chapter 3 image enhancement in the spatial domain
TRANSCRIPT
Chapter 3
Image Enhancement in the Spatial Domain
2
Background
Spatial domain refers to the aggregated of pixels composing an image.
Spatial domain methods are procedures that operate directly on these pixels.
Spatial domain processes will be denoted by the expression
g(x,y)=T [ f(x,y) ]where f(x,y) is the input image, g(x,y) is the processed image, and T is an operator of f
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Some Basic Gray Level Transformations
Discussing gray-level transformation functions. The value of pixels, before and after processing are related by an expression of the form
s = T(r)
Where r is values of pixel before process s is values of pixel after process T is a transformation that maps a pixel value r into a pixel value s.
4
Gray Level Transformations
Image NegativesLog TransformationsExponential TransformationsPower-Law Transformations
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Image Negatives
The negative of an image with gray L levels is given by the expression
s = L – 1 – r
Where r is value of input pixel, and s is value of processed pixel
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7
Log Transformations
The general form of the log transformation shown in
s = c log(1+r)
Where c is constant, and it is assumed that r0
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C = 0.8
C = 1.0
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Exponential Transformations
The general form of the log transformation shown in
s = c exp(r)
Where c is constant, and it is assumed that r0
10
C = 0.8
C = 1.0
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Power-Law Transformations
Power-law transformations have the basic form
Sometime above Equation is written as
crs Where c and are positive constant.
)( rcs
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= 0.5
= 1.0
= 5.0
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Piecewise-Linear Transformation Functions
Translate MappingContrast stretchingGray-level slicingBit-plane slicing
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Translate Mapping
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Contrast stretching
One of the simplest piecewise linear functions is a contrast-stretching transformation.
The idea behind contrast stretching is to increase the dynamic range of the gray levels in the image being processed.
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Contrast stretching
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Contrast stretching
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Gray-level slicing
Highlighting a specific range of gray levels in an image often is desired.
There are several ways of doing level slicing, but most of them are variations of two basic themes.
One approach is to display a high value for all gray levels in the range of interest and a low value for all other gray levels.
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Gray-level slicing
255255 00
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Bit-plane slicing
Instead of highlighting gray-level ranges, highlighting the contribution made to total image appearance by specific bits might be desired.
Suppose that each pixel in an image is represented b y 8 bits. Imagine that the image is composed of eight 1-bit planes, ranging from bit-plane 0, the least significant bit to bit-plane 7, the most significant bit.
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10110011
1
1
0
0
1
1
0
1
Bit-plane 0(least significant)
Bit-plane 7(most significant)
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Bit-plane slicing
Bit-plane 7 Bit-plane 6 Bit-plane 4 Bit-plane 1
Original Image
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Histogram Processing
The histogram of a digital image with gray levels in the range [0,L-1] is a discrete function
kk nrh )(
Where rk is the kth gray level and nk is the number of pixels in the image havinggray level rk
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HISTOGRAM
363622
Image 16x14 = 224 pixels
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0 1 2 3 level
pixels
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Role of Histogram Processing
The role of histogram processing in image enhancement exampleDark imageLight imageLow contrastHigh contrast
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Histogram Processing
Histogram EqualizationHistogram Matching(Specification)Local Enhancement
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Initial part of discussion
Let r represent the gray levels of the image to be enhanced.
Normally, we have used pixel values in the interval [0,L-1], but now we assume that r has been normalized to the interval [0,1], with r=0 representing black and r=1 representing white.
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Histogram Equalization
We focus attention on transformations of the form
10)( rwhererTs
And we assume that the transformation function T(r) satisfies the following conditions:(a) T(r) is single-valued and monotonically increasing in the interval 0 ≤ r ≤ 1; and(b) 0 ≤ T(r) ≤ 1 for 0 ≤ r ≤ 1
That produce a level s for every pixel value r in the original image.
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Reason of Condition
The requirement in (a) that T9r) be single valued is needed to guarantee that the inverse transformation will exist, and the monotonicity condition preserves the increasing order from black to white in the output image.
Condition (b) guarantees that the output gray levels will be in the same range as the input levels.
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Transformation function
T(r)
rk
sk=T(rk)
0 1Level r
Level s
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Note
The inverse transformation from s back to r is denoted
10)(1 rwhererTs
There are some cases that even if T(r) satisfies conditions (a) and (b), it is possible that the corresponding inverse T-1(s) may fail to be single valued.
