c++ - 对音频样本应用 Kiss FFT 并获得 NaN 输出？

Question

标题解释了我的问题。

我想做的很简单：

加载 MP3 曲目（通过 libmpg123）
阅读样本
在样本上应用 Kiss FFT

到目前为止我尝试过的

inline float scale(kiss_fft_scalar val)
{
    int g = 0;
    return val < 0 ? val*(1/32768.0f ) : val*(1/32767.0f);
}

void main()
{
    mpg123_handle *m = NULL;
    int  channels = 0, encoding = 0;
    long rate = 0;
    int err = MPG123_OK;

    err = mpg123_init();        
    m = mpg123_new(NULL, &err);
    mpg123_open(m, "L:\\audio-io\\audio-analysis\\samples\\zero.mp3");
    mpg123_getformat(m, &rate, &channels, &encoding);

    err = mpg123_format_none(m);
    err = mpg123_format(m, rate, channels, encoding);

    // Get 2048 samples
    const int TIME = 2048;

    // 16-bit integer encoded in bytes, hence x2 size
    unsigned char* buffer = new unsigned char[TIME*2];
    size_t done = 0;
    err = mpg123_read(m, buffer, TIME*2, &done);

    short* samples = new short[done/2];
    int index = 0;

    // Iterate 2 bytes at a time
    for (int i = 0; i < done; i += 2)
    {
        unsigned char first = buffer[i];
        unsigned char second = buffer[i + 1];
        samples[index++] = (first | (second << 8));
    }

    // Array to store the calculated data
    int speclen = TIME / 2 + 1;
    float* output = new float[speclen];

    kiss_fftr_cfg config;
    kiss_fft_cpx* spectrum;

    config = kiss_fftr_alloc(TIME, 0, NULL, NULL);
    spectrum = (kiss_fft_cpx*) malloc(sizeof(kiss_fft_cpx) * TIME);

    // Right here...
    kiss_fftr(config, (kiss_fft_scalar*) samples, spectrum);

    for (int i = 0; i < speclen; i++)
    {
        float re = scale(spectrum[i].r) * TIME;
        float im = scale(spectrum[i].i) * TIME;

        output[i] = sqrtf(re*re + im*im);
    }

    return;
}

问题出现在此行，kiss_fftr(config, (kiss_fft_scalar*) samples, spectrum); 其中samples包含音频样本（16 位），并spectrum假设保存输出数据。

函数完成后，调试器窗口中发生了以下情况。

有人能给我一个简单的例子，说明如何在音频（16 位编码）样本上应用 Kiss FFT 函数吗？

score 20 · Accepted Answer

您需要在代码中找到错误。我的测试代码似乎工作得很好。

具有浮点数的复值正向 FFT：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include "kiss_fft.h"

#ifndef M_PI
#define M_PI 3.14159265358979324
#endif

#define N 16

void TestFft(const char* title, const kiss_fft_cpx in[N], kiss_fft_cpx out[N])
{
  kiss_fft_cfg cfg;

  printf("%s\n", title);

  if ((cfg = kiss_fft_alloc(N, 0/*is_inverse_fft*/, NULL, NULL)) != NULL)
  {
    size_t i;

    kiss_fft(cfg, in, out);
    free(cfg);

    for (i = 0; i < N; i++)
      printf(" in[%2zu] = %+f , %+f    "
             "out[%2zu] = %+f , %+f\n",
             i, in[i].r, in[i].i,
             i, out[i].r, out[i].i);
  }
  else
  {
    printf("not enough memory?\n");
    exit(-1);
  }
}

int main(void)
{
  kiss_fft_cpx in[N], out[N];
  size_t i;

  for (i = 0; i < N; i++)
    in[i].r = in[i].i = 0;
  TestFft("Zeroes (complex)", in, out);

  for (i = 0; i < N; i++)
    in[i].r = 1, in[i].i = 0;
  TestFft("Ones (complex)", in, out);

  for (i = 0; i < N; i++)
    in[i].r = sin(2 * M_PI * 4 * i / N), in[i].i = 0;
  TestFft("SineWave (complex)", in, out);

