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因互相关而滞后?

[英]Get lag with cross-correlation?

Let's say have have two signals: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)

I want to shift the first signal to match the second signal.我想移动第一个信号以匹配第二个信号。 I know this can be completed using cross-correlation, as evidenced by Matlab, but how do I accomplish this with SciPy.我知道这可以使用互相关来完成,如 Matlab 所证明的那样,但是我如何使用 SciPy 来完成此操作。

As defined in this Wikipedia article , the lag between signals is given by the argmax of the cross-correlation. 本Wikipedia文章中所定义,信号之间的延迟由互相关的argmax给出。 Consequently, the following code will shift the b_sig to be in phased with a_sig to minimize the error. 因此,以下代码将b_siga_sig移相以使错误最小化。

from scipy.signal import correlate

lag = np.argmax(correlate(a_sig, b_sig))
c_sig = np.roll(b_sig, shift=int(np.ceil(lag)))

See some examples first.先看一些例子。 Assume we are in unit tests class already.假设我们已经在单元测试类中。

# Autocorrelation.
y1 = [1, 1, 0, 0, 1, -1, -1]
corr, lag = cross_corr(y1, y1)
self.assertEqual(lag, 0)

y1 = [1, 1, 0 ,1, -1, -1]
y2 = [1, 0, 1, 0, 0, 2]
corr, lag = cross_corr(y1, y2)
self.assertEqual(lag, -2)

here is my code.这是我的代码。

import numpy as np    

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 = np.correlate(y1, y2, mode='same')
  # The unbiased sample size is N - lag.
  unbiased_sample_size = np.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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