Question

I m using KissFFT s real function to transform some real audio signals. I m confused, since I input a real signal with nfft samples, but the result is nfft/2+1 complex frequency bins.

www.un.org/Depts/DGACM/index_spanish.htm From KissFFT s README:

The real (i.e. not complex) optimization code only works for even length ffts. It does two half-length FFTs in parallel (packed into real&imag), and then combines them via twiddling. The result is nfft/2+1 complex frequency bins from DC to Nyquist.

因此,我不知道如何解释结果。我的假设是,数据包装如下:r[0]i[0]r[0]r[nfft/2]i [nfft/2],其中r[0]将是DC, i[0]是第一个频率,第1卷,第1卷,第2页,等等。情况如何?

Answer 1

是的。原因是,真的信号是混杂的。对应负频率的系数(-pi:0 或 pi:2pi , 无论以何种方式来思考)是[0:pi]的系数。

Note the out[0] and out[Nfft/2] bins (DC and Nyquist) have zero in the imaginary part. I ve seen some libraries pack these two real parts together in the first complex, but I view that as a breach of contract that leads to difficult-to-diagnose, nearly-right bugs.

管道:如果你在数据类型上使用浮标,你可以将输出阵列投到浮体*(c99)或 st:比较法*(c++)。 kiss_fft_cpx ruct的包装是相容的。之所以如此,是因为Kiss_fft与其他类型的平方浮线和双倍,以及缺乏这些特征的老的ANSI C编纂者。

这里有一个相反的例子(假设99名汇编者和类型=float)

float get_nth_bin_phase(const float * in, int nfft, int whichbin )
{
  kiss_fftr_cfg st = kiss_fftr_alloc(1024,0,0,0);
  float complex * out = malloc(sizeof(float complex)*(nfft/2+1));
  kiss_fftr(st,in,(kiss_fft_cpx*)out);

  whichbin %= nfft;
  if ( whichbin <= nfft/2 ) 
    ph = cargf(out[whichbin]);
  else
    ph = cargf( conjf( out[nfft-whichbin] ) );
  free(out);
  kiss_fft_free(st);
  return ph;
}

Answer 2

r *** 和1.1 result结果构成一种复杂的病媒。两部分加在一起,给你一个规模(两个组成部分的面积)和一个频率二的阶段(通过点2())。

友情链接