  return 0;
}

输出：

Zeroes (complex)
 in[ 0] = +0.000000 , +0.000000    out[ 0] = +0.000000 , +0.000000
 in[ 1] = +0.000000 , +0.000000    out[ 1] = +0.000000 , +0.000000
 in[ 2] = +0.000000 , +0.000000    out[ 2] = +0.000000 , +0.000000
 in[ 3] = +0.000000 , +0.000000    out[ 3] = +0.000000 , +0.000000
 in[ 4] = +0.000000 , +0.000000    out[ 4] = +0.000000 , +0.000000
 in[ 5] = +0.000000 , +0.000000    out[ 5] = +0.000000 , +0.000000
 in[ 6] = +0.000000 , +0.000000    out[ 6] = +0.000000 , +0.000000
 in[ 7] = +0.000000 , +0.000000    out[ 7] = +0.000000 , +0.000000
 in[ 8] = +0.000000 , +0.000000    out[ 8] = +0.000000 , +0.000000
 in[ 9] = +0.000000 , +0.000000    out[ 9] = +0.000000 , +0.000000
 in[10] = +0.000000 , +0.000000    out[10] = +0.000000 , +0.000000
 in[11] = +0.000000 , +0.000000    out[11] = +0.000000 , +0.000000
 in[12] = +0.000000 , +0.000000    out[12] = +0.000000 , +0.000000
 in[13] = +0.000000 , +0.000000    out[13] = +0.000000 , +0.000000
 in[14] = +0.000000 , +0.000000    out[14] = +0.000000 , +0.000000
 in[15] = +0.000000 , +0.000000    out[15] = +0.000000 , +0.000000
Ones (complex)
 in[ 0] = +1.000000 , +0.000000    out[ 0] = +16.000000 , +0.000000
 in[ 1] = +1.000000 , +0.000000    out[ 1] = +0.000000 , +0.000000
 in[ 2] = +1.000000 , +0.000000    out[ 2] = +0.000000 , +0.000000
 in[ 3] = +1.000000 , +0.000000    out[ 3] = +0.000000 , +0.000000
 in[ 4] = +1.000000 , +0.000000    out[ 4] = +0.000000 , +0.000000
 in[ 5] = +1.000000 , +0.000000    out[ 5] = +0.000000 , +0.000000
 in[ 6] = +1.000000 , +0.000000    out[ 6] = +0.000000 , +0.000000
 in[ 7] = +1.000000 , +0.000000    out[ 7] = +0.000000 , +0.000000
 in[ 8] = +1.000000 , +0.000000    out[ 8] = +0.000000 , +0.000000
 in[ 9] = +1.000000 , +0.000000    out[ 9] = +0.000000 , +0.000000
 in[10] = +1.000000 , +0.000000    out[10] = +0.000000 , +0.000000
 in[11] = +1.000000 , +0.000000    out[11] = +0.000000 , +0.000000
 in[12] = +1.000000 , +0.000000    out[12] = +0.000000 , +0.000000
 in[13] = +1.000000 , +0.000000    out[13] = +0.000000 , +0.000000
 in[14] = +1.000000 , +0.000000    out[14] = +0.000000 , +0.000000
 in[15] = +1.000000 , +0.000000    out[15] = +0.000000 , +0.000000
SineWave (complex)
 in[ 0] = +0.000000 , +0.000000    out[ 0] = +0.000000 , +0.000000
 in[ 1] = +1.000000 , +0.000000    out[ 1] = +0.000000 , +0.000000
 in[ 2] = +0.000000 , +0.000000    out[ 2] = +0.000000 , +0.000000
 in[ 3] = -1.000000 , +0.000000    out[ 3] = +0.000000 , +0.000000
 in[ 4] = +0.000000 , +0.000000    out[ 4] = +0.000000 , -8.000000
 in[ 5] = +1.000000 , +0.000000    out[ 5] = +0.000000 , +0.000000
 in[ 6] = +0.000000 , +0.000000    out[ 6] = +0.000000 , +0.000000
 in[ 7] = -1.000000 , +0.000000    out[ 7] = +0.000000 , +0.000000
 in[ 8] = +0.000000 , +0.000000    out[ 8] = +0.000000 , +0.000000
 in[ 9] = +1.000000 , +0.000000    out[ 9] = +0.000000 , +0.000000
 in[10] = +0.000000 , +0.000000    out[10] = +0.000000 , +0.000000
 in[11] = -1.000000 , +0.000000    out[11] = +0.000000 , +0.000000
 in[12] = +0.000000 , +0.000000    out[12] = +0.000000 , +8.000000
 in[13] = +1.000000 , +0.000000    out[13] = +0.000000 , +0.000000
 in[14] = +0.000000 , +0.000000    out[14] = +0.000000 , +0.000000
 in[15] = -1.000000 , +0.000000    out[15] = +0.000000 , +0.000000

