Setting the lower order bit plane to zero would have the effect of reducing the number of distinct gray levels by half. This would cause the histogram to become more peaked, with more pixels concentrated in fewer bins.

1. DIGITAL IMAGE PROCESSING
IMAGE ENHANCEMENT
Histogram Processing
by
Dr. K. M. Bhurchandi

2. Histogram Processing
Histogram:
It is a plot of frequency of occurrence of an event.
freq. of
occurrence
0 1 2 3 event
2

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:
3

4. Histogram Processing
 First Method:
 X-axis has gray levels & Y-axis has No. of pixels in each gray
levels.
nk
Gray 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
4

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
0.8
Gray 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
5

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(µk)
0 rk 0 rk
P(µk) P(µk)
0 rk 0 rk
6

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(µk) Bright Image
0 rk 0 rk
P(µk) Low Contrast Image P(µk) Equalized Image
0 rk 0 rk
7

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.
 Normalized histogram is nothing but PDF (Probability Density Function) of
an image.
8

9. Histogram Stretching
1) 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 image
Smin – min gray level of output image
rmax – max gray level of input image
rmin – min gray level of input image
0 min max 0 min max
9

10. Histogram Processing
𝑆 = 𝑇 𝑟 =
𝑆𝑚𝑎𝑥 − 𝑆𝑚𝑖𝑛
𝑟𝑚𝑎𝑥 − 𝑟𝑚𝑖𝑛
𝑟 − 𝑟𝑚𝑖𝑛 + 𝑆𝑚𝑖𝑛
Smax
Smin
rmin rmax
10

11. 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 0 0 50 60 50 20 10 0
11

12. 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 0 0 50 60 50 20 10 0
Soln:- 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) = S
S = (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
12

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 10
60 60
50 50
40 40
30 30
20 20
10 10
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
Original Histogram Stretched Histogram
13

14. Histogram Processing
Ex. 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
14

15. Histogram Processing
Ex. 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 70
Dynamic range increases
BUT
No. of pixels at each gray level remains constant.
15

16. Histogram Equalization
2) 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, the 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.
o It is used to enhance visual content of image.
16

17. Histogram Equalization
nk Normal Histogram
r
nk Equalized Histogram nk Linearly Stretched Histogram
S S
17

18. Histogram Equalization
o 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 ≤ 1
Here, range of r is [0, 1] (Normalized range) instead of [0, 255].
18

19. Histogram Equalization
These 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)
19

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, If reconstruction is required.
 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.
20

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| ----(a)
Prob. Density (Transformed Image) = Prob. Density (Original Image)
x
inv. Slope of transformation
21

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 = (L - 1) ʃ Pr(w) dw
0
diff. wrt. r
ds / dr = (L - 1) Pr(r) ---------(b)
Equating eqn(a) & eqn(b), we get
Ps(s) = 1 / L - 1 ; 0 ≤ S ≤ L - 1
22

23. Histogram Equalization
 A bad Histogram becomes a flat histogram when we find the CDF.
Pr(r) Ps(s)
A 1/L-1
0 L - 1 f 0 L - 1 S
Non-Uniform Function Uniform Function
CDF
Pr(r) dr
23

24. Histogram Equalization
 NOTE: If an image has MxN pixels, then PDF is given by:
Pr(rk) = nk / MN;
Where, nk – no. of pixels with intensity rk
k = 0, 1, 2, …….., L-1
In continuous domain,
r
S = T(r) = (L-1)ʃ Pr(w) dw
0
In discrete domain,
k
Sk = T(rk) = (L - 1)Σ Pr(rj) (CDF)
j=0
24

25. Histogram Equalization
Prob. 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 25

26. Histogram Equalization
Gray Levels
(rk)
No. of Pixels
nk
(PDF)
Pr(rk) = nk/n
(CDF)
Sk = Σ Pk(rk)
(L-1) Sk = 7 x Sk Rounding
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!!!
26

27. Histogram Equalization
Equalized 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 27

28. Histogram Equalization
Prob. 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 28

29. Histogram Equalization
Prob. 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 29

30. Histogram Equalization
Prob. 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 30

31. Histogram Equalization
Prob. 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
31

32. Histogram Equalization
Prob. 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
32

33. Histogram Equalization
Prob. 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.
33

34. 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 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.
34

