EELE 5310: Digital Image Processing

Lecture 2Ch. 3

Eng. Ruba A. Salamah

Rsalamah @ iugaza.Edu

EELE 5310: Digital Image Processing

Lecture 2Ch. 3

Eng. Ruba A. Salamah

Rsalamah @ iugaza.Edu

1

Image Enhancement in the Spatial DomainImage Enhancement in the Spatial Domain

2

Lecture Reading

3.1 Background

3.2 Some Basic Gray Level Transformations Some Basic GrayLevel Transformations

3.2.1 Image Negatives

3.2.2 Log Transformations

3.2.3 Power-Law Transformations

3.2.4 Piecewise-Linear Transformation Functions

Contrast stretching

Gray-level slicing

Bit-plane slicing

3.1 Background

3.2 Some Basic Gray Level Transformations Some Basic GrayLevel Transformations

3.2.1 Image Negatives

3.2.2 Log Transformations

3.2.3 Power-Law Transformations

3.2.4 Piecewise-Linear Transformation Functions

Contrast stretching

Gray-level slicing

Bit-plane slicing3

Process an image so that the result will be moresuitable than the original image for a specificapplication.

The suitableness is up to each application.

A method which is quite useful for enhancing animage may not necessarily be the best approachfor enhancing another images

Principle Objective of Enhancement

Process an image so that the result will be moresuitable than the original image for a specificapplication.

The suitableness is up to each application.

A method which is quite useful for enhancing animage may not necessarily be the best approachfor enhancing another images

4

2 domains

Spatial Domain : (image plane) Techniques are based on direct manipulation of pixels in an

image

Frequency Domain : Techniques are based on modifying the Fourier transform of

an image

There are some enhancement techniques based onvarious combinations of methods from these twocategories.

Spatial Domain : (image plane) Techniques are based on direct manipulation of pixels in an

image

Frequency Domain : Techniques are based on modifying the Fourier transform of

an image

There are some enhancement techniques based onvarious combinations of methods from these twocategories.

5

Good images

For human visual

The visual evaluation of image quality is a highly subjective

process.

It is hard to standardize the definition of a good image.

A certain amount of trial and error usually is required

before a particular image enhancement approach is

selected.

For human visual

The visual evaluation of image quality is a highly subjective

process.

It is hard to standardize the definition of a good image.

A certain amount of trial and error usually is required

before a particular image enhancement approach is

selected.

6

Spatial Domain

Procedures that operatedirectly on pixels.

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

f(f(x,yx,y)) is the input image

g(g(x,yx,y)) is the processed image

TT is an operator on ff definedover some neighborhood of((x,yx,y))

Procedures that operatedirectly on pixels.

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

f(f(x,yx,y)) is the input image

g(g(x,yx,y)) is the processed image

TT is an operator on ff definedover some neighborhood of((x,yx,y))

7

Mask/Filter

Neighborhood of a point (x,y) canbe defined by using asquare/rectangular (commonused) or circular subimage areacentered at (x,y)

The center of the subimage ismoved from pixel to pixel startingat the top left corner.

•(x,y)

Neighborhood of a point (x,y) canbe defined by using asquare/rectangular (commonused) or circular subimage areacentered at (x,y)

The center of the subimage ismoved from pixel to pixel startingat the top left corner.

•

8

Mask Processing or Filtering

Neighborhood is bigger than 1x1 pixel

Use a function of the values of f in a predefinedneighborhood of (x,y) to determine the value of g at(x,y)

The value of the mask coefficients determine thenature of the process.

Used in techniques like: Image Sharpening

Image Smoothing

Neighborhood is bigger than 1x1 pixel

Use a function of the values of f in a predefinedneighborhood of (x,y) to determine the value of g at(x,y)

The value of the mask coefficients determine thenature of the process.

Used in techniques like: Image Sharpening

Image Smoothing9

Point Processing

Neighborhood = 1x1 pixel gg depends on only the value of ff at ((x,yx,y)) TT = gray level (or intensity or mapping) transformation function:

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

rr = gray level of f(f(x,yx,y))ss = gray level of g(g(x,yx,y))

Because enhancement at any point in an image depends only onthe gray level at that point, techniques in this category often arereferred to as point processing.

Neighborhood = 1x1 pixel gg depends on only the value of ff at ((x,yx,y)) TT = gray level (or intensity or mapping) transformation function:

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

rr = gray level of f(f(x,yx,y))ss = gray level of g(g(x,yx,y))

Because enhancement at any point in an image depends only onthe gray level at that point, techniques in this category often arereferred to as point processing.

10

Contrast Stretching

Produce higher contrastthan the original by:

darkening the levelsbelow m in the originalimage

Brightening the levelsabove m in the originalimage

Produce higher contrastthan the original by:

darkening the levelsbelow m in the originalimage

Brightening the levelsabove m in the originalimage

11

Thresholding

Produce a two-level

(binary) image

T(r) is called a

thresholding function.

Produce a two-level

(binary) image

T(r) is called a

thresholding function.

12

3 basic gray-level transformation functions

Linear function Negative and identity

transformations

Logarithm function Log and inverse-log

transformation

Power-law function nth power and nth root

transformations

Negative

Log

nth root

nth power

Linear function Negative and identity

transformations

Logarithm function Log and inverse-log

transformation

Power-law function nth power and nth root

transformationsInput gray level, r

Identity Inverse Log

13

Image Negatives An image with gray level in the

range [[00, L, L--11]]where L =L = 22nn ; n = 1, 2…

Negative transformation :

s = Ls = L –– 11 ––rr

Reversing the intensity levels ofan image.

Suitable for enhancing white orgray detail embedded in darkregions of an image, especiallywhen the black area dominantin size.

