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Image Enhancement in the Spatial Domain

Lec#03: Intensity Transformations

Dr. Ch Vit Nht Anhnhat-anh.che@{hcmut.edu.vn,hotmail.com}

Department of Telecommunications EngineeringFaculty of Electrical and Electronic Engineerings

Ho Chi Minh City, University of Technology

HK I, 2015 - 2016

Dr. Che Viet Nhat Anh Digital Image & Speech Processing DI&SP - 405017 1 / 91

Outline

1 Introduction

2 Histogram

3 Some Basic Intensity Transformation Functions

4 Histogram Processing


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Learning Outcomes

After learning this chapter, you will be able to: Illustrate the effect of power-law & logarithmic histogram transforms Describe the basis of the histogram as a Probability Density

Function (PDF)

Distinguish between the brightness and the contrast of an image Explain the relationship between dynamic range and contrast Explain the effects of histogram stretch, histogram equalization,

histogram matching, histogram modification & distinguish betweenthem


"It makes all the difference whether one sees

darkness through the light or brightness through the

shadows

David Lindsay (Scottish Novelist)


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Outline

1 Introduction- Motivation- Type of operations

- Spatial domain


IntroductionMotivation

Which one looks better?


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Close-by pixel values are difficult to distinguish.

Far apart pixel values are easy to distinguish.

Low contrast: image values concentrated near a narrow range(mostly dark, or mostly bright, or mostly medium values).

Contrast enhancement change the image value distribution tocover a wide range.

Contrast of an image can be revealed by itshistogram.


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IntroductionType of operations



Operation Characterization Generic Complexity/PixelPoint The output value at a specific co-

ordinate is dependent only on the

input value at thatsame coordinate

Constant

Local The output value at a specific co-ordinate is dependent on the input

values in the neighborhood of that

same coordinate.

(2m + 1) (2n + 1)

Global The output value at a specific coor-dinate is dependent on all the valuesin the input image.

M N

Image size = M N; Neighborhood size = (2m + 1) (2n + 1).

The complexity is specified in operations per pixel.


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Domains

Spatial Domain Processing



Spatial domain:Image plane itself, directly process the intensityvalues of the image plane.

Point Operation (Intensity Transformations Function) Gamma correction, window-center correction, histogram

equalization,...

Spatial filters (Neighborhood Operation) Correlation Convolution (Filtering): mean, Gaussian, median, etc. Image gradients,...

Transform domain(Frequency domain, Wavelet domain) Processthe transform coefficients, not directly process the intensity valuesof the image plane.


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Point processing Spatial processing Frequency domainprocessing

Contrast stretching:

Gray level transform

Clipping

Intensity Level slicing

Neighborhood pro-

cessing:

Averaging

Directionalsmoothing

Median filtering

Unsharpmasking

Linear filtering:

Low pass

High-passnonlinearfiltering

Homomorphicfiltering

Histogram techniques:

Equalization

Histogram specification(matching)


IntroductionSpatial domain

Spatial domain refers to the aggregated of pixels composing animage.

Spatial domain methods are procedures that operate directly onthese pixels. Spatial domain processes will be denoted by the

expression:s= g(x0, y0) =T(r) =T{f(x0, y0)} (1.1)

T: an operator defined over some neighborhood of (x0, y0), maybelinear or nonlinear.

N 1(0,0)

r

y

x

x0

M 1

y0 N 1(0,0)

s

y

x

x0

M 1

y0


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Histogram

Histogram of a monochrome image with gray levels in the range[0, L 1]:

A discrete function, h(rk) =nk, k= 0, 1, . . . , L 1. rk: the kth gray level.

nk the number of pixels in the image having gray level rk. Normalized histogram: An estimate of the probability of

occurrence of gray level rk p(rk) =

nk

M N, k= 0, 1, . . . , L 1.

The sum of all components of a normalized histogram is equal to 1.


Example

Sketch the histogram of the following image:

6 6 1 2 50 2 0 5 6

6 7 6 1 0

2 1 0 5 4

0 6 7 1 6

rk 0 1 2 3 4 5 6 7

nk 5 4 3 0 1 3 7 20 1 2 3 4 5 6 7

01234

5

7

rk

nk


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Example

Sketch the normalized histogram of the following image:

6 6 1 2 50 2 0 5 6

6 7 6 1 02 1 0 5 4

0 6 7 1 6

rk 0 1 2 3 4 5 6 7

p(rk) 0.2 0.16 0.12 0 0.04 0.12 0.28 0.08

0 1 2 3 4 5 6 70

0.040.080.120.160.2

0.28

rk

p(rk)


Histogram

The following two images have the same histograms:

The histogram shows the number of pixels that have each pixelvalue, but it does not record where those pixels are located in the

image. Thus, spatial information is discarded. The histogram is unique for a particular image, but different images

could have the same histogram.


