Intensity transformation and spatial filtering

Digital Image Processing

Lecture 3 – Intensity

Transformation& Spatial Filtering

Lecturer: Ha Dai Duong

Faculty of Information Technology

I. Introduction

Spatial Domain vs. Transform Domain

Spatial Domain

Image plane itself, directly process the intensity values of the

image plane

Transform (Frequency) domain

Process the transform coefficients, not directly process the

intensity values of the image plane

Digital Image Processing

2

1

Intensity transformation and spatial filtering

I. Introduction

Spatial Domain Process

g ( x, y ) = T [ f ( x, y )])

f ( x, y ) : input image

g ( x, y ) : output image

T : an operator on f defined over

a neighborhood of point ( x, y )

Digital Image Processing

3

I. Introduction

Spatial Domain Process

Digital Image Processing

4

2

Intensity transformation and spatial filtering

II. Intensity transformation function

Intensity transformation function

s = T (r )

5

Digital Image Processing

II. Intensity transformation function

Some

basic

Functions

Digital Image Processing

intensity

transformation

6

3

Intensity transformation and spatial filtering

II.1. Negative

Image negatives

s = L −1− r

Digital Image Processing

7

II.1. Negative

Small

lesion

Digital Image Processing

8

4

Intensity transformation and spatial filtering

II.2. Log Transform

Log Transformations

s = c log(1 + r )

Digital Image Processing

9

II.2. Log Transform

Digital Image Processing

10

5

Intensity transformation and spatial filtering

II.3. Power – Law

s = cr γ

Digital Image Processing

11

II.3. Power – Law

Digital Image Processing

12

6

Intensity transformation and spatial filtering

II.3. Power – Law

Digital Image Processing

13

II.4. Piecewise-Linear Transform..

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

Digital Image Processing

14

7

Intensity transformation and spatial filtering

II.4. Piecewise-Linear Transform…

Digital Image Processing

15

Digital Image Processing

16

8

Intensity transformation and spatial filtering

II.5. Bit – Plane Slicing

Digital Image Processing

17

II.5. Bit – Plane Slicing

Digital Image Processing

18

9

Intensity transformation and spatial filtering

III. Histogram processing

Histogram Equalization

Histogram Matching

Local Histogram Processing

Using

Histogram

Statistics

Enhancement

Digital Image Processing

for

Image

19

III. Histogram processing

Histogram h(rk ) = nk

rk is the k th intensity value

nk is the number of pixels in the image with intensity rk

nk

MN

nk : the number of pixels in the image of

Normalized histogram p (rk ) =

size M × N with intensity rk

Digital Image Processing

20

10

Intensity transformation and spatial filtering

III. Histogram processing

No. of pixels

6

2

3

3

2

4

2

4

3

3

2

3

5

2

4

2

4

4x4 image

Gray scale = [0,9]

5

4

3

2

1

Gray level

0 1 2 3 4 5 6 7 8 9

Digital Image Processing

histogram

21

III. Histogram processing

Digital Image Processing

22

11

Intensity transformation and spatial filtering

III.1. Histogram Equalization

As the low-contrast image’s 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

23

Digital Image Processing

III.1. Histogram Equalization

Histogram transformation

s = T(r)

s

Where 0 ≤ r ≤ 1

T(r) satisfies

sk= T(rk)

T(r)

0

rk

Digital Image Processing

1

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

and monotonically

increasingly in the interval

0≤r≤1

(b). 0 ≤ T(r) ≤ 1 for

0≤r≤1

r

24

12

Intensity transformation and spatial filtering

III.1. Histogram Equalization

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 won’t cause a

negative image

0 ≤ T(r) ≤ 1 for 0 ≤ r ≤ 1 guarantees that the output gray

levels will be in the same range as the input levels.

