3
$\begingroup$

I am having a time series which shows some kind of periodic behavior looking at the plot. In order to get this seasonal component I am using the rolling mean and use seasonal_decompose of the statsmodels Python library.

sig = df_sensors['S1'].rolling(window=100).mean()[2000:4000].values
decomposed = sm.tsa.seasonal_decompose(sig, freq=4)

As you can see, I am assuming a frequency (freq) of the series of 4 Hz. I am doing this because the FFT of the original data of df_sensors['S1'] shows the following:

enter image description here

So it appears that there are 2 major components. One which is at ~4 Hz and another which is at ~ 175 Hz.

Plotting the result of the decomposition:

decomposed.plot()

shows the following:

enter image description here

As you can see, it is not able to find the actual seasonal part of the signal. You can also see rather long seasons of 4 Hz and the rather fast seasons of 175 Hz in the original signal.

I would like to understand why there was not seasonal component extracted. To the naked eye the behavior looks very seasonal - although there are irregularities.

In the end I would like to try and predict the seasonal behavior but I am a bit stuck here.

Any help would be appreciated and please let me know if you need more information.

$\endgroup$
1
  • $\begingroup$ freq is the cycle length in number of period. That is you need to use freq=500 (based on your plot) or something close to this. $\endgroup$
    – Josef
    Commented Jun 16, 2017 at 14:33

1 Answer 1

6
$\begingroup$

As far as I understand, seasonality is detected but since you had freq set to such a low value (4), the plot fluctuates and covers the whole graph.

I would recommend changing the freq parameter - which should be set as follows: if you assume that some periodic activity is happening every e.g. 24 hours and your measurements are every 10 mins, decomposition_frequency = (60 / 10) * 24, where 60 / 10 = 6 - meaning you have 6 10mins intervals in an hour, and then you multiply it by 24 to make an hour a day.

In your case, if you would like to capture those fluctuations that happen every 500 steps, you could set freq to 500 and see what you get.

$\endgroup$
1
  • $\begingroup$ In that case the what is referred to as freq would be the underlying period of the signal not the frequency? Your solution works for me... $\endgroup$
    – Philipp
    Commented Jun 3, 2020 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.