EE 583 Lecture06

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EE-583: Digital Image Processing

Prepared By:Dr. Hasan Demirel, PhD

Image Enhancement in the Spatial Domain Example 1(PR3.1): Exponentials of the form e- r2, a positive constant, are

useful for constructing smooth gray-level transformation functions. Construct thetransformation functions having the general shapes shown in the following

figures. The constants shown are input parameters, and your proposed

transformations must include them in their specif ications.

A

A/2

L0

s=T(r)

r

(a)

B

B/2

L0

s=T(r)

r

(b)

D

C

L0

s=T(r)

r

(c)

(a)General form of the funct ion :2

)( rAerTs a2/

20 AAe

L aIn Figure (a): solving for :2

0

2

0

/693.0

693.0)5.0ln(

L

L

a

a

Then:2

20

69 3.0

)(

rL

AerTs

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Image Enhancement in the Spatial Domain Example 1(PR3.1):

B

B/2

L0

s=T(r)

r

(b)

(b) General form of the function:

2 2

1

r r

s T( r ) B Be B( e )

a a

In Figure (b): Then:

2

0

2

0

/693.0

693.0)5.0ln(

L

L

a

a

)1()(

2

20

69 3.0r

LeBrTs

2/)1(20 BeB

L a

(c) General form of the function:

D

C

L0

s=T(r)

r

(c)

CeCDrTsr

L

)1)(()(

2

20

693.0

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Image Enhancement in the Spatial Domain Example 2(PR3.3): Propose a set of gray-level-sl icing transformations

capable of producing all the individual bi t planes of an 8-bit monochromeimage.

For Bi t plane 7 (MSB): the following

function can be wri tten

otherwise

rforrT

0

2255)(

7

For Bi t plane 6:

otherwise

rforrT0

)2,mod(2255)(76

For Bi t plane 5:

otherwise

rforrT

0

)2,mod(2255)(

65

Consider mod(x,y)to be the modular division resul ting the remainder of the integer

division x/y.

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For Bi t plane 4:

otherwiserforrT

0)2,mod(2255)(

54

For Bi t plane 3:

otherwise

rforrT

0

)2,mod(2255)(

43

For Bi t plane 2:

otherwiserforrT

0)2,mod(2255)(

32

For Bi t plane 1:

otherwise

rforrT

0

)2,mod(2255)(

21

For Bi t plane 0:

otherwiserforrT

0)2,mod(2255)(

10

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0 50 100 150 200 2500

50

100

150

200

250


Consider Bi t plane 6:6 7255 2 mod( , 2 )

( )0

for rT r

otherwise

r

s

s=T(r)

250=11001110

31=00011111 111=01101111

190=10111110

Note that al l the 1s corresponding to bit

6 are scaled to 255 and al l the bits

corresponding to 0s are scaled down to

0.

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Image Enhancement in the Spatial Domain Example 3(PR3.4): a) What effect would setting to zero the lower-order bit

planes have on the histogram of an image in general? b) What would be the effect on the histogram if we set to zero the higher-

order bit planes instead?

Al l the bits included

a) Removing the low order bit planes would mean the loss of some high frequency detail s.

Fur thermore the image histogram wil l be more sparseas compared with the all 8-bit

plane case.

0 50 100 150 200 250

0

200

400

600

800

1000

0 50 100 150 200 250

0

500

1000

1500

2000

2 of the LSBs removed

This is because, there wil l be no component r epresenting intermediate pixel values

such as 1,2,3,4, 5, 6,7and 9,10,11,12,13,14,15 etc. Instead there wil l be 0and 8and 16 etc.

This would cause the height some of the remaining histogram peaks to increase in general .

Typicall y, less variabil i ty in gray level values wil l reduce contrast.

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Image Enhancement in the Spatial Domain Example 3(PR3.4): b) What would be the eff ect on the histogram if we set

to zero the higher-order bit planes instead?

Al l the bits included

b) Removing the high order bit planes would mean the loss of some very

important DC components away from the image.

0 50 100 150 200 250

0

200

400

600

800

1000

MSB removed0 50 100 150 200 250

0

200

400

600

800

1000

1200

The meaning of this is that the image is much darkerand a lot of

the low f requencycomponents wil l be lost.