The inverse transformation from s back to r is

r = T -1(s) ; 0 ≤ s ≤ 1

25

Digital Image Processing

III.1. Histogram Equalization

Let

pr(r) denote the PDF of random variable r

ps(s) denote the PDF of random variable s

If pr(r) and T(r) are known and T-1(s) satisfies

condition (a) then ps(s) can be obtained using a

formula :

ps(s) = pr(r)

Digital Image Processing

dr

ds

26

13

Intensity transformation and spatial filtering

III.1. Histogram Equalization

A transformation function is a cumulative distribution

function (CDF) of random variable r :

r

s = T ( r ) = ∫ pr ( w )dw

0

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

27

Digital Image Processing

III.1. Histogram Equalization

ds dT ( r )

=

dr

dr

r

⎤

d ⎡

p

(

w

)

dw

=

⎢ r

⎥

dr ⎣ ∫0

⎦

= pr ( r )

p s ( s ) = pr ( r )

Substitute and yield

Digital Image Processing

= pr ( r )

dr

ds

1

pr ( r )

= 1 where 0 ≤ s ≤ 1

28

14

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Ps(s):

As ps(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.

Called ps(s) as a uniform probability density function

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

pr(r)

29

Digital Image Processing

III.1. Histogram Equalization

Discrete Transformation Function

The probability of occurrence of gray level in an image

is approximated by

n

where k = 0 , 1 , ..., L- 1

p r ( rk ) = k

n

The discrete version of transformation

s k = T ( rk ) =

=

Digital Image Processing

k

nj

j=0

n

∑

k

∑

j=0

pr ( rj )

where k

= 0 , 1 , ..., L- 1

30

15

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Thus, an output image is obtained by mapping

each pixel with level rk in the input image into a

corresponding pixel with level sk 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

31

Digital Image Processing

III.1. Histogram Equalization

Example

before

Digital Image Processing

after

Histogram

equalization

32

16

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Example

before

after

Histogram

equalization

The quality is

not improved

much because

the original

image already

has a broaden

gray-level scale

33

Digital Image Processing

III. Histogram processing

No. of pixels

6

2

3

3

2

4

2

4

3

3

2

3

5

2

4

2

4

4x4 image

Gray scale = [0,9]

Digital Image Processing

5

4

3

2

1

Gray level

0 1 2 3 4 5 6 7 8 9

histogram

34

17

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Example

Gray Level

0

1

2

3

4

5

6

7

8

9

No.of pixels

0

0

6

5

4

1

0

0

0

0

0

0

6

11

15

16

16

16

16

16

0

0

0

0

k

∑

n

j=0

k

s=∑

j =0

j

nj

n

sx9

6/

16

3.3

≈3

11 /

15 /

16 /

16/

16/

16/

16/

16

16

16

16

16

16

16

6.1

≈6

8.4

≈8

9

9

9

9

9

35

Digital Image Processing

III.1. Histogram Equalization

Example

No. of pixels

3

6

6

3

8

3

8

6

6

3

6

9

3

8

3

8

Output image

Gray scale = [0,9]

Digital Image Processing

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9

Gray level

Histogram equalization

36

18

Intensity transformation and spatial filtering

III.2. Histogram Matching

Generate a processed image that has a specified

histogram

Let pr ( r ) and pz ( z ) denote the continous probability

density functions of the variables r and z. pz ( z ) is the

specified probability density function.

Let s be the random variable with the probability

r

s = T (r ) = ( L − 1) ∫ pr ( w)dw

0

Define a random variable z with the probability

z

G ( z ) = ( L − 1) ∫ pz (t )dt = s

0

37

Digital Image Processing

III.2. Histogram Matching

r

s = T ( r ) = ( L − 1) ∫ p r ( w ) dw

0

z

G ( z ) = ( L − 1) ∫ p z ( t ) dt = s

0

z = G −1 ( s ) = G −1 [T (r ) ]

Digital Image Processing

38

19

Intensity transformation and spatial filtering

III.2. Histogram Matching

Procedure

1.

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

r

s = ( L − 1) ∫ pr ( w)dw

0

2.

