Lecture 2. Intensity Transformation and Spatial...

88
Lecture 2. Intensity Transformation and Spatial Filtering

Transcript of Lecture 2. Intensity Transformation and Spatial...

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Lecture 2. Intensity Transformation and Spatial Filtering

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Spatial Domain vs. Transform Domain

► Spatial domain image plane itself, directly process the intensity values of

the image plane

►Transform domain process the transform coefficients, not directly process the

intensity values of the image plane

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Spatial Domain Process

( , ) [ ( , )])( , ) : input image( , ) : output image: an operator on defined over

a neighborhood of point ( , )

g x y T f x yf x yg x yT f

x y

=

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Spatial Domain Process

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Spatial Domain Process

Intensity transformation function ( )s T r=

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Some Basic Intensity Transformation Functions

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Image Negatives

Image negatives1s L r= − −

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Example: Image Negatives

Small lesion

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Log Transformations

Log Transformationslog(1 )s c r= +

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Example: Log Transformations

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Power-Law (Gamma) Transformations

s crγ=

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Example: Gamma Transformations

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Example: Gamma Transformations Cathode ray tube (CRT) devices have an intensity-to-voltage response that is a power function, with exponents varying from approximately 1.8 to 2.5

1/2.5s r=

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Example: Gamma Transformations

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Example: Gamma Transformations

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Piecewise-Linear Transformations

►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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Highlight the major blood vessels and study the shape of the flow of the contrast medium (to detect blockages, etc.)

Measuring the actual flow of the contrast medium as a function of time in a series of images

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Bit-plane Slicing

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Bit-plane Slicing

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Bit-plane Slicing

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Histogram Processing

► Histogram Equalization

► Histogram Matching

► Local Histogram Processing

► Using Histogram Statistics for Image Enhancement

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Histogram Processing

Histogram ( )

is the intensity value is the number of pixels in the image with intensity

k kth

k

k k

h r nr kn r

=

Normalized histogram ( )

: the number of pixels in the image of size M N with intensity

kk

k

k

np rMN

nr

=

×

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Histogram Equalization

The intensity levels in an image may be viewed as random variables in the interval [0, L-1]. Let ( ) and ( ) denote the probability density function (PDF) of random variables and .

r sp r p sr s

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Histogram Equalization

( ) 0 1s T r r L= ≤ ≤ −

. T(r) is a strictly monotonically increasing function in the interval 0 -1;. 0 ( ) -1 for 0 -1.

ar L

b T r L r L≤ ≤

≤ ≤ ≤ ≤

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Histogram Equalization

. T(r) is a strictly monotonically increasing function in the interval 0 -1;. 0 ( ) -1 for 0 -1.

ar L

b T r L r L≤ ≤

≤ ≤ ≤ ≤

( ) 0 1s T r r L= ≤ ≤ −

( ) is continuous and differentiable.T r

( ) ( )s rp s ds p r dr=

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Histogram Equalization

0( ) ( 1) ( )

r

rs T r L p w dw= = − ∫

0

( ) ( 1) ( )r

rds dT r dL p w dwdr dr dr

= = − ∫( 1) ( )rL p r= −

( )( ) 1( ) ( )( ) ( 1) ( ) 1

r r rs

r

p r dr p r p rp s L p rdsds Ldr

= = = =− −

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Example

2

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

2 , for 0 r L-1( 1)( )

0, otherwise

Find the transformation function for equalizing the image histogra

r

rLp r

≤ ≤ −=

m.

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Example

0( ) ( 1) ( )

r

rs T r L p w dw= = − ∫

20

2( 1)( 1)

r wL dwL

= −−∫

2

1r

L=

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Histogram Equalization

0

Continuous case:

( ) ( 1) ( )r

rs T r L p w dw= = − ∫

0

Discrete values:

( ) ( 1) ( )k

k k r jj

s T r L p r=

= = − ∑

0 0

1( 1) k=0,1,..., L-1k k

jj

j j

n LL nMN MN= =

−= − =∑ ∑

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Example: Histogram Equalization Suppose that a 3-bit image (L=8) of size 64 × 64 pixels (MN = 4096) has the intensity distribution shown in following table. Get the histogram equalization transformation function and give the ps(sk) for each sk.