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Image Enhancement in the Spatial Domain Example 4(PR3.6): Suppose that a digi tal image is subjected to histogram

equal ization . Show that a second pass of hi stogram equal ization wi l l produceexactl y the same resul t as the first pass?

Let nbe the total number of pixels and let nrjbe the number of pixels in the input image with

in tensity valuerj. Then, the histogram equalization tr ansformation i s:

Since every pixel (and no others) wi th value rkis mapped to value sk, it follows that nsk= nrk

. A

second pass of h istogram equal ization would produce values vkaccording

to the transformation:

k

j

r

k

j

r

kk j

j

nnn

n

rTs00

1

)(

thennnbutn

n

sTv jjj

rs

k

j

s

kk ,,)(0

k

k

j

r

kk sn

nsTv

j 0

)(

Which shows that a second pass of histogram equal ization would yield the same resul t

as the fi rst pass.

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Image Enhancement in the Spatial Domain Example 4(PR3.6):Given an image, the fol lowing histograms and CDF (T(r) graphs can

be obtained.

0 50 100 150 200 250

0

1000

2000

3000

4000

5000

6000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 50 100 150 200 250

0

1000

2000

3000

4000

5000

6000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Histogram of the fi rst pass

H istogram of the 2nd pass

(~uni formly distributed)

T(r )=CDF of the fir st pass T(r )=CDF of the 2nd pass

(~identity transformation)

Note that the I denti ty Transformation does notchange the histogram of the input image

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Image Enhancement in the Spatial Domain Example 6(PR3.22): The three images shown below were blur red using the square

averaging masks of sizes n=23, 25 and 45 respectively. The vertical bars on the lower partof (a) and (c) are blur red, but clear separation exists between them. However, the bars

have merged in in image (b), in spite of the fact that the mask that produced the image is

signi f icantly smaller than the mask produced image (c).Explain this.

(Note that the vertical bars are 5 pixels wide and 20 pixels apart.)

(a) (b) (c)Original image

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Note that the vertical bars are 5 pixels wide and 20 pixels apart.

(a) (b) (c)

The reason why the mask wi th size n=25 producing merged uniform region around the bars is

because of the sizes of the bars and the separation of the bars in the hor izontal dir ection. The

width of each bar is 5 pixels and each bar is separated by 20 pixels.

I n such an environment as the mask moves in the hor izontal dir ection there wil l be 5 black and

20 light gray pixels in each r ow at a time. Th is wil l provide the same average value for each pixel

in the region producing a merged unif orm gray level.

However thi s wil l not be the same in 23 and 45 pixel masks!!

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Image Enhancement in the Spatial Domain Example 7(PR3.24): I n a given application an averaging mask is applied to input

images to reduce noise, and then a Laplacian mask is applied to enhance small detai ls.Would the resul t be the same if the order of these operations were reversed?

Laplacian operation and averaging can be expressed by the fol lowing 3x3 masks:

),(4)1,()1,(),1(),1(),( yxfyxfyxfyxfyxfyxgLaplacian

9

1

1( , )

9 i

i

Averaging h x y f

Both of the two operators are mul tiplying the pixels in a 3x3 neighborhood with constant

numbers and perform the addition. Therefore, these operations are li near operations.

The order of two linear operations does not matter. The resul t would be same in any order.

0 1 0

1 -4 1

0 1 0

1/9 1/9 1/9

1/9 1/9 1/9

1/9 1/9 1/9

Laplacian MaskAveraging Mask

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Image Enhancement in the Frequency Domain Example 8(PR3.25) :Show that the Laplacian operation i s isotropic (invar iant to

rotation). You wil l need the foll owing equations relating coordinates after axis rotationby an angle .

Laplacian operator is defined as:

I f we show that the r ight sides of the first 2 equations are equal than the Laplacian operationis rotation invar iant.

For the rotated Laplacian operator :

Given that:

qq

qq

cossin

sincos

yxy

yxx

We start wi th,

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Image Enhancement in the Frequency Domain Example 8(PR3.25 ):

Taking the parti al derivative of th is expression again wi th r espect to x yields:

Repeat the same operation f or y:

Taking the parti al derivative of th is expression again wi th r espect to yyields:

Adding the 2 expressions for the second der ivatives:

Both sides are equal, Hence Laplacian i s rotational invari ant.