Use the specified PDF and obtain the transformation function

G(z)

z

G ( z ) = ( L − 1) ∫ pz (t )dt = s

0

3.

Mapping from s to z

z = G −1 ( s )

39

Digital Image Processing

III.2. Histogram Matching

Example

Assume an image has a gray level probability density function

pr(r) as shown.

⎧ − 2 r + 2 ;0 ≤ r ≤ 1

pr ( r ) = ⎨

; elsewhere

⎩ 0

Pr(r)

2

1

r

∫

p r ( w ) dw = 1

0

0

1

Digital Image Processing

2

r

40

20

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

We would like to apply the histogram specification with the

desired probability density function pz(z) as shown.

Pz(z)

2

⎧ 2z

pz ( z ) = ⎨

⎩ 0

1

z

0

1

2

∫p

z

;0 ≤ z ≤ 1

; elsewhere

z

( w )dw = 1

0

41

Digital Image Processing

III.2. Histogram Matching

Example

1. Obtain the transformation function T(r)

r

s=T(r)

s = T ( r ) = ∫ pr ( w )dw

0

1

r

= ∫ ( −2 w + 2 )dw

0

= − w 2 + 2w

0

Digital Image Processing

1

r

r

0

= − r + 2r

2

42

21

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

2. Obtain the transformation function G(z)

z

G ( z ) = ∫ ( 2 w )dw

= z2

0

z

0

= z2

43

Digital Image Processing

III.2. Histogram Matching

Example

3. Obtain the inversed transformation function G-1

G ( z ) = T (r )

z 2 = − r 2 + 2r

z = 2r − r 2

We can guarantee that 0 ≤ z ≤1 when 0 ≤ r ≤1

Digital Image Processing

44

22

Intensity transformation and spatial filtering

III.2. Histogram Matching

Procedure in discrete cases

1.

Obtain pr(rj) from the input image and then obtain the values of

sk, round the value to the integer range [0, L-1]

k

sk = T (rk ) = ( L − 1)∑ pr (rj ) =

j =0

2.

( L − 1) k

∑ nj

MN j =0

Use the specified PDF and obtain the transformation function

G(zq), round the value to the integer range [0, L-1].

q

G ( zq ) = ( L − 1)∑ pz ( zi ) = sk

i =0

3.

Mapping from sk to zq

zq = G −1 ( sk )

45

Digital Image Processing

III.2. Histogram Matching

Example

Digital Image Processing

Suppose that a 3-bit image (L=8) of size 64 × 64 pixels (MN =

4096) has the intensity distribution shown in the following

table (on the left). Get the histogram transformation function

and make the output image with the specified histogram, listed

in the table on the right.

46

23

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

Obtain the scaled histogram-equalized values,

s0 = 1, s1 = 3, s2 = 5, s3 = 6, s4 = 7,

s5 = 7, s6 = 7, s7 = 7.

Compute all the values of the transformation function G,

0

G ( z0 ) = 7∑ pz ( z j ) = 0.00

→0

j =0

G ( z1 ) = 0.00 → 0

G ( z3 ) = 1.05 → 1

G ( z5 ) = 4.55 → 5

G ( z7 ) = 7.00 → 7

s0

s2

→0

G ( z4 ) = 2.45 → 2 s1

G ( z2 ) = 0.00

G ( z6 ) = 5.95

s4 s5 s6

→ 6 s3

s7

47

Digital Image Processing

III.2. Histogram Matching

Example

rk

0

1

2

3

4

5

6

7

Digital Image Processing

s0 = 1, s1 = 3, s2 = 5, s3 = 6, s4 = 7,

rk → zq

s5 = 7, s6 = 7, s7 = 7.