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Example: Histogram Equalization

0

0 00

( ) 7 ( ) 7 0.19 1.33r jj

s T r p r=

= = = × =∑ 1→1

1 10

( ) 7 ( ) 7 (0.19 0.25) 3.08r jj

s T r p r=

= = = × + =∑ 3→

2 3

4 5

6 7

4.55 5 5.67 6 6.23 6 6.65 7 6.86 7 7.00 7

s ss ss s

= → = →= → = →= → = →

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Example: Histogram Equalization

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Question

Is histogram equalization always good?

No

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Histogram Matching Histogram matching (histogram specification) — 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

r z

z

p r p zr z p z

s

0

0

ability

( ) ( 1) ( )

Define a random variable with the probability

( ) ( 1) ( )

r

r

z

z

s T r L p w dw

z

G z L p t dt s

= = −

= − =

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Histogram Matching

0

0

( ) ( 1) ( )

( ) ( 1) ( )

r

r

z

z

s T r L p w dw

G z L p t dt s

= = −

= − =

∫∫

[ ]1 1( ) ( )z G s G T r− −= =

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Histogram Matching: Procedure

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

► Use the specified PDF and obtain the transformation function G(z)

► Mapping from s to z

0( 1) ( )

r

rs L p w dw= − ∫

0( ) ( 1) ( )

z

zG z L p t dt s= − =∫

1( )z G s−=

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Histogram Matching: Example

Assuming continuous intensity values, suppose that an image has the intensity PDF

Find the transformation function that will produce an image whose intensity PDF is

2

2 , for 0 -1( 1)( )

0 , otherwiser

r r LLp r

≤ ≤ −=

2

3

3 , for 0 ( -1)( ) ( 1)

0, otherwisez

z z Lp z L

≤ ≤= −

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Histogram Matching: Example

Find the histogram equalization transformation for the input image

Find the histogram equalization transformation for the specified histogram

The transformation function

20 0

2( ) ( 1) ( ) ( 1)( 1)

r r

rws T r L p w dw L dw

L= = − = −

−∫ ∫

2 3

3 20 0

3( ) ( 1) ( ) ( 1)( 1) ( 1)

z z

zt zG z L p t dt L dt s

L L= − = − = =

− −∫ ∫

2

1r

L=

1/321/3 1/32 2 2( 1) ( 1) ( 1)1

rz L s L L rL

= − = − = − −

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Histogram Matching: Discrete Cases

► Obtain pr(rj) from the input image and then obtain the values of sk, round the value to the integer range [0, L-1].

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

► Mapping from sk to zq

0 0

( 1)( ) ( 1) ( )k k

k k r j jj j

Ls T r L p r nMN= =

−= = − =∑ ∑

0( ) ( 1) ( )

q

q z i ki

G z L p z s=

= − =∑

1( )q kz G s−=

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Example: Histogram Matching 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: Histogram Matching

Obtain the scaled histogram-equalized values,

Compute all the values of the transformation function G,

0 1 2 3 4

5 6 7

1, 3, 5, 6, 7,7, 7, 7.

s s s s ss s s= = = = == = =

0

00

( ) 7 ( ) 0.00z jj

G z p z=

= =∑

1 2

3 4

5 6

7

( ) 0.00 ( ) 0.00 ( ) 1.05 ( ) 2.45( ) 4.55 ( ) 5.95( ) 7.00

G z G zG z G zG z G zG z

= == == ==

0→

0→ 0→1→ 2→

6→5→7→

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Example: Histogram Matching

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Example: Histogram Matching

Obtain the scaled histogram-equalized values,

Compute all the values of the transformation function G,

0 1 2 3 4

5 6 7

1, 3, 5, 6, 7,7, 7, 7.

