倒车时，FFT没有得到确切的数据 [英] Not getting the exact data when reversing FFT
问题描述
好吧，我想要实现很简单。在某些随机数据应用的FFT，然后应用在输出反向算法取回输入。我使用 kissFFT 库这一点。
Okay, what I'm trying to achieve is simple. Apply FFT on some random data and then apply the reverse algorithm on the output to get back the input. I'm using kissFFT library for this.
code：
const int fft_siz = 512;
const int inverse = 1;
kiss_fft_cpx* in = (kiss_fft_cpx*)malloc(sizeof(kiss_fft_cpx) * fft_siz);
kiss_fft_cpx* out = (kiss_fft_cpx*)malloc(sizeof(kiss_fft_cpx) * fft_siz);
kiss_fft_cpx* rec = (kiss_fft_cpx*)malloc(sizeof(kiss_fft_cpx) * fft_siz);
kiss_fft_cfg cfg = kiss_fft_alloc(fft_siz, !inverse, NULL, NULL);
kiss_fft_cfg icfg = kiss_fft_alloc(fft_siz, inverse, NULL, NULL);
srand((unsigned int)time(NULL));
for(int i = 0; i < fft_siz; i++)
{
in[i].r = rand() % 256;
in[i].i = rand() % 256;
}
kiss_fft(cfg, in, out);
// scaling
for(int i = 0; i < fft_siz; i++)
{
out[i].r /= fft_siz;
out[i].i /= fft_siz;
}
kiss_fft(icfg, out, rec);
unsigned int count = 0;
for(int i = 0; i < fft_siz; i++)
if(in[i].r != rec[i].r)
{
count++;
printf( "in[%3d].r does not match rec[%3d].r :: %3d :: %f\n",
i, i, count, in[i].r - rec[i].r);
}
else if(in[i].i != rec[i].i)
{
count++;
printf( "in[%3d].i does not match rec[%3d].i :: %3d :: %f\n",
i, i, count, in[i].i - rec[i].i);
}
free(in);
free(out);
free(rec);
free(cfg);
free(icfg);
kiss_fft_cleanup();
输出：
in[ 0]: 71.000000 85.000000 -- out[ 0]: 127.095703 124.541016
in[ 1]: 248.000000 27.000000 -- out[ 1]: -7.083314 0.072701
in[ 2]: 64.000000 18.000000 -- out[ 2]: -3.770610 2.682554
in[ 3]: 6.000000 96.000000 -- out[ 3]: -7.929140 -2.897723
in[ 4]: 98.000000 23.000000 -- out[ 4]: -0.719621 -5.854260
in[ 5]: 250.000000 188.000000 -- out[ 5]: 0.397226 -1.248124
in[ 6]: 231.000000 3.000000 -- out[ 6]: -7.934285 -2.367196
in[ 7]: 6.000000 105.000000 -- out[ 7]: -0.317480 -2.955601
in[ 8]: 172.000000 143.000000 -- out[ 8]: -4.236186 3.911616
in[ 9]: 16.000000 134.000000 -- out[ 9]: -0.162577 -5.353521
in[ 10]: 230.000000 112.000000 -- out[ 10]: -4.703711 7.791993
in[ 11]: 5.000000 26.000000 -- out[ 11]: -2.636305 0.188381
in[ 12]: 16.000000 127.000000 -- out[ 12]: 1.137413 4.576081
in[ 13]: 112.000000 86.000000 -- out[ 13]: 0.978051 -0.408992
in[ 14]: 40.000000 23.000000 -- out[ 14]: 5.231920 -2.347566
in[ 15]: 75.000000 26.000000 -- out[ 15]: 0.009981 -2.091559
note ::count::difference
--------------------------------------------------------
in[ 1].r does not match rec[ 1].r :: 1 :: -0.000031
in[ 3].r does not match rec[ 3].r :: 2 :: -0.000015
in[ 4].i does not match rec[ 4].i :: 3 :: -0.000004
in[ 6].i does not match rec[ 6].i :: 4 :: -0.000008
in[ 7].r does not match rec[ 7].r :: 5 :: -0.000002
in[ 9].r does not match rec[ 9].r :: 6 :: -0.000015
in[ 11].r does not match rec[ 11].r :: 7 :: -0.000015
in[ 12].r does not match rec[ 12].r :: 8 :: -0.000015
in[ 13].i does not match rec[ 13].i :: 9 :: -0.000008
in[ 14].i does not match rec[ 14].i :: 10 :: 0.000008
in[ 15].r does not match rec[ 15].r :: 11 :: -0.000015
调试。如果你去的底部，你会看到有317的不匹配。我也在输出值之间即（[]中河 - REC [] R）的区别或（[]中岛 - REC []。我）。
Debug. If you go to the bottom, you'll see that there's 317 mismatches. I'm also outputting the difference between values i.e. (in[].r - rec[].r) or (in[].i - rec[].i).
什么我展示接下来就是白点重新present的实部和红色的点虚部输入数据。
What I'm showing next is the input data where white dots represent the real part and red dots the imaginary part.
这是紫色与重建白色和红色。
This is the output data of FFT represented in purple along with the reconstructed data in white and red.
注意小的差异？我猜这是关系到浮点precision。
我怎样才能克服这个问题让我用FFT上完全一样的输入数据？
Notice the small difference? I'm guessing this is related to floating point precision. How can I overcome this problem to get the exact same input data that I used FFT on ?
编辑：
我注意到，在我的案件的范围内关闭的错误是一样的东西] 0，0.002]。因此，作为一种解决方法，我圆润重建的数据，得到了良好的效果。但还是......如果我的号码的小数部分为0.0这种方法只适用。 I noticed that the range off error in my case is something like ]0 , 0.002]. So as a workaround, I rounded the reconstructed data and got a good result. But still... This only works if the fractional part of my numbers is 0.0.推荐答案
你的猜测是正确的;浮子具有有限precision，有的输入precision将在此过程中会丢失。如果你想确切的输入，你需要某种arbitrary- precision浮点库中（例如， GNU MP库）。
You're guess is correct; floats have a limited precision, and some of the input precision will be lost in the process. If you want the exact input, you'll need some sort of arbitrary-precision floating-point library (for example, the GNU MP library).
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