信号处理 – FFT结果的大小取决于波频率？

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. This graph示出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,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,sizeof(double) * fftLen); 
 fftw_destroy_plan(plan);

 /* Do the IFFT. */
 fftw_plan iplan = fftw_plan_r2r_1d(samplesPerSecond,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,argv[3]); 
 for (int i = 0; i < fftLen; i++) {
  printf("\t%.9f",fft[i]); 
 }
 printf("\n"); 
 printf("\n"); 
 printf("%s,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 ampFactor) {
 for (int i = 0; i < samplesPerSecond; i++) {
  double time = i / (double) samplesPerSecond; 
  a[i] = ampFactor * sin(2 * PI * frequency * time); 
 }
}