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

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

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

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)

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:

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

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

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

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

Hadamard Transform 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 ] 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

Finding the transform coefficients X = [ 2 1 6 1 ] standard Signal: 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

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.

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

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

Finding the transform coefficients 2 1 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 5 -2 2 -3 ] [ X  new

[ ] [ ] [ ] [ ] ] [ [ ] [ ] Standard Basis: Hadamard Transform: 1 0 1 0 0 0 ] 0 1 0 0 0 1 [ 2 1 6 1 ] coefficients Standard Basis Elements Hadamard Transform: 1 1 [ ] [ ] 1 1 -1 -1 5 -2 2 -3 ] [ /2 /2 -1 1 1 -1 [ ] -1 1 [ ] Hadamard coefficients /2 /2 Hadamard Basis Elements

Part II: Sines and Cosines

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

1 x 1 x 2/2 1 x

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  .

Changing Amplitude: A x Changing Phase: x

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

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)

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()

Part III: Complex Numbers

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

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

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

Number Multiplication Im A*B B A   Re

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

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 ]

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