Negative

Log

nth root

nth power

An image with gray level in therange [[00, L, L--11]]where L =L = 22nn ; n = 1, 2…

Negative transformation :

s = Ls = L –– 11 ––rr

Reversing the intensity levels ofan image.

Suitable for enhancing white orgray detail embedded in darkregions of an image, especiallywhen the black area dominantin size.

Input gray level, r

Identity Inverse Log

14

Image Negatives

Example:1 5 1

5 255 5

1 5 0

254 250 254

250 0 250

254 250 255

Negative

Original

15

Log Transformations c is a constant

and r ≥ 0

Log curve maps a narrowrange of low gray-level valuesin the input image into a widerrange of output levels. Theopposite is true for highervalues.

Used to expand the values ofdark pixels in an image whilecompressing the higher-levelvalues.

Negative

Log

nth root

nth power

s = c log (s = c log (11+r)+r) c is a constantand r ≥ 0

Log curve maps a narrowrange of low gray-level valuesin the input image into a widerrange of output levels. Theopposite is true for highervalues.

Used to expand the values ofdark pixels in an image whilecompressing the higher-levelvalues.

Input gray level, r

Identity Inverse Log

16

Example of Logarithm Image

17

Inverse Logarithm Transformations

Do opposite to theLog Transformations

Used to expand thevalues of high pixelsin an image whilecompressing thedarker-level values.

Negative

Log

nth root

nth power

Do opposite to theLog Transformations

Used to expand thevalues of high pixelsin an image whilecompressing thedarker-level values.Input gray level, r

Identity Inverse Log

18

Power-Law Transformations

s =s = crcr c and γ are positive constants

Power-law curves withfractional values of γ map anarrow range of dark inputvalues into a wider range ofoutput values. The opposite istrue for higher values of inputlevels.

c = γ = 1 Identity function

s =s = crcr c and γ are positive constants

Power-law curves withfractional values of γ map anarrow range of dark inputvalues into a wider range ofoutput values. The opposite istrue for higher values of inputlevels.

c = γ = 1 Identity functionInput gray level, r

Plots of s =s = crcr for various values of γ(c = 1 in all cases)

19

Example 1: Gamma correction

Cathode ray tube (CRT)devices have an intensity-to-voltage response that isa power function, with γvarying from 1.8 to 2.5

The picture will becomedarker.

Gamma correction is doneby preprocessing theimage before inputting itto the monitor with s =s =crcr11//

Monitor

Cathode ray tube (CRT)devices have an intensity-to-voltage response that isa power function, with γvarying from 1.8 to 2.5

The picture will becomedarker.

Gamma correction is doneby preprocessing theimage before inputting itto the monitor with s =s =crcr11//

Monitor

Gammacorrection

= 2.5

=1/2.5 = 0.420

Example 2: MRI

(a) The picture is predominately dark

(b) Result after power-law

transformation with γ = 0.6

(c) transformation with γ = 0.4

(best result)

(d) transformation with γ = 0.3

(washed out look)

under than 0.3 will be reduced tounacceptable level.

a b (a) The picture is predominately dark

(b) Result after power-law

transformation with γ = 0.6

(c) transformation with γ = 0.4

(best result)

(d) transformation with γ = 0.3

(washed out look)

under than 0.3 will be reduced tounacceptable level.

c d

21

Example 3

(a) image has a washed-outappearance, it needs acompression of gray levels needs γ > 1

(b) result after power-lawtransformation with γ = 3.0(suitable)

(c) transformation with γ = 4.0(suitable)

(d) transformation with γ = 5.0(high contrast, the imagehas areas that are too dark,some detail is lost)

a b (a) image has a washed-outappearance, it needs acompression of gray levels needs γ > 1

(b) result after power-lawtransformation with γ = 3.0(suitable)

(c) transformation with γ = 4.0(suitable)

(d) transformation with γ = 5.0(high contrast, the imagehas areas that are too dark,some detail is lost)

dc

22

Piecewise-Linear Transformation Functions

Advantage: The form of piecewise

functions can be arbitrarilycomplex

Disadvantage: Their specification requires

considerably more user input

Advantage: The form of piecewise

functions can be arbitrarilycomplex

Disadvantage: Their specification requires

considerably more user input

23

Contrast Stretching

increase the dynamicrange of the gray levels inthe image

(b) a low-contrast image

(c) result of contraststretching: (r1,s1)=(rmin,0)and (r2,s2) = (rmax,L-1)

(d) result of thresholding

increase the dynamicrange of the gray levels inthe image

(b) a low-contrast image

(c) result of contraststretching: (r1,s1)=(rmin,0)and (r2,s2) = (rmax,L-1)

(d) result of thresholding

24

Gray-level slicing Highlighting a specific range of

gray levels in an image

Display a high value of allgray levels in the range ofinterest and a low value forall other gray levels

(a) transformation highlightsrange [A,B] of gray level andreduces all others to a constantlevel (result in binary image)

(b) transformation highlightsrange [A,B] but preserves allother levels

Highlighting a specific range ofgray levels in an image

Display a high value of allgray levels in the range ofinterest and a low value forall other gray levels

(a) transformation highlightsrange [A,B] of gray level andreduces all others to a constantlevel (result in binary image)

(b) transformation highlightsrange [A,B] but preserves allother levels

25

Bit-plane Slicing

Highlighting the contributionmade to total image appearanceby specific bits

Suppose each pixel is representedby 8 bits

Higher-order bits contain themajority of the visuallysignificant data

Useful for analyzing the relativeimportance played by each bit ofthe image

Bit-plane 7(most significant)

