jls*_*ker 6 signal-processing fft fftw
我对从FFT获得的结果感到困惑,并希望得到任何帮助.
我正在使用FFTW 3.2.2但是与其他FFT实现(在Java中)得到了类似的结果.当我采用正弦波的FFT时,结果的缩放取决于波的频率(Hz) - 具体而言,它是否接近整数.当频率接近整数时,得到的值非常小,当频率在整数之间时,它们的数量级要大一些.该图显示了对应于不同频率的波频率的FFT结果中的尖峰幅度.这是正确的吗？？
我检查了FFT的逆FFT等于原始正弦波乘以样本数,它是.FFT的形状似乎也是正确的.
如果我正在分析单个正弦波,那就不会那么糟糕了,因为无论高度如何,我都可以在FFT中寻找尖峰.问题是我想分析正弦波的总和.如果我正在分析正弦波的总和,例如440 Hz和523.25 Hz,那么只有523.25 Hz的正弦波峰值出现.另一个的尖峰非常小,看起来像是噪音.必须有一些方法来使这项工作,因为在Matlab中它确实有效 - 我在两个频率上得到类似大小的尖峰.如何更改下面的代码以均衡不同频率的缩放？
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <fftw3.h>
#include <cstdio>
using namespace std;
const double PI = 3.141592;
/* Samples from 1-second sine wave with given frequency (Hz) */
void sineWave(double a[], double frequency, int samplesPerSecond, double ampFactor);
int main(int argc, char** argv) {
/* Args: frequency (Hz), samplesPerSecond, ampFactor */
if (argc != 4) return -1;
double frequency = atof(argv[1]);
int samplesPerSecond = atoi(argv[2]);
double ampFactor = atof(argv[3]);
/* Init FFT input and output arrays. */
double * wave = new double[samplesPerSecond];
sineWave(wave, frequency, samplesPerSecond, ampFactor);
double * fftHalfComplex = new double[samplesPerSecond];
int fftLen = samplesPerSecond/2 + 1;
double * fft = new double[fftLen];
double * ifft = new double[samplesPerSecond];
/* Do the FFT. */
fftw_plan plan = fftw_plan_r2r_1d(samplesPerSecond, wave, fftHalfComplex, FFTW_R2HC, FFTW_ESTIMATE);
fftw_execute(plan);
memcpy(fft, fftHalfComplex, sizeof(double) * fftLen);
fftw_destroy_plan(plan);
/* Do the IFFT. */
fftw_plan iplan = fftw_plan_r2r_1d(samplesPerSecond, fftHalfComplex, ifft, FFTW_HC2R, FFTW_ESTIMATE);
fftw_execute(iplan);
fftw_destroy_plan(iplan);
printf("%s,%s,%s", argv[1], argv[2], argv[3]);
for (int i = 0; i < samplesPerSecond; i++) {
printf("\t%.6f", wave[i]);
}
printf("\n");
printf("%s,%s,%s", argv[1], argv[2], argv[3]);
for (int i = 0; i < fftLen; i++) {
printf("\t%.9f", fft[i]);
}
printf("\n");
printf("\n");
printf("%s,%s,%s", argv[1], argv[2], argv[3]);
for (int i = 0; i < samplesPerSecond; i++) {
printf("\t%.6f (%.6f)", ifft[i], samplesPerSecond * wave[i]); // actual and expected result
}
delete[] wave;
delete[] fftHalfComplex;
delete[] fft;
delete[] ifft;
}
void sineWave(double a[], double frequency, int samplesPerSecond, double ampFactor) {
for (int i = 0; i < samplesPerSecond; i++) {
double time = i / (double) samplesPerSecond;
a[i] = ampFactor * sin(2 * PI * frequency * time);
}
}