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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}? • Answer: 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 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 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 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 Hadamard Basis Functions 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 Hadamard Transform: 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 [ 2 -3 ] Hadamard 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 Hadamard Transform: 1 1 5 -2 [ ] [ ] 1 1 /2 1 1 -1 -1 /2 [ 2 -3 ] -1 1 -1 1 [ ] [ ] Hadamard coefficients 1 -1 /2 -1 1 /2 Hadamard 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(2x) 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: • addition/subtraction: • multiplication: • inner Product: • norm: 28 Number Multiplication Im A*B Re 29 The (Co-) Sinusoid • The (Co-)Sinusoid as complex exponential: Or • What about generalization? 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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posted: | 3/24/2011 |

language: | English |

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