正确使用scipy.signal.spectral.lombscargle的方法

Question

我正在参考以下帖子：Using scipy.signal.spectral.lombscargle for period discovery

我意识到答案在某些情况下是正确的。

sin（x）的频率，即1 /（2 * pi）

# imports the numerical array and scientific computing packages
import numpy as np
import scipy as sp
from scipy.signal import spectral

# generates 100 evenly spaced points between 1 and 1000
time = np.linspace(1, 1000, 100)

# computes the sine value of each of those points
mags = np.sin(time)

# scales the sine values so that the mean is 0 and the variance is 1 (the documentation specifies that this must be done)
scaled_mags = (mags-mags.mean())/mags.std()

# generates 1000 frequencies between 0.01 and 1
freqs = np.linspace(0.01, 1, 1000)

# computes the Lomb Scargle Periodogram of the time and scaled magnitudes using each frequency as a guess
periodogram = spectral.lombscargle(time, scaled_mags, freqs)

# returns the inverse of the frequence (i.e. the period) of the largest periodogram value
print "1/2pi = " + str(1/(2*np.pi))
print "Frequency = " + str(freqs[np.argmax(periodogram)] / 2.0 / np.pi)

打印以下内容。很好。我猜。我们将lombscargle结果除以2pi的原因是，我们需要将弧度转换为频率。 （f =弧度/ 2pi）

1/2pi = 0.159154943092
Frequency = 0.159154943092

然而，以下情况似乎出现了问题。

sin的频率（2x），即1 /（pi）

# imports the numerical array and scientific computing packages
import numpy as np
import scipy as sp
from scipy.signal import spectral

# generates 100 evenly spaced points between 1 and 1000
time = np.linspace(1, 1000, 100)

# computes the sine value of each of those points
mags = np.sin(2 * time)

# scales the sine values so that the mean is 0 and the variance is 1 (the documentation specifies that this must be done)
scaled_mags = (mags-mags.mean())/mags.std()

# generates 1000 frequencies between 0.01 and 1
freqs = np.linspace(0.01, 1, 1000)

# computes the Lomb Scargle Periodogram of the time and scaled magnitudes using each frequency as a guess
periodogram = spectral.lombscargle(time, scaled_mags, freqs)

# returns the inverse of the frequence (i.e. the period) of the largest periodogram value
print "1/pi = " + str(1/(np.pi))
print "Frequency = " + str(freqs[np.argmax(periodogram)] / 2.0 / np.pi)

正在印刷以下内容。

1/pi = 0.318309886184
Frequency = 0.0780862900972

似乎不正确。我错过了哪一步？

Answer 1

您理所当然地希望峰值显示在1 / pi，但您测试的最高频率为1 / 2 / pi ...请尝试以下单一更改：

freqs = linspace(0.01, 3, 3000)

现在输出是预期的：

1/pi = 0.318309886184
Frequency = 0.318311478264

但请注意，如果您针对periodogram绘制freqs / 2 / np.pi，则图表如下所示：

因此，对于更复杂的信号，您不能仅仅依靠查找周期图的max来找到主导频率，因为谐波可能会欺骗您。