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Number of complex calculations in FFT and inverse FFT
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[QUOTE="DrClaude, post: 4988550, member: 461323"] Note that the FFT algorithm scales as ##O(N \log_2 N)##, not that the number of operations is exactly ##N \log_2 N## (I don't know if this is relevant to your problem, because I don't know how your teacher presented the material). [B][/B] Why? What is the computer actually calculating? You probably only learned about an algorithm that works for ##2^N## data points. Therefore an FFT on length 5 doesn't make any sense. [/QUOTE]
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Number of complex calculations in FFT and inverse FFT
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