python – 使用scipy fft计算信号的自相关性可以得到直接计算的

我试图使用自相关是功率谱的逆傅里叶变换的属性来计算信号的自相关.但是,当我使用scipy(或numpy)fft执行此操作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,具体而言,fft版本在较大的延迟时间内以较小的负值关闭,这是显然错了.

我的MWE低于输出.我使用fft错了吗？

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()

def autocorrelation(x):
    xp = ifftshift((x - np.average(x))/np.std(x))
    n,= xp.shape
    xp = np.r_[xp[:n//2],np.zeros_like(xp),xp[n//2:]]
    f = fft(xp)
    p = np.absolute(f)**2
    pi = ifft(p)
    return np.real(pi)[:n//2]/(np.arange(n//2)[::-1]+n//2)