r) r S = § Pr(r) dr ; 0 ≤ r ≤ 1 0 diff. wrt. r ds / dr = Pr(r) ---------(b)Equating eqn(a) & eqn(b), we get Ps(s) = *1+ ; 0 ≤ S ≤ 1 i.e. Ps(s) = 1
  • 23. Histogram Equalization A bad Histogram becomes a flat histogram when we find the CDF.Pr(r) Ps(s) 1 1 CDF Pr(r) dr 0 1 f 0 1 S Non-Uniform Function Uniform Function
  • 24. Histogram Equalization NOTE: If ‘n’ is total no. of pixels in an image, then PDF is given by: Pr(rk) = nk / n;In frequency domain, r S = T(r) = § Pr(rk) dr 0In discrete domain, r Sk = T(rk) = Pr(rk) (CDF) k=0
  • 25. Histogram EqualizationProb. 1) Equalize the given histogram Gray Levels (r) 0 1 2 3 4 5 6 7 No. of Pixels 790 1023 850 656 329 245 122 81 nk 1000 900 800 700 600 500 400 300 200 100 0 1 2 3 4 5 6 7 r
  • 26. Histogram EqualizationGray Levels No. of Pixels (PDF) (CDF) (L-1) Sk = 7 x Sk Rounding (rk) nk Pr(rk) = nk/n Sk = Pk(rk) off 0 790 0.19 0.19 1.33 1 1 1023 0.25 0.44 3.08 3 2 830 0.21 0.65 4.55 5 3 656 0.16 0.81 5.67 6 4 329 0.08 0.89 6.23 6 5 245 0.06 0.95 6.65 7 6 122 0.03 0.98 6.86 7 7 81 0.02 1 7 7 n = 4096 1 Equating Gray Levels to No. of Pixels: 0 -> 0 4 -> 0 1 -> 790 5 -> 850 2 -> 0 6 -> 985 3 -> 1023 7 -> 448 Total : 4096 Hence Verified!!!
  • 27. Histogram EqualizationEqualized Histogram: Gray Levels (S) 0 1 2 3 4 5 6 7 No. of Pixels 0 790 0 1023 0 850 985 448 Total: 4096 nk 1000 900 800 700 600 500 400 300 200 100 0 1 2 3 4 5 6 7 r
  • 28. Histogram EqualizationProb. 2) Equalize the given histogram Gray Levels (r) 0 1 2 3 4 5 6 7 No. of Pixels 100 90 50 20 0 0 0 0 nk 100 90 80 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 r
  • 29. Histogram EqualizationProb. 3) Equalize the above histogram twice. Gray Levels (r) 0 1 2 3 4 5 6 7 No. of Pixels 0 0 0 100 0 90 50 20 nk 100 90 80 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 r
  • 30. Histogram EqualizationProb. 3) Equalize the above histogram twice. Gray Levels (r) 0 1 2 3 No. of Pixels 70 20 7 3 nk 100 90 80 70 60 50 40 30 20 10 0 1 2 3 r
  • 31. Histogram EqualizationProb. 1) What effect would setting to zero the lower order bit plane have on the histogram of an image shown? 0 1 2 3 no. of 4 5 6 7 pixels 1 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 no. of levels
  • 32. Histogram EqualizationProb. 1) What effect would setting to zero the lower order bit plane have on the histogram of an image shown? 0 1 2 3 4 5 6 7 8 9 10 11 0000 0001 0010 0011 12 13 14 15 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
  • 33. Histogram EqualizationProb. 1) What effect would setting to zero the lower order bit plane have on the histogram of an image shown? 0 1 2 3 4 5 6 7 8 9 10 11 0000 0001 0010 0011 12 13 14 15 0100 0101 0110 0111 1000 1001 1010 1011 0000 0000 0000 0000 1100 1101 1110 1111 0100 0100 0100 0100 1000 1000 1000 1000 1100 1100 1100 1100 Setting lower order bit plane to zero.
  • 34. Histogram EqualizationProb. 1) What effect would setting to zero the lower order bit plane have on the histogram of an image shown? 0 0 0 0 0 1 2 3 4 4 4 4 4 5 6 7 8 8 8 8 8 9 10 11 0000 0001 0010 0011 12 12 12 12 12 13 14 15 0100 0101 0110 0111 1000 1001 1010 1011 0000 0000 0000 0000 1100 1101 1110 1111 0100 0100 0100 0100 1000 1000 1000 1000 1100 1100 1100 1100 Setting lower order bit plane to zero.