One 8-bit byte

Highlighting the contributionmade to total image appearanceby specific bits

Suppose each pixel is representedby 8 bits

Higher-order bits contain themajority of the visuallysignificant data

Useful for analyzing the relativeimportance played by each bit ofthe image

Bit-plane 0(least significant)

26

Example

An 8-bit fractal image

27

8 bit planes

Bit-plane 7 Bit-plane 6

Bit-plane 5

Bit-plane 4

Bit-plane 3

Bit-plane 5

Bit-plane 4

Bit-plane 3

Bit-plane 2

Bit-plane 1

Bit-plane 0

28

Next lecture Reading

3.3 Histogram processing

3.3.1 Histogram Equalization

3.3.2 Histogram Specification

3.4 Enhancement Using Arithmetic/LogicOperations

3.4.1 Image Subtraction

3.4.2 Image Averaging

3.3 Histogram processing

3.3.1 Histogram Equalization

3.3.2 Histogram Specification

3.4 Enhancement Using Arithmetic/LogicOperations

3.4.1 Image Subtraction

3.4.2 Image Averaging

29

Histogram Processing

Histogram of a digital image with gray levels in therange [0,L-1] is a discrete function

h(rh(rkk) = n) = nkk

Where rk : the kth gray level

nk : the number of pixels in the image having gray level rk

h(rk) : histogram of a digital image with gray levels rk

Histogram of a digital image with gray levels in therange [0,L-1] is a discrete function

h(rh(rkk) = n) = nkk

Where rk : the kth gray level

nk : the number of pixels in the image having gray level rk

h(rk) : histogram of a digital image with gray levels rk

30

Example

2 3 3 2

4 2 4 3 4

5

6

No. of pixels

4 2 4 3

3 2 3 5

2 4 2 4

4x4 image

Gray scale = [0,9]histogram

0 1

1

2

2

3

3

4

4

5 6 7 8 9

Gray level

31

Normalized Histogram

dividing each of histogram at gray level rrkk by the totalnumber of pixels in the image, nn

p(p(rrkk) =) = nnkk / n/ nFor k = 0,1,…,L-1

p(p(rrkk)) gives an estimate of the probability of occurrenceof gray level rrkk

The sum of all components of a normalized histogramis equal to 1

dividing each of histogram at gray level rrkk by the totalnumber of pixels in the image, nn

p(p(rrkk) =) = nnkk / n/ nFor k = 0,1,…,L-1

p(p(rrkk)) gives an estimate of the probability of occurrenceof gray level rrkk

The sum of all components of a normalized histogramis equal to 1

32

Histogram Processing

Basic for numerous spatial domain processingtechniques

Used effectively for image enhancement

Information inherent in histograms also is usefulin image compression and segmentation

Basic for numerous spatial domain processingtechniques

Used effectively for image enhancement

Information inherent in histograms also is usefulin image compression and segmentation

33

Example

Dark image

Components ofhistogram areconcentrated on thelow side of the grayscale.

Bright image

Components ofhistogram areconcentrated on thelow side of the grayscale.

Components ofhistogram areconcentrated on thehigh side of the grayscale.

34

ExampleLow-contrast image

histogram is narrowand centered towardthe middle of thegray scale

High-contrast imagehistogram covers broadrange of the gray scaleand the distribution ofpixels is not too far fromuniform, with very fewvertical lines being muchhigher than the others

35

Histogram Equalization

As the low-contrast image’s histogram is narrow andcentered toward the middle of the gray scale, if wedistribute the histogram to a wider range the quality ofthe image will be improved.

We can do it by adjusting the probability densityfunction of the original histogram of the image so thatthe probability spread equally

As the low-contrast image’s histogram is narrow andcentered toward the middle of the gray scale, if wedistribute the histogram to a wider range the quality ofthe image will be improved.

We can do it by adjusting the probability densityfunction of the original histogram of the image so thatthe probability spread equally

36

sk= T(rk)

s

Histogram transformation

s = T(r)s = T(r) Where 0 ≤ r ≤ 1 T(r) satisfies (a). T(r) is single-

valued andmonotonicallyincreasingly in theinterval 0 ≤ r ≤ 1

(b). 0 ≤ T(r) ≤ 1 for0 ≤ r ≤ 10 1rk r

T(r)

s = T(r)s = T(r) Where 0 ≤ r ≤ 1 T(r) satisfies (a). T(r) is single-

valued andmonotonicallyincreasingly in theinterval 0 ≤ r ≤ 1

(b). 0 ≤ T(r) ≤ 1 for0 ≤ r ≤ 1

٣٧

2 Conditions of T(r)

Single-valued (one-to-one relationship) guarantees thatthe inverse transformation will exist

Monotonicity condition preserves the increasing orderfrom black to white in the output 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) ;(s) ; 00 ss 11

Single-valued (one-to-one relationship) guarantees thatthe inverse transformation will exist

Monotonicity condition preserves the increasing orderfrom black to white in the output 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) ;(s) ; 00 ss 1138

Probability Density Function

The gray levels in an image may be viewed asrandom variables in the interval [0,1]

PDF is one of the fundamental descriptors of arandom variable

The gray levels in an image may be viewed asrandom variables in the interval [0,1]

PDF is one of the fundamental descriptors of arandom variable

39

Applied to Image

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

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

If pprr(r)(r) and T(r)T(r) are known and TT--11(s)(s) satisfies condition(a) then ppss(s)(s) can be obtained using a formula :

The PDF of the transformed variable s is determined by the gray-

level PDF of the input image and by the chosen transformation

function

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

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

If pprr(r)(r) and T(r)T(r) are known and TT--11(s)(s) satisfies condition(a) then ppss(s)(s) can be obtained using a formula :

The PDF of the transformed variable s is determined by the gray-

level PDF of the input image and by the chosen transformation

function

dsdr(r) p(s)p rs =

40

A transformation function of particular importancein DIP is :

Thus:∫==r

r dw)w(p)r(Ts0

Transformation function

)r(p

dw)w(pdrd

dr)r(dT

drds

r

r

r

=

=

=

∫0

A transformation function of particular importancein DIP is :

Thus:

)r(p

dw)w(pdrd

dr)r(dT

drds

r

r

r

=

=

=

∫0

10 where1

1

≤≤=

=

=

s)r(p

)r(p

dsdr)r(p)s(p

rr

rs

Substitute and yield 41

Transformation function

It is important to note that T(r) depends on pr(r),the resulting ps(s) always is uniform, independentof the form of pr(r).

