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CIS 601 Image ENHANCEMENT in the SPATIAL DOMAIN Dr. Rolf Lakaemper.

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Presentation on theme: "CIS 601 Image ENHANCEMENT in the SPATIAL DOMAIN Dr. Rolf Lakaemper."— Presentation transcript:

1 CIS 601 Image ENHANCEMENT in the SPATIAL DOMAIN Dr. Rolf Lakaemper

2 Most of these slides base on the book Digital Image Processing by Gonzales/Woods Chapter 3

3 Introduction Image Enhancement ? enhance otherwise hidden information Filter important image features Discard unimportant image features Spatial Domain ? Refers to the image plane (the ‘natural’ image) Direct image manipulation

4 Remember ? A 2D grayvalue - image is a 2D -> 1D function, v = f(x,y)

5 Remember ? As we have a function, we can apply operators to this function, e.g. T(f(x,y)) = f(x,y) / 2 Operator Image (= function !)

6 Remember ? T transforms the given image f(x,y) into another image g(x,y) f(x,y) g(x,y)

7 Spatial Domain The operator T can be defined over The set of pixels (x,y) of the image The set of ‘neighborhoods’ N(x,y) of each pixel A set of images f1,f2,f3,…

8 Operation on the set of image-pixels 6820 122002010 3410 6100105 Spatial Domain (Operator: Div. by 2)

9 Operation on the set of ‘neighborhoods’ N(x,y) of each pixel 6820 122002010 226 Spatial Domain 68 12200 (Operator: sum)

10 Operation on a set of images f1,f2,… 6820 122002010 Spatial Domain 5510 22034 111330 142202314 (Operator: sum)

11 Operation on the set of image-pixels Remark: these operations can also be seen as operations on the neighborhood of a pixel (x,y), by defining the neighborhood as the pixel itself. The simplest case of operators g(x,y) = T(f(x,y)) depends only on the value of f at (x,y) T is called a gray-level or intensity transformation function Spatial Domain

12 Basic Gray Level Transformations Image Negatives Log Transformations Power Law Transformations Piecewise-Linear Transformation Functions For the following slides L denotes the max. possible gray value of the image, i.e. f(x,y)  [0,L] Transformations

13 Image Negatives: T(f)= L-f Transformations Input gray level Output gray level T(f)=L-f

14 Log Transformations: T(f) = c * log (1+ f) Transformations

15 Log Transformations Transformations InvLogLog

16 Log Transformations Transformations

17 Power Law Transformations T(f) = c*f  Transformations

18 varying gamma (  ) obtains family of possible transformation curves  > 0 Compresses dark values Expands bright values  < 0 Expands dark values Compresses bright values Transformations

19 Used for gamma-correction Transformations

20 Used for general purpose contrast manipulation Transformations

21 Piecewise Linear Transformations Transformations

22 Thresholding Function g(x,y) =L if f(x,y) > t, 0 else t = ‘threshold level’ Piecewise Linear Transformations Input gray level Output gray level

23 Gray Level Slicing Purpose: Highlight a specific range of grayvalues Two approaches: 1. Display high value for range of interest, low value else (‘discard background’) 2. Display high value for range of interest, original value else (‘preserve background’) Piecewise Linear Transformations

24 Gray Level Slicing Piecewise Linear Transformations

25 Bitplane Slicing Extracts the information of a single bitplane Piecewise Linear Transformations

26 BP 7 BP 5 BP 0

27 Exercise: How does the transformation function look for bitplanes 0,1,… ? What is the easiest way to filter a single bitplane (e.g. in MATLAB) ? Piecewise Linear Transformations

28 Histograms Histogram Processing gray level Number of Pixels 1450 3151

29 Histograms Histogram Equalization: Preprocessing technique to enhance contrast in ‘natural’ images Target: find gray level transformation function T to transform image f such that the histogram of T(f) is ‘equalized’

30 Histogram Equalization Equalized Histogram: The image consists of an equal number of pixels for every gray- value, the histogram is constant !

31 Histogram Equalization Example: We are looking for this transformation ! T

32 Histogram Equalization Target: Find a transformation T to transform the grayvalues g1  [0..1] of an image I to grayvalues g2 = T(g1) such that the histogram is equalized, i.e. there’s an equal amount of pixels for each grayvalue. Observation (continous model !): Assumption: Total image area = 1 (normalized). Then: The area(!) of pixels of the transformed image in the gray-value range 0..g2 equals the gray-value g2.

33 Histogram Equalization The area(!) of pixels of the transformed image in the gray- value range 0..g2 equals the gray-value g2.  Every g1 is transformed to a grayvalue that equals the area (discrete: number of pixels) in the image covered by pixels having gray-values from 0 to g1.  The transformation T function t is the area- integral: T: g2 =  0..g1 I da

34 Histogram Equalization Discrete: g1 is mapped to the (normalized) number of pixels having grayvalues 0..g1.

35 Histogram Equalization Mathematically the transformation is deducted by theorems in continous (not discrete) spaces. The results achieved do NOT hold for discrete spaces ! (Why ?) However, it’s visually close.

