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DESCRIPTION
image transforms
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1
Image Transform
•Representation of the signal
•Basis Function
•Extraction of information from the signal
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Role of Transform
White Light? VIBGYOR!
TransformTime domain
Frequency domain
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Transform
Time domain
Frequency domain
Forward Transform
Inverse Transform
Basis Function
Finality of Co-efficients Orthogonality
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Basis Function
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N
0iiicx
x Input signal
Transform coefficient
Basis function
ici
i
NL
0iicx̂
ii ,xc
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Different Image Trasforms
Fourier TransformWalsh TransformHadamard TransformDiscrete Cosine TransformKL TransformWavelet Transform
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Fourier Transform
French Mathematical Physicist
Jean Baptiste Joseph Fourier
Idea in 1809
Flow of heat through objects
1-D Discrete Fourier Transform
]n[x ]k[X
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Forward TransformAnalysis
]k[X ]n[xSynthesis
Inverse Transform
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
2-D DFT
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][nx ),( nmf
][kX ],[ lkF
1-D signal 2-D signal
1-D Spectrum 2-D Spectrum
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
2-D DFT
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Image
Magnitude
Phase
Forward Transform
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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’
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Importance of PhaseOriginalimage OriginalImage
Cameraman image after phase reversal Lenna image after phase reversal
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)
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)
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
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
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
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
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
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
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
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Walsh basis for N=4
24
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
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
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NN
NNN2 HH
HHH
HADAMARD TRANSFORM
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112H
Jacques Salomon Hadamard
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
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