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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