By J. Arndt

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**Example text**

This of course is not much of a trick at all. 50) That is, copy the real data in reversed order to get the transform with the other sign. This technique does not involve an extra pass and should be virtually for free. Obviously for the complex-to-real FTs (C2RFT) one has to negate the imaginary part before the transform. 9: Real valued Fourier transforms 23 All procedures presented here use the following scheme for the real part of the transformed sequence c in the output array a[ ]: = = = ... 51) cn/2 For the imaginary part of the result there are two schemes: Scheme 1 (‘parallel ordering’) is a[n/2 + 1] a[n/2 + 2] a[n/2 + 3] = = = ...

62) we can, for a given sequence, find a transform that is a ‘square root’ of the FT: Compute u+ , u− , v+ and v− . 14: Eigenvectors of the Fourier transform * 35 Then F 0 [a] is the identity and F 1 [a] is the usual FT. The transform F 1/2 [a] is a transform so that F 1/2 F 1/2 [a] = F [a], that is, a ‘square root’ of the FT. The transform F 1/2 [a] is not unique as the expressions ±11/2 and ±i1/2 aren’t. The eigenvectors of the Hartley Transform are u+ u− := a + H [a] := a − H [a] The eigenvalues are ±1, one has H [u+ ] = +1 · u+ and H [u− ] = −1 · u− .

10: Multidimensional FTs 29 For a m-dimensional array ax (where x = (x1 , x2 , x3 , . . , xm ) and xi ∈ 0, 1, 2, . . , Si ) the m-dimensional Fourier transform ck (where k = (k1 , k2 , k3 , . . , km ) and ki ∈ 0, 1, 2, . . k ... x1 =0 x2 =0 where z = eσ 2 π i/n , n = S 1 S 2 . . k where S = (S1 − 1, S2 − 1, . . 56b) x=0 The inverse transform is again the complex conjugate transform. 57) y=0 which shows that the 2-dimensional FT can be obtained by first applying 1-dimensional transforms on the rows and then applying 1-dimensional transforms on the columns.