some problems with FFT

Accelerated Computing CUDA CUDA Programming and Performance
1

Hi,
I need help with Fourier transform. I have success then i work with one array, but I need transform several arrays parallel.
Help me please. I’m sorry, my English not good :unsure:

Thanks!

2

configure your fft plan with batch mode, and then run it… :rolleyes:

3

configure your fft plan with batch mode, and then run it… :rolleyes:

4

Thanks senosy, it’s work External Image

5

Thanks senosy, it’s work External Image

6

Now I have next problem: Inverse of forward cufft not equal to original function,:( my code:

[codebox]// Allocate device memory for signal

cufftComplex* d_signal;

cudaMalloc((void**)&d_signal, sizeof(cufftComplex) * SIZE);

cufftComplex* d_signalout;

cudaMalloc((void**)&d_signalout, sizeof(cufftComplex) * SIZE);

// Copy host memory to device

cudaMemcpy(d_signal, indata, sizeof(cufftComplex) * SIZE, cudaMemcpyHostToDevice);

cufftHandle plan;

/* Create a 1D FFT plan. */

cufftPlan1d(&plan, NX, CUFFT_C2C, BATCH);

/* Use the CUFFT plan to transform the signal in place. */

cufftExecC2C(plan, d_signal, d_signal, CUFFT_FORWARD);

/* Inverse transform the signal in place. */

cufftExecC2C(plan, d_signal, d_signalout, CUFFT_INVERSE);

/* Note:

//(1) Divide by number of elements in data set to get back original data

//(2) Identical pointers to input and output arrays implies in-place

//transformation

*/

cutilSafeCall(cudaMemcpy(outdata, d_signalout, sizeof(cufftComplex) * SIZE,

                         cudaMemcpyDeviceToHost));[/codebox]

Thanks!

7

Now I have next problem: Inverse of forward cufft not equal to original function,:( my code:

[codebox]// Allocate device memory for signal

cufftComplex* d_signal;

cudaMalloc((void**)&d_signal, sizeof(cufftComplex) * SIZE);

cufftComplex* d_signalout;

cudaMalloc((void**)&d_signalout, sizeof(cufftComplex) * SIZE);

// Copy host memory to device

cudaMemcpy(d_signal, indata, sizeof(cufftComplex) * SIZE, cudaMemcpyHostToDevice);

cufftHandle plan;

/* Create a 1D FFT plan. */

cufftPlan1d(&plan, NX, CUFFT_C2C, BATCH);

/* Use the CUFFT plan to transform the signal in place. */

cufftExecC2C(plan, d_signal, d_signal, CUFFT_FORWARD);

/* Inverse transform the signal in place. */

cufftExecC2C(plan, d_signal, d_signalout, CUFFT_INVERSE);

/* Note:

//(1) Divide by number of elements in data set to get back original data

//(2) Identical pointers to input and output arrays implies in-place

//transformation

*/

cutilSafeCall(cudaMemcpy(outdata, d_signalout, sizeof(cufftComplex) * SIZE,

                         cudaMemcpyDeviceToHost));[/codebox]

Thanks!