1991
DOI: 10.1109/78.91199
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A better FFT bit-reversal algorithm without tables

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Cited by 20 publications
(3 citation statements)
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“…This permutation will be referred further in this paper as Fourier Permutation and is sometimes named the Buterfly Permutation [1], or Bit Reversal Permutation [2]- [9] due to the fact that the reversed bit representation of the permuted index of the initial 0-based index (see the following example) is implicitly sorted in ascending order. The Fourier Permutation of the 1-based index [1,2,3,4,5,6,7,8] is [1,5,3,7,2,6,4,8]. The stages leading to this permutation through the explicit calls in (1) are the following: [1,2,3,4,5,6,7,8][1,3,5,7,2,4,6,8][1,5,3,7,2,6,4,8][1,5,3,7,2,6,…”
Section: Introductionmentioning
confidence: 99%
See 1 more Smart Citation
“…This permutation will be referred further in this paper as Fourier Permutation and is sometimes named the Buterfly Permutation [1], or Bit Reversal Permutation [2]- [9] due to the fact that the reversed bit representation of the permuted index of the initial 0-based index (see the following example) is implicitly sorted in ascending order. The Fourier Permutation of the 1-based index [1,2,3,4,5,6,7,8] is [1,5,3,7,2,6,4,8]. The stages leading to this permutation through the explicit calls in (1) are the following: [1,2,3,4,5,6,7,8][1,3,5,7,2,4,6,8][1,5,3,7,2,6,4,8][1,5,3,7,2,6,…”
Section: Introductionmentioning
confidence: 99%
“…The Fourier Permutation of the 1-based index [1,2,3,4,5,6,7,8] is [1,5,3,7,2,6,4,8]. The stages leading to this permutation through the explicit calls in (1) are the following: [1,2,3,4,5,6,7,8][1,3,5,7,2,4,6,8][1,5,3,7,2,6,4,8][1,5,3,7,2,6,4,8], where a left to right reading shows the permutations operated from the first to the last (the innermost) call.…”
Section: Introductionmentioning
confidence: 99%
“…Splitting the summation, applying (58) to π n (j) and π n (j +1), then adjusting missing terms for j = k/2 − 1 and j = k − 1, we obtainV (k, a, b) = k 23) is applied twice. Next, multiplying out terms and using property(23), the above expression simplifies toV (k, a, b) = 2k 1 (π n−1 (b/2)−1)−a k − (π n−1 (a/2)+1)−b k − 2π n−1 (π n−1 (a/2)+1)−b kwhich is equivalent to(37) with a = a, b = b, a = 2π n−1 (π n−1 (a/2)+1)−b, b = 2π n−1 (π n−1 (b/2)−1)−a, and e = 1. PROOF OF LEMMA 14 Assume a, b are both even.…”
mentioning
confidence: 99%