  • 35. Histogram Equalization Prob. 1) What effect would setting to zero the lower order bit plane have on the histogram of an image shown? 0 0 0 0 0 1 2 3 4 4 4 4 4 5 6 7 8 8 8 8 8 9 10 11 12 12 12 12 12 13 14 15 0000 0000 0000 0000 4 0100 0100 0100 0100no. ofpixels 1000 1000 1000 1000 1100 1100 1100 1100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 no. of levels
  • 36. Histogram EqualizationProb. 2) What effect would setting to zero the higher order bit plane have on the histogram of an image shown? Comment on the image. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
  • 37. Histogram EqualizationProb. 3) Given below is the slope transformation & the image. Draw the frequency tables for the original & the new image. Assume Imin=0, Imax = 7 2 3 4 2 5 5 2 4 I 3 6 3 5 Imax 6 3 5 5 Imin mmin mmax µ
  • 38. Frequency Domain2-D Discrete Fourier Transform & Its Inverse:If f(x, y) is a digital image of size M X N then its Fourier Transform is given by: M-1 N-1 F(u, v) = f(x, y) e -j2π(ux/M + vy/N) -----------(1) x=0 y=0 for u = 0, 1, 2, ……., M-1 & v = 0, 1, 2, …….., N-1.Given the transform F(u, v), we can obtain f(x, y) by using Inverse discrete Fourier Transform (IDFT): M-1 N-1 f(x, y) = (1/MN) F(u, v) e j2π(ux/M + vy/N) ------------(2) u=0 v=0 for x = 0, 1, 2, ….M-1 & y = 0, 1, 2, ….., N-1Eqn. (1) & (2) constitute the 2-D discrete Fourier transform pair.
  • 39. Frequency Domain2-D convolution Theorem:The expression of circular convolution is extended for 2-D is given below: M-1 N-1 f(x, y)*h(x, y) = f(m, n)h(x-m, y-n) -----------(3) m=0 n=0 for x = 0, 1, 2, ….M-1 & y = 0, 1, 2, …..N-1The 2-D convolution theorem is given by the expressions f(x, y) ¤ h(x, y) F(u, v) x H(u, v) --------------(4)and conversely f(x, y) x h(x, y) F(u, v) ¤ H(u, v) ---------------(5)
  • 40. Frequency DomainProb 4) Find the convolution of an image x(m, n) & h(m, n) shown in the fig. below: y(p, q) = x(m1, n1) ¤ h(m2, n2) 0 -1 1 1 1 -1 4 -1 1 -1 2x2 (m, n) 0 -1 0 3x3 (m, n)
  • 41. Frequency DomainProb 4) Find the convolution of an image x(m, n) & h(m, n) shown in the fig. below: y(p, q) = x(m1, n1) ¤ h(m2, n2) 0 -1 1 1 1 -1 4 -1 1 -1 origin 2x2 (m2, n2) 0 -1 0 3x3 (m1, n1) p = m1 + m2 -1 q = n1 + n2 -1 = 3 + 2 -1 = 3 + 2 -1 = 4 columns = 4 rows Resultant image dimension is 4x4.
  • 42. Frequency Domain x(m, n) h(m, n)0 -1 1 1 1-1 4 -1 1 -10 -1 0 h(m, -n) h(-m, -n) 1 1 -1 1 -1 1 1 1
  • 43. Frequency Domain x(m, n) h(-m, -n)0 -1 1-1 4 -1 -1 10 -1 0 1 1 y(0, 0) 4
  • 44. Frequency Domain x(m, n) h(1-m, -n)0 -1 1-1 4 -1 -1 10 -1 0 1 1 y(1, 0) 4 -6
  • 45. Frequency Domain x(m, n) h(2-m, -n)0 -1 1-1 4 -1 0 -1 10 -1 0 0 1 1 y(2, 0) 4 -6 1
  • 46. Frequency Domain x(m, n) h(3-m, -n)0 -1 1-1 4 -1 0 0 -1 10 -1 0 0 0 1 1 y(3, 0) 4 -6 1 0
  • 47. Frequency Domain x(m, n) h(-m, 1-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 y(0, 1) 2 4 -6 1
  • 48. Frequency Domain x(m, n) h(-m, 2-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 0 0 -1 y(0, 2) 2 4 -6 1
  • 49. Frequency Domain x(m, n) h(1-m, 1-n)0 -1 1-1 4 -1 -1 10 -1 0 1 1 y(1, 1) -1 2 5 4 -6 1
  • 50. Frequency Domain x(m, n) h(1-m, 2-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 0 0 y(1, 2) -1 0 2 5 4 -6 1
  • 51. Frequency Domain x(m, n) h(2-m, 1-n)0 -1 1-1 4 -1 0 -1 10 -1 0 0 1 1 y(2, 1) -1 0 2 5 -2 4 -6 1
  • 52. Frequency Domain x(m, n) h(2-m, 2-n)0 -1 1 0 -1 1-1 4 -1 0 1 10 -1 0 0 0 0 y(2, 2) -1 0 1 2 5 -2 4 -6 1
  • 53. Frequency Domain x(m, n) h(-1-m, -n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 y(-1, 0) -1 0 1 2 5 -2 -1 4 -6 1
  • 54. Frequency Domain x(m, n) h(-1-m, 1-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 y(-1, 1) -1 0 1 -1 2 5 -2 -1 4 -6 1
  • 55. Frequency Domain x(m, n) h(-1-m, 2-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 0 0 y(-1, 2) 0 -1 0 1 -1 2 5 -2 -1 4 -6 1
  • 56. Frequency Domain x(m, n) h(-1-m, -1-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 y(-1, -1) 0 -1 0 1 -1 2 5 -2 -1 4 -6 1 0
  • 57. Frequency Domain x(m, n) h(-m, -1-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 y(0, -1) 0 -1 0 1 -1 2 5 -2 -1 4 -6 1 0 -1