∫==r

r dw)w(p)r(Ts0∫==r

r dw)w(p)r(Ts0

a random variable scharacterized by

a uniform probabilityfunction

yields

0

1

s

Ps(s)

42

Discrete Transformation Function

The probability of occurrence of gray level inan image is approximated by

The discrete version of transformation

110)( , ..., L-,where knnrp k

kr ==

The probability of occurrence of gray level inan image is approximated by

The discrete version of transformation

∑

∑

=

=

==

==

k

j

j

k

jjrkk

, ..., L-, where knn

)r(p)r(Ts

0

0

110

110)( , ..., L-,where knnrp k

kr ==

43

Histogram Equalization Implementation

Thus, an output image is obtained by mappingeach pixel with level rrkk in the input image into acorresponding pixel with level sskk in the outputimage such that:

Thus, an output image is obtained by mappingeach pixel with level rrkk in the input image into acorresponding pixel with level sskk in the outputimage such that:

∑=

==k

j

jk , ..., L-, where k

nn

s0

110

44

Example

٤٥

Histogram Processing

٤٦

Transformation functions used to equalized the histogram

Example Assume a 10 x 10 image (100 pixels) with the following normalized histogram:

For example, the value 0.1 appears 19 times in the histogram so the number is19/100 = 0.19 in the normalized histogram. Using equation (3), we get:

s1 s2 s3 s4 s5 s6 s7 s8 s9 s100.19 0.37 0.55 0.70 0.83 0.89 0.93 0.95 0.96 1.00

In the histogram equalized image, s5 = 0.83 will have a correspondinghistogram value of (the same as r5) 0.13

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

Assume a 10 x 10 image (100 pixels) with the following normalized histogram:

For example, the value 0.1 appears 19 times in the histogram so the number is19/100 = 0.19 in the normalized histogram. Using equation (3), we get:

s1 s2 s3 s4 s5 s6 s7 s8 s9 s100.19 0.37 0.55 0.70 0.83 0.89 0.93 0.95 0.96 1.00

In the histogram equalized image, s5 = 0.83 will have a correspondinghistogram value of (the same as r5) 0.13

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

47

Histogram Matching (Specification)

Histogram equalization has a disadvantage whichis that it can generate only one type of outputimage.

With Histogram Specification, we can specify theshape of the histogram that we wish the outputimage to have.

It doesn’t have to be a uniform histogram.

٤٨

Histogram equalization has a disadvantage whichis that it can generate only one type of outputimage.

With Histogram Specification, we can specify theshape of the histogram that we wish the outputimage to have.

It doesn’t have to be a uniform histogram.

Let pr(r) denote continuous probability density function of gray-level of inputimage, r

Let pz(z) denote desired (specified) continuous probability density function ofgray-level of output image, z

Let s be a random variable with the property

We need to find z such that:

Thus: s = T(r) = G(z).

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

Assume G-1 exists and satisfies the condition (a) and (b). We can map an input gray levelr to output gray level z

∫==r

r dw)w(p)r(Ts0

sdt)t(p)z(gz

z == ∫0

Consider the continuous domainLet pr(r) denote continuous probability density function of gray-level of inputimage, r

Let pz(z) denote desired (specified) continuous probability density function ofgray-level of output image, z

Let s be a random variable with the property

We need to find z such that:

Thus: s = T(r) = G(z).

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

Assume G-1 exists and satisfies the condition (a) and (b). We can map an input gray levelr to output gray level z

∫==r

r dw)w(p)r(Ts0

sdt)t(p)z(gz

z == ∫0

49

Discrete Formulation

12100

0

−==

==

∑

∑

=

=

L,...,,,knn

)r(p)r(Ts

k

j

j

k

jjrkk

٥٠

12100

−=== ∑=

L,...,,,ks)z(p)z(G k

k

iizk

[ ][ ] 12101

1

−==

=−

−

L,...,,,ksG)r(TGz

k

kk

Histogram Matching Implementation

1. Find histogram of input image, and find histogram equalizationmapping:

2. Specify the desired histogram , and find histogram equalizationmapping:

3. For every value of sk use the stored values of G to find thecorresponding min value zq such that:

∑=

=k

jjrk rps

0)(

1. Find histogram of input image, and find histogram equalizationmapping:

2. Specify the desired histogram , and find histogram equalizationmapping:

3. For every value of sk use the stored values of G to find thecorresponding min value zq such that:

∑=

=k

jjzk zpv

0)(

51

Histogram Matching Implementation

٥٢

• Consider an 8-level image with theshownhistogram

Histogram Matching Example

٥٣

• Match it to the image with thehistogram

1. Equalize the histogram of the input image using transform s =T(r)

Example Continue…

٥٤

2. Equalize the desired histogram v = G(z).

Example Continue…

٥٥

Example Continue…

3. Set v = s to obtain the composite transform

k rk Sk vk zk0 0 1 0 3

٥٦

0 0 1 0 3

1 1 3 0 4

2 2 5 0 5

3 3 6 1 6

4 4 6 2 6

5 5 7 5 7

6 6 7 6 7

7 7 7 7 7

Example

٥٧Image of Mars moon

Image is dominated by large, dark areas,resulting in a histogram characterized bya large concentration of pixels in pixels inthe dark end of the gray scale

