Download FFT: Math Preliminaries

Image Processing
Math Review Towards
Fourier Transform
• Linear Spaces
• Change of Basis
• Cosines and Sines
• Complex Numbers
1
Part I:
Vector Spaces and Basis Vectors
2
Basis Vectors
• A given vector value is represented with
respect to a coordinate system.
• The coordinate system is arbitrarily
chosen.
• 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.
a= 0.5 *v1+1*v2 (0.5 , 1)v
v1
a
• v1, v2 are basis vectors
• The representation of a with
respect to this basis is (0.5,1)
v2
3
Orthonormal 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 …)
….
V2=(0 1)
a
V1=(1 0)
4
Change of Basis
u2
v2
u1
a
v1
• Question: Given a vector av, represented in an
orthonormal basis {vi} , what is the
representation of av in a different orthonormal
basis {ui}?
• Answer:
a u i   a v , u i
a v   a u i  u i
i
where
c, b  cT b   ci  bi 
i
5
Change of Basis: Matrix Form
u2
v2
u1
a
v1
• Defining the new basis as a collection of
columns in a matrix form
U  u1
u2  un 
au  U av
av  U au
T
6
Signal (image) Transforms
1. Basis Functions.
2. Method for finding the transform
coefficients given a signal.
3. Method for finding the signal given
the transform coefficients.
7
The Orthonormal Standard Basis - 1D
Wave Number
0
N = 16
1
2
3
4
5
6
7
8
9
Standard Basis Functions - 1D
8
The Orthonormal Hadamard Basis – 1D
Wave Number
0
N = 16
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
9
Hadamard Transform
Standard Basis
New Basis
Grayscale Image
Transformed Image
Transform Coordinate
spatial Coordinate
Standard Basis:
[ 2 1 6 1 ]standard=
2[ 1 0 0 0 ] + 1[ 0 1 0 0 ] + 6[ 0 0 1 0 ] + 1[ 0 0 0 1 ]
Hadamard Transform:
[ 2 1 6 1 ]standard =
=
5 [ 1 1 1 1 ]/2
+ -2 [ 1 1 -1 -1 ] /2 +
2 [ -1 1 1 -1 ] /2 + -3 [ -1 1 -1 1 ] /2
 [ 5 -2 2 -3 ]
Hadamard
10
Finding the transform coefficients
Signal:
X = [ 2 1 6 1]
standard
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
Hadamard Coefficients:
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
a2 = <X,T2 > = < [ 2 1 6 1 ] , [ -1 1 1 -1 ] > /2 = 2
a3 = <X,T3 > = < [ 2 1 6 1 ] , [ -1 1 -1 1 ] > /2 = -3
Signal:
[ 2 1 6 1]
Standard
 [ 5 -2 2 -3 ]
Hadamard
11
Grayscale
Image
X Coordinate
V Coordinates
Y Coordinate
Transforms: Change of Basis - 2D
Transformed
Image
U Coordinates
The coefficients are arranged in a 2D array.
12
Hadamard Basis Functions
size = 8x8
Black = +1 White = -1
13
Transforms: Change of Basis - 2D
Standard Basis:
[
2
1
6
1
1 0
0 0
0 1
0 0
+ 6
0 0
1 0
0 0
0 1
] [ ] [ ] [ ] [ ]
= 2
+ 1
+ 1
Hadamard Transform:
[
2
1
6
1
] [ ]
= 5