0→3

1→ 4

2→5

3→6

4→7

5→7

6→7

7→7

48

24

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

Digital Image Processing

49

III.2. Histogram Matching

Example

Digital Image Processing

50

25

Digital Image Processing

Lecture 3 – Intensity

Transformation& Spatial Filtering

Lecturer: Ha Dai Duong

Faculty of Information Technology

I. Introduction

Spatial Domain vs. Transform Domain

Spatial Domain

Image plane itself, directly process the intensity values of the

image plane

Transform (Frequency) domain

Process the transform coefficients, not directly process the

intensity values of the image plane

Digital Image Processing

2

1

Intensity transformation and spatial filtering

I. Introduction

Spatial Domain Process

g ( x, y ) = T [ f ( x, y )])

f ( x, y ) : input image

g ( x, y ) : output image

T : an operator on f defined over

a neighborhood of point ( x, y )

Digital Image Processing

3

I. Introduction

Spatial Domain Process

Digital Image Processing

4

2

Intensity transformation and spatial filtering

II. Intensity transformation function

Intensity transformation function

s = T (r )

5

Digital Image Processing

II. Intensity transformation function

Some

basic

Functions

Digital Image Processing

intensity

transformation

6

3

Intensity transformation and spatial filtering

II.1. Negative

Image negatives

s = L −1− r

Digital Image Processing

7

II.1. Negative

Small

lesion

Digital Image Processing

8

4

Intensity transformation and spatial filtering

II.2. Log Transform

Log Transformations

s = c log(1 + r )

Digital Image Processing

9

II.2. Log Transform

Digital Image Processing

10

5

Intensity transformation and spatial filtering

II.3. Power – Law

s = cr γ

Digital Image Processing

11

II.3. Power – Law

Digital Image Processing

12

6

Intensity transformation and spatial filtering

II.3. Power – Law

Digital Image Processing

13

II.4. Piecewise-Linear Transform..

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

Digital Image Processing

14

7

Intensity transformation and spatial filtering

II.4. Piecewise-Linear Transform…

Digital Image Processing

15

Digital Image Processing

16

8

Intensity transformation and spatial filtering

II.5. Bit – Plane Slicing

Digital Image Processing

17

II.5. Bit – Plane Slicing

Digital Image Processing

18

9

Intensity transformation and spatial filtering

III. Histogram processing

Histogram Equalization

Histogram Matching

Local Histogram Processing

Using

Histogram

Statistics

Enhancement

Digital Image Processing

for

Image

19

III. Histogram processing

Histogram h(rk ) = nk

rk is the k th intensity value

nk is the number of pixels in the image with intensity rk

nk

MN

nk : the number of pixels in the image of

Normalized histogram p (rk ) =

size M × N with intensity rk

Digital Image Processing

20

10

Intensity transformation and spatial filtering

III. Histogram processing

No. of pixels

6

2

3

3

2

4

2

4

3

3

2

3

5

2

4

2

4

4x4 image

Gray scale = [0,9]

5

4

3

2

1

Gray level

0 1 2 3 4 5 6 7 8 9

Digital Image Processing

histogram

21

III. Histogram processing

Digital Image Processing

22

11

Intensity transformation and spatial filtering

III.1. Histogram Equalization

As the low-contrast image’s 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

23

Digital Image Processing

III.1. Histogram Equalization

Histogram transformation

s = T(r)

s

Where 0 ≤ r ≤ 1

T(r) satisfies

sk= T(rk)

T(r)

0

rk

Digital Image Processing

1

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

and monotonically

increasingly in the interval

0≤r≤1

(b). 0 ≤ T(r) ≤ 1 for

0≤r≤1

r

24

12

Intensity transformation and spatial filtering

III.1. Histogram Equalization

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 won’t cause a

negative image

0 ≤ T(r) ≤ 1 for 0 ≤ r ≤ 1 guarantees that the output gray

levels will be in the same range as the input levels.

