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Math Review Towards Fourier Transform

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1 Math Review Towards Fourier Transform
Image Processing Math Review Towards Fourier Transform Linear Spaces Change of Basis Cosines and Sines Complex Numbers

2 Vector Spaces and Basis Vectors
Part I: Vector Spaces and Basis Vectors

3 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 The representation of a with respect to this basis is (0.5,1) a v1 v2

4 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=( …) V2=( …) V3=( …) …. V2=(0 1) a V1=(1 0)

5 Change of Basis v2 v1 a u2 u1 Question: Given a vector av, represented in an orthonormal basis {vi} , what is the representation of av in a different orthonormal basis {ui}? Answer:

6 Change of Basis: Matrix Form
v2 v1 a u2 u1 Defining the new basis as a collection of columns in a matrix form

7 Signal (image) Transforms
Basis Functions. Method for finding the transform coefficients given a signal. Method for finding the signal given the transform coefficients.

8 Standard Basis Functions - 1D
The Orthonormal Standard Basis - 1D Wave Number 1 2 3 4 5 6 7 8 9 N = 16 Standard Basis Functions - 1D

9 The Orthonormal Hadamard Basis – 1D
Wave Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 N = 16

10 Hadamard Transform Standard Basis New Basis Grayscale Image
Transformed Image spatial Coordinate Transform Coordinate Standard Basis: [ ]standard= 2 [ ] + 1 [ ] + 6 [ ] + 1 [ ] Hadamard Transform: [ ]standard = = [ ]/ [ ] /2 + 2 [ ] / [ ] /2  [ ] Hadamard

11 Finding the transform coefficients
X = [ ] standard Signal: Hadamard Basis: T0 = [ ] /2 T1 = [ ] /2 T2 = [ ] /2 T3 = [ ] /2 Hadamard Coefficients: a0 = <X,T0 > = < [ ] , [ ] > /2 = 5 a1 = <X,T1 > = < [ ] , [ ] > /2 = -2 a2 = <X,T2 > = < [ ] , [ ] > /2 = 2 a3 = <X,T3 > = < [ ] , [ ] > /2 = -3 Signal: [ ] Standard  [ ] Hadamard

12 Transforms: Change of Basis - 2D
Grayscale Image Transformed Image V Coordinates Y Coordinate X Coordinate U Coordinates The coefficients are arranged in a 2D array.

13 Hadamard Basis Functions
Black = +1 White = -1 size = 8x8

14 Transforms: Change of Basis - 2D
Standard Basis: 1 0 0 0 [ ] 0 1 + 1 + 6 = 2 Hadamard Transform: 1 1 [ ] -1 -1 -1 1 1 -1 -3 +2 -2 = 5 /2 [ ] Hadamard

15 Finding the transform coefficients
6 1 [ ] X = Signal: standard New Basis: [ ]/2 1 1 -1 -1 [ ]/2 1 1 T11 = T12 = [ -1 1 1 -1 ]/2 [ ]/2 -1 1 T21 = T22 = 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 ] [ X  new

16 [ ] [ ] [ ] [ ] ] [ [ ] [ ] Standard Basis: Hadamard Transform: 1 0
1 0 0 0 ] 0 1 0 0 0 1 [ 6 1 ] coefficients Standard Basis Elements Hadamard Transform: 1 1 [ ] [ ] 1 1 -1 -1 ] [ /2 /2 -1 1 1 -1 [ ] -1 1 [ ] Hadamard coefficients /2 /2 Hadamard Basis Elements

17 Part II: Sines and Cosines

18 Wavelength and Frequency of Sine/Cosin
1 sin() 1 cos() 1 x The wavelength of sin(x) is 2 . The frequency is 1/(2) .

19 1 x 1 x 2/2 1 x

20 The wavelength of sin(2x) is 1/ . The frequency is  .
Define K=2 1 x The wavelength of sin(2x) is 1/ . The frequency is  .

21 Changing Amplitude: A x Changing Phase: x

22 } Sine vs Cosine sin(x) = cos(x) with a phase shift of p/2.

23 sin(x) + cos(x) = sin(x) scaled by with a phase shift of p/4.
Sine vs Cosine sin(x) + cos(x) = sin(x) scaled by with a phase shift of p/4. 3 sin(kx) + 4 cos(kx) = sin(kx) with amplitude scaled by 5 and phase shift of 0.3p 3 sin(kx) 4 cos(kx) 5 sin(kx + 0.3p)

24 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: What is the result if a=0? What is the result if b=0? (prove it!) tan()

25 Part III: Complex Numbers

26 Complex Numbers Two kind of representations for a point
Imaginary (a,b) b R 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

27 Conjugate of Z is Z*: Cartesian rep. Polar rep. Imaginary b R  Real a
- R -b

28 Algebraic operations:
addition/subtraction: multiplication: inner Product: norm:

29 Number Multiplication
Im A*B B A Re

30 The (Co-) Sinusoid The (Co-)Sinusoid as complex exponential: Or
What about generalization? S sin(kx) + C cos(kx) = ?

31 S C We saw that : S sin(kx) + C cos(kx) = R sin(kx + )
Using Complex exponentials: Scaling and phase shifting can be represented as a multiplication with and also æ C ö where R = S 2 + C 2 and q = tan - 1 ç ç è S ø Example 6sin(kx) + 4cos(kx) = 1/2 [(4-6i) eikx + (4+6i) e-ikx ]

32 T H E E N D


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