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Fundamental of Image Processing

By

Ashis Kr. Dhara

1

Representation of Image

2

Point Processing

3

Gray-Level Transformation Function

s = T (r)

Contrast stretching

Thresholding

4

Image Negative

s = L - 1 - r

5

Basic Gray Level Transformation

6

Log Transformations

s = c log(1+r)

Pixel values dynamic range=[0 - 1.5×106]

7

Power-Law Transformations

s = cr

s = r2.5

r = [1 10 20 30 40 210 220 230 240 250 255]

s( = 2.5) = [0 0 0 1 2 157 176 197 219 243 255]

s( =.4) = [28 70 92 108 122 236 240 245 249 253 255] 8


c=1

= 0.4 c=1

= 0.3

c=1

= 0.6

9


c=1

= 3

c=1

= 5

c=1

= 4

10

Contrast Stretching

Objective: Increase the dynamic range of the gray levels in the image

Causes for poor image -Poor illumination -Wrong lens aperture -Wrong shutter speed -Lack of dynamic range in the imaging sensor

11

Histogram The histogram of a digital image with gray levels in the range [0, L-1] is a discrete function h(rk) = nk, where rk is the kth gray level and nk is the number of pixels in the image having gray level rk.

Normalized Histogram Dividing each value of the histogram by the total number of pixels in the image, denoted by n. p(rk) = nk /n. Normalized histogram provide useful image statistics.

Histogram Processing

12

Histogram Processing

13

Histogram equalization is used to enhance image contrast and gray-level detail by spreading the histogram of the original image. s = T ( r ) 0 r 1, where r and s are normalized pixel intensities

Conditions for the transformation (a) T( r ) is single-valued and monotonically increasing in the interval 0 r 1 (b) 0 T ( r ) 1 for 0 r 1

1

Histogram Equalization

14

Objective of histogram equalization Transform the histogram function of the original image pr(r) to a uniform histogram function. ps(s) = 1 0 s 1

Equalizer

Transformation pr(r) ps(s)

r

r dwwprTs

0

)()(

k

j

jrkk rprTs0

)()(

Continuous case Discrete case


15

0

1

2

3

:

:

128

129

130

:

:

253

254

255

0

0

:

0

:

:

0.004

0.15

0.05

:

:

0.005

0.006

0.004

pr(rk) k

T (rk)

0

0

:

0

0

0

0.004

0.154

0.204

:

:

0.989

0.994

1

sk

0

0

0

0

:

:

10

39

52

:

:

252

253

255

0

0.004

0.008

0.011

:

:

0.5

0.505

0.51

:

:

0.992

0.996

1

rk

/255 *255

Input image Output image

16


17


15

1

8

0.5

4

0.25

12

0.75

0.4

0.2

0.3

0.1

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0

0

0

0

0.09

0.06

0.1

0.15

0.3

0.18

0.05

0.07

0

0

0

0

0

0

0

0

0.09

0.15

0.25

0.4

0.7

0.88

0.93

1

1

1

1

1

rk pr(rk) Normalized sk

0

0

0

0

1

2

4

6

11

13

14

15

15

15

15

15

sk

>>plot(sk)

Input image size 100 100

18

Is histogram equalization a good approach to enhance the below image?

Histogram Matching

19

Histogram Matching method generates a processed image that has a specified histogram.

Input image

pr(r)

T (r)

r

r dttp

0

)( s

Desired

Output image

pz(z)

G (z)

z

z dttp

0

)( v

s v, since both will have

approximately uniform pdf

z = G-1[T( s )] s = G[T( r )]

Histogram Matching

20

Histogram Matching

21

Histogram Matching

22

0.6

0.12

0.08

0.0

:

0.01

0.005

0

0.6

0.72

0.8

0.8

:

0.9

0.95

1

153

184

204

204

:

230

242

255


23

1) Modify the histogram of the image to obtain pz(z). 2) Find the transformation function G (z) using the modified histogram in step 1. 3) Find the inverse G-1 (z) 4) Enhanced image is

obtained by applying G-1 to the pixels of the histogram-equalized image shown in Fig. 3.21.c

1 G (z)

2 G-1 (z)

Histogram Matching

24

Arithmetic/logic operations involving images are performed on a pixel-by-pixel basis between two or more images. Arithmetic Operations Addition, Subtraction, Multiplication, and Division Logic Operations AND, OR, NOT

Enhancement Using Arithmetic/Logic Op.

25

AND

OR

Logic operations are performed on the binary representation of the pixel intensities

Enhancement Using AND and OR Logic Op.

