正确使用方法scipy.signal.光谱.lombscarg公司

Question

我指的是以下帖子：Using scipy.signal.spectral.lombscargle for period discovery

我知道在某些情况下给出的答案是正确的。在

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)

打印以下内容。很好。我想是吧。我们将lombscargle结果与2pi分开的原因是，我们需要将弧度转换为频率。（f=弧度/2pi）

^{pr2}$

然而，下面的情况似乎出了问题。在

sin（2x）的频率，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)

下面正在打印。在

1/pi = 0.318309886184
Frequency = 0.0780862900972

似乎不正确。我错过了哪一步？在

Answer 1

您理所当然地期望峰值出现在1 / pi，但是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)

现在输出是预期的：

^{pr2}$

不过，请注意，如果您将periodogram与freqs / 2 / np.pi对应，则图形如下所示：

所以对于更复杂的信号，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。在