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ippsFFT related - need number of operations

summa1234
Beginner
‎01-15-2005 02:21 AM
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Hi,
I could not find implementation details about ippsFFT. Basically I want to know how many numerical operations should I substitute for the ippsFFT.

I am aware that for a N element long array the FFT has order of Nlog2(N) operations and if it is radix 2 implementation it goes upto 5N log2(N). Please confirm.

So similarly for a N element long array, how many operations should I substitute for "ippsFFT"?
My guess based on some measurements is 2N log2(N). Can anyone please confirm this for me?
Any help in this regards will be highly appreciated.
Thanks,
~SA
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Vladimir_Dudnik
Employee
‎01-18-2005 04:53 PM
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Hi,
5Nlog2(N) is estimation for usual implementation of RADIX2 algorithm. With using SSE and RADIX4 algorithm we get closer to 2Nlog2(N)
Regards,
Vladimir
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