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I know the efficiency for an $N$ point 2-radix FFT is $N\log_2(N)$ but assuming $k\leq N$, what if you were looking for the efficiency of calculating $k$ positions of the FFT? Would the efficiency be $k\log_2(N)$?

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No.

The FFT is pretty much an "all or nothing" algorithm. There is almost no efficiency gain in reducing the number of desired outputs.

For $k \ll N$ the Goertzel algorithm is typically the better choice. See https://en.wikipedia.org/wiki/Goertzel_algorithm. The break even point is quite small. Maybe $k = 4\cdot log_2(N)$ or thereabouts.

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  • $\begingroup$ Thank you! I think that was the algorithm I was looking for. $\endgroup$ – digger1235 Mar 19 '18 at 13:20

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