本文介绍了互相关是否滞后?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
假设有两个信号:
import numpy
dt = 0.001
t_steps = np.arange(0, 1, dt)
a_sig = np.sin(2*np.pi*t_steps*4+5)
b_sig = np.sin(2*np.pi*t_steps*4)
我想移动第一个信号以匹配第二个信号.我知道这可以使用互相关来完成,正如Matlab所证明的那样,但是我如何使用SciPy做到这一点.
I want to shift the first signal to match the second signal. I know this can be completed using cross-correlation, as evidenced by Matlab, but how do I accomplish this with SciPy.
推荐答案
请先查看一些示例.假设我们已经在单元测试课中.
See some examples first. Assume we are in unit tests class already.
# Autocorrelation.
y1 = [1, 1, 0, 0, 1, -1, -1]
corr, lag = util.cross_corr(y1, y1)
self.assertEqual(lag, 0)
y1 = [1, 1, 0 ,1, -1, -1]
y2 = [1, 0, 1, 0, 0, 2]
corr, lag = util.cross_corr(y1, y2)
self.assertEqual(lag, -2)
这是我的代码.
def cross_corr(y1, y2):
"""Calculates the cross correlation and lags without normalization.
The definition of the discrete cross correlation is in:
https://www.mathworks.com/help/matlab/ref/xcorr.html
Args:
y1, y2: Should have the same length.
Returns:
max_corr: Maximum correlation without normalization.
lag: The lag in terms of the index.
"""
if len(y1) != len(y2):
raise ValueError('The lengths of the inputs should be the same.')
y1_auto_corr = np.dot(y1, y1) / len(y1)
y2_auto_corr = np.dot(y2, y2) / len(y1)
corr = correlate(y1, y2, mode='same')
# The unbiased sample size is N - lag.
unbiased_sample_size = correlate(
np.ones(len(y1)), np.ones(len(y1)), mode='same')
corr = corr / unbiased_sample_size / np.sqrt(y1_auto_corr * y2_auto_corr)
shift = len(y1) // 2
max_corr = np.max(corr)
argmax_corr = np.argmax(corr)
return max_corr, argmax_corr - shift
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