Connecting spatial and frequency domains for the quaternion Fourier transform

Transcription

1 Connecting spatial and frequency domains for the quaternion Fourier transform Ghent University (joint work with N. De Schepper, T. Ell, K. Rubrecht and S. Sangwine) MOIMA, Hannover, June, 2016

2 Direct formulas Quaternions qft Convolution

3 Quaternions qft Convolution Outline Direct formulas Quaternions qft Convolution

4 Quaternions qft Convolution The quaternion algebra H is defined over R with three imaginary units i, j and k satisfying ij = ji = k i 2 = j 2 = k 2 = 1. Every quaternion can be written as q = q 0 + q 1 i + q 2 j + q 3 k, q 0,..., q 3 R A pure quaternion q has q 0 = 0 and satisfies q 2 = (q q q 2 3).

5 Quaternions qft Convolution Quaternionic Fourier transform: Let µ, ν H be pure quaternions with µ 2 = ν 2 = 1. Define for h L 1 (R 2 ; H) the left qft as F µ,ν (h)(y 1, y 2 ) := (2π) 1 R 2 e µx 1y 1 e νx 2y 2 h(x 1, x 2 )dx 1 dx 2,

6 Quaternions qft Convolution History: µ = j and ν = k by Ell (1992) general µ, ν and application to image processing: [Sangwine, Ell] discrete versions [Pei et al.] T.A. Ell, Hypercomplex Spectral Transformations, Ph.D. Thesis. University of Minnesota, June S. J. Sangwine, Fourier transforms of colour images using quaternion, or hypercomplex, numbers. Electron. Lett. 32 (1996), S-C. Pei, J-J. Ding and J-H. Chang, Efficient implementation of quaternion Fourier transform, convolution, and correlation by 2-D complex FFT. IEEE Trans. Signal Process. 49 (2001), T.A. Ell and S.J. Sangwine, Hypercomplex Fourier Transforms of Color Images, IEEE Trans. Image Process. 16 (2007),

7 Quaternions qft Convolution Basic problem: Identify a color image with quaternion function h(x 1, x 2 ) = R(x 1, x 2 )i + G(x 1, x 2 )j + B(x 1, x 2 )k Applying a mask g(x 1, x 2 ) to h(x 1, x 2 ) amounts to computing g h Our aim: = use qft for this

8 Quaternions qft Convolution Convolution for the qft Ordinary convolution f g(x) := f (y)g(x y)dy, R 2 with x = (x 1, x 2 ) and y = (y 1, y 2 ) No convolution theorem for the qft: F µ,ν (f g) F µ,ν (f )F µ,ν (g) A formula is available, but complicated. See: Bujack R, Scheuermann G and Hitzer E, A general geometric Fourier transform convolution theorem. Adv. appl. Clifford alg. 23 (2013),

9 Quaternions qft Convolution Alternatively, based on an idea of Mustard, one may define convolution for the qft as follows which clearly satisfies f q g(x) := F ν, µ (F µ,ν (f )F µ,ν (g)) F µ,ν (f q g) = F µ,ν (f )F µ,ν (g). However, only useful if there is connection f q g f g D. Mustard, Fractional convolution. J. Austral. Math. Soc. Ser. B 40 (1998),

10 Quaternions qft Convolution Direct formula: There is a finite expansion: f q g = F ν, µ (F µ,ν (f )F µ,ν (g)) = i (a i f φ i ) (b i g ψ i ), a i, b i H with: f φ (x) := f (( 1) φ 1 x 1, ( 1) φ 2 x 2 ) with φ 1, φ 2 {0, 1} Note that the explicit formula depends on the geometric position of µ and ν

11 Outline Direct formulas Quaternions qft Convolution

12 Aim: find reverse expansion f g = i (c i f φ i ) q (d i g ψ i ) c i, d i H. where φ i, ψ i denote suitable argument-reflections. This is essentially a combinatorial problem. Consequence: F µ,ν (f g) = i F µ,ν (c i f φ i )F µ,ν (d i g ψ i ), c i, d i H.

13 The classic convolution theorem in md Consider f, g ordinary functions and F the ordinary FT F(f g(x))(u)

14 The classic convolution theorem in md Consider f, g ordinary functions and F the ordinary FT F(f g(x))(u) m = (2π) m 2 e ix ku k (f g)(x)dx definition of FT

15 The classic convolution theorem in md Consider f, g ordinary functions and F the ordinary FT F(f g(x))(u) = (2π) m 2 = (2π) m 2 m e ix ku k (f g)(x)dx definition of FT m e ix ku k f (x y)g(y)dydx Fubini R m

16 The classic convolution theorem in md Consider f, g ordinary functions and F the ordinary FT F(f g(x))(u) = (2π) m 2 = (2π) m 2 = (2π) m 2 R m m e ix ku k (f g)(x)dx definition of FT m e ix ku k f (x y)g(y)dydx Fubini R m m e ix ku k f (x y)g(y)dydx x z + y