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Histogram

The information inherent in histograms also is quite useful in otherimage processing applications:

Image compression Segmentation

Histograms are simple to calculate in software & also lendthemselves to economic hardware implementations, thus makingthem a popular tool for real-time image processing.

The shape of the histogram of an image does provide usefulinformation about the possibility for contrast enhancement.

The role of histogram processing in image enhancement: Dark image

Light image Low contrast High contrast


Histogram

Dark image

0 63 127 193 2550

2

4

6

8102

rk

p(rk

)


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Histogram

Light image

0 63 127 193 2550

2

4

6

8102

rk

p(rk

)


Histogram

Low contrast

0 63 127 193 2550

2

4

6

8102

rk

p(rk

)


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Histogram

High contrast

0 63 127 193 2550

2

4

6

8102

rk

p(rk

)


Histogram

Dynamic range: The range of pixel values, defined as the difference between the

maximum (rmax) the minimum (rmin) pixel values found in theimage, ignoring any obvious outliers.

It can be expressed either as the difference in pixel values or dB:

Dynamic range of image= 20 log10(rmax rmin) (2.1)


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Outline

3 Some Basic Intensity Transformation Functions- Point operations- Identity- Image Negatives- Gamma Transformations- Log Transformations- Piecewise-Linear Transformations- Bit Plane Slicing


Some Basic Intensity Transformation FunctionsPoint operations

g depends only onthe value off at (x0, y0). T becomes a gray-level(also called an intensity or mapping) transformation function of theform:

s= g(x0, y0) =T(r) =T{f(x0, y0)} (3.1)

Point operations arezero memory operations.

N 1(0,0)

r

y

x

x0

M 1

y0 N 1(0,0)

s

y

x

x0

M 1

y0


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Some Basic Intensity Transformation FunctionsPoint operations


Some Basic Intensity Transformation FunctionsIdentity

No change on pixel values, s= T(r) =r.


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Some Basic Intensity Transformation FunctionsImage Negatives

Contrast reversal: negative, s= T(r) =L 1 r.

It produces the equivalent of a photographic negatives.


Some Basic Intensity Transformation FunctionsGamma Transformations

Power-Law transformations: s= cr or s= cr, c, > 0.

Values ofsuch that 0< 1 have the opposite behavior to above (darken theintensity).

c= = 1gives the identity transformation.

Gamma correction is an application.


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Plots of the equation s= r for various values of. All curves werescaled to fit in the range shown.




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Some Basic Intensity Transformation FunctionsLog Transformations

Maps a narrow range of low intensity values in input to a wideroutput range.

The opposite is true for high intensity input values.

Compresses the dynamic range of images with large variations inpixel values. (The dynamic range of the image data may be verylarge -> Dynamic range can be compresses via the logarithmictransformation).




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Some Basic Intensity Transformation FunctionsPiecewise-Linear Transformations: Contrast Stretching

Expands the range of intensity levels in an image so that it spansthe full intensity range of the recording medium or display device.


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Low contrast images can result from poor illumination. Reduce intensity: (0, 0)(r1, s1), (r2, s2)(L 1, L 1). Increase intensity: The slope of the transformation is chosen greater

than unity in the region of stretch.

s= T(r) =

s1

r1r 0 rr1

s2 s1r2 r1

(r r1) + s1 r1 < rr2

L 1 s2L 1 r2

(r r2) + s2 r2 < rL 1

(3.2)




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Some Basic Intensity Transformation FunctionsPiecewise-Linear Transformations: Contrast Compressing


Some Basic Intensity Transformation FunctionsPiecewise-Linear Transformations: Intensity-level Slicing

Highlighting a specific range of intensities in an image.

Two main approaches: Display in one value (e.g. white) all the values in the range of interest

and all others in another value (e.g black all other intensities)

binary image. Brighten or darkens the desired range of intensity levels in the imageunchanged.


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This transformationhighlights intensity range[A, B]and reduces all othersintensities to a lower level.

This transformationhighlights intensity range[A, B] and preserves all othersintensities to a lower level.




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Some Basic Intensity Transformation FunctionsBit Plane Slicing

Pixels are digital numbers composed of bits.

Suppose that each pixel in an image is represented by 8 bitsb7b6b5b4b3b2b1b0. Imagine that the image is composed of eight 1-bit

planes, ranging from bit-plane 0, the Least Significant Bit(Lowest-order bit, b0) to bit-plane 7, the Most Significant Bit(Highest-order bit, b7).