The inverse transformation from s back to r is

r = T -1(s) ; 0 ≤ s ≤ 1

25

Digital Image Processing

III.1. Histogram Equalization

Let

pr(r) denote the PDF of random variable r

ps(s) denote the PDF of random variable s

If pr(r) and T(r) are known and T-1(s) satisfies

condition (a) then ps(s) can be obtained using a

formula :

ps(s) = pr(r)

Digital Image Processing

dr

ds

26

13

Intensity transformation and spatial filtering

III.1. Histogram Equalization

A transformation function is a cumulative distribution

function (CDF) of random variable r :

r

s = T ( r ) = ∫ pr ( w )dw

0

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

27

Digital Image Processing

III.1. Histogram Equalization

ds dT ( r )

=

dr

dr

r

⎤

d ⎡

p

(

w

)

dw

=

⎢ r

⎥

dr ⎣ ∫0

⎦

= pr ( r )

p s ( s ) = pr ( r )

Substitute and yield

Digital Image Processing

= pr ( r )

dr

ds

1

pr ( r )

= 1 where 0 ≤ s ≤ 1

28

14

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Ps(s):

As ps(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.

Called ps(s) as a uniform probability density function

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

pr(r)

29

Digital Image Processing

III.1. Histogram Equalization

Discrete Transformation Function

The probability of occurrence of gray level in an image

is approximated by

n

where k = 0 , 1 , ..., L- 1

p r ( rk ) = k

n

The discrete version of transformation

s k = T ( rk ) =

=

Digital Image Processing

k

nj

j=0

n

∑

k

∑

j=0

pr ( rj )

where k

= 0 , 1 , ..., L- 1

30

15

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Thus, an output image is obtained by mapping

each pixel with level rk in the input image into a

corresponding pixel with level sk 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

31

Digital Image Processing

III.1. Histogram Equalization

Example

before

Digital Image Processing

after

Histogram

equalization

32

16

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Example

before

after

Histogram

equalization

The quality is

not improved

much because

the original

image already

has a broaden

gray-level scale

33

Digital Image Processing

III. Histogram processing

No. of pixels

6

2

3

3

2

4

2

4

3

3

2

3

5

2

4

2

4

4x4 image

Gray scale = [0,9]

Digital Image Processing

5

4

3

2

1

Gray level

0 1 2 3 4 5 6 7 8 9

histogram

34

17

Intensity transformation and spatial filtering

III.1. Histogram Equalization

Example

Gray Level

0

1

2

3

4

5

6

7

8

9

No.of pixels

0

0

6

5

4

1

0

0

0

0

0

0

6

11

15

16

16

16

16

16

0

0

0

0

k

∑

n

j=0

k

s=∑

j =0

j

nj

n

sx9

6/

16

3.3

≈3

11 /

15 /

16 /

16/

16/

16/

16/

16

16

16

16

16

16

16

6.1

≈6

8.4

≈8

9

9

9

9

9

35

Digital Image Processing

III.1. Histogram Equalization

Example

No. of pixels

3

6

6

3

8

3

8

6

6

3

6

9

3

8

3

8

Output image

Gray scale = [0,9]

Digital Image Processing

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9

Gray level

Histogram equalization

36

18

Intensity transformation and spatial filtering

III.2. Histogram Matching

Generate a processed image that has a specified

histogram

Let pr ( r ) and pz ( z ) denote the continous probability

density functions of the variables r and z. pz ( z ) is the

specified probability density function.

Let s be the random variable with the probability

r

s = T (r ) = ( L − 1) ∫ pr ( w)dw

0

Define a random variable z with the probability

z

G ( z ) = ( L − 1) ∫ pz (t )dt = s

0

37

Digital Image Processing

III.2. Histogram Matching

r

s = T ( r ) = ( L − 1) ∫ p r ( w ) dw

0

z

G ( z ) = ( L − 1) ∫ p z ( t ) dt = s

0

z = G −1 ( s ) = G −1 [T (r ) ]

Digital Image Processing

38

19

Intensity transformation and spatial filtering

III.2. Histogram Matching

Procedure

1.

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

r

s = ( L − 1) ∫ pr ( w)dw

0

2.

