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using ippsDFTInit_C with using ippsFFTInit

Hello.

it is confusing the difference between DFTInit and FFTInit

for example:

{

    int *sizeSpec, *sizeInit, *sizeBuff;

    IppDFTSpec_C_64fc *pSpecDfr, *pSpecFft;

    Ipp8u *pSpecFft, *pInit, *pBuff;

#ifdef USE_DFT

    ippsDFTGetSize_C_64fc(16, IPP_FFT_NODIV_BY_ANY, ippAlgHintNone, &sizeSpec, &sizeInit, &sizeBuff);

#else

    ippsFFTGetSize_C_64fc(4, IPP_FFT_NODIV_BY_ANY, ippAlgHintNone, &sizeSpec, &sizeInit, &sizeBuff);

#endif

    pSpec = ippsMalloc_8u(sizeSpec); pInit = ippsMalloc_8u(sizeInit); pBuff = ippsMalloc_8u(sizeBuff);

#ifdef USE_DFT

    ippsDFTInit_C_64fc(10, IPP_FFT_NODIV_BY_ANY, ippAlgHintNone, pSpec , pInit );

#else

    ippsFFTInit_C_64fc(&pSpecFft, 4, IPP_FFT_NODIV_BY_ANY, ippAlgHintNone, pSpec, pInit ); // here we provide additional argument

#endif

    // using fft or dft converting functions

}

Did I making right the initialization for FFt and DFT?

Why the interface is different for DFT and FFT? I mean why for FFT I need to provide additional argument for Init function? Why it can be done in the same way for DFT and FFT? It is confusing.

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1 Reply
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Beginner
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I found link http://software.intel.com/en-us/articles/how-to-use-intel-ipp-s-1d-fourier-transform-functions that answer on my first question.

But the second question remains relevant. Why interface for DFT and FFT initialization functions is different? It is only confuse. May be it can be changed in next versions of IPP?

Thank you,

Itzhak

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