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Fundamental of random variable
PDF (probability density function) is the probability of each element
CDF (cumulative distribution function) is summation of the probability of the element that value less than or equal this element
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The PDF (probability density function) is denoted by p(x)
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34
CDF
The CDF (cumulative density function) is denoted by P(x)
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cdf
1/6
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Relation between PDF and CDF
PDF can find from this equation
CDF can find from this equation
)()]([
)( xPdx
xPdxp
nx
xx
n dxxpxxP0
)()(
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Idea of Histogram Equalization
The gray levels in an image may be viewed as random variables in the interval [0,1].
ds
drrpsp rs )()(
Let pr(r) denote the pdf of random variable r and ps(s) denote the pdf of random variable s; if pr(r) and T(r) are known and T-1(s) satisfies condition (a), the formula should be
r
r dwwprTs0
)()(
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Idea of Histogram Equalization
By Leibniz’s rule that the derivative of a definite integral with respect to its upper limit is simply the integrand evaluated at that limit.
)(
)(
)(
0
rp
dwwpdr
d
dr
rdT
dr
ds
r
r
r
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Idea of Histogram Equalization
Substituting into the first equaltion
10,1
)(
1)(
)()(
s
rprp
ds
drrpsp
rr
rs
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Idea of Histogram Equalization
The probability of occurrence of gray level rk in an image is approximated by
1,...,2,1,0)( Lkn
nrp k
kk
1,...,2,1,0
)()(
0
0
Lkn
n
rprTs
k
j
j
k
jjrkk
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Histogram Equalization
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Example forHistogram Equalization
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Example forHistogram Equalization
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Example forHistogram Equalization(2)
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จ"าน่วน่พ$กเซลู(nj)
0 30 30 0.0037 0.0256
1 50 80 0.0098 0.0683
2 100 180 0.0220 0.1537
3 1500 1680 0.2050 1.4241
4 2300 3980 0.4854 3.3976
5 4000 7980 0.9732 6.8122
6 200 8180 0.9976 6.9830
7 20 8200 1.0000 7.0000
k
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45
Histogram Specification
can called “Histogram Matching”Histogram equalization automatically
determines a transformation function that seeks to produce an output image that has a uniform histogram.
In particular, it is useful sometimes to be able to specify the shape of the histogram that we wish the processed image to have.
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Meaning:Histogram SpecificationIn this notation, r and z denote the gray
levels of the input and output (processed) images, respectively.
We can estimate pr(r) from the given image
While pz(z) is the specified probability density function that we wish the output image to have
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Meaning:Histogram SpecificationLet s be a random variable with the
property
Suppose that we define a random variable z with property
Then two equations imply G(z)=T(r)
)1()()(0
r
r dwwprTs
)2()()(0
sdttpzGz
z
)3()]([)( 11 rTGsGz
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Procedure:Histogram Specification
1) Use Eq.(1) to obtain the transformation function T(r)
2) Use Eq.(2) to obtain the transformation function G(z)
3) Obtain the inverse transformation function G-
1
4) Obtain the output image by applying Eq.(3) to all the pixels in the input image
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Specified histogram
Input Image
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Result form Histogram Specification
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Local Enhancement
Normally, Transformation function based on the content of an entire image.
Some cases it is necessary to enhance details over small areas in an image.
The histogram processing techniques are easily adaptable to local enhancement.
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Local Enhancement
Pixel-to-pixel translation Nonoverlapping region
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Local Equalization
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Enhancement Using Arithmetic/Logic OperationsAre performed on a pixel-by-pixel basis
between two or more imagesLogic operations are concerned with the ability
to implement the AND, OR, and NOT logic operators because these three operators are functionally complete.
Arithmetic operations are concerned about +,-,*, / and so on (arithmetic operators)
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Image Subtraction
The difference between two images f(x,y) and h(x,y) expressed as
g(x,y) = f(x,y) – h(x,y)
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Image Averaging
Consider a noisy image g(x,y) formed by the addition of noise to original image f(x,y)
The objective of this procedure is to reduce the noise content.
),(),(),( yxyxfyxg noise is where η(x,y)
Where the assumption is that at every pair of coordinates(x,y) the noise is uncorrelated and has zero average value.
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Image Averaging
Let there are K different noisy imagesIf an image is formed by
averaging K different noisy images
K
ii yxg
Kyxg
1
),(1
),(
),( yxg
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Image Averaging (Gray Scale)
2images
5images
10images
20images
1 image
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Image Averaging (Color Image)
Average image
(1) (2) (3)