s s s s ss s s= = = = == = =

0

00

( ) 7 ( ) 0.00z jj

G z p z=

= =∑

1 2

3 4

5 6

7

( ) 0.00 ( ) 0.00 ( ) 1.05 ( ) 2.45( ) 4.55 ( ) 5.95( ) 7.00

G z G zG z G zG z G zG z

= == == ==

0→

0→ 0→1→ 2→

6→5→7→

s0

s2 s3

s5 s6 s7

s1

s4

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Example: Histogram Matching

0 1 2 3 4

5 6 7

1, 3, 5, 6, 7,7, 7, 7.

s s s s ss s s= = = = == = =

01234567

kr

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Example: Histogram Matching

0 31 42 53 64 75 76 77 7

k qr z→

→→→→→→→→

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Example: Histogram Matching

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Example: Histogram Matching

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Example: Histogram Matching

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Example: Histogram Matching

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Local Histogram Processing

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

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Local Histogram Processing: Example

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Using Histogram Statistics for Image Enhancement

1

0( )

L

i ii

m r p r−

=

=∑1

0( ) ( ) ( )

Ln

n i ii

u r r m p r−

=

= −∑1

2 22

0( ) ( ) ( )

L

i ii

u r r m p rσ−

=

= = −∑

1 1

0 0

1 ( , )M N

x yf x y

MN

− −

= =

= ∑∑

[ ]1 1

2

0 0

1 ( , )M N

x yf x y m

MN

− −

= =

= −∑∑

Average Intensity

Variance

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Using Histogram Statistics for Image Enhancement

1

0

Local average intensity

( )

denotes a neighborhood

xy xy

L

s i s ii

xy

m r p r

s

=

=∑

12 2

0

Local variance

( ) ( )xy xy xy

L

s i s s ii

r m p rσ−

=

= −∑

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Using Histogram Statistics for Image Enhancement: Example

0 1 2

0 1 2

( , ), if and ( , )

( , ), otherwise

: global mean; : global standard deviation0.4; 0.02; 0.4; 4

xy xys G G s G

G G

E f x y m k m k kg x y

f x y

mk k k E

σ σ σ

σ

≤ ≤ ≤=

= = = =

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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

( , ) ( , ) ( , )a b

s a t bg x y w s t f x s y t

=− =−

= + +∑ ∑

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Spatial Filtering

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Spatial Correlation

The correlation of a filter ( , ) of size with an image ( , ), denoted as ( , ) ( , )

w x y m nf x y w x y f x y

×

( , ) ( , ) ( , ) ( , )a b

s a t bw x y f x y w s t f x s y t

=− =−

= + +∑ ∑

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Spatial Convolution

The convolution of a filter ( , ) of size with an image ( , ), denoted as ( , ) ( , )

w x y m nf x y w x y f x y

×

( , ) ( , ) ( , ) ( , )a b

s a t bw x y f x y w s t f x s y t

=− =−

= − −∑ ∑

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Smoothing Spatial Filters

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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Spatial Smoothing Linear Filters

The general implementation for filtering an M N imagewith a weighted averaging filter of size m n is given

( , ) ( , ) ( , )

( , )

where 2 1

a b

s a t ba b

s a t b

w s t f x s y tg x y

w s t

m a

=− =−

=− =−

××

+ +=

= +

∑ ∑

∑ ∑, 2 1.n b= +

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Two Smoothing Averaging Filter Masks

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Example: Gross Representation of Objects

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Order-statistic (Nonlinear) Filters

— Nonlinear — Based on ordering (ranking) the pixels contained in the

filter mask — Replacing the value of the center pixel with the value

determined by the ranking result E.g., median filter, max filter, min filter

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Example: Use of Median Filtering for Noise Reduction

