lecture 03,04,05 - intensity transformation and spatial filtering.pdf

8/12/2019 Lecture 03,04,05 - Intensity transformation and Spatial Filtering.pdf

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Intensity transformation and spatial filtering

2

Digital Image Processing 3

I.Introduction Spatial Domain Process

( , ) [ ( , )])

( , ) : input image

( , ) :output image

: an operator on defined over

a neighborhood of point ( , )

g x y T f x y

f x y

g x y

T f

x y

=

Digital Image Processing

4

I.Introduction

Spatial Domain Process


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II.IntensitytransformationfunctionIntensity transformation function

( )s T r=


6

II.Intensitytransformationfunction

Some basic intensity transformation

Functions


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II.1.NegativeImage negatives

1s L r=


8

II.1.Negative

Smalllesion


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II.2.LogTransformLog Transformations

log(1 )s c r= +


10

II.2.LogTransform


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II.3.Power Law

s cr=


12

II.3.Power Law


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II.3.Power Law


14

II.4.Piecewise-LinearTransform..

Contrast Stretching Expands the range of intensity levels in an image so

that it spans the full intensity range of the recording

medium or display device

Intensity-level Slicing Highlighting a specific range of intensities in an image

often is of interest


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II.4.Piecewise-LinearTransform


16


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II.5.Bit PlaneSlicing


18

II.5.Bit PlaneSlicing


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III.Histogramprocessing Histogram Equalization

Histogram Matching

Local Histogram Processing

Using Histogram Statistics for Image

Enhancement


20

III.Histogramprocessing

Histogram ( )

is the intensity value

is the number of pixels in the image with intensity

k k

th

k

k k

h r n

r k

n r

=

Normalized histogram ( )

: the number of pixels in the image of

size M N with intensity

kk

k

k

np r

MN

n

r

=


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4x4 image

Gray scale = [0,9]histogram

0 1

1

2

2

3

3

4

4

5

5

6

6

7 8 9

No. of pixels

Gray level

2 3 3 2

4 2 4 3

3 2 3 5

2 4 2 4


22



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III.1.HistogramEqualization As the low-contrast images histogram is narrow

and centered toward the middle of the gray scale,

if we distribute the histogram to a wider range the

quality of the image will be improved.

We can do it by adjusting the probability density

function of the original histogram of the image so

that the probability spread equally


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III.1.HistogramEqualization

Histogram transformation

0 1rk

sk= T(rk)

s

r

T(r)

s = T(r)s = T(r)

Where 0 r 1

T(r) satisfies (a). T(r) is single-valued

and monotonicallyincreasingly in the interval0 r 1

(b). 0 T(r) 1 for0 r 1


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III.1.HistogramEqualization 2 conditions of T(r)

Single-valued (one-to-one relationship) guarantees that

the inverse transformation will exist

Monotonicity condition preserves the increasing order from

black to white in the output image thus it wont cause a

negative image

0 T(r) 1 for 0 r 1 guarantees that the output graylevels will be in the same range as the input levels.

The inverse transformation from s back to r is

r = Tr = T --11(s) ; 0(s) ; 0 ss 11


26


Let

pprr(r)(r) denote the PDF of random variable r

ppss(s(s)) denote the PDF of random variable s

Ifpprr(r)(r) and T(r)T(r) are known and TT--11(s)(s) satisfiescondition (a) thenppss(s(s)) can be obtained using a

formula :

ds

dr(r)p(s)p rs =


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III.1.HistogramEqualization A transformation function is a cumulative distribution

function (CDF) of random variable r :

CDF is an integral of a probability function (always positive) is the

area under the function Thus, CDF is always single valued and monotonically increasing

Thus, CDF satisfies the condition (a)

We can use CDF as a transformation function

==r

r dw)w(p)r(Ts0


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)r(p

dw)w(pdr

d

dr

)r(dT

dr

ds

r

r

r

=

=

=

0

10where1

1

=

=

=

s

)r(p)r(p

ds

dr)r(p)s(p

r

r

rs

Substitute and yield


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III.1.HistogramEqualization Ps(s):

Asps(s) is a probability function, it must be zero outside

the interval [0,1] in this case because its integral over all

values of s must equal 1.

