image enhancement – introduction processing & machine vision/y. gao 3 image enhancement --...

Image Processing & Machine Vision/Y. Gao 1

Image Enhancement – Introduction

1. Introduction

1.1 What is image enhancement?--- Modify the intensities of pixels in an image so that it

can be more suitable for a specific application.

Different enhancement process suits different application.

An enhancement method is good for an application but maybe bad for another application.



1.2 Categories of methods for enhancement process.

Spatial domain: process pixels in the image plane directly.

Frequency domain: modify the Fourier transform of an image.


Image Enhancement -- Introduction

Mathematically, the enhancement process in the spatial domain can be described by the transform function

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

f(x,y) --- input image,

g(x,y) --- output image,

T --- a transform

For a single pixel, we have s = T (r)

r --- grey-level value for an input pixel

s --- grey-level value for an output pixel



Grey-level value transform graph

We use a graph to denote the transform function s = T (r) .

r

sT(r)

r

sT(r)

r

s

T(r)

No change image brighter image darker



Two simple examples for contrast enhancement

x



For the enhancement in the frequency domain, according to the convolution theorem we have

),( yxh ),( yxg),( yxf

),( vuH),( vuF ),( vuG



)],(),([ ),(

),(),(),(),(),(),(

1- vuFvuHyxg

vuFvuHvuGyxfyxhyxg

=

=∗=

F

)],([ ),()],([ ),()],([ ),(

yxgvuGyxhvuHyxfvuF

=== F

FF


Image Enhancement -- outline

1. Introduction

2. Transforms

3. Histogram processing

4. Arithmetic/logic operations

5. Filtering

1) Spatial filters

i. Smoothing

ii. Sharpening

2) Frequency-domain filters

i. Smoothing

ii. Sharpening


Image Enhancement -- Transformations

2. Transformations

2.1. Image negatives

0

T(r)s

r

L-1

rLrTs −−== 1)(

L-1



2.2. Contrast stretching

For a low-contrast image, we feel it uncomfortable and sometimes can not see details clearly.



Mathematically, We have following transform for stretching contrast.

srrs

srrrss

rs

rLL

r

r

222

2

1112

12

1

1

)()1()1(

)()()(

+−⋅−−−−

+−⋅−−

⋅

{)( == rTs12

1

0

2

1

−<≤<≤<≤

Lrrrrrrr



Special cases:

When

ssrr

ssrrsrsr

L

2121

2121

2211

and

1 and 0,

and

≤≤

−===

== ----------------- No change

--- Thresholding, which produces a binary image

--- General, adjusting contrast



2.3. Compression of dynamic range

When the dynamic range of the grey-level in an image far exceedsthe capability of the display device, it needs to be compressed to display it better.

A good transform function (Log transform) is as follow,

)1log( rcs +⋅= c is a scaling constant



T(r)

s

r



For example, an image with dynamic range

So

[ ]105.2,0[],0 6×=R

4.6)1log( 105.2 6 ≈+ ×When the dynamic range of the display device is [0,255], we have

404.6

255 ≈=c

The transform result is shown as follows



2.4. Gray-level slicingSometimes need to highlight a specific range of gray levels in an image.

Two approaches:1. A high value for all gray levels in the range of

interest and a low value for all other gray levels.2. Brighten the desired range of gray levels but

preserve all other parts including background.



2.5. Bit-plane slicingSuppose we use one byte ( 8 bits ) to store the gray level valuefor each pixel, as follows

bbbbbbbb 01234567

For all pixels in an image we take all bits in the same position, which makes a bit-plane.



Usually higher order bit-plane contains much more visual data, Bit-plane is the most significant for the image but Bit-plane is the least significant

b7

b0


Image Enhancement -- Histogram

3. Histogram Processing

3.1. What is the histogram of an image?

The distribution of the pixel grey level is very important for an image display.

For a limited grey-level domain [0, L-1] we hope there is anuniform distribution of the pixel grey level instead of non-uniform distribution.

Based on the distribution of the pixel grey level we define the image histogram.



For a digital image with gray levels in the range [0, L-1], we define a discrete function

rnr kk

k np )( =

is the number of the pixels with the gray levelnk rkis the total number of pixels in the image.

We call the histogram of a digital image.

Four special histograms are as follows.

)(rkp

n

is the kth gray level.



)(r kp

r k

Dark image

)(r kp

r k

Bright image

)(r kp

r k

High-contrast image

)(r kp

r k

Low-contrast image



Compute Histogram H of gray-level image I

procedure histogram (I,H) {

“Initialize all bins to zero”

for i:=0 to MaxVal

H[i]:=0;

“Compute values”

for L:=0 to MaxRow

for P:=0 to MaxCol {

grayval:= I[L,P];

H[grayval]:= H[grayval] + 1;

}

}



3.2. Histogram equalization

Sometimes we need to convert a low-contrast image into a high-contrast image for displaying it better.

We can do it by extending its histogram ---- histogram equalization

For our discussion, we suppose the gray level r is a continuous quantity and normalized with the range [0,1].

