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I have a fabric-based strain sensor and would like to find out what other recommended models can be used in the hysteresis modelling.

I have tried using the cubic function for loading and unloading phase, but it wasn't accurate in the when I sub the unloading phase model into my calculation to find the strain using my resistance value.

Appreciate some advice on how I model the unloading phase better. Thank you. 

 ListLinePlot[{Strainsensor50}, PlotLegends -> {"50%"}, 
 PlotLabel -> "normalize strain-resistance graph", 
 PlotRange -> {-0.1, 1.5}, AxesLabel -> {"(Li-Lo)/Lo", "(Ri-Ro)/Ro"}, 
 PlotStyle -> Red]

enter image description here

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  • $\begingroup$ Many times hysteresis loops are fit with separate functions with "time" being the predictor. Also it looks like you just have a single cycle (i.e., no replication over multiple cycles). Can your data in terms of {time, (Li-Lo)/Lo, (Ri-Ro)/Ro}} be made available? $\endgroup$
    – JimB
    Sep 24, 2019 at 2:59
  • $\begingroup$ Hi JimB, the data in terms of {time, (Li-Lo)/Lo, (Ri-Ro)/Ro}} are the mean of 30cycles over 3 different sensors. The file are too large to be uploaded here. $\endgroup$
    – Boajj04
    Sep 24, 2019 at 3:54
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    $\begingroup$ OK. Do the plots of (Li-Lo)/Lo and (Ri-Ro)/Ro over time suggest a reasonable model for each? Or (because of the differential-equation tag) do you mean a differential equation when you say "model" ? And if you want to model without time, how do you know if you're in the loading or unloading phase? $\endgroup$
    – JimB
    Sep 24, 2019 at 4:17
  • $\begingroup$ Maybe my comment about how you might know whether you're in the loading or unloading phase might sound silly (at first) because one can look at the figure and probably have a reasonable idea about that. However, fitting curves to data isn't about "I'll know it when I see it" but rather having a well-defined estimation process to determine the phase with some measure of precision associated with splitting up the data into two parts. (Admittedly, I usually deal with biological data which is never as "smooth" as what you show.) $\endgroup$
    – JimB
    Sep 24, 2019 at 4:40
  • $\begingroup$ Thank you JimB, that data you see is a filtered data. I have also used one estimation through a fourth-order polynomial and it didn't work out too. $\endgroup$
    – Boajj04
    Sep 24, 2019 at 5:49

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