Each bit plane is a binary image



Highlight the contribution made to total image appearance byspecific bits

Higher-order bits usually contain most of the significant visualinformation.

Lower-order bits contain subtle details.

Useful for compression.

We have to use bit get and bit set to extract 8 images:

b7b6b5b4b3b2b1b0

136 11 124 67 7

33 17 3

0 1 1

0 1 1

1 1 1

0 0 0

1 0 0

0 1 0

0 1 0

0 1 1

0 0 1

0 0 0

0 0 0

1 0 0

0 0 0

0 0 1

0 0 0

0 0 0

0 1 0

0 0 0

1 1 0

1 0 0

0 0 0

1 0 0

0 0 0

0 0 0


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Reconstruction is obtained by:


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Outline

4 Histogram Processing- Histogram Equalization- Histogram matching or Histogram specification- Local Histogram Processing- Using Histogram Statistics for Image Enhancement


Histogram ProcessingHistogram Equalization

Consider for a momentcontinuous intensity values, & let thevariable r represent the gray levels of the image to be enhanced,that is r[0, L 1], with r = 0 representingblack, & r=L 1representingwhite.

For any r satisfying the aforementioned conditions, we focusattention on transformations of the form:

s= T(r) 0 r L 1 (4.1)

Let pr(r), ps(s)denote the Probability Density Function (PDF) ofrandom variables r and s.

0 rL 1

pr(r)

0 sL 1

1

L 1

ps(s)


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Transforms an image with an arbitrary histogram to one with a flathistogram.

The transformation function T(r) satisfies the following conditions:

T(r)is amonotonically increasing functionin the interval0 r L 1.

0 s= T(r) L 1 for 0 r L 1.

T(r)is continuous and differentiable.

0 rr1 r2 L 1

L 1

s1

s

0 rL 1

L 1

s1

s

r1



0 rL 1

L 1

s

0 rdr L 1

pr(r)

s

ps(s)

ds

ps(s)ds= pr(r)drds= 1

ps(s)pr(r)dr = (L 1)pr(r)dr


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Continuous case:

s= T(r) = (L 1) r

0

pr(w)dw (4.2)

w: a dummy variable of integration. pr(r)ishigh, T(r)has asteep slope, dswill bewide, causing ps(s)to

belowto keep ps(s)ds= pr(r)dr. pr(r)islow, T(r)has ashallow slope, dswill benarrow, causing

ps(s)to behighto keep ps(s)ds= pr(r)dr.


Example

Suppose that the (continuous) intensity values in an image have theProbability Density Function (PDF)

pr(r) =

2r

(L 1)2 for0 r L 1

0 otherwise

Find the transformation function for equalizing the image histogram.

s= T(r) = (L 1)

r0

pr(w)dw= (L 1)

r0

2w

(L 1)2dw=

r2

L 1


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Example

The histograms of two images are illustrated below. Sketch atransformation function for each image that will make the image has abetter contrast. Use the axis provided below to sketch yourtransformation functions.

r20 128 255

1

108

pr(r)

0 r100 200 255

pr(r)

r20 128 255

255

s

0 r100 200 255

128

s

255



Discrete values: For k= 0, 1, . . . , L 1

sk = T(rk) = (L 1)k

j=0

pr(rj) (4.3)

= (L 1)k

j=0

nj

M N=

L 1

M N

kj=0

nj (4.4)

The histogram equalization process consists of four steps:1 Find the running sum of the histogram values.2 Normalize the values from step 1 by dividing by the total number of

pixels.

3 Multiply the values from step 2 by the maximum gray level value andround.

4 Map the gray-level values to the result from step 3 using one-to-onecorrespondence.


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Example

Suppose that a 3-bit image (L= 8 intensity levels) of size 64 64 pixels(M N= 4096) has the intensity distribution shown in following table.

rk 0 1 2 3 4 5 6 7

nk 790 1023 850 656 329 245 122 81

1 Sketch the original histogram.

2 Get & sketch the histogram equalization transformation function.

3 Give the ps(sk)for each sk. Sketch the equalized histogram.


Example

1 Sketch the original histogram.

rk 0 1 2 3 4 5 6 7pr(rk) 0.193 0.25 0.208 0.16 0.08 0.06 0.03 0.02

0 1 2 3 4 5 6 7 rk

pr(rk)

