Image Enhancement in the Spatial Domain1
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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.
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