What time step is assumed in the function lyapunovExponent?

Hi All,

I'm analysing a measured time series for the Lyapunov exponent using the function "lyapunovExponent". I wonder what value for the timestep delta t is assumed in this function, (this is not clear from the documentation provided). I guess it is '1', so to convert the result to a physical meaningfull answer I have to rescale the numerical results obtained with the reciprocal time step (e.g. 0.01 s ), right?

kind regards

Francois Clemens

Answer 1

[Star Strider]

From the documentation:

lyapExp = lyapunovExponent(X,fs) estimates the Lyapunov exponent of the uniformly sampled time-domain signal X using sampling frequency fs.

So the time step would be 1/fs.

I do not have the Predictive Maintenance Toolbox, however this is the usual conversion between the sampling interval and the sampling frequency.

Answer 2

[Francois Clemens]

Hi Star Rider,

you're right, this will mean that the phsical timestep is defined by the time interval the measured data, since this info is not passed into the functionas you can feed it with a simgle series of data wthout time-info.

regards

Francois

[Star Strider]

Thank you.