I am trying to use Kepler Data for Eclipsing Binaries to estimate time period, and then other parameters such as mass, eccentricity, semi-major axis, distance, etc. of the system. I want to write code in MATLAB which will use FFT. The available data has the following columns:

bjd(date) | phase | raw_flux | raw_err | corr_flux | corr_err | dtr_flux | dtr_err

Does the data require any cleaning? If I want to take the FFT of the data to find time period, which data should I use?

  • $\begingroup$ Unless you're trying to practice with FFT or something to that effect, your proposed method to find the period is overkill. You can get that directly from the BJD and the phase data. No need to FFT the flux data. $\endgroup$
    – zephyr
    Commented Oct 26, 2016 at 18:33

1 Answer 1


For this, you should create a lightcurve, a graph of brightness over time, to view the data. For Kepler data, the bjd(date) column is the time in BJD. The dtr_flux stands for detrended flux, meaning that it should contain "cleaned" data.

First, I would plot the dtr_flux column over bjd(time) to obtain a lightcurve. Depending on the type of eclipsing binary, you should see a minimum and maximum dip occurring periodically. The data also contains a "phase" column, meaning the data has been "folded" to allow you to plot each cycle of the eclipsing binary on top of each other. See http://www.southampton.ac.uk/~sdc1g08/BinningFolding.html

After that, you can take the FFT if you wish. Remember that since this is an eclipsing binary with both maximum and minumum brightness dips, the FFT will have a spike both at P and P/2.

  • $\begingroup$ So I got the plot of bjd vs dtr_flux as imgur.com/gallery/3ZFNW. I tried taking the FFT of this, but all the values are coming imaginary except the max value which is coming as 6.0836e+04. What is the problem here? $\endgroup$
    – ruskin23
    Commented Oct 27, 2016 at 3:52
  • $\begingroup$ @ruskin23 That is to be expected. It sounds to me like you should go back and study exactly what an FFT is and how it works to get a better understanding of your results. What's more, you should probably disclude fitting the gap in your data as that will throw off your results to some degree. $\endgroup$
    – zephyr
    Commented Oct 27, 2016 at 14:22
  • $\begingroup$ The FFT assumes that the samples used are sampled at regular intervals. It will not work with irregularly sampled data. $\endgroup$ Commented Oct 28, 2016 at 17:00

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