Differences between FFTW and CUFFT Output

Question

In char, which I posted below, I compare the results of IFFT execution in FFTW and CUFFT.

What are the possible reasons for this? Are there really so many rounding errors?

Here is the relevant code snippet:

cufftHandle plan; cufftComplex *d_data; cufftComplex *h_data; cudaMalloc((void**)&d_data, sizeof(cufftComplex)*W); complex<float> *temp = (complex<float>*)fftwf_malloc(sizeof(fftwf_complex) * W); h_data = (cufftComplex *)malloc(sizeof(cufftComplex)*W); memset(h_data, 0, W*sizeof(cufftComplex)); /* Create a 1D FFT plan. */ cufftPlan1d(&plan, W, CUFFT_C2C, 1); if (!reader->getData(rowBuff, row)) return 0; // copy from read buffer to our FFT input buffer memcpy(indata, rowBuff, fCols * sizeof(complex<float>)); for(int c = 0; c < W; c++) h_data[c] = make_cuComplex(indata[c].real(), indata[c].imag()); cutilSafeCall(cudaMemcpy(d_data, h_data, W* sizeof(cufftComplex), cudaMemcpyHostToDevice)); cufftExecC2C(plan, d_data, d_data, CUFFT_INVERSE); cutilSafeCall(cudaMemcpy(h_data, d_data,W * sizeof(cufftComplex), cudaMemcpyDeviceToHost)); for(int c = 0; c < W; c++) temp[c] =(cuCrealf(h_data[c]), cuCimagf(h_data[c])); //execute ifft plan on "indata" fftwf_execute(ifft); ... //dump out abs() values of the first 50 temp and outdata values. Had to convert h_data back to a normal complex 

ifft was defined as follows:

 ifft = fftwf_plan_dft_1d(freqCols, reinterpret_cast<fftwf_complex*>(indata), reinterpret_cast<fftwf_complex*>(outdata), FFTW_BACKWARD, FFTW_ESTIMATE); 

and to generate the graph, I unloaded h_data and outdata after fftw_execute W - this is the line width of the image processed by me.

Anything obvious?

Answer 1

So, it looks like CUFFT returns the real and imaginary parts, and FFTW only the real ones. The cuCabsf () function, which comes through the complex CUFFT library, forces it to give me a multiple of sqrt (2) when I have both parts of the complex

Aside - I could never get exactly the corresponding results in the intermediate stages between FFTW and CUFFT. If you are performing both IFFT and FFT, you should get something close.