  • 58. Frequency Domain x(m, n) h(1-m, -1-n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 y(1, -1) 0 -1 0 1 -1 2 5 -2 -1 4 -6 1 0 -1 1
  • 59. Frequency Domain x(m, n) h(2-m, -1-n)0 -1 1 0 -1 1-1 4 -1 0 1 10 -1 0 y(2, -1) 0 -1 0 1 -1 2 5 -2 -1 4 -6 1 0 -1 1 0
  • 60. Frequency Domain x(m, n) h(-m, -n)0 -1 1 -1 1-1 4 -1 1 10 -1 0 y(m, n) 0 -1 0 1 -1 2 5 -2 -1 4 -6 1 0 -1 1 0
  • 61. Frequency DomainHomomorphic Filtering:Illumination-Reflectance model for an image f(x, y) is given by: f(x, y) = i(x, y)r(x, y) ------------(1) where, i(x, y) – illumination r(x, y) – reflectanceIt is used to develop a frequency domain procedure for improving image appearance by : (i) simultaneous intensity range compression and (ii) contrast enhancement.Eqn (1) cant be directly operated on the frequency components of illumination & reflectanceThis is because F*f(x, y)+ ≠ F*i(x, y)]F[r(x, y)]However, if we definez(x, y) = In f(x, y) = ln i(x, y) + ln r(x, y)
  • 62. Frequency DomainThen F{z(x, y)} = F{ln f(x, y)} = F{ln i(x, y)} + F{ln r(x, y)}OrZ(u, v) = Fi(u, v) + Fr(u, v)Where Fi(u, v) & Fr(u, v) are FT of ln i(x, y) & r(x, y) resp.We can filter Z(u, v) by using a filter H(u, v) so thatS(u, v) = H(u, v)Z(u, v) = H(u, v) Fi(u, v) + H(u, v) Fr(u, v)Filtered image in spatial domain is s(x, y) = F-1{S(u, v)} = F-1{H(u, v) Fi(u, v)} + F-1{H(u, v) Fr(u, v)}
  • 63. Frequency DomainDefining i’(x, y) = F-1{H(u, v) Fi(u, v)}and r’(x, y) = F-1{H(u, v) Fr(u, v)}We can express it in form s(x, y) = i’(x, y) + r’(x, y)Finally, we reverse the process by taking the exponential of the filtered result to form the output image: g(x, y) = es(x, y) = ei’(x, y) er’(x, y) = i0(x, y) r0(x, y)Where, i0(x, y) & r0(x, y) are the illumination and reflectance components of the output (processed) image.
  • 64. Frequency Domainf(x, y) g(x, y) ln DFT H(u, v) (DFT)-1 expThis method is based on special case of a class of systems known as Homomorphic systems.
  • 65. Question Bank of Unit-3Q.1) Explain Histogram Processing?Q.2) Why an ideal LPF not suitable for smoothing operation? Explain how the Gaussian LPF overcomes the drawback of ideal LPF.Q.3) Write in short about Minimum Mean Square Error (Wiener) filtering.Q.4) Suppose F(u, v) is a 2D Fourier transform of an image f(x, y) and a new function G(u, v) is obtained as G(u, v) = f(u/2, v/2). What would be the image g(x, y) after taking the Inverse Fourier transform of G(u, v)?Q.5) Write a short note on Histogram Equalization.Q.6) Write a short note on Bit Plane Slicing.Q.7) Write a short note on Order Statistics Filter.
  • 66. Question Bank of Unit-3Q.8) The Gray level histogram of an image is given below: Gray Level 0 1 2 3 4 5 6 7Frequency of occurrence 400 700 1350 2500 3000 1500 550 0Compute the Gray level histogram of the output image obtained by enhancing the input by histogram equalization technique. [7m]Q.9) What is Histogram Matching? Explain. [5m]Q.10) What are the advantages and disadvantages of median filter? Obtain output image by applying 3 x 3 median filter on the following image. [8m] [ 2 4 15 0] [ 3 5 2 6] [11 0 2 10] [ 6 16 0 2]
  • 67. Question Bank of Unit-3Q.11) Write a short note on different Gradient filters.Q.12) Explain the following terms in significant details in reference to image enhancement in spatial domain: (i) Contrast Stretching (ii) Histogram Equalization (iii) Bit Plane Slicing (iv) The LaplacianQ.13) Explain Homomorphic filtering in frequency domain. [5m]Q.14) The gray level histogram of an image is given below:Compute the gray histogram of the output image obtained by enhancing the input by the histogram equalization technique. [9m] Gray Level 0 1 2 3 4 5 6 7 Freq. of Occurrence 123 78 281 417 639 1054 816 688
  • 68. Question Bank of Unit-3Q.15) Explain the principle objective of image enhancement process in spatial domain process.Q.16) Explain image subtraction, image averaging and image smoothing through smoothing linear filter.Q.17) A short note on Median Filtering.
  • 69. End of Unit-3