Calledps(s) as a uniform probability densitya uniform probability density functionfunction

ps(s) is always a uniform, independent of the form of

pr(r)


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Discrete Transformation Function

The probability of occurrence of gray level in an image

is approximated by

The discrete version of transformation

=

=

==

==

k

j

j

k

j

jrkk

, ..., L-,where kn

n

)r(p)r(Ts

0

0

110

110 , ..., L-,where kn

n)r(p kkr ==


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III.1.HistogramEqualization Thus, an output image is obtained by mapping

each pixel with level rrkk in the input image into a

corresponding pixel with level sskk in the output

image

In discrete space, it cannot be proved in general

that this discrete transformation will produce the

discrete equivalent of a uniform probability density

function, which would be a uniform histogram


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Example before after Histogramequalization


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III.1.HistogramEqualization Example

before after Histogramequalization

The quality is

not improvedmuch becausethe originalimage alreadyhas a broadengray-level scale


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4x4 image

Gray scale = [0,9]histogram

0 1

1

2

2

3

3

4

4

5

5

6

6

7 8 9

No. of pixels

Gray level

2 3 3 2

4 2 4 3

3 2 3 5

2 4 2 4


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III.1.HistogramEqualization Example

Gray Level 0 1 2 3 4 5 6 7 8 9

No.of pixels 0

16

16/

16

9s x 9

0 0 6 5 4 1 0 0 0

0 0 6 11 15 16 16 16 16

0 06 /

16

11 /

16

15 /

16

16 /

16

16/

16

16/

16

16/

16

0 03.3

3

6.1

6

8.4

89 9 9 9

=

k

j

jn0

=

=k

j

j

n

ns

0


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Example

Output image

Gray scale = [0,9]

Histogram equalization

0 1

1

2

2

3

3

4

4

5

5

6

6

7 8 9

No. of pixels

Gray level

3 6 6 3

8 3 8 6

6 3 6 9

3 8 3 8


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III.2.HistogramMatching Generate a processed image that has a specified

histogram

Let ( ) and ( ) denote the continous probability

density functions of the variables and . ( ) is the

specified probability density function.

Let be the random variable with the prob

r z

z

p r p z

r z p z

s

0

0

ability ( ) ( 1) ( )

Define a random variable with the probability

( ) ( 1) ( )

r

r

z

z

s T r L p w dw

z

G z L p t dt s

= =

= =


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III.2.HistogramMatching

0

0

( ) ( 1) ( )

( ) ( 1) ( )

r

r

z

z

s T r L p w dw

G z L p t dt s

= =

= =

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


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III.2.HistogramMatching Procedure

1. Obtain pr(r) from the input image and then obtain the values of s

2. Use the specified PDF and obtain the transformation function

G(z)

3. Mapping from s to z

0( 1) ( )

r

rs L p w dw=

0( ) ( 1) ( )

z

zG z L p t dt s= =1( )z G s=


40


Example

Assume an image has a gray level probability density functionpr(r) as shown.

0 1 2

1

2

Pr(r)

+=

elsewhere;0

1r;022 r)r(p r

10

=r

r dw)w(p

r


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III.2.HistogramMatching Example

We would like to apply the histogram specification with thedesired probability density function pz(z) as shown.

0 1 2

1

2

Pz(z)

z

=elsewhere;0

1z;02z)z(pz

10

=z

z dw)w(p


42


Example

0 1

1

s=T(r)

rrr

ww

dw)w(

dw)w(p)r(Ts

r

r

r

r

2

2

22

2

0

2

0

0

+=

+=

+=

==

1. Obtain the transformation function T(r)


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III.2.HistogramMatching Procedure in discrete cases

1. Obtain pr(rj) from the input image and then obtain the values ofsk, round the value to the integer range [0, L-1]

2. Use the specified PDF and obtain the transformation functionG(zq), round the value to the integer range [0, L-1].

3. Mapping from sk to zq

0 0

( 1)( ) ( 1) ( )

k k

k k r j j

j j

Ls T r L p r n

MN= =

= = =

0

( ) ( 1) ( )q

q z i k

i

G z L p z s=

= =

1( )q kz G s=


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ExampleSuppose that a 3-bit image (L=8) of size 64 64 pixels (MN =4096) has the intensity distribution shown in the followingtable (on the left). Get the histogram transformation functionand make the output image with the specified histogram, listedin the table on the right.