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Image Enhancement in the Spatial Domain Example 9(PR4.7): What is the source of nearl y per iodic bright points in the

hor izontal axi s of the spectrum in the following f igur e.

The near ly per iodic bri ght points in the frequency spectrum corr esponds to the peri odic barsas well as

the repeating boxes, letters and cir cles in the hor izontal dir ection.

l

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Image Enhancement in the Spatial Domain Example 10(PR4.6) -a): Prove the validity of the foll owing Equation.

TransformFourierthedenotesNvMuFyxf yx [.],)2/,2/(])1)(,([

1 12

0 0

1 M N j (ux / M vy / N )

x y

F( u,v ) f ( x, y )eMN

1j( x y ) x y j( x y )e ( ) e cos( ( x y )) j sin( ( x y ))

1 12 2 2

0 0

12 2

M Nj (( u M / ) x / M ( v N / ) y / N )

x y

F( u M / ,v N / ) f ( x, y )e

MN

1 12 1 2 1 2

0 0

1 M N j (( ux / M / x ) ( vy / N / y ))

x y

f ( x, y )eMN

)//(2)//(2)())2/1/()2/1/((2 )1( NvyMuxjyxNvyMuxjyxjyNvyxMuxj eeee

always zero

1 or -1depending on the addition of x+y. I f (x+y) is even then 1, otherwise -1.

1 12

0 0

11

M Nx y j ( ux / M vy / N )

x y

f ( x, y )( ) e

MN

EE 583 Di i l I P i

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Image Enhancement in the Spatial Domain Example 10(PR4.6) -b): Prove the validity of the following 2 Equations.

),(),( 00)//(2 00 vvuuFeyxf NyvMxuj

)//(21

0

1

0

)//(2)//(2 0000 ),(1

]),([ NvyMuxjM

x

N

y

NyvMxujNyvMxujeeyxf

MNeyxf

)//(2

0000),(),(

NvyMuxjevuFyyxxf

Simi lar ly, it can be shown that : ),(]),([ 00)//(2 00 yyxxfevuF NvyMuxj

Thi s is the Translation property of the 2D Four ier transform:

When x0=u0=M /2 and x0=u0=N/2, then )2/,2/()1)(,( NvMuFyxf yx

vuvuFNyMxf )1)(,()2/,2/(

1

0

1

0

)/][/]([2 00),(1 M

x

N

y

NyvvMxuujeyxf

MN

),( 00 vvuuF

EE 583 Di it l I P i

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Image Enhancement in the Spatial Domain Example 11(PR4.9): Consider the images shown below. The image on the r ight i s

obtained by (a) mult iplying the image on the lef t by (-1)x+y

; (b) computing the DFT; (c)taking the complex conjugate of the transform; (d) computing the inverse DFT; and

(e) mul tiplying the real part of the resul t by (-1)x+y. Explain (mathematically) why the

image on the right appears as it does.

The complex conjugate simply changes j toj in the inverse tr ansform, so

the image on the r ight is given by:

1

0

1

0

)//(2),()*],([M

u

N

v

NvyMuxjevuFvuF

Which simply mir rors f(x,y) about the origin, thus producing the image on

the right

1

0

1

0

)/)(/)((2),(M

u

N

v

NyvMxujevuF

),( yxf

http://localhost/Hasan%20Demirel/Desktop/EE%20583/clear%0d%0ax=imread('DIP.jpg');%20x=double(x);%20y=x;%0d%0afx=fft2(x);%0d%0aifx=ifft2(conj(fx));%0d%0aimshow(ifx,%5b%5d);%0d%0a%0d%0a

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Image Enhancement in the Spatial Domain Example 12(PR4.14): Suppose that you form a low pass fi l ter that averages the

four immediate neighbors of a poin t (x,y), but excludes the point itself . (a) F ind the equivalent f il ter H (u,v) in the fr equency domain .

