快速傅里叶变换信号重建的奇怪结果

Question

我有一些数据，如下图所示，我有兴趣找到它的一些傅立叶级数系数。

r = np.array([119.80601628, 119.84629291, 119.85290735, 119.45778804,
   115.64497439, 105.58519852, 100.72765819, 100.04327702,
   100.08590518, 100.35824977, 101.58424993, 105.47976376,
   112.27556007, 117.07679226, 118.99998888, 119.60458086,
   119.78624424, 119.83022022, 119.36116943, 115.72323767,
   106.58946834, 101.19479124, 100.11537349, 100.13313755,
   100.41846106, 101.42255377, 104.33650237, 109.73625492,
   115.14763728, 118.24665037, 119.35359999, 119.68061835])

z = np.array([-411.42980545, -384.98596279, -358.13032372, -330.89578468,
   -303.39129113, -275.76248957, -248.24478443, -221.07069838,
   -194.33260984, -168.05271807, -142.19357982, -116.62090103,
    -91.15354178,  -65.56745626,  -39.65284757,  -13.29632162,
     13.54374939,   40.84929432,   68.50496394,   96.33720787,
    124.08525182,  151.36802193,  177.98791952,  204.0805317 ,
    229.85399128,  255.44727674,  281.02166554,  306.75399703,
    332.74638285,  359.05528646,  385.74336711,  412.8189858 ])

plt.plot(z, r, label='data')
plt.legend()

​

然后我计算平均采样周期，因为在Z变量中它不是常数：

l = []
for i in range(32-1):
    l.append(z[i]-z[i+1])
Ts = np.mean(l)

然后计算快速傅立叶变换:从scipy.fftpack导入快速傅立叶变换

rf = scipy.fftpack.fft(r)

对于信号的重建，则：

fs = 1/Ts

amp = np.abs(rf)/r.shape[0]
n = r.shape[0]
s = 0
for i in range(n//2):

    phi = np.angle(rf[i], deg=False)
    a   = amp[i]
    k   = i*fs/n

    s += a*np.cos(2*np.pi*k  *(z)  +phi)

plt.plot(z, s, label='fft result')
plt.plot(z, r, label='data')
plt.legend()

​

​

然而，就振幅和频率而言，结果都很奇怪。

Answer 1

复数谱是范围为(-fMax/2，…，+fMax/2)的对称谱。你只用到了光谱的右手边的正值部分。这意味着，你的重建信号只包含一半的频谱频率。因为频谱是对称的，所以您所要做的就是将计算出的绝对值加倍。然而，有一个重要的例外。直流值幅度不能加倍。