[
1 1
1 1
-1 1
-1 1
-2
+2
-3
1 1 /2
-1 -1 /2
1 -1 /2
-1 1 /2
5
-2
2
-3
[ ] [ ]
]
[ ]
Hadamard
14
Finding the transform coefficients
X =
Signal:
[
2
1
6
1
]
standard
New Basis:
1 1
T11 =
1 1
1 1
T12 =
-1 -1
-1 1
T21 =
1 -1
-1 1
T22 =
-1 1
[ ]/2
[ ]/2
[ ]/2
[ ]/2
Signal:
X = a11T11 +a12T12 + a21T21 + a22T22
New Coefficients:
a11 =
a12 =
a21 =
a22 =
<X,T11 > =
<X,T12 > =
<X,T21 > =
<X,T22 > =
X 
sum(sum(X.*T11)) =
sum(sum(X.*T12)) =
sum(sum(X.*T21)) =
sum(sum(X.*T22)) =
[
5
2
-2
-3
]
5
-2
2
-3
new
15
Standard Basis:
[
2
1
6
1
1 0
0 0
0 1
0 0
0 0
1 0
0 0
0 1
[ ] [ ]
]
coefficients
[ ] [ ]
Standard
Basis Elements
Hadamard Transform:
[
5
-2
2
-3
]
coefficients
1 1
1 1 /2
1 1
-1 -1 /2
-1 1
1 -1 /2
-1 1
-1 1 /2
[ ] [ ]
Hadamard
[ ] [ ]
Hadamard
Basis Elements
16
Part II:
Sines and Cosines
17
Wavelength and Frequency of
Sine/Cosin
1
sin()

cos()
1
sinx 
1
x
2
– The wavelength of sin(x) is 2 .
– The frequency is 1/(2) .
18
sinx 
1
x
2
sin2x 
1
x
2/2
sinkx 
1
x
2 / k
19
– Define K=2
sin2πωx 
1
x
1
ω
– The wavelength of sin(2x) is 1/ .
– The frequency is  .
20
– Changing Amplitude:
A sin2πωx 
A
x
– Changing Phase:
A sin2πωx  φ

2
x
21
Sine vs Cosine
sin(x) = cos(x) with a phase shift of /2.
}
/2
sin(x) + cos(x) = ?
22
Sine vs Cosine
sin(x) + cos(x) = sin(x) scaled by 2 with
a phase shift of /4.
3 sin(kx) + 4 cos(kx) = sin(kx) with amplitude
scaled by 5 and phase shift of 0.3
3 sin(kx)
4 cos(kx)
5 sin(kx + 0.3)
23
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   
b
where R  a  b and   tan  
a
2
2
• What is the result if a=0?
• What is the result if b=0?
1
(prove it!)
tan()
24
Part III:
Complex Numbers
25
Complex Numbers
Imaginary
(a,b)
b
The Complex Plane

Real
a
• Two kind of representations for a point
(a,b) in the complex plane
– The Cartesian representation:
Z  a  ib
where i 2  1
– The Polar representation:
Z  Reiθ
(Complex exponential)
• Conversions:
– Polar to Cartesian:
Reiθ  R cosθ  iR sinθ
– Cartesian to Polar
a  ib  a  b e
2
2
i tan 1  b / a 
26
• Conjugate of Z is Z*:
– Cartesian rep.
a  ib 
– Polar rep.
Re 

iθ 
 a  ib
 Re  iθ
Imaginary
a  ib  Reiθ
b
R

-
a
Real
R
-b
a  ib  Reiθ
27
Algebraic operations:
• addition/subtraction:
(a  ib)  (c  id)  (a  c)  i(b  d)
• multiplication:
a  ib c  id   ac  bd  ibc  ad
Aeiα Beiβ  ABei α β 
• inner Product:
(a  ib), (c  id )  (a  ib)* (c  id )  (a  ib)(c  id )
• norm:
a  ib  a  ib  a  ib   a 2  b2

2
Re
iθ 2
 Re
 Re
iθ 
iθ
 Reiθ Reiθ  R 2
28
Number Multiplication
Im
A*B


i
Re
i
Ae Be  ABe
i    
29
The (Co-) Sinusoid
e  cos  i sin
i
• The (Co-)Sinusoid as complex exponential:
e ix  e ix
cosx  
2
e ix  e ix
sin x  
2i
Or
cosx   Reale

sinx   Imag e 
ix
ix
• What about generalization?
S sin(kx) + C cos(kx) = ?
30
We saw that :
S sin(kx) + C cos(kx) = R sin(kx + )
where
1  C 
2
2




R
tan 
S C and
S 
Using Complex exponentials:
Scaling and phase shifting can be
represented as a multiplication with Z  Re i
R sin kx     Imag ( R e e )
i
ikx
 Imag (Z eikx )
and also
R sin kx     21i ( R ei eikx  R ei eikx )
 (Z e  Z e
1
2i
ikx
*
ikx
)
31
THE END
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