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Radix 2 FFT Complexity is N Log N

Putting together the length $ N$ DFT from the $ N/2$ length-$ 2$ DFTs in a radix-2 FFT, the only multiplies needed are those used to combine two small DFTs to make a DFT twice as long, as in Eq.$ \,$ (A.1). Since there are approximately $ N$ (complex) multiplies needed for each stage of the DIT decomposition, and only $ \lg N$ stages of DIT (where $ \lg N$ denotes the log-base-2 of $ N$ ), we see that the total number of multiplies for a length $ N$ DFT is reduced from $ {\cal O}(N^2)$ to $ {\cal O}(N\lg N)$ , where $ {\cal O}(x)$ means ``on the order of $ x$ ''. More precisely, a complexity of $ {\cal O}(N\lg N)$ means that given any implementation of a length-$ N$ radix-2 FFT, there exist a constant $ C$ and integer $ M$ such that the computational complexity $ {\cal C}(N)$ satisfies

$\displaystyle {\cal C}(N) \leq C N \lg N
$

for all $ N>M$ . In summary, the complexity of the radix-2 FFT is said to be ``N log N'', or $ {\cal O}(N\lg N)$ .


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``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8.
Copyright © 2014-04-06 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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