在音频采样应用吻FFT和获得NaN的输出？(Applying Kiss FFT on audio s

2019-08-08 08:41发布

站内文章 / 后端开发
88 0 0

标题解释了我的问题。

我所试图做的是非常简单的：

  • 加载MP3曲目（通过libmpg123）
  • 阅读样本
  • 对样品申请吻FFT

我至今尝试过

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;
}

发生在这条线的问题kiss_fftr(config, (kiss_fft_scalar*) samples, spectrum); 其中samples包含音频样本（16位），和spectrum是假设以保持输出数据。

该功能完成后，这里发生的事情在调试器窗口。

有人可以给我如何在音频（16位编码）样品申请吻FFT功能的一个简单的例子？

Answer 1:

你需要找到在你的代码的bug（一个或多个）。 我的测试代码，似乎工作得很好。

复数正向FFT浮筒： 

#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

实值FFT向前花车： 

#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


Answer 2:

当我第一次开始寻找这个答案我一直在想，为什么-8.0在虚部而不是真正的部分转动起来。 这是同时重新阅读FFT的，我才意识到我一直在思考幅度的印刷物品。

所以我调整了在复杂的代码的答案改变的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)));

其生产出的大小以及答案。 

...
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

我也发挥周围改变用于循环产生的正弦波的频率。 

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;

并且只要我住以1.0的倍数而根据奈奎斯特频率16/2 = 8结果从箱转移到箱相当不错。 课程设置频率，以分数值看到它的大小跨箱，不实施窗口函数，我们得到泄漏扩散。 如果你还在用FFT的挣扎像我玩的这样的代码，你可以看到所有的结果，在一个屏幕上一会儿，事情开始变得清晰。

最后感谢阿列克谢的答案的投票它帮助我开始使用吻FFT。



Answer 3:

试试这个： 

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


文章来源: Applying Kiss FFT on audio samples and getting NaN output?
标签： c++ audio kissfft
举报

收藏的人(0)

Ta的文章 更多文章
登录 后发表评论
0条评论
还没有人评论过~