用scipy fft计算信号的自相关与直接计算得到的结果不同

Question

我正在尝试使用自相关是功率谱的逆傅立叶变换的属性来计算信号的自相关。然而，当我使用scipy (或numpy) fft来做这件事，并与自相关函数的直接计算进行比较时，我得到了错误的答案，具体地说，fft版本在大延迟时间内稳定在一个小的负值，这显然是错误的。

下面是我的MWE和输出。我是不是用错了fft？

import numpy as np
import matplotlib.pyplot as pl
from scipy.fftpack import fft, ifft


def autocorrelation(x) :
    xp = (x - np.average(x))/np.std(x)
    f = fft(xp)
    p = np.absolute(f)**2
    pi = ifft(p)
    return np.real(pi)[:len(xp)/2]/(len(xp))

def autocorrelation2(x):
    maxdelay = len(x)/5
    N = len(x)
    mean = np.average(x)
    var = np.var(x)
    xp = (x - mean)/np.sqrt(var)
    autocorrelation = np.zeros(maxdelay)
    for r in range(maxdelay):
        for k in range(N-r):
            autocorrelation[r] += xp[k]*xp[k+r]
        autocorrelation[r] /= float(N-r)
    return autocorrelation


def autocorrelation3(x):
    xp = (x - np.mean(x))/np.std(x)
    result = np.correlate(xp, xp, mode='full')
    return result[result.size/2:]/len(xp)

def main():
    t = np.linspace(0,20,1024)
    x = np.exp(-t**2)
    pl.plot(t[:200], autocorrelation(x)[:200],label='scipy fft')
    pl.plot(t[:200], autocorrelation2(x)[:200],label='direct autocorrelation')
    pl.plot(t[:200], autocorrelation3(x)[:200],label='numpy correlate')
    pl.legend()
    pl.show()


if __name__=='__main__':
    main()

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