fftw c2c:转换后的真实数据中缺少对称性 [英] fftw c2c: missing symmetry in transformed real data
问题描述
最近我在使用 fftw 和它的 c2c 转换方面遇到了一些问题(参见:带有 fftw 库的 3d c2c fft).当我发现我在使用 fftw lib 时遇到的问题时，我创建了一个新问题，以便以更具体的方式讨论这种情况.由于我正在对真实数据进行复杂到复杂的转换，因此我在傅立叶空间中的转换数据应该是对称的:F[n] = con(F[N-n])
recently I faced some problems concerning the use of fftw and it's c2c transformation (see: 3d c2c fft with fftw library). As I located my problems in the use of the fftw lib I created a new Question in order to discus this situation in a more concrete way. Since I am doing a complex to complex transform with real data my transformed data in fourier space is supposed to be symmetric: F[n] = con(F[N-n])
现在我对小块测试数据进行了一些转换，以检查转换后的数据是否具有这种对称性.对于 1D 变换，每件事都按预期工作，但对于更高维度，我得到了真正意想不到的结果.
Now I did some transformations with small blocks of test-data to check the transformed data for this symmetry. For 1D transform at every things worked as expected, but for higher dimensions I got real unexpected results.
我使用 fftwf_plan_dft_2d 将 8x8 灰度图像转换为傅立叶空间，复数结果为:
I am using fftwf_plan_dft_2d to transform a 8x8 grayscale image into fourier space and the complex result is given by:
n
0 real 7971 imag 0
1 real -437.279 imag -802.151
2 real -289 imag -566
3 real -182.721 imag 15.8486
4 real 31 imag 0
5 real -182.721 imag -15.8486
6 real -289 imag 566
7 real -437.279 imag 802.151
8 real -1499.79 imag -315.233
9 real 182.693 imag -74.5563
10 real 55.9239 imag -12.8234
11 real -84.7868 imag -9.10052
12 real -14.4264 imag 211.208
13 real 289.698 imag 214.723
14 real 452.659 imag -246.279
15 real 1136.35 imag -763.85
16 real 409 imag -134
17 real -141.865 imag 42.6396
18 real -33 imag 122
19 real 129.075 imag -49.7868
20 real 1 imag -150
21 real 109.865 imag -84.6396
22 real 95 imag -142
23 real -841.075 imag -92.2132
24 real -108.207 imag -89.2325
25 real -127.213 imag 28.8995
26 real -36.6589 imag -8.27922
27 real -74.6934 imag 43.4437
28 real 70.4264 imag 29.2082
29 real -88.3545 imag -81.8499
30 real -127.924 imag -190.823
31 real 230.302 imag 8.7229
32 real -53 imag 0
33 real -73.1127 imag -22.8578
34 real -85 imag -82
35 real -10.8873 imag 51.1421
36 real -65 imag 0
37 real -10.8873 imag -51.1421
38 real -85 imag 82
39 real -73.1127 imag 22.8578
40 real -108.207 imag 89.2325
41 real 230.302 imag -8.7229
42 real -127.924 imag 190.823
43 real -88.3545 imag 81.8499
44 real 70.4264 imag -29.2082
45 real -74.6934 imag -43.4437
46 real -36.6589 imag 8.27922
47 real -127.213 imag -28.8995
48 real 409 imag 134
49 real -841.075 imag 92.2132
50 real 95 imag 142
51 real 109.865 imag 84.6396
52 real 1 imag 150
53 real 129.075 imag 49.7868
54 real -33 imag -122
55 real -141.865 imag -42.6396
56 real -1499.79 imag 315.233
57 real 1136.35 imag 763.85
58 real 452.659 imag 246.279
59 real 289.698 imag -214.723
60 real -14.4264 imag -211.208
61 real -84.7868 imag 9.10052
62 real 55.9239 imag 12.8234
63 real 182.693 imag 74.5563
抱歉提供这么长的数据列表，但它显示了我的问题.
Sorry for this long list of data, but it shows my problem.
例如对于 F[3]=-182.721 + 15.8486i 我期望 F[64-3] = F[61] = -182.721 - 15.8486i，但正如你所看到的，它是 -84.7868 + 9.10052i.相反，F[3] 的共轭位于索引 5.其他对也是如此.
For example for F[3]=-182.721 + 15.8486i I expected F[64-3] = F[61] = -182.721 - 15.8486i, but as you can see it is -84.7868 + 9.10052i. Instead the conjugate of F[3] is located at index 5. Same thing for other pairs.
如果有系统我找不到.
完整代码如下:
QImage image("/Users/wolle/Desktop/wolf.png");
int w = image.width();
int h = image.height();
int size = w * h;
cl_float *rawImage = imageToRaw(image); // converts a QImage into an rgb array [0..255]
fftwf_complex *complexImage = (fftwf_complex*) fftwf_malloc(sizeof(fftwf_complex) * size);
fftwf_complex *freqBuffer = (fftwf_complex*) fftwf_malloc(sizeof(fftwf_complex) * size);
// real data to complex data
for (int i = 0; i < size; i++)
{
complexImage[i][0] = (float)rawImage[i];
complexImage[i][1] = 0.0f;
}
fftwf_plan forward = fftwf_plan_dft_2d(w, h, complexImage, freqBuffer, FFTW_FORWARD, FFTW_ESTIMATE);
fftwf_execute(forward);
for (int y = 0; y < h; y++)
{
for (int x = 0; x < w; x++)
{
int gid = y * w + x;
qDebug() << gid << "real" << freqBuffer[gid][0] << "imag" << freqBuffer[gid][1];
}
}
我希望得到一些帮助.:-D
I would appreciate some Help. :-D
问候
狼
推荐答案
对于二维傅立叶变换，当 x 为实数时，FFT(x) 是共轭对称的仍然成立.但那是两个维度.所以索引 16*x+y 处的 (x,y) 元素应该是索引 16*(16-x mod 16)+(16-y mod 16) 处的 (16-x,16-y) 元素的共轭, 当 y 不为 0 时为 272-16*xy mod 256.
For a 2D Fourier transform, it is still true that when x is real FFT(x) is conjugate-symmetric. But that's in two dimensions. So the (x,y) element at index 16*x+y should be the conjugate of the (16-x,16-y) element at index 16*(16-x mod 16)+(16-y mod 16), which when y isn't 0 is 272-16*x-y mod 256.
但我认为虽然您说的是 16x16，但实际上您的意思是 8x8.所以 (x,y) 在 8*x+y 与 (8-x,8-y) 在 8*(8-x mod 8) + (8-y mod 8) 是共轭的.
BUT I think that although you said 16x16 you actually meant 8x8. So (x,y) at 8*x+y is conjugate to (8-x,8-y) at 8*(8-x mod 8) + (8-y mod 8).
特别是，例如，当 x=0 时，共轭元素是 y 和 8-y——例如，包括你发现的 3 和 5.
In particular, e.g., when x=0 the conjugate elements are y and 8-y -- including, for instance, 3 and 5, as you found.
(当 x=0 或 y=0 时，上面的8-y mod 8"之类的意思是 0.)
(When x=0 or y=0, things like "8-y mod 8" above mean 0.)
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