带有浮点数的实值正向 FFT：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include "kiss_fftr.h"

#ifndef M_PI
#define M_PI 3.14159265358979324
#endif

#define N 16

void TestFftReal(const char* title, const kiss_fft_scalar in[N], kiss_fft_cpx out[N / 2 + 1])
{
  kiss_fftr_cfg cfg;

  printf("%s\n", title);

  if ((cfg = kiss_fftr_alloc(N, 0/*is_inverse_fft*/, NULL, NULL)) != NULL)
  {
    size_t i;

    kiss_fftr(cfg, in, out);
    free(cfg);

    for (i = 0; i < N; i++)
    {
      printf(" in[%2zu] = %+f    ",
             i, in[i]);
      if (i < N / 2 + 1)
        printf("out[%2zu] = %+f , %+f",
               i, out[i].r, out[i].i);
      printf("\n");
    }
  }
  else
  {
    printf("not enough memory?\n");
    exit(-1);
  }
}

int main(void)
{
  kiss_fft_scalar in[N];
  kiss_fft_cpx out[N / 2 + 1];
  size_t i;

  for (i = 0; i < N; i++)
    in[i] = 0;
  TestFftReal("Zeroes (real)", in, out);

  for (i = 0; i < N; i++)
    in[i] = 1;
  TestFftReal("Ones (real)", in, out);

  for (i = 0; i < N; i++)
    in[i] = sin(2 * M_PI * 4 * i / N);
  TestFftReal("SineWave (real)", in, out);

  return 0;
}

输出：

Zeroes (real)
 in[ 0] = +0.000000    out[ 0] = +0.000000 , +0.000000
 in[ 1] = +0.000000    out[ 1] = +0.000000 , +0.000000
 in[ 2] = +0.000000    out[ 2] = +0.000000 , +0.000000
 in[ 3] = +0.000000    out[ 3] = +0.000000 , +0.000000
 in[ 4] = +0.000000    out[ 4] = +0.000000 , +0.000000
 in[ 5] = +0.000000    out[ 5] = +0.000000 , +0.000000
 in[ 6] = +0.000000    out[ 6] = +0.000000 , +0.000000
 in[ 7] = +0.000000    out[ 7] = +0.000000 , +0.000000
 in[ 8] = +0.000000    out[ 8] = +0.000000 , +0.000000
 in[ 9] = +0.000000    
 in[10] = +0.000000    
 in[11] = +0.000000    
 in[12] = +0.000000    
 in[13] = +0.000000    
 in[14] = +0.000000    
 in[15] = +0.000000    
Ones (real)
 in[ 0] = +1.000000    out[ 0] = +16.000000 , +0.000000
 in[ 1] = +1.000000    out[ 1] = +0.000000 , +0.000000
 in[ 2] = +1.000000    out[ 2] = +0.000000 , +0.000000
 in[ 3] = +1.000000    out[ 3] = +0.000000 , +0.000000
 in[ 4] = +1.000000    out[ 4] = +0.000000 , +0.000000
 in[ 5] = +1.000000    out[ 5] = +0.000000 , +0.000000
 in[ 6] = +1.000000    out[ 6] = +0.000000 , +0.000000
 in[ 7] = +1.000000    out[ 7] = +0.000000 , +0.000000
 in[ 8] = +1.000000    out[ 8] = +0.000000 , +0.000000
 in[ 9] = +1.000000    
 in[10] = +1.000000    
 in[11] = +1.000000    
 in[12] = +1.000000    
 in[13] = +1.000000    
 in[14] = +1.000000    
 in[15] = +1.000000    
SineWave (real)
 in[ 0] = +0.000000    out[ 0] = +0.000000 , +0.000000
 in[ 1] = +1.000000    out[ 1] = +0.000000 , +0.000000
 in[ 2] = +0.000000    out[ 2] = +0.000000 , +0.000000
 in[ 3] = -1.000000    out[ 3] = +0.000000 , +0.000000
 in[ 4] = +0.000000    out[ 4] = +0.000000 , -8.000000
 in[ 5] = +1.000000    out[ 5] = +0.000000 , +0.000000
 in[ 6] = +0.000000    out[ 6] = +0.000000 , +0.000000
 in[ 7] = -1.000000    out[ 7] = +0.000000 , +0.000000
 in[ 8] = +0.000000    out[ 8] = +0.000000 , +0.000000
 in[ 9] = +1.000000    
 in[10] = +0.000000    
 in[11] = -1.000000    
 in[12] = +0.000000    
 in[13] = +1.000000    
 in[14] = +0.000000    
 in[15] = -1.000000