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Obtain the scaled histogram-equalized values,

Compute all the values of the transformation function G,

0 1 2 3 4

5 6 7

1, 3, 5, 6, 7,

7, 7, 7.

s s s s s

s s s

= = = = =

= = =

0

0

0

( ) 7 ( ) 0.00z jj

G z p z

=

= =

1 2

3 4

5 6

7

( ) 0.00 ( ) 0.00

( ) 1.05 ( ) 2.45

( ) 4.55 ( ) 5.95

( ) 7.00

G z G z

G z G z

G z G z

G z

= =

= =

= =

=

0

0 01 2

65

7

s0

s2 s3

s5 s6 s7

s1

s4


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Example 0 1 2 3 4

5 6 7

1, 3, 5, 6, 7,

7, 7, 7.

s s s s s

s s s

= = = = =

= = =

01

2

3

4

5

6

7

kr

0 3

1 4

2 5

3 6

4 7

5 7

6 7

7 7

k qr z


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50


Example


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III.3.LocalHistogramProcessingDefine a neighborhood and move its center from pixel to pixel

At each location, the histogram of the points in the

neighborhood is computed. Either histogram equalization or

histogram specification transformation function is obtained

Map the intensity of the pixel centered in the neighborhood

Move to the next location and repeat the procedure


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III.3.LocalHistogramProcessing


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III.4.UsingHistogramStatistics

1

0

( )L

i i

i

m r p r

=

= 1

0

( ) ( ) ( )L

n

n i i

i

u r r m p r

=

=

12 2

2

0

( ) ( ) ( )L

i i

i

u r r m p r

=

= =

1 1

0 0

1( , )

M N

x y

x yMN

= =

=

[ ]1 1

2

0 0

1( , )

M N

x y

f x y mMN

= =

=

Average Intensity

Variance


54


1

0

Local average intensity

( )

denotes a neighborhood

xy xy

L

s i s i

i

xy

m r p r

s

=

=

12 2

0

Local variance

( ) ( )xy xy xy

L

s i s s i

i

r m p r

=

=


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0 1 2

0 1 2

( , ), if and( , )

( , ), otherwise

: global mean; : global standard deviation

0.4; 0.02; 0.4; 4

xy xys G G s G

G G

E f x y m k m k kg x y

f x y

m

k k k E

=

= = = =


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IV.UsingArithmetic/LogicOperators

Arithmetic/Logic operations perform on pixel by

pixel basis between two or more images, except

NOT operation which perform only on a single

image

Logic operation performs on gray-level images,

the pixel values are processed as binary

numbers:

Light represents a binary 1, and

Dark represents a binary 0


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IV.1.LogicOperators


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IV.2.ImageSubtraction

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

Extraction of the differences between images


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IV.3.ImageAveraging Consider a noisy image g(x,y) formed by the addition of noise (x,y)

to an original image f(x,y)

g(x,y) = f(x,y) + (x,y) If noise has zero mean and be uncorrelated then it can be shown that

if

),( yxg = image formed by averagingK different noisy images

=

=K

i

i yxgK

yxg1

),(1

),(


60

IV.3.ImageAveraging

Then

),(2

),(2 1

yxyxg

K

=

= variances of g and ),(2

),(2 , yxyxg

if K increase, it indicates that the variability (noise) of the pixel ateach location (x,y) decreases.


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V.SpatialFiltering

Example

M=3, N=3

3x3

Or M=5, N=5

5x5


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V.1.SmoothingFilters

Smoothing filters are used for blurring and for noise reduction

Blurring is used in removal of small details and bridging of small

gaps in lines or curves

Smoothing spatial filters include linear filters and nonlinear filters.


