Fourier Transform 2D

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Kristopher Anderson
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1 Image Processing - Lesson 8 Fourier Transform 2D Discrete Fourier Transform - 2D Continues Fourier Transform - 2D Fourier Properties Convolution Theorem Eamples = + + +

2 The 2D Discrete Fourier Transform For an image f(,y) =..N-, y=..m-, there are two-indices basis functions B u.v (,y): B u,v (,y) = MN e u v y 2πi + N M u=..n-, M=..M- The inner product of 2 functions (in 2D) is defined similarly to the D case : F ( u, v ) = f (, y ),B (, y ) u,v = = N = M y = f * (, y ) B (, y ) u,v 2

3 V u=-2, v=2 u=-, v=2 u=, v=2 u=, v=2 u=2, v=2 u=-2, v= u=-, v= u=, v= u=, v= u=2, v= U u=-2, v= u=-, v= u=, v= u=, v= u=2, v= u=-2, v=- u=-, v=- u=, v=- u=, v=- u=2, v=- u=-2, v=-2 u=-, v=-2 u=, v=-2 u=, v=-2 u=2, v=-2 3

4 The 2D Discrete Fourier Transform Image f(,y) =,,...,N- y=,,...,m- The 2D Discrete Fourier Transform (DFT) is defined as: F(u,v) N M = = y f(,ye ) 2πi ( u /N+ v y /M) u =,, 2,..., N- v =,, 2,..., M- Matlab: F=fft2(f); The Inverse Discrete Fourier Transform (IDFT) is defined as: N f(,y) = MN M u= v Fu,v ( ) e 2πi ( u /N + v y /M) y =,, 2,..., N- =,, 2,..., M- Matlab: f=ifft2(f); 4

5 Fourier Transform - Image u v 5

6 Fourier Image - Eample Original Fourier Image = F(u,v) Shifted Fourier Image Shifted Log Fourier Image = log(+ F(u,v) ) 6

7 Visualizing the Fourier Transform Image using Matlab Routines F(u,v) is a Fourier transform of f(,y) and it has comple entries. F = fft2(f); In order to display the Fourier Spectrum F(u,v) Reduce dynamic range of F(u,v) by displaying the log: D = log(+abs(f)); Cyclically rotate the image so that F(,) is in the center: Eample: D = fftshift(d); F(u) = Display in Range([..]): log(+ F(u) ) = log(+ F(u) )/.462 = fftshift(log(+ F(u) ) =

8 Visualizing the Fourier Image - Eample Original F(u,v) log( + F(u,v) ) fftshift(log( + F(u,v) )) 8

9 Properties of The Fourier Transform Linearity: F ~ [ α f] = αf ~ [ f] Distributive (additivity): [ f + f ] = F ~ [ f ] [ ] F ~ + F ~ 2 f2 DC (average): Parseval ( ) f(,y) F, = y e (, y) F( u, v) f = y 2 u v 2 9

10 ~ Distributive: F{ f + g} = F{ f} + F{ g} ~ ~ f() F() g() G() f+g F()+G()

11 Parseval s Theorem One more characteristic: ΣΣ f(,y) 2 = ΣΣ F(u,v) 2 y u v v y a u

12 Properties of The Fourier Transform Symmetric: If f(,y) is real then, * ( ) = F ( u, v) thus F( u,v ) = F( u, v) F u,v Cyclic: if f(,y) is discrete ( u,v ) = F( u+ N,v ) = F( u,v + M) = F( u+ N,v M) F + 2

13 Cyclic and Symmetry of the Fourier Transform - D Eample F(u) cycle -N/2 N/2 N- N 3

14 Separability Properties: Cont. ( ) = f(,y) F u,v = y f y vy N e u vy 2πi + N M u N 2πi 2πi 2πi N (,y) e e = F(,v) e = u Thus, performing a 2D Fourier Transform is equivalent to performing 2 D transforms:. Perform D transform on EACH column of image f(,y), obtaining F(,v). 2. Perform D transform on EACH row of F(,v), obtaining F(u,v). Higher Dimensions: Fourier in any dimension can be performed by applying D transform on each dimension. 4

15 Eample - Separability 2D Image Fourier Spectrum 5

16 Image Transformations Translation: u vy ~ 2π i + [ ( )] ( ) N M F f, y y = F u, v e The Fourier Spectrum remains unchanged under translation: F ( u, v ) = F ( u, v ) e 2πi u N + vy M Rotation: Rotation of f(,y) by θ rotation of F(u,v) by θ Scaling: ~ F ab [ ( )] f a, b y = F, u a v b 6

17 Eample - Translation D Image real(f(u)) imag(f(u)) F(u) Translated Differences:

18 Change of Scale- D: if ~ F a { f ( ) } = F( ) then F{ f ( a) } = F ~ a f() F() f() F() f() F() 8

19 Change of Scale f() F() f(2).5 F(/2) f(/2) 2 F(2) 9

20 Eample - Rotation 2D Image 2D Image - Rotated Fourier Spectrum Fourier Spectrum 2

21 Fourier Transform Eamples Image Domain Frequency Domain 2

22 Fourier Transform Eamples Image Domain Frequency Domain 22

23 Fourier Transform Eamples Image Domain Frequency Domain 23

24 Fourier Transform Eamples Image Domain Frequency Domain 24

25 Why do we need representation in the frequency domain? Problem in Frequency Space Relatively easy solution Solution in Frequency Space Fourier Transform Inverse Fourier Transform Original Problem Difficult solution Solution of Original Problem 25

