2-Point Operations Mar31 April2

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Bernd Girod: EE368 Digital Image Processing Point Operations no. 1

What is an image?

Ideally, we think of an image as a 2-dimensional light

intensity function,f(x,y), wherex andy are spatial

coordinates, andfat (x,y) is related to the brightness orcolor of the image at that point.

In practice, most images are defined over a rectangle.

Continous in amplitude (continous-tone)

Continous in space: no pixels!


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Digital Images and Pixels

A digital image is the representation of a continuous image

f(x,y)by a 2-d array of discrete samples. The amplitude ofeach sample is quantized to be represented by a finite

number of bits.

Each element of the 2-d array of samples is called a pixelor pel (from picture element)

Think of pixels as point samples, without extent.


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Image Resolution

200x200 100x100 50x50 25x25

These images were produced by simply picking every n-th sample

horizontally and vertically and replicating that value nxn times.

We can do better

prefiltering before subsampling to avoid aliasing

Smooth interpolation


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Color Components

RedR(x,y) Green G(x,y) BlueB(x,y)

Monochrome image

R(x,y) = G(x,y) = B(x,y)


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Different numbers of gray levels

256 32 16

8 4 2

Contouring


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How many gray levels are required?

Contouring is most visible for a ramp

Digital images typically are quantized to 256 gray levels.

32 levels

64 levels

128 levels

256 levels


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130129128127126125

Brightness discrimination experiment

Can you see the circle?

Visibility threshold

I

I+ I

I I KWeber

12%Weber fractionWebers Law

Note: I is luminance,

measured in cd m2


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Contrast with 8 Bits According to Webers Law

Assume that the luminance difference between two

successive representative levels is just at visibility threshold

For

Typical display contrast

Cathode ray tube 100:1

Print on paper 10:1

Suggests uniform perception in the log(I) domain(Fechners Law)

Imax

Imin

= 1+ KWeber( )

255

KWeber = 0.010.02Imax

Imin

=13156


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log vs. -predistortion

Similar enough for most practical applications

U

I

U~ I1

U ~ log(I)

Imax

Imin

=100


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Photographic film

measures film contrast General purpose films: = -0.7 . . . -1.0

High-contrast films: = -1.5 . . . -10

Lower speed films tend to have higher absolute

slope -

logEEis exposure

densityd

shoulder

toe

linear region

I = I0 10d

= I0 10 logE+d0( )

= I0 10d0

E

Luminance

Hurter & Driffield curve (H&D curve)for photographic negative

d00

2.0

1.0


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Intensity Scaling

Original image Scaled image

f x,y( ) a f x,y( )

Scaling in the -domain is equivalent to scaling in the linear luminancedomain

. . . same effect as adjusting camera exposure time.

I~ a f x,y( )( )

= a

f x,y( )( )


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Adjusting

Original image increased by 50%

f x,y( ) a f x,y( )( )

with =1.5

. . . same effect as using a different photographic film . . .


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Changing gradation by -adjustment

Original ramp 0

Scaled ramp 20

Scaled ramp 0.50

Scaling chosen to

approximately preservebrightness of mid-gray


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Histograms

Distribution of gray-levels can be judged by measuring a

histogram:

For B-bit image, initialize 2B

counters with 0

Loop over all pixelsx,y

When encountering gray levelf(x,y)=i, increment counter #

Histogram can be interpreted as an estimate of theprobability density function (pdf) of an underlying random

process.

You can also use fewer, larger bins to trade off amplitude

resolution against sample size.


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Example histogram

gray level

#pixels

Cameramanimage


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Histogram equalization

Idea: find a non-linear transformation

to be applied to each pixel of the input imagef(x,y), suchthat a uniform distribution of gray levels in the entire range

results for the output image g(x,y). Analyse ideal, continuous case first, assuming

T(f)is strictly monotonically increasing, hence, there

exists

Goal: pdfpg(g)= const. over the range

0 f 1

f = T1

g( )

g = T f( )

0 g 1

0 g 1


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Histogram equalization for continuous case

From basic probability theory

Consider the transformation function

Then . . .

pg g( ) = pf f( )df

dg

f=T1 g( )

g = T f( ) = pf ( )0

f

d 0 f 1

dg

df= pf f( )

pg g( ) = pf f( )df

dg

f=T

1g( )

= pf f( )1

pf f( )

f=T

1 g( )

= 1 0 g 1


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Histogram equalization for discrete case

Now,fonly assumes discrete amplitude values

with probabilities

Discrete approximation of

The resulting values gkare in the range [0,1] and need tobe scaled and rounded appropriately.

f0 , f1,, fL1

P0 =n

0

n P

1=

n1

n P

L1=

nL1

n

g = T f( ) = pf ( )0

f

d

gk =

T fk( ) = Pi

i=0

k


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Histogram equalization example

Original image Pout Poutafter histogram equalization


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Original image Pout . . . after histogram equalization

gray level

#pixels

gray level

#pixels


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Original imageCameraman

Cameraman

after histogram equalization


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Original image Cameraman . . . after histogram equalization

gray level

#

pixels

gray level

#pixels


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Original image Moon Moonafter histogram equalization


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Original image Moon . . . after histogram equalization

gray level

#

pixels

gray level

#

pixels


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Adaptive Histogram Equalization

Original Global histogram Tiling

8x8 histograms

Tiling

32x32 histograms


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Adaptive histogram equalization

Original image Tire Tireafter

equalization ofglobal histogram

Tireafter

adaptive histogram equalization8x8 tiles


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Point Operations Combining Images

Image averaging for noise reduction

Combination of different exposure for high-dynamic rangeimaging

Image subtraction for change detection

Accurate alignment is always a requirement


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http://www.cambridgeincolour.com/tutorials/noise-reduction.htm

1 image 2 images 4 images


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Take N aligned images

Average image:

Mean squared error vs. noise-free imageg

f1x,y( ),f2 x,y( ),,fN x,y( )

f x,y( ) =1

Nf

ix,y( )

i=1

N

E f g( )2

= E

1

Nf

ii

g

2

= E

1

Ng+ n

i( )i

g

2

= E 1N

ni

i

2

= 1

N2E n

i2

{ }i= 1

NE n2{ }

E ni

{ } =E n{ }iprovidedE n

inj{ } = 0i,j


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Image subtraction

Find differences/changes between 2 mostly identical images

Example from IC manufacturing:defect detection in photomasks

by die-to-die comparison


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Where is the Defect?

Image A (no defect) Image B (w/ defect)


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Absolute Difference Between Two Images

w/o alignment w/ alignment


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Digital Subtraction Angiography

- +

Contrast

enhancement

http://www.isi.uu.nl/Research/Gallery/DSA/

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2-Point Operations Mar31 April2