Image Equalization

Result imageafter histogram

equalization

٥٨

Result imageafter histogram

equalizationTransformation functionfor histogram equalization Histogram of the result image

Histogram Equalization

Solve the problem

Since the problem with thetransformation function of thehistogram equalization wascaused by a large concentrationof pixels in the original image withlevels near 0

A reasonable approach is tomodify the histogram of thatimage so that it does not havethis property

٥٩

Since the problem with thetransformation function of thehistogram equalization wascaused by a large concentrationof pixels in the original image withlevels near 0

A reasonable approach is tomodify the histogram of thatimage so that it does not havethis property

Histogram Specification

(1) the transformationfunction G(z) obtainedfrom the specifiedhistogram.

(2) the inversetransformation G-1(s)

Histogram Specification

٦٠

(1) the transformationfunction G(z) obtainedfrom the specifiedhistogram.

(2) the inversetransformation G-1(s)

Result image and its histogram

٦١

Original image The output image’s histogramAfter applingthehistogrammatching

Notes

Histogram specification is a trial-and-errorprocess

There are no rules for specifying histograms, andone must resort to analysis on a case-by-casebasis for any given enhancement task.

٦٢

Histogram specification is a trial-and-errorprocess

There are no rules for specifying histograms, andone must resort to analysis on a case-by-casebasis for any given enhancement task.

Readings

From the book the following sections:

3.4: Enhancement Using Arithmetic/LogicOperations

3.4.1: Image Subtraction

3.4.2 :Image Averaging

From the book the following sections:

3.4: Enhancement Using Arithmetic/LogicOperations

3.4.1: Image Subtraction

3.4.2 :Image Averaging

63

Enhancement using Arithmetic/LogicOperations Arithmetic/Logic operations are performed on pixel by

pixel basis between two or more images

Except NOT operation which perform only on a singleimage.

We need only be concerned with the ability to implementthe AND, OR, and NOT which are functionally complete.

Pixel values are processed as strings of binary numbers.

NOT operation = negative transformation

٦٤

Arithmetic/Logic operations are performed on pixel bypixel basis between two or more images

Except NOT operation which perform only on a singleimage.

We need only be concerned with the ability to implementthe AND, OR, and NOT which are functionally complete.

Pixel values are processed as strings of binary numbers.

NOT operation = negative transformation

Example of AND Operation The AND and OR operations are used for masking; for selecting

subimages in an image. ROI processing. The AND and OR operations are used for masking; for selecting

subimages in an image. ROI processing.

original image AND imagemask

result of ANDoperation

65

Example of OR Operation

original image OR imagemask

result of ORoperation

٦٦

Image Subtraction

The difference between two images f(x, y) and h(x, y):

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

is obtained by computing the difference between all pairsof corresponding pixels from f and h.

The key usefulness of subtraction is the enhancement ofdifferences between images.

The difference between two images f(x, y) and h(x, y):

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

is obtained by computing the difference between all pairsof corresponding pixels from f and h.

The key usefulness of subtraction is the enhancement ofdifferences between images.

67

Image Subtraction

68

Mask mode radiography

Note:• the background is dark because it doesn’t change much in bothimages.• the difference area is bright because it has a big change

٦٩

Note

The values in a difference image can range from –255 to255, so some sort of scaling is required.

One method is to add 255 to every pixel and then divide by2.

Another method is the min_max normalization:

Where minA and maxA are the min and max values of thediff. image, and new_maxA = 255, and new_minA= 0.

thus

The values in a difference image can range from –255 to255, so some sort of scaling is required.

One method is to add 255 to every pixel and then divide by2.

Another method is the min_max normalization:

Where minA and maxA are the min and max values of thediff. image, and new_maxA = 255, and new_minA= 0.

thus

70

Image Averaging

consider a noisy image g(x,y) formed by theaddition of noise η(x,y) to an original imagef(x,y)

g(x,y) = f(x,y) + η(x,y)

consider a noisy image g(x,y) formed by theaddition of noise η(x,y) to an original imagef(x,y)

g(x,y) = f(x,y) + η(x,y)

71

Image Averaging if noise has zero mean and is uncorrelated then it can be

shown that if

image formed by averaging K different noisy images.

Then: and

Thus if K increases, the variability (noise) of the pixel ateach location (x,y) decreases in g(x,y).

This means that g(x,y) approaches f(x, y) as the number ofnoisy images used in the averaging process increases.

if noise has zero mean and is uncorrelated then it can beshown that if

image formed by averaging K different noisy images.

Then: and

Thus if K increases, the variability (noise) of the pixel ateach location (x,y) decreases in g(x,y).

This means that g(x,y) approaches f(x, y) as the number ofnoisy images used in the averaging process increases.