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 0100
1000 1000 1000 1000
1100 1100 1100 1100
no. of
pixels
no. of levels
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
35

36. Histogram Equalization
Prob. 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
36

37. Histogram Equalization
Prob. 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
37

38. Frequency Domain
2-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) Σ Σ 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-1
Eqn. (1) & (2) constitute the 2-D discrete Fourier transform pair.
38
Σ Σ

39. Frequency Domain
2-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) = Σ Σ 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-1
The 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)
39

40. Frequency Domain
Prob 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)
40

41. Frequency Domain
Prob 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.
41

42. Frequency Domain
x(m, n) h(m, n)
0 -1 1 1 1
-1 4 -1 1 -1
0 -1 0
h(m, -n) h(-m, -n)
1 1
-1 1 -1 1
1 1
42

43. Frequency Domain
x(m, n) h(-m, -n)
0 -1 1
-1 4 -1 -1 1
0 -1 0 1 1
y(0, 0)
4
43

44. Frequency Domain
x(m, n) h(1-m, -n)
0 -1 1
-1 4 -1 -1 1
0 -1 0 1 1
4 -6
y(1, 0)
44

45. Frequency Domain
x(m, n) h(2-m, -n)
0 -1 1
-1 4 -1 0 -1 1
0 -1 0 0 1 1
4 -6 1
y(2, 0)
45

46. Frequency Domain
x(m, n) h(3-m, -n)
0 -1 1
-1 4 -1 0 0 -1 1
0 -1 0 0 0 1 1
4 -6 1 0
y(3, 0)
46

47. Frequency Domain
x(m, n) h(-m, 1-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0
2
4 -6 1
y(0, 1)
47

48. Frequency Domain
x(m, n) h(-m, 2-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0 0 0
-1
2
4 -6 1
y(0, 2)
48

49. Frequency Domain
x(m, n) h(1-m, 1-n)
0 -1 1
-1 4 -1 -1 1
0 -1 0 1 1
y(1, 1) -1
2 5
4 -6 1
49

50. Frequency Domain
x(m, n) h(1-m, 2-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0 0 0
y(1, 2) -1 0
2 5
4 -6 1
50

51. Frequency Domain
x(m, n) h(2-m, 1-n)
0 -1 1
-1 4 -1 0 -1 1
0 -1 0 0 1 1
y(2, 1) -1 0
2 5 -2
4 -6 1
51

52. Frequency Domain
x(m, n) h(2-m, 2-n)
0 -1 1 0 -1 1
-1 4 -1 0 1 1
0 -1 0 0 0 0
y(2, 2) -1 0 1
2 5 -2
4 -6 1
52

53. Frequency Domain
x(m, n) h(-1-m, -n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0
y(-1, 0) -1 0 1
2 5 -2
-1 4 -6 1
53

54. Frequency Domain
x(m, n) h(-1-m, 1-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0
y(-1, 1) -1 0 1
-1 2 5 -2
-1 4 -6 1
54

55. Frequency Domain
x(m, n) h(-1-m, 2-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0 0 0
y(-1, 2) 0 -1 0 1
-1 2 5 -2
-1 4 -6 1
55

56. Frequency Domain
x(m, n) h(-1-m, -1-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0
y(-1, -1) 0 -1 0 1
-1 2 5 -2
-1 4 -6 1
0
56

57. Frequency Domain
x(m, n) h(-m, -1-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0
y(0, -1) 0 -1 0 1
-1 2 5 -2
-1 4 -6 1
0 -1
57

58. Frequency Domain
x(m, n) h(1-m, -1-n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0
y(1, -1) 0 -1 0 1
-1 2 5 -2
-1 4 -6 1
0 -1 1
58

59. Frequency Domain
x(m, n) h(2-m, -1-n)
0 -1 1 0 -1 1
-1 4 -1 0 1 1
0 -1 0
y(2, -1) 0 -1 0 1
-1 2 5 -2
-1 4 -6 1
0 -1 1 0
59

60. Frequency Domain
x(m, n) h(-m, -n)
0 -1 1 -1 1
-1 4 -1 1 1
0 -1 0
y(m, n) 0 -1 0 1
-1 2 5 -2
-1 4 -6 1
0 -1 1 0
60