FFT and IFFT with FFTW

Question

I am trying to do ifft and fft on a floating point array. However the outcome is the same for both. Do you have any idea? Why are results the same, even though I use FFTW_FORWARD for one and FFTW_BACKWARD for another one?

  int N=16;
  fftwf_complex in[N], out[N];
  fftwf_plan p1, q;

  /* prepare a cosine wave */
  for (i = 0; i < N; i++) {
    in[i][0] = cos(3 * 2*M_PI*i/N);
    in[i][1] = 0;
  }

  /* forward Fourier transform, save the result in 'out' */
  p1 = fftwf_plan_dft_1d(N, in, out, FFTW_FORWARD, FFTW_ESTIMATE);
  fftwf_execute(p1);
  for (i = 0; i < N; i++)
  cout << out[i][0] << endl;
  fftwf_destroy_plan(p1);

  printf("\nInverse transform:\n");
  q = fftwf_plan_dft_1d(N, in, out, FFTW_BACKWARD, FFTW_ESTIMATE);
  fftwf_execute(q);
  for (i = 0; i < N; i++)
  cout << out[i][0] << endl;
  fftwf_destroy_plan(q);

Answer 1

You're only displaying the real parts of the output bins, and ignoring the imaginary components. It just so happens that the real parts match, but the imaginary components are different (they are actually complex conjugates):

#include <iostream>
#include <cmath>
#include "fftw3.h"

using namespace std;

int main()
{
    int N=16;
    fftwf_complex in[N], out[N];
    fftwf_plan p1, q;

    for (int i = 0; i < N; i++) {
        in[i][0] = cos(3 * 2*M_PI*i/N);
        in[i][1] = 0;
    }

    p1 = fftwf_plan_dft_1d(N, in, out, FFTW_FORWARD, FFTW_ESTIMATE);
    fftwf_execute(p1);
    for (int i = 0; i < N; i++)
        cout << out[i][0] << " + j" << out[i][1] << endl; // <<<
    fftwf_destroy_plan(p1);

    printf("\nInverse transform:\n");
    q = fftwf_plan_dft_1d(N, in, out, FFTW_BACKWARD, FFTW_ESTIMATE);
    fftwf_execute(q);
    for (int i = 0; i < N; i++)
        cout << out[i][0] << " + j" << out[i][1] << endl; // <<<
    fftwf_destroy_plan(q);

    return 0;
}

Compile and run:

$ g++ -Wall fftwf.cpp -lfftw3f && ./a.out

3.67394e-16 + j0
1.19209e-07 + j7.34788e-16
-3.67394e-16 + j0
8 + j-7.34788e-16
3.67394e-16 + j0
2.38419e-07 + j7.34788e-16
-3.67394e-16 + j0
1.19209e-07 + j-7.34788e-16
3.67394e-16 + j0
1.19209e-07 + j7.34788e-16
-3.67394e-16 + j0
2.38419e-07 + j-7.34788e-16
3.67394e-16 + j0
8 + j7.34788e-16
-3.67394e-16 + j0
1.19209e-07 + j-7.34788e-16

Inverse transform:
3.67394e-16 + j0
1.19209e-07 + j-7.34788e-16
-3.67394e-16 + j0
8 + j7.34788e-16
3.67394e-16 + j0
2.38419e-07 + j-7.34788e-16
-3.67394e-16 + j0
1.19209e-07 + j7.34788e-16
3.67394e-16 + j0
1.19209e-07 + j-7.34788e-16
-3.67394e-16 + j0
2.38419e-07 + j7.34788e-16
3.67394e-16 + j0
8 + j-7.34788e-16
-3.67394e-16 + j0
1.19209e-07 + j7.34788e-16

It's interesting to note that the FFT and IFFT are mathematically almost identical. They are often both implemented as a single routine, with a flag indicating direction (forward or inverse). Typically this flag just affects the sign of the imaginary part of the twiddle factors.