Image Enhancement and Filtering

8/2/2019 Image Enhancement and Filtering

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EE4H, M.Sc 0407191Computer Vision

Dr. Mike Spann

[email protected]://www.eee.bham.ac.uk/spannm
mailto:[email protected]://www.eee.bham.ac.uk/spannmhttp://www.eee.bham.ac.uk/spannmmailto:[email protected]


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Introduction Images may suffer from the following degradations:

Poor contrast due to poor illumination or finitesensitivity of the imaging device

Electronic sensor noise or atmospheric disturbancesleading to broad band noise

Aliasing effects due to inadequate sampling

Finite aperture effects or motion leading to spatial


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IntroductionWe will consider simple algorithms for image

enhancement based on lookup tables

Contrast enhancement

We will also consider simple linear filteringalgorithms

Noise removal


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Histogram equalisation In an image of low contrast, the image has grey

levels concentrated in a narrow band

Define the grey level histogram of an image h(i)where : h(i)=number of pixels with grey level=I

For a low contrast image, the histogram will be

concentrated in a narrow band The full greylevel dynamic range is not used


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Histogram equalisation

h i( )

i


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Histogram equalisation Can use a sigmoid lookup to map input to output grey

levels

A sigmoid functiong(i) controls the mapping frominput to output pixel

Can easily be implemented in hardware for maximumefficiency


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Histogram equalisationh i( )

g i( )

h i' ( )

h i h g i' ( ) ( ( ))1

g i i( ) exp 11


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Histogram equalisation controls the position of maximum slope controls the slope

Problem - we need to determine the optimumsigmoid parameters and for each image Abetter method would be to determine the best

mapping function from the image data


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Histogram equalisationA general histogram stretching algorithm is defined in

terms of a transormationg(i)

We require a transformation g(i) such that from anyhistogram h(i) :

constant)()(')(:

jgijjhih


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Histogram equalisation Constraints (NxNx 8 bit image)

No crossover in grey levels after transformation

2)(' Nihi

i i g i g i1 2 1 2 ( ) ( )


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Histogram equalisationAn adaptive histogram equalisation algorithm can be

defined in terms of the cumulative histogram H(i) :

H i( ) = number of pixels with grey levels i

i

jjhiH

0)()(


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Histogram equalisation Since the required h(i) is flat, the required H(i) is a

ramp:

h(i) H(i)


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Histogram equalisation Let the actual histogram and cumulative

histogram be h(i) and H(i)

Let the desired histogram and desired cumulativehistogram be h(i) and H(i)

Let the transformation beg(i)

H g i N g i' ( ( )) ( ) 2255

( ' ( ) , ' ( ) ) H N H 255 0 02


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Histogram equalisation Sinceg(i)is an ordered transformation

i i g i g i1 2 1 2 ( ) ( )H g i H i

N g i' ( ( )) ( )

( ) 2255

g iH i

N( )

( ) 255 2


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Histogram equalisationWorked example, 32 x 32 bit image with grey levels

quantised to 3 bits

g iH i

( )( ) 7

1024

h i h jj i g j

' ( ) ( ): ( )


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Histogram equalisationi h i( ) H i( ) g i( )

0 197 197 1.351 -

1 256 453 3.103 1972 212 665 4.555 -

3 164 829 5.676 256

4 82 911 6.236 -

5 62 993 6.657 2126 31 1004 6.867 246

7 20 1024 7.07 113

h i' ( )


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0 1 2 3 4 5 6 7

0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Original histogram

Stretched histogram


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0.00

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i

h(i)

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i

h(i)


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i

h(i)

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i

h(i)


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Histogram equalisation ImageJ demonstration

http://rsb.info.nih.gov/ij/signed-applet
http://rsb.info.nih.gov/ij/signed-applet/http://rsb.info.nih.gov/ij/signed-applet/http://rsb.info.nih.gov/ij/signed-applet/http://rsb.info.nih.gov/ij/signed-applet/


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Image Filtering Simple image operators can be classified as 'pointwise'

or 'neighbourhood' (filtering) operators

Histogram equalisation is a pointwise operation

More general filtering operations use neighbourhoodsof pixels


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Image Filtering

(x,y) (x,y)

Input image Output image

(x,y) (x,y)

pointwisetransformation

neighbourhoodtransformation

Input image Output image


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Image Filtering Examples of filters:

g x y h f x y h f x y

h f x y

( , ) ( , ) ( , )

..... ( , )

1 2

9

1 1 1

1 1

g x y median

f x y f x y

f x y( , )

( , ), ( , )

..... ( , )

1 1 1

1 1


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Linear filtering and convolution Example

3x3 arithmetic mean of an input image (ignoringfloating point byte rounding)

(x,y) (x,y)

Input image f(x,y) Output image g(x,y)

(x-1,y-1)

(x+1,y+1)


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Linear filtering and convolution

(x,y) (x,y)


Image point

Filter mask point


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Linear filtering and convolutionWe can define the convolution operator

mathematically

Defines a 2D convolution of an imagef(x,y) with a filterh(x,y)

g x y h x y f x x y y

f x x y y

yx

yx

( , ) ( ' , ' ) ( ' , ' )

( ' , ' )

''

''

1

1

1

1

1

1

1

11

9


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Linear filtering and convolution Example convolution with a Gaussian filter kernel

determines the width of the filter and hence theamount of smoothing

g x yx y

g x g y

g xx

( , ) exp(( )

)

( ) ( )

( ) exp( )

2 2

2

2

2

2

2


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0.00

0.20

0.40

0.60

0.80

1.00

-6 -4 -2 0 2 4 x

g(x)