score 3 · Accepted Answer

当我第一次开始查看这个答案时，我一直想知道为什么 -8.0 出现在虚数部分而不是实数部分。在重新阅读有关 FFT 的印刷文章时，我意识到我一直在考虑幅度。

所以我调整了复杂代码中的答案以更改 printf 如下

for (i = 0; i < N; i++)
    printf(" in[%02i]=%+f, %+f  out[%02i]=%+f, %+f M[%02i]=%+f\n",
         i, in[i].r, in[i].i,
         i, out[i].r, out[i].i,
         i, sqrt((out[i].r * out[i].r) + (out[i].i * out[i].i)));

这也产生了一个显示幅度的答案。

...
SineWave (complex)
 in[00]=+0.000000, +0.000000  out[00]=+0.000000, +0.000000 M[00]=+0.000000
 in[01]=+1.000000, +0.000000  out[01]=+0.000000, +0.000000 M[01]=+0.000000
 in[02]=+0.000000, +0.000000  out[02]=+0.000000, +0.000000 M[02]=+0.000000
 in[03]=-1.000000, +0.000000  out[03]=+0.000000, +0.000000 M[03]=+0.000000
 in[04]=-0.000000, +0.000000  out[04]=-0.000000, -8.000000 M[04]=+8.000000
 in[05]=+1.000000, +0.000000  out[05]=+0.000000, -0.000000 M[05]=+0.000000
 in[06]=+0.000000, +0.000000  out[06]=+0.000000, -0.000000 M[06]=+0.000000
 in[07]=-1.000000, +0.000000  out[07]=+0.000000, -0.000000 M[07]=+0.000000
 in[08]=-0.000000, +0.000000  out[08]=+0.000000, +0.000000 M[08]=+0.000000
 in[09]=+1.000000, +0.000000  out[09]=+0.000000, +0.000000 M[09]=+0.000000
 in[10]=+0.000000, +0.000000  out[10]=+0.000000, +0.000000 M[10]=+0.000000
 in[11]=-1.000000, +0.000000  out[11]=+0.000000, +0.000000 M[11]=+0.000000
 in[12]=-0.000000, +0.000000  out[12]=-0.000000, +8.000000 M[12]=+8.000000
 in[13]=+1.000000, +0.000000  out[13]=+0.000000, -0.000000 M[13]=+0.000000
 in[14]=+0.000000, +0.000000  out[14]=+0.000000, -0.000000 M[14]=+0.000000
 in[15]=-1.000000, +0.000000  out[15]=+0.000000, -0.000000 M[15]=+0.000000

我还尝试改变生成正弦波的 for 循环中的频率。

float freq;
...
freq = 6.0;
for (i = 0; i < N; i++)
    in[i].r = sin(2 * M_PI * freq * i / N), in[i].i = 0;

只要我保持 1.0 的倍数并且在奈奎斯特频率 16/2 = 8 下，结果从 bin 到 bin 的变化非常好。当然，将频率设置为小数值会看到其幅度分布在各个 bin 中，并且在不应用窗口函数的情况下，我们会得到泄漏。如果你还在为 FFT 苦苦挣扎，就像我在玩这样的代码，你可以在一个屏幕上看到所有结果一段时间，事情开始变得更加清晰。

最后感谢 Alexey 的回答，它帮助我开始使用 Kiss FFT。

score -1 · Accepted Answer

-1

试试这个：

in[i].r = sin(2 * M_PI * freq * (i / N*1.00)), in[i].i = 0;

于 2017-11-05T10:52:47.073 回答

c++ - 对音频样本应用 Kiss FFT 并获得 NaN 输出？

3 回答 3

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