image enhancement in the spatial domain1

8/2/2019 Image Enhancement in the Spatial Domain1

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1

Image Enhancement in the

Spatial Domain


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Principle Objective ofEnhancement

Process an image so that the result will bemore suitable than the original image for a

specific application.

The suitableness is up to each application.

A method which is quite useful for

enhancing an image may not necessarily bethe best approach for enhancing anotherimages


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2 domains

Spatial Domain : (image plane)

Techniques are based on direct manipulation of pixelsin an image

Frequency Domain : Techniques are based on modifying the Fourier

transform of an image

There are some enhancement techniques based

on various combinations of methods from thesetwo categories.


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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 goodimage.

For machine perception The evaluation task is easier.

A good image is one which gives the best machinerecognition results.

A certain amount of trial and error usually isrequired before a particular image enhancementapproach is selected.


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Spatial Domain

Procedures that operatedirectly on pixels.

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

where f(x,y) is the input image

g(x,y) is the processedimage

Tis an operator onfdefined over someneighborhood of (x,y)


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Point Processing

Neighborhood = 1x1 pixel

g depends on only the value offat (x,y)

T= gray level (or intensity or mapping)

transformation functions = T(r)

Where

r = gray level off(x,y)

s = gray level ofg(x,y)


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3 basic gray-leveltransformation functions

Linear function

Negative and identitytransformations

Logarithm function Log and inverse-log

transformation

Power-law function

nth power and nth roottransformations

Input gray level, r

Negative

Lognth root

Identity

nth power

Inverse Log


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Identity function

Output intensities areidentical to inputintensities.

Is included in thegraph only forcompleteness.

Input gray level, r

Negative

Lognth root

Identity

nth power

Inverse Log


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Image Negatives

An image with gray level in therange [0, L-1]whereL = 2n; n = 1, 2

Negative transformation :s = L1r

Reversing the intensity levelsof an image.

Suitable for enhancing whiteor gray detail embedded indark regions of an image,especially when the black areadominant in size.Input gray level, r

Negative

Lognth root

Identity

nth power

Inverse Log


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Example of Negative Image

Original image Negative Image : gives abetter vision to analyzethe image


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Log Transformations

s = c log (1+r)

c is a constantand r 0

Log curve maps a narrow

range of low gray-levelvalues in the input imageinto a wider range ofoutput levels.

Used to expand thevalues of dark pixels in animage while compressingthe higher-level values.

Input gray level, r

Negative

Lognth root

Identity

nth power

Inverse Log


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Example of Logarithm Image

Result after apply the log

transformation with c = 1,range = 0 to 6.2Fourier Spectrum withrange = 0 to 1.5 x 106


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Inverse LogarithmTransformations

Do opposite to the Log Transformations

Used to expand the values of high pixelsin an image while compressing the darker-level values.


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Power-Law Transformations

s = cr c and are positive

constants

Power-law curves withfractional values of mapa narrow range of darkinput values into a widerrange of output values,

with the opposite beingtrue for higher values ofinput levels.

c = = 1 Identityfunction

Input gray level, rPlots ofs = cr

for various values of (c = 1 in all cases)


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Another example : MRI

(a) a magnetic resonance image ofan upper thoracic human spinewith a fracture dislocation and

spinal cord impingement The picture is predominately dark

An expansion of gray levels aredesirable needs < 1

(b) result after power-law

transformation with = 0.6, c=1

(c) transformation with = 0.4

(best result)

(d) transformation with = 0.3

(under acceptable level)

a b

c d


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Effect of decreasing gamma

When the is reduced too much, theimage begins to reduce contrast to thepoint where the image started to have very

slight wash-out look, especially in thebackground


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Another example

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

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

(c) transformation with = 4.0

(suitable)(d) transformation with = 5.0

(high contrast, the image hasareas that are too dark,some detail is lost)

a b

c d


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Piecewise-LinearTransformation Functions

Advantage:

The form of piecewise functions can bearbitrarily complex

Disadvantage:

Their specification requires considerably moreuser input


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Contrast Stretching

Produce highercontrast than theoriginal by

darkening the levelsbelow m in the originalimage

Brightening the levels

above m in the originalimage


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Thresholding

Produce a two-level(binary) image


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Contrast Stretching

increase the dynamic range of thegray levels in the image

(b) a low-contrast image : resultfrom poor illumination, lack ofdynamic range in the imagingsensor, or even wrong setting of alens aperture of image acquisition

(c) result of contrast stretching:(r1,s1) = (rmin,0) and (r2,s2) =

(rmax,L-1)

(d) result of thresholding


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Gray-level slicing

Highlighting a specificrange of gray levels in animage Display a high value of all

gray 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

constant level (b) transformation highlights

range [A,B] but preserves allother levels


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Bit-plane slicing

Highlighting the contributionmade to total imageappearance by specific bits

Suppose each pixel is

represented by 8 bits Higher-order bits contain the

majority of the visuallysignificant data

Useful for analyzing therelative importance playedby each bit of the image

Bit-plane 7(most significant)

Bit-plane 0(least significant)

One 8-bit byte


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Example

The (binary) image for bit-plane 7 can be obtainedby processing the inputimage with a thresholding

gray-level transformation. Map all levels between 0

and 127 to 0

Map all levels between 129and 255 to 255

An 8-bit fractal image


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8 bit planes

Bit-plane 7 Bit-plane 6

Bit-plane 5

Bit-plane 4

Bit-plane 3

Bit-plane 2

Bit-plane 1

Bit-plane 0


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Histogram Processing

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

h(rk

) = nk

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 levelsrk


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Example

2 3 3 2

4 2 4 3

3 2 3 5

2 4 2 4

4x4 image

Gray scale = [0,9]histogram

0 11

2

2

3

3

4

4

5

5

6

6

7 8 9

No. of pixels

Gray level


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Normalized Histogram

dividing each of histogram at gray levelrk by thetotal number of pixels in the image,n

p(rk

) = nk

/ n For k = 0,1,,L-1

p(rk)gives an estimate of the probability ofoccurrence of gray levelrk

The sum of all components of a normalizedhistogram is equal to 1


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Histogram Processing

Basic for numerous spatial domainprocessing techniques

Used effectively for image enhancement

Information inherent in histograms also isuseful in image compression andsegmentation


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Example

rk

h(rk) or p(rk)

Dark image

Bright image

Components of

histogram areconcentrated on thelow side of the grayscale.