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Original Noisy

Filtered

=1.5

Filtered

=3.0


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Linear filtering and convolution ImageJ demonstration

http://rsb.info.nih.gov/ij/signed-applet/http://rsb.info.nih.gov/ij/signed-applet/http://rsb.info.nih.gov/ij/signed-applet/http://rsb.info.nih.gov/ij/signed-applet/


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Linear filtering and convolution The inverse DFT is defined by

f x yN

F u v jN

ux vyy

N

x

N

( , ) ( , )exp( ( ))

1 220

1

0

1

x y N , .. 0 1


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Linear filtering and convolutionx

y

f(x,y)

(0.0)

(N-1,N-1)

v

u(0,0)

(N-1,N-1)

F(u,v)

DFT IDFT


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log( ( , ) )1 F u v


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Linear filtering and convolution F(u,v) is the frequency content of the image at spatial

frequency position(u,v) Smooth regions of the image contribute low frequency

components to F(u,v)Abrupt transitions in grey level (lines and edges)

contribute high frequency components toF(u,v)


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Linear filtering and convolutionWe can compute the DFT directly using the formula

An N point DFT would require N2 floating pointmultiplications per output point

Since there are N2 output points , the computationalcomplexity of the DFT is N4

N4=4x109 for N=256

Bad news! Many hours on a workstation


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Linear filtering and convolution The FFT algorithm was developed in the 60s for

seismic exploration

Reduced the DFT complexity to 2N2log2N

2N2log2N~106 for N=256

A few seconds on a workstation


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Linear filtering and convolution The filtering interpretation of convolution can be

understood in terms of the convolution theorem

The convolution of an imagef(x,y) with a filter h(x,y) isdefined as:

g x y h x y f x x y y

f x y h x yy

M

x

M

( , ) ( ' , ' ) ( ' , ' )

( , )* ( , )''

0

1

0

1


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(x,y)


(x,y)

Filter mask h(x,y)


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Linear filtering and convolution Note that the filter mask is shifted and inverted prior

to the overlap multiply and add stage of theconvolution

Define the DFTs off(x,y),h(x,y), andg(x,y) asF(u,v),H(u,v) and G(u,v)

The convolution theorem states simply that :

G u v H u v F u v( , ) ( , ) ( , )


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DFT

IDFT


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Linear filtering and convolution Frequency domain implementation of

convolution Imagef(x,y)NxNpixels

Filter h(x,y) MxM filter mask points UsuallyM


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Linear filtering and convolutionInput image f(x,y) Output image g(x,y)

(x,y)

Filter mask h(x,y)

(x',y')

x' = x modulo N

y' = y modulo N


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Linear filtering and convolutionWe can evaluate the computational complexity of

implementing convolution in the spatial andspatial frequency domains

Nx Nimage is to be convolved with anMx Mfilter Spatial domain convolution requiresM2 floating point

multiplications per output point orN2 M2 in total

Frequency domain implementation requires 3x(2N2

log 2N) +N2 floating point multiplications ( 2 DFTs +

1 IDFT +N2 multiplications of the DFTs)


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Linear filtering and convolution Example 1,N=512,M=7

Spatial domain implementation requires 1.3 x 107floating point multiplications

Frequency domain implementation requires 1.4 x 107floating point multiplications

Example 2,N=512,M=32 Spatial domain implementation requires 2.7 x 108

floating point multiplications Frequency domain implementation requires1.4 x 107

floating point multiplications


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Linear filtering and convolution For smaller mask sizes, spatial and frequency

domain implementations have about the samecomputational complexity

However, we can speed up frequency domaininterpretations by tessellating the image into sub-blocks and filtering these independently

Not quite that simple we need to overlap the filteredsub-blocks to remove blocking artefacts

Overlap and add algorithm


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Linear filtering and convolutionWe can look at some examples of linear filters

commonly used in image processing and theirfrequency responses

In particular we will look at a smoothing filter and afilter to perform edge detection


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Linear filtering and convolution Smoothing (low pass) filter

Simple arithmetic averaging

Useful for smoothing images corrupted by additivebroad band noise

H H3 5

1

9

1 1 1

1 1 11 1 1

1

25

1 1 1 1 1

1 1 1 1 1

1 1 1 1 11 1 1 1 1

1 1 1 1 1

etc


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Linear filtering and convolutionh x( ) H u( )

Spatial domain Spatial frequency domain

ux


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Linear filtering and convolution Edge detection filter

Simple differencing filter used for enhancing edged

Has a bandpass frequency response

H

1 0 1

1 0 1

1 0 1


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Linear filtering and convolution ImageJ demonstration

http://rsb.info.nih.gov/ij/signed-applethttp://rsb.info.nih.gov/ij/signed-applethttp://rsb.info.nih.gov/ij/signed-applethttp://rsb.info.nih.gov/ij/signed-applet


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Linear filtering and convolutionf x( )

p

x

f x( )* ( )1 0 1

p

x


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Linear filtering and convolutionWe can evaluate the (1D) frequency response of

the filter h(x)={1,0,-1 } from the DFT definition

H u h xjux

N

ju

Nju

N

ju

N

ju

N

jju

N

u

N

x

N

( ) ( )exp( )

exp( )

exp( ) exp( ) exp( )

exp( ) sin( )

2

14

2 2 2

22 2

0

1


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Linear filtering and convolution The magnitude of the response is therefore:

This has a bandpass characteristic

H u

u

N( ) sin( )

2


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H u( )

u


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Image Enhancement and Filtering

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