Download Basis Vectors Orthonormal Basis Vectors

Image Processing
Part I:
Vector Spaces and Basis Vectors
Math Review Towards
Fourier Transform
• Linear Spaces
• Change of Basis
• Cosines and Sines
• Complex Numbers
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Basis Vectors
Orthonormal Basis Vectors
• A given vector value is represented with
respect to a coordinate system.
• A coordinate system is defined by a set
of linearly independent vectors forming
the system basis.
• Any vector value is represented as a
linear sum of the basis vectors.
• If the basis vectors are mutually
orthogonal and are unit vectors, the
vectors form an orthonormal basis.
• Example:
The standard basis is orthonormal:
V1=(1 0 0 0 …)
V2=(0 1 0 0 …)
V3=(0 0 1 0 …)
….
a= 0.5 *v1+1*v2≡ (0.5 , 1)v
V2=(0 1)
v1
a
• v1, v2 are basis vectors
• The representation of a with
respect to this basis is (0.5,1)
(a1 ,a2)
a2
a
v2
a1
3
V1=(1 0)
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1
Change of Basis
u2
Signal (Image) Transforms
(a1 ,a2)v = (a3 ,a4)u
v2
av
1. Basis Functions.
u1
a4
a3
2. Method for finding the transform
coefficients given a signal.
v1
Given a vector av, represented in orthonormal
basis {vi} , what is the representation of av in
a different orthonormal basis {ui}?
3. Method for finding the signal given
the transform coefficients.
a u (i) = a v , u i
a v = ∑ a u(i) u i
i
where
c, b = c T b = ∑ c(i )b (i )
i
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The Orthonormal Standard Basis
The Orthonormal Hadamard Basis
Wave Number
0
Wave Number
0
N = 16
1
N = 16
2
1
3
2
4
3
5
4
6
5
7
6
8
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9
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10
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11
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Standard Basis Functions
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Hadamard Transform
Standard Basis
New Basis
Grayscale Image
Transformed Image
spatial Coordinate
Finding the transform coefficients
Signal:
X = [ 2 1 6 1]
standard
Hadamard Basis:
Transform Coordinate
T0 = [ 1 1 1 1 ] /2
T1 = [ 1 1 -1 -1 ] /2
T2 = [ 1 -1 -1 1 ] /2
Standard Basis:
T3 = [ 1 -1 1 -1 ] /2
[ 2 1 6 1 ]standard=
Hadamard Coefficients:
2[ 1 0 0 0 ] + 1[ 0 1 0 0 ] + 6[ 0 0 1 0 ] + 1[ 0 0 0 1 ]
a0 = <X,T0 > = < [ 2 1 6 1 ] , [ 1 1 1 1 ] > /2 = 5
a1 = <X,T1 > = < [ 2 1 6 1 ] , [ 1 1 -1 -1 ] > /2 = -2
Hadamard Transform:
a2 = <X,T2 > = < [ 2 1 6 1 ] , [ 1 -1 -1 1 ] > /2 = -2
[ 2 1 6 1 ]standard =
=
a3 = <X,T3 > = < [ 2 1 6 1 ] , [ 1 -1 1 -1 ] > /2 =
5 [ 1 1 1 1 ]/2
+ -2 [ 1 1 -1 -1 ] /2 +
Signal:
-2 [ 1 -1 -1 1 ] /2 + 3 [ 1 -1 1 -1 ] /2
≡
[ 5 -2 -2 3 ]
3
[ 2 1 6 1]
Standard
≡ [ 5 -2 -2 3 ]
Hadamard
Hadamard
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Transforms: Change of Basis –
2D Images
Reconstructing the Image from the transform
coefficients
Y = [ 5 -2 -2 3 ]
Hadamard
Y Coordinate
Hadamard Basis:
T0 = [ 1 1 1 1 ] /2
T1 = [ 1 1 -1 -1 ] /2
T2 = [ 1 -1 -1 1 ] /2
T3 = [ 1 -1 1 -1 ] /2
Grayscale
Image
X Coordinate
V Coordinates
Transform:
Transformed
Image
U Coordinates
Reconstruction;
Σ Y(i),Ti =
The coefficients are arranged in a 2D array.
5 [ 1 1 1 1 ]/2
+ -2 [ 1 1 -1 -1 ] /2 +
-2 [ 1 -1 -1 1 ] /2 + 3 [ 1 -1 1 -1 ] /2
=
[ 2 1 6 1 ]standard
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Transforms: Change of Basis
Hadamard Basis Functions
Standard Basis:
[
2
1
6
1
] [ ] [ ] [ ] [ ]
1 0
= 2
0 0
+ 1
0 1
+ 6
0 0
0 0
0 0
+ 1
1 0
0 1
Hadamard Transform:
[
2
1
6
1
] [ ]
≡
size = 8x8
1 1
= 5
1 1 /2
[
5
3
-2
-2
+ -2
]
[ ]+ [ ]
1 1
-1 -1 /2
-2
1 -1
-1 1 /2
+3
[ ]
1 -1
1 -1 /2
Hadamard
White = +1 Black = -1
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Finding the transform coefficients
X =
Signal:
[
2
1
6
1
]
Standard Basis:
standard
[ ] [ ]
New Basis:
[ ]/2
[ ]/2
T11 =
1 1
1 1
1 1
T21 =
-1 -1
[
[ ]/2
[ ]/2
T12 =
T22 =
1 -1
1 -1
2
1
6
1
]
coefficients
1 -1
1 0
0 1
0 0
0 0
[ ] [ ]
0 0
0 0
1 0
0 1
Standard
-1 1
Basis Elements
Signal:
X = a11T11 +a12T12 + a21T21 + a22T22
Hadamard Transform:
New Coefficients:
a11 =
a12 =
a21 =
a22 =
<X,T11 > =
<X,T21 > =
<X,T22 > =
<X,T12 > =
X ≡
sum(sum(X.*T11)) =
sum(sum(X.*T21)) =
sum(sum(X.*T22)) =
sum(sum(X.*T12)) =
[
5
3
-2
-2
]
5
-2
-2
3
5
3
-2
-2
[
]
coefficients
[ ] [ ]
Hadamard
1 1
1 -1
1 1 /2
1 -1 /2
[ ] [ ]
1 1
1 -1
-1 -1 /2
-1 1 /2
Hadamard
new
Basis Elements
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Continuous images/signals f(x):
Part II:
Sines and Cosines
1) The number of Basis Elements Bi is ∞.