72

Example

a) original image b) image corrupted by

additive Gaussiannoise with zero meanand a standarddeviation of 64 graylevels.

c). -f). results ofaveraging K = 8, 16,64 and 128 noisyimages

a) original image b) image corrupted by

additive Gaussiannoise with zero meanand a standarddeviation of 64 graylevels.

c). -f). results ofaveraging K = 8, 16,64 and 128 noisyimages

a bc de f

٧٣

Example

74

Reading

3.5: Basics of Spatial Filtering

3.6:Smoothing Spatial Filters 3.6.1: Smoothing Linear Filters

3.6.2: Order-Statistics Filters

3.7: Sharpening Spatial Filters

3.7.2 The Laplacian Operator

3.7.3 Use of First Derivatives for Enhancement

3.8: Combining Spatial Enhancement Methods

3.5: Basics of Spatial Filtering

3.6:Smoothing Spatial Filters 3.6.1: Smoothing Linear Filters

3.6.2: Order-Statistics Filters

3.7: Sharpening Spatial Filters

3.7.2 The Laplacian Operator

3.7.3 Use of First Derivatives for Enhancement

3.8: Combining Spatial Enhancement Methods

75

Spatial Filtering

r s t

u v w

x y z

Origin x

FilterSimple 3*3

Neighbourhood e

a b c

d e f

g h iOriginal Image

Pixels

*

y Image f (x, y)

eprocessed = v*e +r*a + s*b + t*c +u*d + w*f +x*g + y*h + z*i

Simple 3*3Neighbourhood e

3*3 Filter

Original ImagePixels

The above is repeated for every pixel in theoriginal image to generate the processed image

Spatial Filtering

use filter (can also be called asmask/kernel/template or window)

the values in a filter subimage are referred to ascoefficients, rather than pixel.

our focus will be on masks of odd sizes, e.g.3x3, 5x5,…

use filter (can also be called asmask/kernel/template or window)

the values in a filter subimage are referred to ascoefficients, rather than pixel.

our focus will be on masks of odd sizes, e.g.3x3, 5x5,…

77

Spatial Filtering

78

Spatial Filtering Process

simply move the filter mask from point to point in animage.

at each point (x,y), the response of the filter at that

point is calculated using a predefined relationship.

Or:

simply move the filter mask from point to point in animage.

at each point (x,y), the response of the filter at that

point is calculated using a predefined relationship.

Or:

79

Linear Filtering

Linear Filtering of an image f of size MxN filtermask of size mxn is given by the expression

To generate a complete filtered image thisequation must be applied for x = 0, 1, 2, … , M-1and y = 0, 1, 2, … , N-1

Linear Filtering of an image f of size MxN filtermask of size mxn is given by the expression

To generate a complete filtered image thisequation must be applied for x = 0, 1, 2, … , M-1and y = 0, 1, 2, … , N-1

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

80

Note An important consideration in implementing neighborhood

operations for spatial filtering is the issue of what happenswhen the center of the filter approaches the border of theimage.

There are several ways to handle this situation: limit the excursions of the center of the mask to be at a distance

no less than (n-1)2 pixels from the border. filter all pixels only with the section of the mask that is fully

contained in the image. Other approaches include “padding” the image by adding rows

and columns of 0’s (or other constant gray level), or paddingby replicating rows or columns.

An important consideration in implementing neighborhoodoperations for spatial filtering is the issue of what happenswhen the center of the filter approaches the border of theimage.

There are several ways to handle this situation: limit the excursions of the center of the mask to be at a distance

no less than (n-1)2 pixels from the border. filter all pixels only with the section of the mask that is fully

contained in the image. Other approaches include “padding” the image by adding rows

and columns of 0’s (or other constant gray level), or paddingby replicating rows or columns.

81

Smoothing Spatial Filters

Image sensors and transmission channels mayproduce certain type of noise characterized byrandom and isolated pixels with out-of-rangegray levels, which are either much lower orhigher than the gray levels of the neighboringpixels.

One way to get rid of this kind of noise is to usethe average of a small local region in the imageso that the out-of-range gray levels can besuppressed.

Image sensors and transmission channels mayproduce certain type of noise characterized byrandom and isolated pixels with out-of-rangegray levels, which are either much lower orhigher than the gray levels of the neighboringpixels.

One way to get rid of this kind of noise is to usethe average of a small local region in the imageso that the out-of-range gray levels can besuppressed.

82

Smoothing Spatial Filters

Used for image blurring and for noise reduction Blurring is used in preprocessing steps, such as removal of small details from an image prior to

object extraction bridging of small gaps in lines or curves

noise reduction can be accomplished byblurring with a linear filter and also by anonlinear filter

Used for image blurring and for noise reduction Blurring is used in preprocessing steps, such as removal of small details from an image prior to

object extraction bridging of small gaps in lines or curves

noise reduction can be accomplished byblurring with a linear filter and also by anonlinear filter

83

Two Types of Smoothing filters

Smoothing Linear Filter

Order-Statistics Filters

Smoothing Linear Filter

Order-Statistics Filters

84

Smoothing Linear Filters

These filters are called averaging filters or lowpassfilters.

Replacing the value of every pixel in an image bythe average of the gray levels in the neighborhoodwill reduce the “sharp” transitions in gray levels.

sharp transitions random noise in the image edges of objects in the image

Thus, smoothing can reduce noises (desirable) andblur edges (undesirable)

These filters are called averaging filters or lowpassfilters.

Replacing the value of every pixel in an image bythe average of the gray levels in the neighborhoodwill reduce the “sharp” transitions in gray levels.

sharp transitions random noise in the image edges of objects in the image

Thus, smoothing can reduce noises (desirable) andblur edges (undesirable)

85

3x3 Smoothing Linear Filters

box filter weighted average

the center is the most important and otherpixels are inversely weighted as a function oftheir distance from the center of the mask

86

General form : smoothing mask

Filter of size mxn (m and n odd)

∑ ∑

∑ ∑

−= −=

−= −=

++= a

as

b

bt

a

as

b

bt

tsw

tysxftswyxg

),(

),(),(),(

∑ ∑

∑ ∑

−= −=

−= −=

++= a

as

b

bt

a

as

b

bt

tsw

tysxftswyxg

),(

),(),(),(

summation of all coefficient of the mask

87

Example

88

Example a). original image 500x500 pixel b). - f). results of smoothing with

square averaging filter masks ofsize 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 beblended with the background.

a). original image 500x500 pixel b). - f). results of smoothing with

square averaging filter masks ofsize 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 beblended with the background.