Components ofhistogram areconcentrated on thehigh side of the gray

scale.


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Example

Low-contrast image

High-contrast image

histogram is narrow

and centered towardthe middle of thegray scale

histogram covers broad

range of the gray scaleand the distribution ofpixels is not too far fromuniform, with very fewvertical lines being much

higher than the others


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Histogram Equalization

As the low-contrast images histogram is narrowand centered toward the middle of the grayscale, 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 sothat the probability spread equally


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Example

before after Histogramequalization


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Example

before after Histogramequalization

The quality isnot improved

much becausethe originalimage alreadyhas a broadengray-level scale


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Example

2 3 3 2

4 2 4 3

3 2 3 5

2 4 2 4

4x4 image

Gray scale = [0,9]histogram

0 11

2

2

3

3

4

4

5

5

6

6

7 8 9

No. of pixels

Gray level


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Gray

Level(j)0 1 2 3 4 5 6 7 8 9

No. ofpixels

0 0 6 5 4 1 0 0 0 0

0 0 6 11 15 16 16 16 16 16

0 0

6

/

16

11

/

16

15

/

16

16

/

16

16

/

16

16

/

16

16

/

16

16

/

16

s x 9 0 03.3

3

6.1

6

8.4

89 9 9 9 9

k

j

jn0

k

j

j

n

ns

0


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Example

3 6 6 3

8 3 8 6

6 3 6 9

3 8 3 8

Output image

Gray scale = [0,9]Histogram equalization

0 11

2

2

3

3

4

4

5

5

6

6

7 8 9

No. of pixels

Gray level


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Histogram Matching(Specification)

Histogram equalization has a disadvantage

which is that it can generate only one type ofoutput image.

With Histogram Specification, we can specifythe shape of the histogram that we wish theoutput image to have.

It doesnt have to be a uniform histogram


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Consider the continuous domainLetpr(r) denote continuous probability density

function of gray-level of input image, rLetpz(z) denote desired (specified) continuousprobability density function of gray-level ofoutput image, z

Let s be a random variable with the property

r

r dw)w(p)r(Ts0

Where w is a dummy variable of integrationHistogram equalization


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Next, we define a random variable z with the property

s = T(r) = G(z)

We can map an input gray level r to output gray level z

thus

sdt)t(p)z(g

z

z 0

Where t is a dummy variable of integrationHistogram equalization

Therefore, z must satisfy the conditionz = G

-1

(s) = G-1

[T(r)]

Assume G-1 exists and satisfies the condition (a) and (b)


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Procedure Conclusion

1. Obtain the transformation function T(r) bycalculating the histogram equalization of theinput image

2. Obtain the transformation function G(z) bycalculating histogram equalization of thedesired density function

r

r dw)w(p)r(Ts0

sdt)t(p)z(G

z

z

0


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Procedure Conclusion

3. Obtain the inversed transformationfunction G-1

4. Obtain the output image by applying theprocessed gray-level from the inversed

transformation function to all the pixels inthe input image

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


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Example

Assume an image has a gray level probability densityfunction pr(r) as shown.

0

1

2

12

Pr(r)

elsewhere;0

1r;022r)r(pr

1

0

r

r dw)w(p

r


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Example

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

0

1

2

12

Pz(z)

z

elsewhere;0

1z;02z)z(pz

1

0

z

z dw)w(p


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Step 1:

0 1

1s=T(r)

rrr

ww

dw)w(

dw)w(p)r(Ts

r

r

r

r

2

2

22

2

0

2

0

0

Obtain the transformation function T(r)

One to onemappingfunction


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Step 2:

2

0

2

0

2 zzdw)w()z(Gz

z

Obtain the transformation function G(z)


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Step 3:

2

22

2

2

rrz

rrz

)r(T)z(G

Obtain the inversed transformation function G-1

We can guarantee that 0 z 1 when 0 r 1


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Discrete formulation

12100

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

k

k

i izk

12100

0

L,...,,,knn

)r(p)r(Ts

k

j

j

k

j

jrkk

12101

1

L,...,,,ksG

)r(TGz

k

kk


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Example

Image 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


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Image Equalization

Result imageafter histogram

equalizationTransformation function

for histogram equalization Histogram of the result imageThe histogram equalization doesnt make the result image look better thanthe original image. Consider the histogram of the result image, the neteffect of this method is to map a very narrow interval of dark pixels intothe upper end of the gray scale of the output image. As a consequence, the

output image is light and has a washed-out appearance.


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Histogram Equalization

Histogram Specification

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


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Histogram Specification

(1) the transformationfunction G(z) obtained

from

(2) the inversetransformation G-1(s)

1210

0

L,...,,,k

s)z(p)z(G k

k

i

izk


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Result image and its histogram

Original image

The output images histogramNotice that the outputhistograms low end hasshifted right toward thelighter region of the gray

scale as desired.

After appliedthe histogram

equalization


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Note

Histogram specification is a trial-and-errorprocess

There are no rules for specifying

histograms, and one must resort toanalysis on a case-by-case basis for anygiven enhancement task.

image enhancement in the spatial domain1

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