Correct way to use scipy.signal.spectral.lombscargle

Question

I'm refering to the following post : Using scipy.signal.spectral.lombscargle for period discovery

I realize the answer given correct for certain case.

Frequency for sin(x), which is 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)

The following is printed. Is fine. I guess. The reason we divide the lombscargle result with 2pi is that, we need to convert radian to frequency. (f = radian / 2pi)

1/2pi = 0.159154943092
Frequency = 0.159154943092

However, thing seems goes wrong for the following case.

Frequency for sin(2x), which is 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)

The following is being printed.

1/pi = 0.318309886184
Frequency = 0.0780862900972

Seem incorrect. Any step that I had missed out?

Answer 1

You are rightfully expecting the peak to show up at 1 / pi, but the highest frequency you are testing is 1 / 2 / pi... Try the following single change :

freqs = linspace(0.01, 3, 3000)

and now the output is the expected:

1/pi = 0.318309886184
Frequency = 0.318311478264

Note, though, that if you plot periodogram against freqs / 2 / np.pi, the graph looks like this:

So for a more complicated signal, you cannot rely on just looking for the max of periodogram to find the dominant frequency, because the harmonics may fool you.