f (x) =
∫ aiBi(x) di
i
2) The dot product:
∫ f(x)Bi(x) dx
<f(x),Bi(x)> =
x
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Wavelength and Frequency of
Sine/Cosin
sin(x )
1
1
x
sin(θ
θ)
θ
cos(θ
θ)
2π
1
sin(2x )
1
sin(x )
x
1
2π/2
x
sin(kx )
2π
1
x
– The wavelength of sin(x) is 2π .
2π / k
– The frequency is 1/(2π) .
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– Changing Amplitude:
– Define K=2πω
A sin(2πωx )
A
sin(2πωx )
1
x
x
1
ω
– Changing Phase:
A sin(2πωx + φ )
– The wavelength of sin(2πωx) is 1/ω .
– The frequency is ω .
x
ϕ
2πω
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Sine vs Cosine
Sine vs Cosine
sin(x) = cos(x) with a phase shift of π/2.
sin(x) + cos(x) = sin(x) scaled by 2 with
a phase shift of π/4.
}
3 sin(kx) + 4 cos(kx) = sin(kx) with amplitude
scaled by 5 and phase shift of 0.3π
π/2
sin(x) + cos(x) = ?
3 sin(kx)
4 cos(kx)
5 sin(kx + 0.3π)
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Part III:
Complex Numbers
Combining Sine and Cosine
If we add a Sine wave to a Cosine wave with
the same frequency we get a scaled and shifted
(Co-) Sine wave with the same frequency:
a sin (kx ) + b cos (kx ) = R sin (kx + φ )
b
where R = a 2 + b 2 and φ = tan −1  
a
(prove it!)
What is the result if a=0?
What is the result if b=0?
tan(φ)
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Complex Numbers
Imaginary
• Conjugate of Z is Z*:
(a,b)
b
– Cartesian rep. (a + ib)* = a - ib
The Complex Plane
θ
a
Imaginary
• Two kind of representations for a point
(a,b) in the complex plane
R
where i 2 = −1
θ
– The Polar representation:
Z = Re
iθ
a + ib = Reiθ
b
– The Cartesian representation:
Z = a + ib
(Reiθ)* = Re-iθ
– Polar rep.
Real
-θ
(Complex exponential)
a
Real
R
-b
• Conversions:
– Polar to Cartesian:
Re = R cos(θ ) + iR sin(θ )
– Cartesian to Polar
a + ib = a 2 + b 2 ei tan
a − ib = Re − iθ
iθ
−1
(b / a )
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Algebraic operations:
The (Co-) Sinusoid
eiθ = cos(θ) + i sin(θ)
• addition/subtraction:
(a + ib) + (c + id) = (a + c) + i(b + d)
• The (Co-)Sinusoid as complex exponential:
• multiplication:
cos(x ) = Real(eix )
(a + ib )(c + id ) = (ac − bd ) + i(bc + ad)
sin(x ) = Imag(eix )
Aeiα Beiβ = ABei (α + β )
• inner Product:
Or
e ix + e −ix
2
ix
e − e −ix
sin (x ) =
2i
( a + ib), (c + id ) = (a + ib )* (c + id ) = (a − ib )( c + id )
cos(x ) =
<Aeiα,Beiβ> = Ae-iαBeiβ = ABei(β-α)
• norm:
a + ib = (a + ib ) (a + ib ) = a 2 + b 2
∗
2
Reiθ = (Reiθ ) Reiθ = Re − iθ Reiθ = R 2
2
∗
• What about generalization?
S sin(kx) + C cos(kx) = ?
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Scaling and phase shifting can be
represented as a multiplication with Z = Re iθ
cos(kx)
Rcos(kx+θ)
sin(kx)
Rsin(kx+θ)
eikx
Rei(kx+θ)
We saw that :
S sin(kx) + C cos(kx) = R sin(kx + θ)
where R =
2
S +C
2
−1
and θ = tan
C
(S)
R sin (kx + θ) = Imag( R e iθ e ikx )
= Imag( Z e ikx )
= Reiθ eikx
R sin (kx + θ ) = 21i ( R eiθ eikx − R e −iθ e −ikx )
= Zeikx
= 21i ( Z eikx − Z * e −ikx )
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The 1D Continuous
Fourier Transform
S sin(kx) + C cos(kx) = R sin(kx + θ)
The Continuous Fourier Transform finds the
F(ω) given the (cont.) signal f(x):
R sin (kx + θ) = Imag(R e e )
iθ
ikx
= Imag( Z eikx )
F (ω ) =
∫
f(x) e − i 2 πω x dx
x
R sin (kx + θ ) =
1
2i
( R eiθ eikx − R e −iθ e −ikx )
Bω(x)=ei2πωx is a complex wave function for
each ω .
= 21i ( Z eikx − Z * e −ikx )
The Inverse Continuous Fourier Transform
composes a signal f(x) given F(ω):
f(x) =
∫ F (ω ) e
i 2 πω x
dω
ω
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