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By Arndt J.

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The eigenvectors are pairwise perpendicular. 107) we can, for a given sequence, find a transform that is the ‘square root’ of the FT: Simply compute u+ , u− , v+ , v− . 108) F 0 [a] is the identity, F 1 [a] is the (usual) FT, F 1/2 [a] (which is not unique) is a transform so that F 1/2 F 1/2 [a] = F [a], that is, a ‘square root’ of the FT. The eigenvectors of the Hartley Transform are u+ := a + H [a] (with eigenvalue +1) and u+ := a − H [a] (with eigenvalue −1). 2) x=0 where negative indices τ − x must be understood as n + τ − x, it is a cyclic convolution.

This means that one chooses a shape of the matrix so that the rows are as big as possible subject to the constraint that they have to fit into main memory, which in turn means there are R = α rows, leading to an optimal seek count of K = 2 α + 4 α2 . If one seek takes 10 milliseconds then one has for α = 16 (probably quite a big FFT) a total of K · 10 = 1056 · 10 milliseconds or approximately 10 seconds. With a RAM workspace of 64 Megabytes4 the CPU time alone might be in the order of several minutes.

Apply a (length C) FFT on each column (transposed row). 3. Multiply each matrix element (index r, c) by exp(±2 π i r c/n). 4. Apply a (length R) FFT on each row (transposed column). e. g. in unit strides). In radix 2 (or 2n ) algorithms one even has skips of powers of 2, which is particularly bad on computer systems that use direct mapped cache memory: One piece of cache memory is responsible for caching addresses that lie apart by some power of 2. TBD: move cache discussion to appendix With an ‘usual’ FFT algorithm one gets 100% cache misses and therefore a memory performance that corresponds to the access time of the main memory, which is very long compared to the clock of modern CPUs.

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Algorithms for programmers ideas and source code by Arndt J.


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