٨٩

Example

original image result after smoothingwith 15x15 averaging mask

result of thresholding

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

90

Order-Statistics Filters (Nonlinear Filters) The response is based on ordering (ranking) the

pixels contained in the image area encompassedby 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 maskMedian example: (10,20,20,20,15,20,20,25,100)

rank to:( 10,15,20,20,20,20,20,25,100)Median=20

The response is based on ordering (ranking) thepixels contained in the image area encompassedby 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 maskMedian example: (10,20,20,20,15,20,20,25,100)

rank to:( 10,15,20,20,20,20,20,25,100)Median=20

91

Median Filters

Replaces the value of a pixel by the median of the graylevels in the neighborhood of that pixel (the originalvalue of the pixel is included in the computation of themedian)

Quite popular because for certain types of random noise(impulse noiseimpulse noise salt and pepper noisesalt and pepper noise) , they provideprovideexcellent noiseexcellent noise--reduction capabilitiesreduction capabilities, with consideringless blurring than linear smoothing filters of similar size.less blurring than linear smoothing filters of similar size.

Replaces the value of a pixel by the median of the graylevels in the neighborhood of that pixel (the originalvalue of the pixel is included in the computation of themedian)

Quite popular because for certain types of random noise(impulse noiseimpulse noise salt and pepper noisesalt and pepper noise) , they provideprovideexcellent noiseexcellent noise--reduction capabilitiesreduction capabilities, with consideringless blurring than linear smoothing filters of similar size.less blurring than linear smoothing filters of similar size.

92

Median Filters

Forces the points with distinct gray levels to be more liketheir neighbors.

Isolated clusters of pixels that are light or dark withrespect to their neighbors, and whose area is less thann2/2 (one-half the filter area), are eliminated by an n x nmedian filter.

Eliminated = forced to have the value equal the medianintensity of the neighbors.

Larger clusters are affected considerably less

Forces the points with distinct gray levels to be more liketheir neighbors.

Isolated clusters of pixels that are light or dark withrespect to their neighbors, and whose area is less thann2/2 (one-half the filter area), are eliminated by an n x nmedian filter.

Eliminated = forced to have the value equal the medianintensity of the neighbors.

Larger clusters are affected considerably less

93

Example : Median Filters

94

Sharpening Spatial Filters To highlight fine details in an image or to

enhance detail that has been blurred, either inerror or as a natural effect of a particular methodof image acquisition.

Blurring can be done in spatial domain by pixelaveraging in a neighborhood, since averaging isanalogous to integration, thus, we can guess thatthe sharpening must be accomplished by spatialspatialdifferentiation.differentiation.

To highlight fine details in an image or toenhance detail that has been blurred, either inerror or as a natural effect of a particular methodof image acquisition.

Blurring can be done in spatial domain by pixelaveraging in a neighborhood, since averaging isanalogous to integration, thus, we can guess thatthe sharpening must be accomplished by spatialspatialdifferentiation.differentiation. 95

Derivative operator

The strength of the response of a derivativeoperator is proportional to the degree ofdiscontinuity of the image at the point at whichthe operator is applied.

Thus, image differentiation enhances edges and other discontinuities (noise)

The strength of the response of a derivativeoperator is proportional to the degree ofdiscontinuity of the image at the point at whichthe operator is applied.

Thus, image differentiation enhances edges and other discontinuities (noise)

96

First-order derivative

a basic definition of the first-order derivative of aone-dimensional function f(x) is the difference

)()1( xfxfxf −+=

∂∂

97

Second-order derivative

similarly, we define the second-order derivativeof a one-dimensional function f(x) is thedifference

similarly, we define the second-order derivativeof a one-dimensional function f(x) is thedifference

)(2)1()1(2

2

xfxfxfx

f −−++=∂∂

98

Derivative operator

99

First and Second-order derivative of f(x,y)

When we consider an image function of twovariables, f(x,y), at which time we will dealingwith partial derivatives along the two spatial axes.

When we consider an image function of twovariables, f(x,y), at which time we will dealingwith partial derivatives along the two spatial axes.

yyxf

xyxf

yxyxf

∂∂+

∂∂=

∂∂∂=∇ ),(),(),(f

2

2

2

22 ),(),(

yyxf

xyxff

∂∂+

∂∂=∇(linear operator)

Laplacian operator

Gradient operator

100

Discrete Form of Laplacian

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

2

yxfyxfyxfx

f −−++=∂∂

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

2

yxfyxfyxfy

f −−++=∂∂

from

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

2

yxfyxfyxfy

f −−++=∂∂

yield,

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

yxfyxfyxfyxfyxff

−−+++−++=∇

101

Result Laplacian mask

102

Laplacian mask implemented an extensionof diagonal neighbors

103

Other implementations of Laplacian masks

give the same result, but we have to keep in mind that when combining(add / subtract) a Laplacian-filtered image with another image.

104

Effect of Laplacian Operator As it is a derivative operator, It highlights gray-level discontinuities in an image

Tends to produce images that have Grayish edge lines and other discontinuities, all

superimposed on a dark, featureless background.

١٠٥

As it is a derivative operator, It highlights gray-level discontinuities in an image

Tends to produce images that have Grayish edge lines and other discontinuities, all

superimposed on a dark, featureless background.

Correct the effect of featurelessbackground

Easily by adding the original and Laplacianimage.

Be careful with the Laplacian filter used

Easily by adding the original and Laplacianimage.