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V.1.SmoothingFilters Linear filters

The general implementation for filtering an M N image

with a weighted averaging filter of size m n is given

( , ) ( , )

( , )( , )

where 2 1

a b

s a t b

a b

s a t b

w s t f x s y t

g x yw s t

m a

= =

= =

+ +

=

= +

, 2 1.n b= +


66


Linear filters

Averaging Filter Masks


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V.1.SmoothingFilters Example

Averaging Filter Masks

a). original image 500x500 pixel

b). - f). results of smoothing with

square averaging filter masks of

size n = 3, 5, 9, 15 and 35,

respectively.

Note:

Big mask is used to eliminate small objects

from an image.

The size of the mask establishes the

relative size of the objects that will be

blended with the background.

a b

c d

e f


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original image result after smoothing with15x15 averaging mask

result of thresholding

we can see that the result after smoothing and thresholding, theremains are the largest and brightest objects in the image.

Example


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V.1.SmoothingFilters Order-Statistics Filters (Nonlinear Filters)

The response is based on ordering (ranking) the pixels

contained in the image area encompassed by the filter

Example

Median filter : R = median{zk |k = 1,2,,n x n}

Max filter : R = max{zk |k = 1,2,,n x n}

Min filter : R = min{zk |k = 1,2,,n x n}

Note: n x n is the size of the mask


70


Median Filter Median: X= {x1,x2,, x2b+1}, x is called the median of X if x

greater than or equal b elements and less than or equal b other

elements in X;

For example: X={3,2,3,2,3,4,5,6,5} (b=4) -> Median is 3

Peplaces the value of a pixel by the median of the gray levels in

the neighborhood of that pixel (the original value of the pixel is

included in the computation of the median)

Quite popular because for certain types of random noise (impulseimpulse

noisenoise salt and pepper noisesalt and pepper noise) , they provide excellent noiseprovide excellent noise--

reduction capabilitiesreduction capabilities, with considering less blurring than linearless blurring than linear

smoothing filters of similar sizesmoothing filters of similar size


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V.1.SmoothingFilters Median Filter

Example


72

V.2.SharpeningSpatialFilters

Foundation

Laplacian Operator

Unsharp Masking and Highboost Filtering

Using First-Order Derivatives for Nonlinear Image

Sharpening - The Gradient


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Laplace Operator

V.3.MethodbasedonLaplaceOperator

The second-order isotropic derivative operator is the Laplacian

for a function (image) f(x,y)2 2

2

2 2

f ff

x y

= +

2

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

ff x y f x y f x y

x

= + +

22

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

f x y f x y f x yy

= + +

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

- 4 ( , )

f f x y f x y f x y f x y

f x y

= + + + + +


76

Masks



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Image sharpening in the way of using the

Laplacian:

where g(x,y) sharpened image



78

Example



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V.4. Unsharp Masking and Highboost

Filtering


80


Filtering


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Filtering


82


Filtering Example


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Filtering Example


84

V.5. Method based on First-Order

Derivatives(Gradient)

For function ( , ), the gradient of at coordinates ( , )

is defined as

grad( ) x

y

f x y f x y

f

g xf f

fg

y

=

2 2

The of vector , denoted as ( , )

( , ) mag( ) x y

magnitude f M x y

M x y f g g

= = +


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2 2

The of vector , denoted as ( , )

( , ) mag( ) x y

magnitude f M x y

M x y f g g

= = +

( , ) | | | |x y

x y g g +

z1 z2 z3

z4 z5 z6

z7 z8 z9

8 5 6 5( , ) | | | |x y z z z z= +


86



z1 z2 z3

z4 z5 z6

z7 z8 z9

9 5 8 6

Roberts Cross-gradient Operators

( , ) | | | |M x y z z z z +

7 8 9 1 2 3

3 6 9 1 4 7

Sobel Operators

( , ) | ( 2 ) ( 2 ) |

| ( 2 ) ( 2 ) |

M x y z z z z z z

z z z z z z

+ + + +

+ + + + +


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88



Example


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V.6. Combining Spatial Enhancement

Methods

Solve:

1. Laplacian to highlight fine detail

2. Gradient to enhance prominent edges

3. Gray-level transformation to increase the

dynamic range of gray levels


90


Methods

Exampe


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Methods

Exampe

lecture 03,04,05 - intensity transformation and spatial filtering.pdf

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