And the enhancement transform function is s=T( r );



We also suppose that T( r ) satisfy the following two conditions,

(1) T( r ) is a single-valued and monotonically increasing for rbetween [0,1];

(2) for 1 )( 0 ≤≤ rT 1 0 ≤≤ r

Condition (1) preserves the order in the gray scale;Condition (2) guarantees a mapping with the range [0,1];



An example satisfying above conditions is as follow,

And the inverse transform can be denoted r 1 s 0 )(1 ≤≤= − sT



Assume the probability density functions of the gray level for theoriginal image and transformed image are respectively

Set

So,

)( and )( sr pp sr

1r0 )()(0

≤≤== ∫ dwwrTs r

rp

)(rdrds pr=

−=

=dsdrrpp r

T sr

ss )(

)(1

)( Q

10 1)( )(

1)()(1

≤≤==∴

−=

ssrp

rppr

rT sr

s



It means that the probability density of the gray level for thetransformed image is uniform

For the digital image, we have

So,

1r0 )()(0

≤≤== ∫ dwwrTs r

rp

1,...1,0 and 1 0 )( −=≤≤= Lkn rnrp kk

kr

1,.......,1,0 and 10

)()(00

−=≤≤

=== ∑∑==

Lkrn

T rpnrs j

k

jr

k

j

jkk



rk

17

67

57

47

37

27

10

7

6

5

4

3

2

1

0

=

=

=

=

=

=

=

=

rr

r

r

r

r

rr

nk

811222453296568501023790

02.003.006.008.016.021.025.019.0

nnrp k

kr=)( sk

00.198.095.089.081.065.044.019.0

1

1000

nk

500

rk

1000

nk

500

sk1



3.3. Histogram Matching (specification)

Sometimes we need to convert an image into another image with a specific gray level distribution.

Let and be the original and desired probabilitydensity functions respectively.

Firstly, we do the equalization transform for both of them.

)(rpr )(zpz



)()(0

dwwrTs r

rp∫==

We think s and v are the same. So,

substitute s with v,

)()(0

dwwzGv z

zp∫==

))(()(z )(

11

1

rTsvz

GGG

−−

−

==∴=Q



The procedure can be summarized as follows,(1) Equalize the original image using

(2) Equalize the desired image using

(3) Apply the inverse transformation functionwe can get the value z.

)()(0

dwwrTs r

rp∫==

)()(0

dwwzGv z

zp∫== )(1 sz G−=



The procedure above is also described as the following graph,



3.4. Local enhancement

When histogram equalization and histogram matching are usedto all sub-images in an image, we call it local enhancement.

Take the neighborhood of a point, for example 5x5, do theenhancement for all points.

An example is as follow.


Image Enhancement – Arithmetic/Logic

4.1. Image Subtraction

The result is that the different parts between two image are kept but the same parts between them are removed (become dark).

A typical application is the medical X-ray image test for a specificbody area of the patient.

We can do it by injecting some dye into the parts to be tested to let them differ from others.

),(),(),( yxhyxfyxg −=



4.2. Image Averaging

Assume a noisy image can be expressed as the following

is an original image

is the noise

And the noise is uncorrelated and has zero average value.

),),(),( yxyxfyxg (+= η

),(),(yxyxf

η



When we acquire a set of noisy images , we compute)},({ yxgithe average value

When

),(1),(

y)](x,),([1),(1),(

1

11

yxM

yxf

yxfM

yxM

yxg

M

ii

M

ii

M

iig

∑

∑∑

=

==

+=

+==

η

η

),(),( 0),(1 ,1

yxfyxgyxM

MM

i→∴→∞→ ∑

=ηι


Image Enhancement – Filtering

5. Filtering

Removing or intensifying some components in an image by usingenhancement processing is called filtering.

Using spatial masks to process an image is called spatial filteringUsing the Fourier transform to process an image is called Frequency domain filtering.



5.1. Spatial Filtering

We need to know that

1) A great change of gray level between the pixels usually stands for some obvious contours in an image and it corresponds to the high frequency components in an image.

2) Averaging the pixel grey levels in an image will let a digitalimage become blurring.

3) Intensifying the high frequency components in an image willlet a digital image have obvious contours.



An example

High frequency components

Low frequency components



5.1.1. Mask

Applying a mask to an image is a basic concept in image processing – convolution.

(1) What is a mask: A mask is a set of pixel positions with values called weights. A 3x3 mask is as follows.

1 w2 w3

w4 w5 w6

w7 w8 w9

21 1

242

1 2 1

1 w2 w3

w4 w5 w6

w7 w8 w9

11 1

111

1 1 1

w1 w2 w3

w4 w5 w6

w7 w8 w9



(2) How to use a mask: For any point (x,y) in the image, align the center of the mask to it. We have

w1 w2 w3

w4 w5 w6

w7 w8 w9

z1 z 2 z3

z 4 z5 z6

z7 z8 z9

zwyxf ii

inew ∑==

9

1),(

(x,y)



5.1.2. Smoothing filters

For low-pass filtering --- called smooth processing

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1 1 11 1

1 1 1

1 1 1

1 1 1

×251

×91

1 1 11 1


Effects of mask sizes




Median filter – less blurring and effective in removing salt and pepper noise.



5.1.3. Sharpening filters

For High-pass filtering

w

1− 1− 1− 1− 1−

1− 1− 1− 1−1− 1− 1− 1− 1−

1− 1− 1− 1−1−

1−

w1− 1−

1− 1− 1−

1− 1− ×251

×91

1− 1− 1− 1−1−



Other forms of filters

image enhancement – introduction processing & machine vision/y. gao 3 image enhancement --...

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