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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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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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