# Image Processing

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```					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, v2 are basis vectors
v1      a         • 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
a        u1

v1
• Question: Given a vector av, represented in an
orthonormal basis {vi} , what is the
representation of av in a different orthonormal
basis {ui}?

5
Change of Basis: Matrix Form

u2
v2
a        u1

v1
• Defining the new basis as a collection of
columns in a matrix form

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

Standard Basis                               New Basis
Grayscale Image                          Transformed Image

spatial Coordinate                     Transform 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 ]

[ 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

T0 = [ 1 1 1 1 ] /2
T1 = [ 1 1 -1 -1 ] /2
T2 = [ -1 1 1 -1 ] /2
T3 = [ -1 1 -1 1 ] /2

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
Transforms: Change of Basis - 2D

V Coordinates
Y Coordinate

Grayscale                      Transformed
Image                             Image

X Coordinate                    U Coordinates

The coefficients are arranged in a 2D array.

12

Black = +1 White = -1
size = 8x8

13
Transforms: Change of Basis - 2D

Standard Basis:
2   1            1 0               0 1              0 0           0 0
[   6   1   ] [ ] [ ] [ ] [ ]
= 2
0 0
+ 1
0 0
+ 6
1 0
+ 1
0 1

2   1
[   6   1   ] [ ]
= 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


14
Finding the transform coefficients

2    1
Signal:          X =
[   6    1   ]   standard

New Basis:
1 1                        1 1
T11 =[ ]/2
1 1
T12 =   [ ]/2
-1 -1
-1 1
T21 =[ ]/2
1 -1
T22 =
-1 1
[ ]/2
-1 1

Signal:
X = a11T11 +a12T12 + a21T21 + a22T22

New Coefficients:

a11 =   <X,T11 > =   sum(sum(X.*T11)) =           5
a12 =   <X,T12 > =   sum(sum(X.*T12)) =          -2
a21 =   <X,T21 > =   sum(sum(X.*T21)) =           2
a22 =   <X,T22 > =   sum(sum(X.*T22)) =          -3

5   -2
X 
[   2   -3   ]    new

15
Standard Basis:

1 0        0 1

2   1
[ ] [ ]
0 0        0 0

[   6   1    ]               0 0         0 0
coefficients
[ ] [ ]
1 0         0 1
Standard

Basis Elements

1 1

5   -2
[ ] [ ]
1 1 /2
1 1
-1 -1 /2
[   2   -3   ]              -1 1       -1 1
[ ] [ ]

coefficients                  1 -1 /2   -1 1 /2

Basis Elements
16
Part II:
Sines and Cosines

17
Wavelength and Frequency of
Sine/Cosin
1
sin()


1
cos()

1

x

– The wavelength of sin(x) is 2 .
– The frequency is 1/(2) .

18
1

x

1

x

2/2

1

x

19
– Define K=2

1

x

– The wavelength of sin(2x) is 1/ .
– The frequency is  .

20
– Changing Amplitude:

A

x

– Changing Phase:

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          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:

(prove it!)
• What is the result if a=0?
• What is the result if b=0?

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:

– The Polar representation:
(Complex exponential)

• Conversions:
– Polar to Cartesian:

– Cartesian to Polar
26
• Conjugate of Z is Z*:

– Cartesian rep.

– Polar rep.

Imaginary

b

R


Real
a
-

R
-b

27
Algebraic operations:

• multiplication:

• inner Product:

• norm:

28
Number Multiplication

Im

A*B



Re

29
The (Co-) Sinusoid

• The (Co-)Sinusoid as complex exponential:

Or

S sin(kx) + C cos(kx) = ?

30
We saw that :

S sin(kx) + C cos(kx) = R sin(kx + )

C 
where   R = S 2 + C 2 and  = tan-1 
S
 

Using Complex exponentials:

Scaling and phase shifting can be
represented as a multiplication with

and also

31
THE END

32

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