Be careful with the Laplacian filter used

∇+∇−

=),(),(),(),(

),(2

2

yxfyxfyxfyxf

yxg

if the center coefficientof the Laplacian mask isnegative

if the center coefficientof the Laplacian mask ispositive

106

107

Example

a). image of the North poleof the moon

b). Laplacian-filtered imagewith

c). Laplacian image scaledfor display purposes

d). image enhanced byaddition with original image

a). image of the North poleof the moon

b). Laplacian-filtered imagewith

c). Laplacian image scaledfor display purposes

d). image enhanced byaddition with original image

1 1 1

1 -8 1

1 1 1

١٠٨

Mask of Laplacian + addition

To simply the computation, we can create a maskwhich do both operations, Laplacian Filter andAddition the original image.

)]1,()1,(),1(),1([),(5)],(4)1,()1,(

),1(),1([),(),(

−+++−++−=

−−+++−++−=

yxfyxfyxfyxfyxfyxfyxfyxf

yxfyxfyxfyxg

)]1,()1,(),1(),1([),(5)],(4)1,()1,(

),1(),1([),(),(

−+++−++−=

−−+++−++−=

yxfyxfyxfyxfyxfyxfyxfyxf

yxfyxfyxfyxg

0 -1 0

-1 5 -10 -1 0

109

Example

١١٠

Unsharp Masking and High-Boost Filtering

),(),(),( yxfyxfyxfs −=

sharpened image = original image – blurred imagesharpened image = original image – blurred image

This operation is known as unsharp masking

where fs(x, y) denotes the sharpened imageobtained by unsharp masking, and f(x, y) is ablurred version of f(x, y).

111

High-boost filtering

A slight further generalization of unsharp masking iscalled high-boost filtering.

A high-boost filtered image, fhb (x,y) is defined at anypoint (x, y) as:

where A ≥ 1.

A slight further generalization of unsharp masking iscalled high-boost filtering.

A high-boost filtered image, fhb (x,y) is defined at anypoint (x, y) as:

where A ≥ 1.

112

High-boost filtering

If we use Laplacian filter to create sharpenimage fs(x,y) with addition of original image

If we use Laplacian filter to create sharpenimage fs(x,y) with addition of original image

∇+∇−

=),(),(),(),(

),(2

2

yxfyxfyxfyxf

yxfs

113

High-boost filtering

yields

∇+∇−

=),(),(),(),(

),(2

2

yxfyxAfyxfyxAf

yxfhb

if the center coefficientof the Laplacian mask isnegative

∇+∇−

=),(),(),(),(

),(2

2

yxfyxAfyxfyxAf

yxfhb

if the center coefficientof the Laplacian mask ispositive

114

High-boost Masks

A ≥ 1 if A = 1, it becomes “standard” Laplacian

sharpening115

Example

One of the principal applications of boostfiltering is when the input image is darker thandesired. By varying the boost coefficient, itgenerally is possible to obtain an overall increasein average gray level of the image, thus helping tobrighten the final result.

One of the principal applications of boostfiltering is when the input image is darker thandesired. By varying the boost coefficient, itgenerally is possible to obtain an overall increasein average gray level of the image, thus helping tobrighten the final result.

116

Example

117

Gradient Operator

first derivatives are implemented using themagnitude of the gradientmagnitude of the gradient.

∂∂∂∂

=

=∇

yfxf

GG

y

xf

21

22

2122 ][)f(

∂∂+

∂∂=

+=∇=∇

yf

xf

GGmagf yx

21

22

2122 ][)f(

∂∂+

∂∂=

+=∇=∇

yf

xf

GGmagf yx

the magnitude becomes nonlinearyx GGf +≈∇

commonly approx.

118

Gradient Mask

simplest approximation, 2x2z1 z2 z3

z4 z5 z6

z7 z8 z9

)(and)( 5658 zzGzzG yx −=−=

212

562

582

122 ])()[(][ zzzzGGf yx −+−=+=∇

5658 zzzzf −+−≈∇119

Gradient Mask here

Roberts cross-gradient operators, 2x2

z1 z2 z3

z4 z5 z6

z7 z8 z9

)(and)( 6859 zzGzzG yx −=−= )(and)( 6859 zzGzzG yx −=−=

212

682

592

122 ])()[(][ zzzzGGf yx −+−=+=∇

6859 zzzzf −+−≈∇

120

Gradient Mask

Sobel operators, 3x3

z1 z2 z3

z4 z5 z6

z7 z8 z9

)2()2()2()2(

741963

321987

zzzzzzGzzzzzzG

y

x

++−++=++−++=

)2()2()2()2(

741963

321987

zzzzzzGzzzzzzG

y

x

++−++=++−++=

yx GGf +≈∇

121

Note

the summation of coefficients in all masks equals0, indicating that they would give a response of 0in an area of constant gray level.

the summation of coefficients in all masks equals0, indicating that they would give a response of 0in an area of constant gray level.

122

Example

123

Example of Combining SpatialEnhancement Methods

want to sharpen theoriginal image andbring out more skeletaldetail.

problems: narrowdynamic range of graylevel and high noisecontent makes theimage difficult toenhance

want to sharpen theoriginal image andbring out more skeletaldetail.

problems: narrowdynamic range of graylevel and high noisecontent makes theimage difficult toenhance

١٢٤

Example of Combining SpatialEnhancement Methods

solve :

1. Laplacian to highlight fine detail2. gradient to enhance prominent edges3. gray-level transformation to increase the

dynamic range of gray levels

solve :

1. Laplacian to highlight fine detail2. gradient to enhance prominent edges3. gray-level transformation to increase the

dynamic range of gray levels

125

126

127