Use the specified PDF and obtain the transformation function

G(z)

z

G ( z ) = ( L − 1) ∫ pz (t )dt = s

0

3.

Mapping from s to z

z = G −1 ( s )

39

Digital Image Processing

III.2. Histogram Matching

Example

Assume an image has a gray level probability density function

pr(r) as shown.

⎧ − 2 r + 2 ;0 ≤ r ≤ 1

pr ( r ) = ⎨

; elsewhere

⎩ 0

Pr(r)

2

1

r

∫

p r ( w ) dw = 1

0

0

1

Digital Image Processing

2

r

40

20

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

We would like to apply the histogram specification with the

desired probability density function pz(z) as shown.

Pz(z)

2

⎧ 2z

pz ( z ) = ⎨

⎩ 0

1

z

0

1

2

∫p

z

;0 ≤ z ≤ 1

; elsewhere

z

( w )dw = 1

0

41

Digital Image Processing

III.2. Histogram Matching

Example

1. Obtain the transformation function T(r)

r

s=T(r)

s = T ( r ) = ∫ pr ( w )dw

0

1

r

= ∫ ( −2 w + 2 )dw

0

= − w 2 + 2w

0

Digital Image Processing

1

r

r

0

= − r + 2r

2

42

21

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

2. Obtain the transformation function G(z)

z

G ( z ) = ∫ ( 2 w )dw

= z2

0

z

0

= z2

43

Digital Image Processing

III.2. Histogram Matching

Example

3. Obtain the inversed transformation function G-1

G ( z ) = T (r )

z 2 = − r 2 + 2r

z = 2r − r 2

We can guarantee that 0 ≤ z ≤1 when 0 ≤ r ≤1

Digital Image Processing

44

22

Intensity transformation and spatial filtering

III.2. Histogram Matching

Procedure in discrete cases

1.

Obtain pr(rj) from the input image and then obtain the values of

sk, round the value to the integer range [0, L-1]

k

sk = T (rk ) = ( L − 1)∑ pr (rj ) =

j =0

2.

( L − 1) k

∑ nj

MN j =0

Use the specified PDF and obtain the transformation function

G(zq), round the value to the integer range [0, L-1].

q

G ( zq ) = ( L − 1)∑ pz ( zi ) = sk

i =0

3.

Mapping from sk to zq

zq = G −1 ( sk )

45

Digital Image Processing

III.2. Histogram Matching

Example

Digital Image Processing

Suppose that a 3-bit image (L=8) of size 64 × 64 pixels (MN =

4096) has the intensity distribution shown in the following

table (on the left). Get the histogram transformation function

and make the output image with the specified histogram, listed

in the table on the right.

46

23

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

Obtain the scaled histogram-equalized values,

s0 = 1, s1 = 3, s2 = 5, s3 = 6, s4 = 7,

s5 = 7, s6 = 7, s7 = 7.

Compute all the values of the transformation function G,

0

G ( z0 ) = 7∑ pz ( z j ) = 0.00

→0

j =0

G ( z1 ) = 0.00 → 0

G ( z3 ) = 1.05 → 1

G ( z5 ) = 4.55 → 5

G ( z7 ) = 7.00 → 7

s0

s2

→0

G ( z4 ) = 2.45 → 2 s1

G ( z2 ) = 0.00

G ( z6 ) = 5.95

s4 s5 s6

→ 6 s3

s7

47

Digital Image Processing

III.2. Histogram Matching

Example

rk

0

1

2

3

4

5

6

7

Digital Image Processing

s0 = 1, s1 = 3, s2 = 5, s3 = 6, s4 = 7,

rk → zq

s5 = 7, s6 = 7, s7 = 7.

0→3

1→ 4

2→5

3→6

4→7

5→7

6→7

7→7

48

24

Intensity transformation and spatial filtering

III.2. Histogram Matching

Example

Digital Image Processing

49

III.2. Histogram Matching

Example

Digital Image Processing

50

25

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