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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(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:


   • 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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