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1 Image Transform Representation of the signal Basis Function Extraction of information from the signal

Image Transform [Compatibility Mode]

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Page 1: Image Transform [Compatibility Mode]

1

Image Transform

•Representation of the signal

•Basis Function

•Extraction of information from the signal

Page 2: Image Transform [Compatibility Mode]

2

Role of Transform

White Light? VIBGYOR!

TransformTime domain

Frequency domain

Page 3: Image Transform [Compatibility Mode]

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Transform

Time domain

Frequency domain

Forward Transform

Inverse Transform

Page 4: Image Transform [Compatibility Mode]

Basis Function

Finality of Co-efficients Orthogonality

4

Page 5: Image Transform [Compatibility Mode]

Basis Function

5

N

0iiicx

x Input signal

Transform coefficient

Basis function

ici

i

NL

0iicx̂

ii ,xc

Page 6: Image Transform [Compatibility Mode]

6

Different Image Trasforms

Fourier TransformWalsh TransformHadamard TransformDiscrete Cosine TransformKL TransformWavelet Transform

Page 7: Image Transform [Compatibility Mode]

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Fourier Transform

French Mathematical Physicist

Jean Baptiste Joseph Fourier

Idea in 1809

Flow of heat through objects

Page 8: Image Transform [Compatibility Mode]

1-D Discrete Fourier Transform

]n[x ]k[X

8

Forward TransformAnalysis

]k[X ]n[xSynthesis

Inverse Transform

Page 9: Image Transform [Compatibility Mode]

1-D DFT (cont..)

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1N

0n

knN2j

1Nto0k,e]n[x]k[X

x[n] represents the signal

X[k] represents the spectrum

Page 10: Image Transform [Compatibility Mode]

2-D DFT

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][nx ),( nmf

][kX ],[ lkF

1-D signal 2-D signal

1-D Spectrum 2-D Spectrum

Page 11: Image Transform [Compatibility Mode]

2-D DFT

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1

0

N

n][kX

1

0

N

m

1

0

N

n),( nmf mk

Nj

e2

nlN

je

2),( lkF

][nxkn

Nj

e2

Page 12: Image Transform [Compatibility Mode]

2-D DFT

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Image

Magnitude

Phase

Forward Transform

Page 13: Image Transform [Compatibility Mode]

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Importance of Phase

Read Image ‘A’ Read Image ‘B’

Fourier Transform Fourier Transform

Magnitude Phase Magnitude Phase

Inverse Fourier Transform Inverse Fourier Transform

Reconstructed Image ‘A1’ Reconstructed Image ‘B1’

Page 14: Image Transform [Compatibility Mode]

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Importance of PhaseOriginalimage OriginalImage

Cameraman image after phase reversal Lenna image after phase reversal

Page 15: Image Transform [Compatibility Mode]

Application of Fourier Transform in Image Processing

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Image Filtering

h(m,n)f(m,n) g(m,n)

f(m,n) Input image

h(m,n) Impulse response of the filter

g(m,n) Output image

g(m,n) = f(m,n) * h(m,n)

Page 16: Image Transform [Compatibility Mode]

Application of Fourier Transform in Image Processing

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f(m,n) DFT F(k,l) H(k,l) IDFTg(m,n)

F(k,l) Spectrum of input signal

H(k,l) Filter functiong(m,n) Output image

Convolution in time domain = Multiplication in frequency domain

G(k,l)

Page 17: Image Transform [Compatibility Mode]

Walsh Transform

Nlog m 2

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1m

0i

)k(b)n(b i1mi)1(N1)k,n

sh basis function:

n time index

k frequency index

N Order

Walsh

Page 18: Image Transform [Compatibility Mode]

Walsh basis for N=4

Nlog m 2

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24log m 2

ep 1: To compute ‘m’

ep 2: Substituting N=4 and m=2 in g(n,k)

1m

0i

)k(b)n(b i1mi)1(N1)k,n(g

12

0

)(12)()1(41),(

i

kibnibkng

1

)()( 1)1(41),( kbnb iikng

Page 19: Image Transform [Compatibility Mode]

Walsh basis for N=4

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ep 3: Substituting n=0 and k=0 in g(n,k)

1

0

)()( 1)1(41),(

i

kbnb iikng

1

0

)0(1)0()1(41)0,0(

iibibg

Page 20: Image Transform [Compatibility Mode]

Walsh basis for N=4

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)0()0()0()0( 0110 )1()1(41)0,0( bbbbg

p 4: Substituting i = 0 and i = 1 in g(0,0)

)0(0b Zeroth bit position in the binary value of zero

)0(1b First bit position in the binary value of zero

0 0 0

Zeroth bit positionFirst bit position

binary representation

Page 21: Image Transform [Compatibility Mode]

Walsh basis for N=4

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Step 5: Substitute and in g(0,0)0)0(0 b 0)0(1 b

41)1()1(

41)0,0( 00 g

To compute g(2,1)

1

0

)()( 1)1(41),(

i

kbnb iikng

1

0

)1()2( 1)1(41)1,2(

i

bb iig

Page 22: Image Transform [Compatibility Mode]

Walsh basis for N=4

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Substituting i=0 and i=1 in the expression of g(2,1)

1

0

)1(1)2()1(41)1,2(

iibibg

)1()2()1()2( 0110 )1()1(41)1,2( bbbbg

)2(0b Zeroth bit position in the binary value of 2

)1(1b First bit position in the binary value of 1

)2(1b First bit position in the binary value of 2

h b h b l f

Page 23: Image Transform [Compatibility Mode]

Walsh basis for N=4

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0

12

3

0

01

ecimal number Binary representation

Zeroth bit position

First bit position

)2(0b = 0)1(1b = 0)2(1b = 1)1(0b = 1

)1(0b)2(1b)1(1b)2(0b )1()1(41)1,2(g

41)1()1(

41)1,2(g 10

0

0

1

11

Page 24: Image Transform [Compatibility Mode]

Walsh basis for N=4

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41

41

41

41

41

41

41

41

41

41

41

41

41

41

41

41

)k,n(

No sign change

One sign change

Three sign changes

Two sign changes

Page 25: Image Transform [Compatibility Mode]

Walsh Basis for N=4

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41

41

41

41

41

41

41

41

41

41

41

41

41

41

41

41

)k,n(

n=1 and k=2

n=1 0 1

k=2 1 0

k=2 0 1

original

reversal

0 1

0 1No. of overlap is odd

No. of overlap of one is odd then sign is negative

Page 26: Image Transform [Compatibility Mode]

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NN

NNN2 HH

HHH

HADAMARD TRANSFORM

11

112H

Jacques Salomon Hadamard

Page 27: Image Transform [Compatibility Mode]

ISCRETE COSINE TRANSFORM

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Nln

Nkmnmflklk

N

m

N

n 212cos

212cos),()()(],

1

0

1

0

02

01

)(kif

N

kifNk

02

01

)(lif

N

lifNl