正确使用scipy.signal.spectral.lombscargle

Question

我指的是下面的帖子：Using scipy.signal.spectral.lombscargle for period discovery

我意识到在某些情况下给出的答案是正确的。

sin(x)的频率为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)

将打印以下内容。很好。我想是的。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，对于下面的情况，事情似乎出了问题。

sin(2x)的频率为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)

正在打印以下内容。

1/pi = 0.318309886184
Frequency = 0.0780862900972

似乎是不正确的。我遗漏了什么步骤吗？

Answer 1

您理所当然地希望峰值出现在1 / pi上，但您测试的最高频率是1 / 2 / pi……尝试以下单个更改：

freqs = linspace(0.01, 3, 3000)

现在的输出是预期的：

1/pi = 0.318309886184
Frequency = 0.318311478264

但请注意，如果您绘制periodogram与freqs / 2 / np.pi的关系图，则该图如下所示：

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因此，对于更复杂的信号，你不能仅仅依靠寻找周期图的max来找到主频，因为谐波可能会欺骗你。