Lecture 03,04,05 intensity transformation and spatial filtering

Digital Image Processing 19 Histogram Equalization Histogram Matching Local Histogram Processing Using Histogram Statistics for Image Enhancement III... We can do it by adjusti

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Digital Image Processing

Lecture 3 – Intensity Transformation& Spatial Filtering

Lecturer: Ha Dai DuongFaculty of Information Technology

Process the transform coefficients, not directly process the

intensity values of the image plane

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Digital Image Processing 3

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Intensity transformation function

s T r = ( )

II Intensity transformation function

Functions

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Log Transformations log(1 )

II.2 Log Transform

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II.3 Power – Law

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II.4 Piecewise-Linear Transform

Contrast Stretching

Expands the range of intensity levels in an image so

that it spans the full intensity range of the recording

medium or display device

Intensity-level Slicing

Highlighting a specific range of intensities in an image

often is of interest

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II.5 Bit – Plane Slicing

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Histogram Equalization

Histogram Matching

Local Histogram Processing

Using Histogram Statistics for Image

Enhancement

III Histogram processing

Histogram ( )

is the intensity value

is the number of pixels in the image with intensity

k k th

: the number of pixels in the image of

size M N with intensity

k k

r

=

×

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As the low-contrast image’s histogram is narrow

and centered toward the middle of the gray scale,

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 so

that the probability spread equally

III.1 Histogram Equalization

s = T(r)

Where 0 ≤ r ≤ 1

T(r) satisfies

(a) T(r) is single-valued and monotonically increasingly in the interval

0 ≤ r ≤ 1

(b) 0 ≤ T(r) ≤ 1 for

0 ≤ r ≤ 1

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2 conditions of T(r)

Single-valued (one-to-one relationship) guarantees that

the inverse transformation will exist

Monotonicity condition preserves the increasing order from

black to white in the output image thus it won’t cause a

negative image

0 ≤ T(r) ≤ 1 for 0 ≤ r ≤ 1 guarantees that the output gray

levels will be in the same range as the input levels

The inverse transformation from s back to r is

r = T -1 (s) ; 0 ≤s ≤1

Let

p r (r) denote the PDF of random variable r (r)

p s (s) ) denote the PDF of random variable s

If pr(r) (r) and T(r) are known and T(r) T-1(s) satisfies (s)

condition (a) then ps(s) ) can be obtained using a

formula :

ds

dr (r) p (s)

ps = r

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function (CDF) of random variable r :

CDF is an integral of a probability function (always positive) is the

area under the function

Thus, CDF is always single valued and monotonically increasing

Thus, CDF satisfies the condition (a)

We can use CDF as a transformation function

p

dw ) w ( p dr

d

dr

) r ( dT

dr

ds

r

r r

where 1

ds

dr ) r ( p ) s ( p

r r

r s

Substitute and yield

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Ps(s):

As p s (s) is a probability function, it must be zero outside

the interval [0,1] in this case because its integral over all

values of s must equal 1.

Called p s (s) as a uniform probability density function

p s (s) is always a uniform, independent of the form of

p r (r)

Discrete Transformation Function

The probability of occurrence of gray level in an image

k j

j r k

k

, , ,

where k n

n

) r ( p )

r ( T s

0

11

0

11

0 where k , , , L- n

n ) r (

k

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Thus, an output image is obtained by mapping

each pixel with level rk in the input image into a

corresponding pixel with level sk in the output

image

In discrete space, it cannot be proved in general

that this discrete transformation will produce the

discrete equivalent of a uniform probability density

function, which would be a uniform histogram

Example before after Histogram

equalization

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Example

equalization

The quality is not improved much because the original image already has a broaden gray-level scale

III Histogram processing

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Generate a processed image that has a specified

histogram

Let ( ) and ( ) denote the continous probability

density functions of the variables and ( ) is the

specified probability density function

Let be the random variable with the prob

z

r z p z s

0

ability ( ) ( 1) ( )

Define a random variable with the probability

( ) ( 1) ( )

r r

z z

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1 Obtain pr(r) from the input image and then obtain the values of s

2 Use the specified PDF and obtain the transformation function

1r

;0 2

2 r

) r (

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Example

We would like to apply the histogram specification with the

desired probability density function pz(z) as shown

1 z

;0

2 z )

z (

dw ) w (

dw ) w ( p ) r ( T s

r r

2 2

2

0 2

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Example

2 0

( )

2 Obtain the transformation function G(z)

III.2 Histogram Matching

Example

2

2 2

) ( )

(

r r z

r r

z

r T z

3 Obtain the inversed transformation function G-1

We can guarantee that 0 ≤ z ≤1 when 0 ≤ r ≤1

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Procedure in discrete cases

1 Obtain pr(rj) from the input image and then obtain the values of

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

2 Use the specified PDF and obtain the transformation function

G(zq), round the value to the integer range [0, L-1].

