对称快速傅里叶变换

Question

我在R中有值s。我只是使用函数fft(s)来计算快速傅里叶变换。然后我将值打印到 其中x很简单1,...length(fft(s))。我看到这里有一种强烈的对称性 - 为什么会这样？我期待类似的东西 我做错了什么？

这是第一张图的MWE：

df <- structure(list(value = c(3, 1537.68157138987, 531.727745627154, 
8.50387522397935, 5.40309722205537, 2.30232661295167, 0.420000953675226, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -4.73469069995451, -829.997919240523, 
-3919.09161844073, -6460.79922752307, -6507.49592868557, -4849.27485532219, 
-1674.36434175734, -83.1428571428571, -2.55813257187214, -261.182084182236, 
-2915.2762208206, -5668.7235841082, -6990.36364128962, -6997.7273121356, 
-6211.86050115633, -4378.77236521268, -1421.9056880109, -30.3635181412285, 
-3.58181865471396, -2408.88296333223, -3106.95655095245, -6482.32762645408, 
-6111.41765122797, -1821.04333977064, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 41.499896933588, 998.907991416867, 3778.2752099702, 
6543, 7005.95412518948, 6995.72726353206, 6293.82005765438, 3269.99904633924, 
592.910420911179, 2, 1378.82116693398, 4391.17937491872, 6822.45459767001, 
6220.908446583, 3447.42762806772, 581, 10.2727745627154, 873.056274992053, 
3248.72737676457, 3776.64174236464, 3779.59937375451, 2711.30950879224
)), .Names = "value", row.names = c(NA, -182L), class = "data.frame")
fft_res <- fft(df$value)
ggplot(data.frame(x = 1:length(fft_res), value = Mod(fft_res)), aes(x,value)) + geom_line()

Answer 1

对称性是应用FFT的基本属性 到实值数据（fft()返回正负频率的值）。例如，

x = 1:8
fft(x)
## [1] 36+0.000000i -4+9.656854i -4+4.000000i -4+1.656854i -4+0.000000i
## [6] -4-1.656854i -4-4.000000i -4-9.656854i

Mod(fft(x))
## [1] 36.000000 10.452504  5.656854  4.329569
## [5]  4.000000  4.329569  5.656854 10.452504

以下是相应的频率 （可能有一种更简单的方法......）

freqs = seq(0,1/2,by=1/8)
freqs = c(freqs,-rev(freqs[-c(1,5)]))
## [1]  0.000  0.125  0.250  0.375  0.500
## [6] -0.375 -0.250 -0.125

因此对于实值输入，您只需要保留 上半年的产出。还有更多 用于实值的高效FFT算法 输入，但我不知道是否有任何 实现它们的R包。

对称性不再适用于复值输入：

set.seed(101); fft(x+rnorm(8)*1i)
# [1] 36.000000+ 1.756632i
# [2] -5.501913+ 8.349288i
# [3] -2.375197+ 4.040887i
# [4] -2.914446+ 1.690809i
# [5] -4.000000- 1.899475i
# [6] -5.085554- 1.622900i
# [7] -5.624803- 3.959113i
# [8] -2.498087-10.964420i