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Sharpening Spatial Filters

► Foundation ► Laplacian Operator ► Unsharp Masking and Highboost Filtering ► Using First-Order Derivatives for Nonlinear Image

Sharpening — The Gradient

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Sharpening Spatial Filters: Foundation

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

► The second-order derivative of f(x) as the difference

( 1) ( )f f x f xx∂

= + −∂

2

2 ( 1) ( 1) 2 ( )f f x f x f xx

∂= + + − −

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Sharpening Spatial Filters: Laplace Operator

The second-order isotropic derivative operator is the Laplacian for a function (image) f(x,y)

2 22

2 2

f ffx y

∂ ∂∇ = +

∂ ∂2

2 ( 1, ) ( 1, ) 2 ( , )f f x y f x y f x yx

∂= + + − −

∂2

2 ( , 1) ( , 1) 2 ( , )f f x y f x y f x yy

∂= + + − −

∂2 ( 1, ) ( 1, ) ( , 1) ( , 1)

- 4 ( , )f f x y f x y f x y f x y

f x y∇ = + + − + + + −

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Sharpening Spatial Filters: Laplace Operator

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Sharpening Spatial Filters: Laplace Operator

Image sharpening in the way of using the Laplacian:

2

2

( , ) ( , ) ( , )

where, ( , ) is input image,

( , ) is sharpenend images,-1 if ( , ) corresponding to Fig. 3.37(a) or (b)

and 1 if either of the other two filters is us

g x y f x y c f x y

f x yg x yc f x y

c

= + ∇

= ∇= ed.

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Unsharp Masking and Highboost Filtering

► Unsharp masking Sharpen images consists of subtracting an unsharp (smoothed)

version of an image from the original image e.g., printing and publishing industry

► Steps

1. Blur the original image

2. Subtract the blurred image from the original

3. Add the mask to the original

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Unsharp Masking and Highboost Filtering

Let ( , ) denote the blurred image, unsharp masking is

( , ) ( , ) ( , )Then add a weighted portion of the mask back to the original ( , ) ( , ) * ( , )

mask

mask

f x y

g x y f x y f x y

g x y f x y k g x y

= −

= + 0k ≥

when 1, the process is referred to as highboost filtering.k >

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Unsharp Masking: Demo

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Unsharp Masking and Highboost Filtering: Example

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Image Sharpening based on First-Order Derivatives

For function ( , ), the gradient of at coordinates ( , )is defined as

grad( ) x

y

f x y f x y

fg xf f

fgy

∂ ∂ ∇ ≡ ≡ = ∂ ∂

2 2

The of vector , denoted as ( , )

( , ) mag( ) x y

magnitude f M x y

M x y f g g

= ∇ = +Gradient Image

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Image Sharpening based on First-Order Derivatives

2 2

The of vector , denoted as ( , )

( , ) mag( ) x y

magnitude f M x y

M x y f g g

= ∇ = +

( , ) | | | |x yM x y g g≈ +

z1 z2 z3

z4 z5 z6

z7 z8 z9

8 5 6 5( , ) | | | |M x y z z z z= − + −

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Image Sharpening based on First-Order Derivatives

z1 z2 z3

z4 z5 z6

z7 z8 z9

9 5 8 6

Roberts Cross-gradient Operators( , ) | | | |M x y z z z z≈ − + −

7 8 9 1 2 3

3 6 9 1 4 7

Sobel Operators( , ) | ( 2 ) ( 2 ) |

| ( 2 ) ( 2 ) |M x y z z z z z z

z z z z z z≈ + + − + +

+ + + − + +

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Image Sharpening based on First-Order Derivatives

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Example

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Example: Combining Spatial Enhancement Methods Goal: Enhance the image by sharpening it and by bringing out more of the skeletal detail

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Example: Combining Spatial Enhancement Methods Goal: Enhance the image by sharpening it and by bringing out more of the skeletal detail