Example Suppose that a 3-bit image (L=8) of size 64 × 64 pixels (MN = 4096) has the intensity distribution shown in the following

table (on the left) Get the histogram transformation function and make the output image with the specified histogram, listed

in the table on the right.

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Example

Obtain the scaled histogram-equalized values,

Compute all the values of the transformation function G,

→7

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Example

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Define a neighborhood and move its center from pixel to pixel

At each location, the histogram of the points in the

neighborhood is computed Either histogram equalization or

histogram specification transformation function is obtained

Map the intensity of the pixel centered in the neighborhood

Move to the next location and repeat the procedure

III.3 Local Histogram Processing

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1 0( )

L

i i i

=

1 0

Local average intensity

( ) denotes a neighborhood

L

i xy

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( , ), if and ( , )

Arithmetic/Logic operations perform on pixel by

pixel basis between two or more images, except

NOT operation which perform only on a single

image

Logic operation performs on gray-level images,

the pixel values are processed as binary

numbers:

Light represents a binary 1, and

Dark represents a binary 0

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IV.2 Image Subtraction

g(x,y) = f(x,y) – h(x,y)

Extraction of the differences between images

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Consider a noisy image g(x,y) formed by the addition of noise η(x,y)

to an original image f(x,y)

g(x,y) = f(x,y) + η(x,y)

If noise has zero mean and be uncorrelated then it can be shown that

if

) ,

K different noisy images

y x

g

1

) , (

1 ) , (

IV.3 Image Averaging

),(

2)

,(

y x y

x g

= variances of g and η)

,(

2)

,(

2g x y , σ η x y

σ

if K increase, it indicates that the variability (noise) of the pixel at

each location (x,y) decreases

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) , ( )}

, (

)}

, (

(output after averaging)

= original image f(x,y)

V Spatial Filtering

A spatial filter consists of (a) a neighborhood, and (b) a predefined

operation

Linear spatial filtering of an image of size MxN with a filter of size

mxn is given by the expression

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Smoothing filters are used for blurring and for noise reduction

Blurring is used in removal of small details and bridging of small

gaps in lines or curves

Smoothing spatial filters include linear filters and nonlinear filters.

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The general implementation for filtering an M N image

with a weighted averaging filter of size m n is given

( , )

( , ) where 2 1

V.1 Smoothing Filters

Averaging Filter Masks

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Averaging Filter Masks

a) original image 500x500 pixel

b) - f) results of smoothing with

square averaging filter masks of

The size of the mask establishes the

relative size of the objects that will be

blended with the background.

original image result after smoothing with

15x15 averaging mask result of thresholding

we can see that the result after smoothing and thresholding, the

remains are the largest and brightest objects in the image

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Order-Statistics Filters (Nonlinear Filters)

The response is based on ordering (ranking) the pixels

contained in the image area encompassed by the filter

Median filter : R = median{zk |k = 1,2,…,n x n}

Max filter : R = max{zk |k = 1,2,…,n x n}

Min filter : R = min{zk |k = 1,2,…,n x n}

Note: n x n is the size of the mask

Median Filter

Median: X= {x1,x2,…, x2b+1}, x is called the median of X if x

greater than or equal b elements and less than or equal b other

elements in X;

For example: X={3,2,3,2,3,4,5,6,5} (b=4) -> Median is 3

Peplaces the value of a pixel by the median of the gray levels in

the neighborhood of that pixel (the original value of the pixel is

included in the computation of the median)

Quite popular because for certain types of random noise ( impulse

noise > salt and pepper noise ) , they provide excellent

noise-reduction capabilities , with considering less blurring than linear

smoothing filters of similar size

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Unsharp Masking and Highboost Filtering

Using First-Order Derivatives for Nonlinear Image

Sharpening - The Gradient

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The first-order derivative of a one-dimensional function f(x) is the

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Laplace Operator

The second-order isotropic derivative operator is the Laplacian

for a function (image) f(x,y)

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Image sharpening in the way of using the

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Filtering

V.4 Unsharp Masking and Highboost

Filtering

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Filtering

V.4 Unsharp Masking and Highboost

Filtering

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Derivatives (Gradient)

V.5 Method based on First-Order

Derivatives (Gradient)

Example

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Methods

dynamic range of gray levels

V.6 Combining Spatial Enhancement

Methods

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Methods

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