17 The classic convolution theorem in md Consider f, g ordinary functions and F the ordinary FT F(f g(x))(u) = (2π) m 2 = (2π) m 2 = (2π) m 2 = (2π) m 2 R m R m m e ix ku k (f g)(x)dx definition of FT m e ix ku k f (x y)g(y)dydx Fubini R m m e ix ku k f (x y)g(y)dydx m e i(z k+y k )u k f (z)g(y)dzdy x z + y identifying FTs

18 The classic convolution theorem in md Consider f, g ordinary functions and F the ordinary FT F(f g(x))(u) = (2π) m 2 = (2π) m 2 = (2π) m 2 = (2π) m 2 = F(f )F(g) R m R m m e ix ku k (f g)(x)dx definition of FT m e ix ku k f (x y)g(y)dydx Fubini R m m e ix ku k f (x y)g(y)dydx m e i(z k+y k )u k f (z)g(y)dzdy x z + y identifying FTs

19 Extra steps for a general strategy: Problem is situated in rearrangement of last step. General strategy 1. Usual preparatory steps 2. Reordering of the exponentials m k=1 e i k (z k +y k )u k 3. Split of the function f (z) 4. Separating factors in y and z 5. Reduction in terms 6. Identification of the qfts

20 Roots of 1 For quaternions we denote i 1 = µ, i 2 = ν µ 2 = 1 µ = a 1 i + b 1 j + c 1 k, a b c 1 2 = 1 for µ H and a 1, b 1, c 1 R likewise ν = a 2 i + b 2 j + c 2 k [µ, ν] = 0 µ = ±ν {µ, ν} = a 1 a 2 b 1 b 2 c 1 c 2 R

21 Subdivision in three cases {µ, ν} = a 1 a 2 b 1 b 2 c 1 c 2 R It became clear early on that anticommutation (orthogonality) simplifies a lot. Distinguish 3 cases 1. there is just one root µ = ±ν 2. the roots µ and ν are anticommuting {µ, ν} = 0 3. the roots µ and ν are general linearly independent roots.

22 Let us consider the second case {µ, ν} = 0 After the preparatory steps we obtain: R 2 R 2 e µ(z 1+y 1 )u 1 e ν(z 2+y 2 )u 2 f (z)g(y)dzdy We want to swap e µy 1u 1 e νz 2u 2 e µy 1u 1 e νz 2u 2 = 1 ( (e νz 2 u 2 + e νz 2u 2 )e µy 1u 1 + (e νz 2u 2 e νz 2u 2 )e µy ) 1u 1 2

23 Conclusions from the example: Swapping increases the number of terms. It changes the sign in the exponentials. After substitution this sign change is captured by argument reflections f (φ 1,φ 2 ) (z) := f (( 1) φ 1 z 1, ( 1) φ 2 z 2 ) with φ 1, φ 2 {0, 1} e.g. f (0,1) (z 1, z 2 ) = f (z 1, z 2 ) It gives rise to a prefactor. Swapping with the function f introduces products of the roots: f = 1 2 (f + µf µ) (f µf µ) Further reduction by summing terms with equal signs in exponentials

24 The final result: Theorem Direct formulas Let f and g be quaternion functions on R 2, and F µ,ν the left qft with µ and ν anticommuting roots of 1. The classical convolution can then be expressed as 4f g = f q g + f q g (0,1) + f q g (1,0) + f q g (1,1) νf q νg + νf q νg (0,1) νf q νg (1,0) + νf q νg (1,1) µf (0,1) q µg µf (0,1) q µg (0,1) + µf (0,1) q µg (1,0) + µf (0,1) q µg (1,1) + νµf (0,1) q µνg νµf (0,1) q µνg (0,1) νµf (0,1) q µνg (1,0) + νµf (0,1) q µνg (1,1) For general a = {µ, ν} extra terms in a and a 2.

25 Formulas look complicated However, in practice often additional symmetry E.g. mask symmetric w.r.t. to x 1 : f (x 1, x 2 ) = f ( x 1, x 2 ) or f (1,0) = f This reduces number of terms drastically

26 Conclusion: Ordinary convolution f g(x) := f (y)g(x y)dy R 2 Difficult convolution theorem New convolution product f q g = (a i f φ i ) (b i g ψ i ) i Easy convolution theorem f g = i (c i f φ i ) q (d i g ψ i ) F µ,ν (f q g) = F µ,ν (f )F µ,ν (g) Bujack R, De Bie H, De Schepper N and Scheuermann G, Convolution products for hypercomplex Fourier transforms J. Math. Imaging Vision 48 (2014), De Bie H, De Schepper N, Ell T, Rubrecht K., Sangwine S., Connecting spatial and frequency domains for the quaternion Fourier transform Applied Mathematics and Computation 271 (2015),

27 Further results and outlook Direct formulas available for wide class of hypercomplex transforms much more complicated Applications in color image processing Design of filters, LTI systems, etc.