0.250.2

0.150.1

0.05


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Example

2 Get & sketch the histogram equalization transformation function.

rk 0 1 2 3 4 5 6 7k

j=0

nj 790 1813 2663 3319 3648 3893 4015 4096

sk = L 1

M N

kj=0

nj 1.33 3.08 4.55 5.67 6.03 6.65 6.86 7

sk 1 3 5 6 6 7 7 7

0 r1 2 3 4 5 6 7

s

1.33

3.08

4.555.67

7

0 r1 2 3 4 5 6 7

s

1

3

5677


Example

3 Give the ps(sk)for each sk. Sketch the equalized histogram.

sk 0 1 2 3 4 5 6 7nk 0 790 0 1023 0 850 985 448

ps(sk) 0 0.19 0 0.25 0 0.20 0.24 0.12

0 1 2 3 4 5 6 7 sk

ps(sk)

0.250.2

0.150.1

0.05


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0 63 127 193 2550

2

4

6

8102

rk

p(rk

)

0 63 127 193 2550

2

4

6

8102

sk

p

(sk

)


0 63 127 193 2550

2

4

6

8102

rk

p(rk

)

0 63 127 193 2550

2

4

6

8102

sk

p(sk

)


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0 63 127 193 2550

2

4

6

8102

rk

p(rk

)

0 63 127 193 2550

2

4

6

8102

sk

p(sk

)


0 63 127 193 2550

2

4

6

8102

rk

p(rk

)

0 63 127 193 2550

2

4

6

8102

sk

p(

sk

)


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Histogram

Transformation function for histogram equalization:



The above procedure is for grayscale images

For color images: Apply separately the procedure to each of the three channels

will yield in a dramatic change in the color balance NOT RECOMMENDED unless you have a good reason for doing

that.

Recommended method Convert the image into another color space such as the Lab color

space OR HSL/HSV color space. Apply the histogram equalization on the Luminance channel Convert back into the RGB color space.


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Histogram ProcessingHistogram equalization

Can contrast stretching achieve similar result as histogramequalization?

If it can, why histogram equalization then? Histogram equalization (HE) results are similar to contrast stretching

but offer the advantage of full automation, since HE automaticallydetermines a transformation function to produce a new image with auniform histogram.

Histogram equalization method: Only generates one result: an image with approximately uniform

histogram (without any flexibility). Enhancement may not be achieved as desired.


Histogram ProcessingHistogram matching or Histogram specification

What if the desired histogram is not flat?

Histogram specification:Transform an image according to aspecified gray-level histogram

Let pr(r)and pz(z)denote the continuous probability density

functions of the variables r and z. pz(z)is the specified probability density function or specifyparticular histogram shapes pz(z)capable of highlighting certaingray-level ranges.

Obtain the transformation function for transformation ofr to z.

r

pr(r)

z

pz(z)

Histogram

Matching


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Let sbe the random variable with the probability:

s= T(r) = (L 1) r

0

pr(w)dw (4.5)

Define a random variable z with the probability

G(z) = (L 1)

z0

pz(t)dt= s (4.6)

The inverse of the above transform is

z =G1(s) =G1(T(r)) (4.7)

r

pr(r)

sHistogram

Equalization

G(z) z

pz(z)

Histogram

Equalization


Example

Assuming continuous intensity values, suppose that an image has theintensity PDF:

pr(r) =

2r

(L 1)2 for0 r L 1

0 otherwise

Find the transformation function that will produce an image whoseintensity PDF is:

pz(z) =

3z2

(L 1)3 for0 z L 1

0 otherwise


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Example

1 Find the histogram equalization transformation for the input image

s= T(r) = (L 1)

r0

pr(w)dw= (L 1)

r0

2w

(L 1)2dw=

r2

L 1

2 Find the histogram equalization transformation for the specifiedhistogram

G(z) = (L 1)

z0

pz(t)dt= (L 1)

z0

3t2

(L 1)3dt=

z3

(L 1)2 =s

3 The transformation function

z=

(L 1)2s1/3

=

(L 1)2 r2

L 1

1/3

=[

(L 1)r2]1/3



Discrete Cases: Obtain pr(rj) from the input image then obtain the values ofsk,

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

sk =T(rk) = (L 1)

kj=0

pr(rj) =

(L 1)

M N

kj=0

nj (4.8)

Use the specified PDF and obtain the transformation function G(zq),round the value to the integer range [0, L 1]:

G(zq) = (L 1)

qi=0

pz(zi) =sk (4.9)

Find the smallest value ofzq so that G(zq)is closest to sk:

zq =G1(sk) (4.10)


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Example

Suppose that a 3-bit image (L = 8) of size 64 64 pixels (M N= 4096)has the intensity distribution shown in the following table (on the left).Get the histogram transformation function and make the output imagewith the specified histogram, listed in the table at the bottom.

