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Computational cost of FFT with IPP

alzhykhar
Beginner
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Hi!

I wonder is there a definitive way (thumb rules :)) to tell how
computational time (Tc) for FFT using IPP depends on block size.
The Good Old Rule says

Tc ~ k*N*log(N)

for N-point FFT, where N is power of 2, and k being a
proportionality constant that depends on the particular FFT
implementation. Here one does believe that the time to compute
FFT is (proportional to) the number of operations (mul's + add's)
only.

I made some benchmarking to figure out how Intel FFT routine
ippsFFTFwd_RToPerm_32f() behaves in dependence on FFT order
and got some strange (wrong?) results. For FFT order in region
from 6 (block_size=64) up to 15 (block_size=32768) there is not
any dependency at all: the time is the same!


Any comments would be appreciated.

Thanks,
Albert


PS. My environment:
- Pentium 4 (2.40GHz, 512Mb RAM)
- Linux-2.6.11 (Mandrake-10.1)
- Intel IPP v4.1
- gcc-3.4.1
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2 Replies
Vladimir_Dudnik
Employee
298 Views
Hi,
please look at this post, might be it is the info you are looking for
Regards,
Vladimir
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borix
Beginner
298 Views
Hi Albert, what do you mean "the same"? total time or per butterfly? total time for different fft orders could be the same only in the case the function does nothing and returns with a bad status, please checkthe status codes, then let us know. Thank you
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