(b) Show that H (u,v) is a lowpass fi l ter.

a) The spatial average is given by:

Then, using the following property:

Where the H(u ,v) is the f il ter function. We get the foll owing transfer function:

)//(2

0000),(),(

NvyMuxjevuFyyxxf


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Image Enhancement in the Spatial Domain

a) The H(u,v) Fil ter function can be centered by:

Example 12(PR4.14): Suppose that you form a low pass fi l ter that averages the

four immediate neighbors of a poin t (x,y), but excludes the point itself . (a) F ind the equivalent f il ter H (u,v) in the fr equency domain .

(b) Show that H (u,v) is a lowpass fi l ter.

b)Consider one variable for convenience. As u ranges from 0 to M , the value of cos(2[u-M/2]/M)starts at -1, peaks at 1 when u = M /2 (the center of the fi lter ) and then decreases to -1 again when u = M .

Thus, we see that the ampli tude of the fi lter decreases as a function of distance from the ori gin of the

centered fi l ter, which is the character istic of a lowpass fi lter.

A simil ar argument is easil y carr ied out when consider ing both vari ables simul taneously.


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Image Enhancement in the Spatial Domain Example 8-PR4.19: Deri ve the frequency domain f il ter that corresponds to the

Laplacian operator in the spatial domain.

Consider theLaplacian mask given. Then,

Where

The H(u,v) F il ter function can be centered by:


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Image Enhancement in the Spatial Domain Example 8-PR4.22: The two four ier spectra shown are of the same image. The spectrum

of the left corresponds to the ori ginal image, and the spectrum on the ri ght was obtained after theimage was padded wi th zeros.

(a) Explain the diff erence of the overall contrast.

(b) Explain the signi fi cant increase in signal strength along the vertical and the hor izontal axes of

the spectrum shown on the right.

(a)Padding with zero increases the size but r educes the average

gray level of image. The average gray level of the padded image is

less than the original image. F(0,0) in the padded image is less than

F (0,0) of the original image.Al l the others away fr om the or igin are

less in the padded image than the original image. Th is produces a

nar rower range of values hence a lower contrast spectrum i n the

padded image.

(b)Padding with 0s introduces significant discontinu iti es at the bordersof the original image. Thi s

process in troduces strong hor izontal and vertical edgeswhere the image ends abruptly and then continues

with o values. These sharp transit ions correspond to the strength of the spectrum along the hori zontal and

verti cal axes.

EE 583: Digital Image Processing

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Image RestorationRestoration in the presence of Noise: Only-Spatial Filtering

Example 1-PR5.10: Given the two subimages below. The sub image on theleft i s the resul t of using ari thmetic mean f il ter of size 3x3. The other

subimage is the resul t of using the geometr ic mean f i l ter of the same size.

a) Why the subimage obtained with geometr ic f il ter ing is less blur red. Hint you

can start your analysis with 1-D step edge profi le of the image.

b) Explain why the black components on the ri ght image are thicker.

a) Lets consider the mathematical expressions of the ari thmetic and geometr ic

mean fi l ters:

xySts

tsgmn

yxf,

),(1

),( mn

Sts xy

tsgyxf

1

,

),(),(


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Example 1-PR5.10 :

a) I f we take a rough estimate of 1-D

Step function profi le before the fi lter ing:

xySts

tsg

mn

yxf,

),(1

),(

mn

Sts xy

tsgyxf

1

,

),(),(

6 6 6 6 0 0 6 6 6 6

Before fi lteri ng:

After fil tering:

6 6 6 4 2 2 4 6 6 6

Ar ithmetic mean fi ltering:

6 6 6 6 0 0 6 6 6 6

Before fi lteri ng:

After fi ltering:

6 6 6 0 0 0 0 6 6 6

Geometr ic mean fi l ter ing:


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Example 1-PR5.10 :

a) The resul ting 1-D profi les clearly indicates that the ari thmetic mean fi lter

produces smoother/blurred transition and

b) The geometri c fi lter ing increases the thicknessof the black components.

6 6 6 6 0 0 6 6 6 6

Before fi lteri ng:

After fi ltering:

6 6 6 4 2 2 4 6 6 6

Ar ithmetic mean fi ltering:

6 6 6 6 0 0 6 6 6 6

Before fi lteri ng:

After fi ltering:

6 6 6 0 0 0 0 6 6 6

Geometr ic mean fi l ter ing:

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EE 583 Lecture06