rk 0 1 2 3 4 5 6 7nk 790 1023 850 656 329 245 122 81

zk 0 1 2 3 4 5 6 7pz(zk) 0 0 0 0.15 0.2 0.3 0.2 0.15


Example

1 Obtain the scaled histogram-equalized values:

rk 0 1 2 3 4 5 6 7k

j=0nj 790 1813 2663 3319 3648 3893 4015 4096

sk =T(rk) 1.33 3.08 4.55 5.67 6.03 6.65 6.86 7

sk 1 3 5 6 6 7 7 72 Compute all the values of the transformation function G

zq 0 1 2 3 4 5 6 71

j=0pz(zj) 0 0 0 0.15 0.35 0.65 0.85 1

G(zq) 0 0 0 1.05 2.45 3.55 5.95 7

G(zq) 0 0 0 1 2 5 6 7


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Example

3 Relate the two mapping above to build a lookup table for the overall mapping. Specifically, for each input level k, find an output levelqso that G(zq)best matches sk.

rk 0 1 2 3 4 5 6 7

sk 1 3 5 6 6 7 7 7G(zq) 0 0 0 1 2 5 6 7

zq 0 1 2 3 4 5 6 7

nq 0 0 0 790 1023 850 985 448

pz(zq) 0 0 0 0.19 0.25 0.21 0.24 0.11

0 1 2 3 4 5 6 7 zq

pz(zq)

0.150.2

0.3

0.20.15

0 1 2 3 4 5 6 7 zq

pz(zq)

0.190.25

0.11



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Issues with Histogram specification/matching: No rule for specifying an optimal histogram. Each given enhancement task needs to be analyzed on a case-by-case

basis. Histogram specification is somehow a trial-and-error process.


Histogram ProcessingLocal Histogram Processing

The histogram processing methods mentioned up to now are globaltransformation where:

Function is designed according to the gray-level distribution over anentire image

Global transformation methods may not be suitable for enhancing

details over small areas Where number of pixels in these small areas may have negligible

influence on designing the global transformation function

To enhance details over small areas in an image Procedure:

Define a neighborhood (e.g. N8) Move it from pixel to pixel. For every pixel:

Histogram computed for the neighborhood. Transfer function computed for Histogram sHistogram specification

or Histogram specification. Applied on Centre Pixel.


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Histogram ProcessingLocal Histogram Processing


Histogram ProcessingUsing Histogram Statistics for Image Enhancement

Global statistics: Themean (average) intensity is given by

m=

L1

i=0

ripr(ri) = 1

M N

M1

x=0

N1

y=0

f(x, y) (4.11)

Thevariance of the intensities is given by

2 =u2(r) =

L1i=0

(ri m)2pr(ri) =

1

M N

M1x=0

N1y=0

[f(x, y) m]2

(4.12) The nth moment of the intensity variable r is

un(r) =L1

i=0

(ri m)npr(ri) (4.13)


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Mean gives the average brightness of the image.

Variance 2 and its square root the standard deviation gives thedeviation of intensities on average from the mean value (average

contrast).

= 14.3 = 31.6 = 49.2



Local Statistics: Sxy: a neighborhood (subimage) of specific sizecentered at (x, y):

Local average intensity:

mSxy =

L1

i=0

ripSxy(ri) (4.14)

Local variance:

2Sxy =L1i=0

(ri mSxy)2pSxy(ri) (4.15)


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The statistical parameters can

be used in various ways. Enhance the background

filament.

Enhance details in dark areaswhile leaving light areaunchanged.

Define rules to chose the

candidate pixels that need tobe enhanced.



A pixel at point (x, y)is considered if: mSxy k0mG, where k0 is a positive constant less than 1.0, and mG

is global mean. Sxy k2G, where G is the global standard deviation and k2 is a

positive constant. Also need to put a lower limit on standard deviation to avoid

distorting areas which dont have details, i.e., k1G Sxy , withk1 < k2.

A pixel that meets all above conditions is processed simply bymultiplying it by a specified constant, E, to increase or decrease thevalue of its gray level relative to the rest of the image.

The values of pixels that do not meet the enhancement conditions

are left unchanged.

g(x, y) =

{ Ef(x, y) ifmSxy < k0mG and k1G < Sxy < k2Gf(x, y) otherwise

(4.16)Dr. Che Viet Nhat Anh Digital Image & Speech Processing DI&SP - 405017 90 / 91

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di&sp#-03-405017

Documents

image values

image enhancement

input image

n image size

output value

histogram equalization

histogram matching

effects of histogram