26

Problem: The pixel intensities in the difference image can range from -255 to 255. Solutions: 1) Add 255 to every pixel and then divide by 2. 2) Add the minimum value of the pixel intensity in the difference image to every pixel and then divide by 255/Max. Max is the maximum pixel value in the modified difference image.

Enhancement Using Arithmetic Op. _ SUB Mask Mode Radiography

27

Spatial Filtering

28

Spatial filtering are filtering operations performed on the pixel intensities of an image and not on the frequency components of the image.

a

as

b

bt

tysxftswyxg ),(),(),(

a = (m - 1) / 2 b = (n - 1) / 2

Basics of Spatial Filtering - Linear

29

Response, R, of an m n mask at any point (x, y)

mn

i

iizwR1

Special consideration is given when the center of the filter approach the boarder of the image.

Basics of Spatial Filtering

30

2

),(

2

),(

1

1

),(),(

),(1

),(

),(),(),(

yxyxg

k

i

i

k

yxfyxgE

yxgk

yxg

yxyxfyxg

Original image Noise with zero mean

Enhancement Using Arithmetic Op. Averaging

31

Gaussian Noise

mean = 0

variance = 64

K = 8

K = 32

K = 16

K = 128

f(x,y) g(x,y)

+ -

x


32

Difference images between original image and images obtained from averaging.

K = 8

K = 16

K = 32

K = 128

Notice the mean and variance of the difference images decrease as K increases.


33

Nonlinear spatial filters operate on neighborhoods, and the mechanics of sliding a mask past an image are the same as was just outlined. In general however, the filtering operation is based conditionally on the values of the pixel in the neighborhood under consideration, and they do not explicitly use coefficients in the sum-of products manner described previously.

Example Computation for the median is a nonlinear operation.

Nonlinear of Spatial Filtering

34

Smoothing filters are used - Noise reduction - Smoothing of false contours - Reduction of irrelevant detail Undesirable side effect of smoothing filters - Blur edges

Weighted

average

Box

filter

Weighted average filter reduces blurring in the

smoothing process.

Smoothing Spatial Filtering - Linear Averaging (low-pass) Filters

35

n = filter size

n = 3

n = 5 n = 9

n = 15

n = 35

Smoothing Spatial Filtering _ Linear Averaging (low-pass) Filters

36

filter size

n = 15

Thrsh = 25% of

highest intensity

Smoothing Spatial Filtering Averaging & Threshold

37

Order-statistics filters are nonlinear spatial filters whose response is based on ordering (ranking) the pixels contained in the image area encompassed by the filter, and then replacing the value of the center pixel with the value determined by the ranking result.

3 3 Median filter [10 125 125 135 141 141 144 230 240] = 141

3 3 Max filter [10 125 125 135 141 141 144 230 240] = 240

3 3 Min filter [10 125 125 135 141 141 144 230 240] = 10

Median filter eliminates isolated clusters of pixels that are light or dark with respect to their neighbors, and whose area is less than n2/2.

Smoothing Spatial Filtering Order Statistic Filters

38

n = 3

Average

filter

n = 3

Median

filter

Order Statistic Filters

39

The principal objective of sharpening is to highlight fine detail in an image or to enhance detail that has been blurred.

9

19

1izImage Blurred Image

The derivatives of a digital function are defined in terms of differences.

Sharpening Spatial Filters

z

1 z2

z6

z8

z4

z7

z3

z9

z5

40

First and Second Derivatives

The signs of the derivatives would be reversed for an edge that transitions from light to dark

41

Requirements for digital derivative First derivative 1) Must be zero in flat segment 2) Must be nonzero along ramps. 3) Must be nonzero at the onset of a gray-level step or ramp

Second derivative 1) Must be zero in flat segment 2) Must be zero along ramps. 3) Must be nonzero at the onset and end of a gray-level step or ramp

)(2)1()1(

)()1(

2

2

xfxfxfx

f

xfxfx

f


42


43

Comparing the response between first- and second-ordered derivatives: 1) First-order derivative produce thicker edge 2) Second-order derivative have a stronger response to fine detail, such as thin lines and isolated points. 3) First-order derivatives generally have a stronger response to a gray-level step {2 4 15} 4) Second-order derivatives produce a double response at step changes in gray level.

In general the second derivative is better than the first derivative for image enhancement. The principle use of first derivative is for edge extraction.


44

First derivatives in image processing are implemented using the magnitude of the gradient.

yx

t

GGy

f

x

fmagf

y

f

x

f

5.022

)( f

f z1 z2

z6

z8

z4

z7

z3

z9

z5

Roberts operator

Gx = (z9-z5) and Gy = (z8 - z6) Sobel operator

Gx = (z3+2z6 +z9) - (z1+2z4+z7) and

Gy = (z7+2z8+z9) - (z1+2z2+z3)

Use of First Derivative for Edge Extraction Gradient

45

Ideal and Ramp Edges

because of optics, sampling, image acquisition imperfection

46

Gradient Operator

first derivatives are implemented using the magnitude of the gradient.

y

fx

f

G

G

y

xf

21

22

21

22 ][)f(

y

f

x

f

GGmagf yx

the magnitude becomes nonlinear yx GGf

commonly approx.