26 The Convolution Theorem g = f * h g = f h implies implies G = F H G = F * H Convolution in one domain is multiplication in the other and vice versa 26

27 The Convolution Theorem ~ { ( ) ( )} ~ { ( )} ~ F f g = F f F{ g( ) } and likewise ~ { ( ) ( )} ~ { ( )} ~ F f g = F f F{ g( ) } 27

28 Convolution Theorem - Eample f() g() f * g () F(u) G(u) F(u)G(u) F - [F(u) G(u)]

29 Convolution Theorem - 2D Eample f(,y) g(,y) f * g (,y) F(u,v) G(u,v) F(u,v) G(u,v) F - [F(u,v) G(u,v)] 29

30 Eample: What is the Fourier Transform of: h() - f() f() h() = * F() H() =. F() F() = 3

31 Eample: What is the Fourier Transform of the Dirac Function? δ ( ) = if = otherwise Proof : Consider any function f() f ( ) * δ ( ) = f ( ) F ( u) F[ δ ( )] = F( u) * = F [ δ()] = 3

32 What is the Fourier Transform of the Dirac Function? Answer II: f() F(u) What is the Fourier Transform of an image with constant gray value? 32

33 Sampling The Image Sampling a function f() with impulse train of cycle T produces replicas in the frequency domain with cycle /T: f() F^ F() f() * F() F^ T /T f() F^ F() -/2T /2T 33

34 Sampling the Transform Sampling a function F() with impulse train of cycle S produces replicas in the image domain with cycle /S: f() ^ F - F() * ^ F - /S S f() ^ F - F() 34

35 Sampling Image & Transform Sampling both f() with impulse train of cycle T and F() with impulse train of cycle S: f() F() F^ * F^ T /T F^ -/2T /2T S = NT S /T 35

36 f() Undersampling the Image F^ F() f() F^ F() * f() T /T F^ F() -/2T /2T f() F() F^ T /T f() F^ F() -/2T /2T 36

37 Critical Sampling If the maimal frequency of f() is ma, it is clear from the above replicas that ma should be smaller that /2T. Alternatively: T > 2 ma Nyquist Theorem: If the maimal frequency of f() is ma the sampling rate should be larger than 2 ma in order to fully reconstruct f() from its samples. If the sampling rate is smaller than 2 ma overlapping replicas produce aliasing. F() -/2T /2T 37

38 Optimal Interpolation It is possible to fully reconstruct f() from its samples: f() ^ F - F() * f() ^ F - -/2T /2T F() -/2T /2T f() ^ F - F() -/2T /2T 38

39 Optimal Interpolation- Eample 39

40 Optimal Interpolation- Eample Image FFT Duplicate Zero FFT - 4

41 Image Domain Frequency Domain 4

42 Fast Fourier Transform - FFT F(u) = N N = f()e 2πiu N O(n 2 ) operations u =,, 2,..., N- F(u) = N N/ 2 = f(2)e 2πiu2 2πiu(2 N/ 2 N N + N = f(2 + )e + ) = 2 N / 2 N / 2 even = f(2)e 2πiu N / 2 + e 2πiu N N / 2 odd N / 2 f(2 + )e = 2πiu N / 2 Fourier Transform of of N/2 even points Fourier Transform of of N/2 odd points All sampling points Even sampling points Odd sampling points The Fourier transform of N inputs, can be performed as 2 Fourier Transforms of N/2 inputs each + one comple multiplication and addition for each value i.e. O(N). Note, that only N/2 different transform values are obtained for the N/2 point transforms. 42

43 F N (u) = 2 N / 2 N / 2 = f(2)e 2πiu N / 2 + e 2πiu N N / 2 N / 2 f(2 + ) e = 2πiu N / 2 F N (u) 2πiu e N o = FN / 2(u) + e FN / 2(u) 2 2πiu' 2πi(u+ N / 2) 2πiu For u' = u+n/2 : N N N πi N e = e = e e = e obtain : 2πiu 2πiu e N o FN (u) = FN / 2(u) + e FN / 2(u) 2 2πiu N e N o FN (u + ) = F (u) e FN / 2(u) 2 N / 2 2 For u =,, 2,..., N/2- Thus: only one comple multiplication is needed for two terms. F N / F N / F N / 4 e o Calculating 2(u) and 2(u) is done recursively by e o calculating F (u) and (u). N / 4 43

44 F() F() F(2) F(3) F(4) F(5) F(6) F(7) F() F(2) F(4) F(6) F() F(3) F(5) F(7) F() F(4) F(2) F(6) F() F(5) F(3) F(7) F() F() F(2) F(3) F(4) F(5) F(6) F(7) 2-point transform 4-point transform FFT FFT : O(nlog(n)) operations FFT of NN Image: O(n 2 log(n)) operations 44

45 Frequency Enhancement Image FFT Transform Filtered New Image FFT - Filtered Image Demo 45

&& && F( u)! "{ f (x)} = f (x)e # j 2$ u x. f (x)! " #1. F(u,v) = f (x, y) e. f (x, y) = 2D Fourier Transform. Fourier Transform - review.

&& && F( u)! { f (x)} = f (x)e # j 2$ u x. f (x)! #1. F(u,v) = f (x, y) e. f (x, y) = 2D Fourier Transform. Fourier Transform - review. 2D Fourier Transfor 2-D DFT & Properties 2D Fourier Transfor 1 Fourier Transfor - review 1-D: 2-D: F( u)! "{ f (x)} = f (x)e # j 2$ u x % & #% dx f (x)! " #1 { F(u) } = F(u)e j 2$ u x du F(u,v) = f (x,