吻FFT bin幅度

时间:2011-05-31 16:23:28

标签: fft amplitude kissfft

我花了不少时间研究FFT。我特别感兴趣的是使用KISSFFT,因为它是一个非常便携的C实现。

我仍然不清楚如何将i [x]和r [x]转换为频率仓的幅度。因此创建了一个签名的int 16版本的sin。我有512个我的sin波样本。我希望看到一个Bin有数据,其余的为零。不是这样......

这是我的代码......

- (IBAction)testFFT:(id)sender{
NSLog(@"testFFT");

static double xAxis = 0;
static int sampleCount = 0;
static double pieSteps;
static double fullSinWave = 3.14159265*2;
static double sampleRate = 44100;
static double wantedHz = 0;
int octiveOffset;
char * globalString = stringToSend;
SInt16 dataStream[512];

// Notes: ioData contains buffers (may be more than one!)
// Fill them up as much as you can. Remember to set the size value in each buffer to match how
// much data is in the buffer.
for (int j = 0; j < 512; j++) {
    wantedHz = 1000;
    pieSteps = fullSinWave/(sampleRate/wantedHz);
    xAxis += pieSteps; 
    dataStream[j] = (SInt16)(sin(xAxis) * 32768.0);
    NSLog(@"%d) %d", j, dataStream[j]);
}

kiss_fft_cfg mycfg = kiss_fft_alloc(512,0,NULL,NULL);
kiss_fft_cpx* in_buf = malloc(sizeof(kiss_fft_cpx)*512);
kiss_fft_cpx* out_buf = malloc(sizeof(kiss_fft_cpx)*512);
for (int i = 0;i < 512;i++){
    in_buf[i].r = dataStream[i];
    in_buf[i].i = dataStream[i];
}    
kiss_fft(mycfg,in_buf, out_buf);
for (int i = 0;i < 256;i++){
    ix = out_buf[i].i;
    rx = out_buf[i].r;
    printfbar(sqrt(ix*ix+rx*rx)););
}

}

我得到的结果看起来像这样......


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2 个答案:

答案 0 :(得分:2)

首先进行一些编程更改:

xAxis += pieSteps;
if (xAxis >= fullSinWave)
  xAxis -= fullSinWave; //wrap x back into 0-2pi period

将有助于减少数字错误。

in_buf[i].r = dataStream[i];
in_buf[i].i = 0;

会将输入缓冲区设置为sin(x),之前您已将其设置为sin(x) + j*sin(x),其中j = sqrt(-1)

wantedHz = 1000;移出循环看起来更好。

还有一个更基本的问题:你设置wantedHz = 1000。采样率为44.1 kHz时,这相当于44100 points/sec * (1/1000) sec/cycle = 44.1 points/cycle。使用512点的缓冲区,您将在缓冲区中获得11.6个正弦波周期。非整数周期导致leakage

在开始之前,尝试设置wantedHz = 12*44100.0/512以在缓冲区中准确地给出12个周期。你应该在变换中看到两个尖峰:一个在索引12处,一个在索引511-12处。

您会看到两个峰值的原因是sin(w_0*x)的转换为j*{-delta(w-w_0) - delta(w+w_0)}。也就是说,在变换的虚部中,您在w_0和-w_0处获得脉冲函数。他们在这些地方的原因是变换从0变为2 * pi。

执行此操作后,返回wantedH = 1000,在缓冲区中给出非整数个循环。您应该看到一个宽大的帐篷形状的结果,以11和511-11为中心。您应该将dataStream乘以窗口函数(Hann是好的)以减少此效果的影响。

答案 1 :(得分:2)

我也一直在与这个库作斗争,这段代码可以帮助你测试。这是我在互联网上阅读的代码混合,看看在项目中包含它是否有趣。工作良好。它写入一个文件,其中包含原始信号的波形和值,FFT和反向FFT也只是为了测试。它是用VS2010编译的

#include "kiss_fft.h"
#include "tools\kiss_fftr.h"
#include <stdio.h>
#include <conio.h>
#define numberOfSamples 1024

int main(void)
{
    struct KissFFT
    {
            kiss_fftr_cfg forwardConfig;
            kiss_fftr_cfg inverseConfig;
            kiss_fft_cpx* spectrum;
            int numSamples;
            int spectrumSize;
    } fft;

    static double dospi = 3.14159265*2;
    static double sampleRate = 44100;
    static double wantedHz = 0;
    int j,i,k;
    float dataStream[numberOfSamples];
    float dataStream2[numberOfSamples];
    float mags[numberOfSamples];
    FILE * pFile;

    //Frequency to achive
    wantedHz = 9517;

    fft.forwardConfig = kiss_fftr_alloc(numberOfSamples,0,NULL,NULL);
    fft.inverseConfig = kiss_fftr_alloc(numberOfSamples,1,NULL,NULL);
    fft.spectrum = (kiss_fft_cpx*)malloc(sizeof(kiss_fft_cpx) * numberOfSamples);
    fft.numSamples = numberOfSamples;
    fft.spectrumSize = numberOfSamples/2+1;


    pFile = fopen ("c:\\testfft.txt","w");
    //filling the buffer data with a senoidal wave of frequency -wantedHz- and printing to testing it
    for (j = 0; j < numberOfSamples; j++) {

            dataStream[j] = 32768*(sin(wantedHz*dospi*j/sampleRate));
            //Draw the wave form 
            for (k=-64;k<(int)(dataStream[j]/512);k++)  fprintf(pFile," ");
            fprintf(pFile,"*\n");
    }

    //spectrum
    kiss_fftr(fft.forwardConfig, dataStream, fft.spectrum);
    //inverse just to testing
    kiss_fftri( fft.inverseConfig, fft.spectrum, dataStream2 );
    for(i=0;i<fft.spectrumSize;i++) {
        mags[i] = hypotf(fft.spectrum[i].r,fft.spectrum[i].i);
        fprintf(pFile,"[Sample %3d] ORIGINAL[%6.0f]   -SPECTRUM[%5dHz][%11.0f]",i,dataStream[i],i*(int)sampleRate/numberOfSamples,mags[i]);
        dataStream2[i] = dataStream2[i] / (float)fft.numSamples;
        fprintf(pFile,"     -INVERSE[%6.0f]\n",dataStream2[i]);
    }
    //end

    //free and close
    fclose (pFile);
    kiss_fft_cleanup();   
    free(fft.forwardConfig);
    free(fft.inverseConfig);
    free(fft.spectrum);

    getch();
    return 0;

}