47

2

2

2

22

y

f

x

ff

)],(4)1,()1,(),1(),1([

),(2)1,()1,(

),(2),1(),1(

2

2

2

2

2

yyfyxfyxfyxfyxff

yxfyxfyxfy

f

yxfyxfyxfx

f

2nd Derivative _ Laplacian

48

2

2

2

22

y

f

x

ff

Isotropic filter response is independent of the direction of the discontinuities in the image to which the filter is applied.

Use of 2nd Derivative for Enhancement Laplacian

49

f(x,y) = [90, 100] 0 1

1

1

1

0

0

0

-4

… 0 0 0 0 …

… 0 10 0 0 …

… 10 -40 10 0 …

… 0 10 0 0 …

… 0 0 0 0 …


50

),(),(

),(),(),(

2

2

yxfyxf

yxfyxfyxg

If the center coefficient of the laplacian mask is negative

1 1

1

1

1

1

1

1

-8


51

g(x,y) = f(x,y) -

0 1

1

1

1

0

0

0

-4

),(2 yxf

=

)],(4)1,()1,(),1(),1([2 yxfyxfyxfyxfyxff


52

High-boost filtering is used when the original image is blurred and dark.

),(),( 2 yxfyxAffhb A > 1

Un-sharp Masking and High-boost Filtering

53

(b)

Un-sharp Masking and High-boost Filtering

54

Laplacian

2

2

2

22 ),(),(

y

yxf

x

yxff

(linear operator)

Laplacian operator

)],(4)1,()1,(

),1(),1([2

yxfyxfyxf

yxfyxff

Laplacian masks

55

Laplacian of Gaussian (LOG)

Laplacian combined with smoothing as a precursor to find edges via zero-crossing.

2

2

2)(

r

erh

where r2 = x2+y2, and is the standard deviation

2

2

24

222 )(

r

er

rh

56

Edge detection

58

Edge Detection

Edges characterize boundaries of objects in image

A fundamental problem in image processing

Edges are areas with strong intensity contrasts

A jump in intensity from one pixel to the next

Edge detected image

Reduces significantly the amount of data,

Filters out useless information,

Preserves the important structural properties

59

Our goal is to extract a “line drawing” representation from an image

Useful for recognition: edges contain shape information

invariance

60

Need of Edge Detection

Digital artists use it to create image outlines.

The output of an edge detector can be added back to an original image to enhance the edges

Edge detection is often the first step in image segmentation

Edge detection is also used in image registration by alignment of two images that may have been acquired at separate times or from different sensors

61

Gradient Masks

62

Sobel

Canny LOG

63

Most of these partial derivative operators are sensitive to noise, Use of these masks produces thick edges or boundaries, Gives spurious edge pixels due to noise.

Problems with Edge Detection Methods

To overcome the effect of noise, smoothing operation is performed before edge detection

Edge Detection

Input Image

Edge Map

Edge Detection

Input Image

Edge Map

Smoothing operation

64

Smoothing based Edge Detection

Two operators which use smoothing

Marr-Hildreth operator

Laplacian of Gaussian function (LOG)

Follows 2-Operations

Smoothing

Applying Laplacian operator

Or generate the combined mask of LOG

Canny Edge Detector

65

Canny Edge Detection

Normally, edge operators use one threshold for whole image

Sobel output

Sobel output

66

Sobel

Canny LOG

67

Sobel

Canny LOG

68

Canny Edge Detector (J. Canny’

1986):

An "optimal" edge detector means:

Good detection - the algorithm should mark as many real edges in the image as possible.

Good localization - edges marked should be as close as possible to the edge in the real image.

Canny edge detector uses two threshold values to detect weak and strong edges

69

Frequency Domain Filtering

70

2/122

1

0

1

0

)(2

1

0

1

0

)(2

),(),(),(

),(),(

),(1

),(

vuIvuRvuF

evuFyxf

eyxfMN

vuF

M

u

N

v

yN

vx

M

uj

M

x

N

y

N

yv

M

xuj

Fourier Spectrum

Discrete Two-Dimensional Fourier Transform and Its Inverse

71

1

0

1

0

),(1

)0,0(M

x

N

y

yxfMN

F

F(0,0) is the average intensity of an image


72

Use Matlab to generate the above figures


73

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