12

所以我试图用 C 编写离散傅立叶变换来处理真正的 32 位浮点 wav 文件。它一次读取 2 帧（每个通道一个，但出于我的目的，我假设它们都是相同的，所以我使用帧 [0]）。此代码应该通过使用频率 20、40、60、...、10000 探测输入文件来写出输入文件的幅度谱。我在输入帧上使用 Hanning 窗口。如果可以的话，我想避免使用复数。当我运行它时，它给了我一些非常奇怪的幅度（其中大部分都非常小，并且与正确的频率无关），这让我相信我在计算中犯了一个根本性的错误。有人可以对这里发生的事情提供一些见解吗？这是我的代码：

int windowSize = 2205;
int probe[500];
float hann[2205];
int j, n;
// initialize probes to 20,40,60,...,10000
for (j=0; j< len(probe); j++) {
    probe[j] = j*20 + 20;
    fprintf(f, "%d\n", probe[j]);
}
fprintf(f, "-1\n");
// setup the Hann window
for (n=0; n< len(hann); n++) {
    hann[n] = 0.5*(cos((2*M_PI*n/(float)windowSize) + M_PI))+0.5;
}

float angle = 0.0;
float w = 0.0; // windowed sample
float realSum[len(probe)]; // stores the real part of the probe[j] within a window
float imagSum[len(probe)]; // stores the imaginary part of probe[j] within window
float mag[len(probe)]; // stores the calculated amplitude of probe[j] within a window
for (j=0; j<len(probe);j++) {
    realSum[j] = 0.0;
    imagSum[j] = 0.0;
    mag[j] = 0.0;
}

n=0; //count number of samples within current window
framesread = psf_sndReadFloatFrames(ifd,frame,1);
totalread = 0;
while (framesread == 1){
    totalread++;

    // window the frame with hann value at current sample
    w = frame[0]*hann[n];

    // determine both real and imag product values at sample n for all probe freqs times the windowed signal
    for (j=0; j<len(probe);j++) {
        angle = (2.0 * M_PI * probe[j] * n) / windowSize;
        realSum[j] = realSum[j] + (w * cos(angle));
        imagSum[j] = imagSum[j] + (w * sin(angle));
    }
    n++;
    // checks to see if current window has ended
    if (totalread % windowSize == 0) {
        fprintf(f, "B(%f)\n", totalread/44100.0);
        printf("%f breakpoint written\n", totalread/44100.0);
        for (j=0; j < len(mag); j++) { // print out the amplitudes 
            realSum[j] = realSum[j]/windowSize;
            imagSum[j] = imagSum[j]/windowSize;
            mag[j] = sqrt(pow((double)realSum[j],2)+pow((double)imagSum[j],2))/windowSize;
            fprintf(f, "%d\t%f\n", probe[j], mag[j]);
            realSum[j] = 0.0;
            imagSum[j] = 0.0;
        }
        n=0;
    }
    framesread = psf_sndReadFloatFrames(ifd,frame,1);
}
4

2 回答 2

1

我认为错误在于角度的计算。每个样本的角度增量取决于采样频率。像这样的东西（你似乎有 44100Hz）：

angle = (2.0 * M_PI * probe[j] * n) / 44100;

您的采样窗口将包含一个完整的循环，用于您的最低探测频率 20Hz。如果将 n 循环到 2205，则该角度将为 2*M_PI。您看到的可能是混叠，因为您的参考频率为 2205Hz，所有高于 1102Hz 的频率都混叠为较低的频率。

于 2013-02-14T12:12:18.777 回答
0

使用下面的代码 - 只是稍微重组以编译和创建一个假样本，我没有得到全零。我已将输出调用更改为最后：

fprintf(f, "%d\t%f\n", probe[j], mag[j] );

至

if (mag[j] > 1e-7)
    fprintf(f, "%d\t%f\n", probe[j], mag[j] * 10000);

这只是使查看非零数据变得更容易。也许唯一的问题是理解比例因子？请注意我如何伪造输入以生成纯音作为测试用例。

#include <math.h>

#include <stdio.h>

#define M_PI 3.1415926535

#define SAMPLE_RATE 44100.0f

#define len(array) (sizeof array/sizeof *array)


unsigned psf_sndReadFloatFrames(FILE* inFile,float* frame,int framesToRead)
{
    static float counter = 0;   
    float frequency = 1000;
    float time = counter++;
    float phase = time/SAMPLE_RATE*frequency;
    *frame = (float)sin(phase);
    return counter < SAMPLE_RATE;
}

void discreteFourier(FILE* f)                    
{
    FILE* ifd = 0;
    float frame[1];
    int windowSize = 2205;
    int probe[500];
    float hann[2205];


    float angle = 0.0;
    float w = 0.0; // windowed sample
    float realSum[len(probe)]; // stores the real part of the probe[j] within a window
    float imagSum[len(probe)]; // stores the imaginary part of probe[j] within window
    float mag[len(probe)]; // stores the calculated amplitude of probe[j] within a window

    int j, n;

    unsigned framesread = 0;
    unsigned totalread = 0;

    for (j=0; j<len(probe);j++) {
        realSum[j] = 0.0;
        imagSum[j] = 0.0;
        mag[j] = 0.0;
    }

    // initialize probes to 20,40,60,...,10000
    for (j=0; j< len(probe); j++) {
        probe[j] = j*20 + 20;
        fprintf(f, "%d\n", probe[j]);
    }
    fprintf(f, "-1\n");
    // setup the Hann window
    for (n=0; n< len(hann); n++) 
    {
        hann[n] = 0.5*(cos((2*M_PI*n/(float)windowSize) + M_PI))+0.5;
    }
    n=0; //count number of samples within current window
    framesread = psf_sndReadFloatFrames(ifd,frame,1);
    totalread = 0;
    while (framesread == 1){
        totalread++;

        // window the frame with hann value at current sample
        w = frame[0]*hann[n];

        // determine both real and imag product values at sample n for all probe freqs times the windowed signal
        for (j=0; j<len(probe);j++) {
            angle = (2.0 * M_PI * probe[j] * n) / windowSize;
            realSum[j] = realSum[j] + (w * cos(angle));
            imagSum[j] = imagSum[j] + (w * sin(angle));
        }
        n++;
        // checks to see if current window has ended
        if (totalread % windowSize == 0) {
            fprintf(f, "B(%f)\n", totalread/SAMPLE_RATE);
            printf("%f breakpoint written\n", totalread/SAMPLE_RATE);
            for (j=0; j < len(mag); j++) { // print out the amplitudes 
                realSum[j] = realSum[j]/windowSize;
                imagSum[j] = imagSum[j]/windowSize;
                mag[j] = sqrt(pow((double)realSum[j],2)+pow((double)imagSum[j],2))/windowSize;
                if (mag[j] > 1e-7)
                    fprintf(f, "%d\t%f\n", probe[j], mag[j] * 10000);
                realSum[j] = 0.0;
                imagSum[j] = 0.0;
            }
            n=0;
        }
        framesread = psf_sndReadFloatFrames(ifd,frame,1);
    }
}
于 2012-12-22T00:08:23.560 回答