c++ - 使用 FFTW 库 C++ 计算 FFT 和 IFFT

Question

我正在尝试计算 FFT，然后计算 IFFT，只是为了尝试是否可以返回相同的信号，但我不确定如何完成它。这就是我进行 FFT 的方式:

    plan = fftw_plan_r2r_1d(blockSize, datas, out, FFTW_R2HC, FFTW_ESTIMATE);
    fftw_execute(plan);

Answer 1

这是一个例子。它做了两件事。首先，它准备一个输入数组 in[N] 作为余弦波，其频率为 3，幅度为 1.0，并对其进行傅里叶变换。因此，在输出中，您应该在 out[3] 处看到一个峰值，在 out[N-3] 处看到另一个峰值。由于余弦波的幅度为 1.0，因此在 out[3] 和 out[N-3] 处得到 N/2。

其次，它将数组 out[N] 逆傅立叶变换回 in2[N]。在适当的规范化之后，您可以看到 in2[N] 与 in[N] 相同。

#include <stdlib.h>
#include <math.h>
#include <fftw3.h>
#define N 16
int main(void) {
  fftw_complex in[N], out[N], in2[N]; /* double [2] */
  fftw_plan p, q;
  int i;

  /* 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' */
  p = fftw_plan_dft_1d(N, in, out, FFTW_FORWARD, FFTW_ESTIMATE);
  fftw_execute(p);
  for (i = 0; i < N; i++)
    printf("freq: %3d %+9.5f %+9.5f I\n", i, out[i][0], out[i][1]);
  fftw_destroy_plan(p);

  /* backward Fourier transform, save the result in 'in2' */
  printf("\nInverse transform:\n");
  q = fftw_plan_dft_1d(N, out, in2, FFTW_BACKWARD, FFTW_ESTIMATE);
  fftw_execute(q);
  /* normalize */
  for (i = 0; i < N; i++) {
    in2[i][0] *= 1./N;
    in2[i][1] *= 1./N;
  }
  for (i = 0; i < N; i++)
    printf("recover: %3d %+9.5f %+9.5f I vs. %+9.5f %+9.5f I\n",
        i, in[i][0], in[i][1], in2[i][0], in2[i][1]);
  fftw_destroy_plan(q);

  fftw_cleanup();
  return 0;
}

这是输出:

freq:   0  -0.00000  +0.00000 I
freq:   1  +0.00000  +0.00000 I
freq:   2  -0.00000  +0.00000 I
freq:   3  +8.00000  -0.00000 I
freq:   4  +0.00000  +0.00000 I
freq:   5  -0.00000  +0.00000 I
freq:   6  +0.00000  -0.00000 I
freq:   7  -0.00000  +0.00000 I
freq:   8  +0.00000  +0.00000 I
freq:   9  -0.00000  -0.00000 I
freq:  10  +0.00000  +0.00000 I
freq:  11  -0.00000  -0.00000 I
freq:  12  +0.00000  -0.00000 I
freq:  13  +8.00000  +0.00000 I
freq:  14  -0.00000  -0.00000 I
freq:  15  +0.00000  -0.00000 I

Inverse transform:
recover:   0  +1.00000  +0.00000 I vs.  +1.00000  +0.00000 I
recover:   1  +0.38268  +0.00000 I vs.  +0.38268  +0.00000 I
recover:   2  -0.70711  +0.00000 I vs.  -0.70711  +0.00000 I
recover:   3  -0.92388  +0.00000 I vs.  -0.92388  +0.00000 I
recover:   4  -0.00000  +0.00000 I vs.  -0.00000  +0.00000 I
recover:   5  +0.92388  +0.00000 I vs.  +0.92388  +0.00000 I
recover:   6  +0.70711  +0.00000 I vs.  +0.70711  +0.00000 I
recover:   7  -0.38268  +0.00000 I vs.  -0.38268  +0.00000 I
recover:   8  -1.00000  +0.00000 I vs.  -1.00000  +0.00000 I
recover:   9  -0.38268  +0.00000 I vs.  -0.38268  +0.00000 I
recover:  10  +0.70711  +0.00000 I vs.  +0.70711  +0.00000 I
recover:  11  +0.92388  +0.00000 I vs.  +0.92388  +0.00000 I
recover:  12  +0.00000  +0.00000 I vs.  +0.00000  +0.00000 I
recover:  13  -0.92388  +0.00000 I vs.  -0.92388  +0.00000 I
recover:  14  -0.70711  +0.00000 I vs.  -0.70711  +0.00000 I
recover:  15  +0.38268  +0.00000 I vs.  +0.38268  +0.00000 I