36 Histogram Equalization Conclusion: The transformation function that yields an image having an equalized histogram is the integral of the histogram of the source-image The discrete integral is given by the cumulative sum, MATLAB function: cumsum() The function transforms an image into an image, NOT a histogram into a histogram ! The histogram is just a control tool ! In general the transformation does not create an image with an equalized histogram in the discrete case !

37 Operations on a set of images Operation on a set of images f1,f2,… 6820 122002010 5510 22034 111330 142202314 (Operator: sum)

38 Operations on a set of images Logic (Bitwise) Operations AND OR NOT

39 Operations on a set of images The operators AND,OR,NOT are functionally complete: Any logic operator can be implemented using only these 3 operators

40 Operations on a set of images Any logic operator can be implemented using only these 3 operators: ABOp 001 011 100 110 Op= NOT(A) AND NOT(B) OR NOT(A) AND B

41 Operations on a set of images Image 1 AND Image 2 1239 7364 1111 2222 1011 2220 (Operator: AND)

42 Operations on a set of images Image 1 AND Image 2: Used for Bitplane-Slicing and Masking

43 Operations on a set of images Exercise: Define the mask-image, that transforms image1 into image2 using the OR operand 1239 7364 25527 37 (Operator: OR)

44 Operations Arithmetic Operations on a set of images 1239 7364 1111 2222 23410 9586 (Operator: +)

45 Operations Exercise: What could the operators + and – be used for ?

46 Operations (MATLAB) Example: Operator – Foreground-Extraction

47 Operations (MATLAB) Example: Operator + Image Averaging

48 Part 2 CIS 601 Image ENHANCEMENT in the SPATIAL DOMAIN

49 Histograms So far (part 1) : Histogram definition Histogram equalization Now: Histogram statistics

50 Histograms Remember: The histogram shows the number of pixels having a certain gray-value number of pixels grayvalue (0..1)

51 Histograms The NORMALIZED histogram is the histogram divided by the total number of pixels in the source image. The sum of all values in the normalized histogram is 1. The value given by the normalized histogram for a certain gray value can be read as the probability of randomly picking a pixel having that gray value

52 Histograms What can the (normalized) histogram tell about the image ?

53 Histograms 1. The MEAN VALUE (or average gray level) M =  g g h(g) 1*0.3+2*0.1+3*0.2+4*0.1+5*0.2+6*0.1= 2.6 0.3 0.2 0.1 0.0 1 2 3 4 5 6

54 Histograms The MEAN value is the average gray value of the image, the ‘overall brightness appearance’.

55 Histograms 2. The VARIANCE V =  g (g-M) 2 h(g) (with M = mean) or similar: The STANDARD DEVIATION D = sqrt(V)

56 Histograms VARIANCE gives a measure about the distribution of the histogram values around the mean. 0.3 0.2 0.1 0.0 0.3 0.2 0.1 0.0 V1 > V2

57 Histograms The STANDARD DEVIATION is a value on the gray level axis, showing the average distance of all pixels to the mean 0.3 0.2 0.1 0.0 0.3 0.2 0.1 0.0 D1 > D2

58 Histograms VARIANCE and STANDARD DEVIATION of the histogram tell us about the average contrast of the image ! The higher the VARIANCE (=the higher the STANDARD DEVIATION), the higher the image’s contrast !

59 Histograms Example: Image and blurred version

60 Histograms Histograms with MEAN and STANDARD DEVIATION M=0.73 D=0.32M=0.71 D=0.27

61 Histograms Exercise: Design an autofocus system for a digital camera ! The system should analyse an area in the middle of the picture and automatically adjust the lens such that this area is sharp.

62 Histograms In between the basics… …histograms can give us a first hint how to create image databases:

63 Feature Based Coding Determine a feature-vector for a given image Compare images by their feature-vectors Two operations need to be defined: a mapping of shape into the feature space and a similarity of feature vectors. Where are the histograms ? RepresentationFeature ExtractionVector Comparison

64 Feature Based Coding Determine a feature-vector for a given image Compare images by their feature-vectors Two operations need to be defined: a mapping of shape into the feature space and a similarity of feature vectors. HERE ! Question: how can we compare histograms (vectors) ? RepresentationHISTOGRAMHistogram Comp.

65 Vector Comparison,

66

67

68 What’s the meaning of the Cosine Distance with respect to histograms ? i.e.: what’s the consequence of eliminating the vector’s length information ?

69 Vector Comparison More Vector Distances: Quadratic Form Distance Earth Movers Distance Proportional Transportation Distance …

70 Vector Comparison Histogram Intersection (non symmetric): d(h1,h2) = 1 -  min(h1,h2 ) /  h1 Ex.: What could be a huge drawback of image comparison using histogram intersection ? iiiii

71 Histograms Exercise: Outline an image database system, using statistical ( histogram ) information

72 Histograms Discussion: Which problems could occur if the database consists of the following images ?

73 Histograms

74 Spatial Filtering End of histograms. And now to something completely different …


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