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Use FindFit with an exponential function?

Mor Bo
Posted 7 years ago

I have been for several days now trying to FIndFit some data to a decaying e function:

k[t_] := \[Omega] E^(-((R t)/J))

This function describes a flywheel that starts with an initial velocity and falls to zero. I have a rough estimate of what the friction and the inertia (R and J respectively) of the system should be, but need to verify.

I have several datasets that I'm using to fit with and known initial/boundry conditions, this one for example:

data = {{0.04, 350.73}, {0.44, 342.26}, {0.84, 334.38}, {1.24, 
  326.29}, {1.64, 318.2}, {2.04, 310}, {2.44, 302.28}, {2.84, 
  294.72}, {3.24, 287.09}, {3.64, 280.08}, {4.04, 272.48}, {4.44, 
  265.03}, {4.84, 258.31}, {5.24, 251.15}, {5.64, 244.37}, {6.04, 
  237.75}, {6.44, 230.89}, {6.84, 225.03}, {7.24, 218.29}, {7.64, 
  212.28}, {8.04, 206.17}, {8.44, 199.7}, {8.84, 193.95}, {9.24, 
  188.05}, {9.64, 182.1}, {10.04, 176.14}, {10.44, 170.77}, {10.84, 
  165.5}, {11.24, 159.8}, {11.64, 154.4}, {12.04, 148.86}, {12.44, 
  143.87}, {12.84, 138.62}, {13.24, 133.58}, {13.64, 128.43}, {14.04, 
  123.22}, {14.44, 118.98}, {14.84, 113.85}, {15.24, 109.16}, {15.64, 
  104.38}, {16.04, 99.54}, {16.44, 95.2}, {16.84, 90.84}, {17.24, 
  86.46}, {17.64, 82}, {18.04, 77.62}, {18.44, 73.43}, {18.84, 
  69.11}, {19.24, 65.09}, {19.64, 61.05}, {20.04, 57.17}, {20.44, 
  53.06}, {20.84, 48.88}, {21.24, 45.03}, {21.64, 41.29}, {22.04, 
  37.4}, {22.44, 33.91}, {22.84, 30.4}, {23.24, 26.59}, {23.64, 
  23.02}, {24.04, 19.78}, {24.44, 16}, {24.84, 12.34}, {25.24, 
  9.28}, {25.64, 6.17}, {26.04, 4.1}};

plot2 = ListPlot[data, ImageSize -> Medium, PlotRange -> All, 
  PlotTheme -> "Scientific"]

correct

Nothing particularly special.

Whereby omega is equal to 350.73 and the duration of the decay is 26.04 seconds.

I'm trying to find the values for R and J, which should be in a specific range, which you can see in the following FindFit[].

fitparams = 
 Join[FindFit[
    data, {k[t] /. \[Omega] -> 350.73, {0.0001 <= R <= 0.0005, 
      0.0013 <= J <= 0.0018}}, {R, J}, t] // Flatten, {\[Omega] -> 350.73}]

With a result of:

{R\[Rule]0.00012762656668736746`,J\[Rule]0.0016248495392962767`,\
\[Omega]\[Rule]350.73`}

When trying to plot these to visualize the fit, I get the following plotting error(s):

In the image below, the function k[t] approaches zero at 26s like expected, however now the Listplot crosses the x axis just above 20 (and does not terminate at 26.04s like it should) and falls into the negative Y axis...which can't be possible, since there are no negative data points or even ones below 4, for that matter.

Show[Plot[{k[t] /. fitparams}, {t, 0, 26.04}], 
 ListPlot[Style[data, Red]], PlotTheme -> "Scientific"]

weird 1

If I switch the order of the plots, now the ListPlot plots correctly, and the function approaches 50 instead of Zero...which is completely wrong. 

   Show[ListPlot[Style[data, Red]], 
     Plot[{k[t] /. fitparams}, {t, 0, 26.04}], 
     PlotTheme -> "Scientific"]

weird 2

Both the function with calculated R and J and the data display/plot individually how I expect them to, However together with Show[] do not, regardless of order.

Why are the plots not aligning correctly when plotted together, but individually are fine? It's impossible to verify anything.

I greatly appreciate the help!
POSTED BY: Mor Bo
5 Replies
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POSTED BY: Sjoerd Smit
Jim Baldwin
Jim Baldwin, Retired
Posted 7 years ago

Please edit your code. Both time and sdata are undefined. Also, you mention that there are "plotting errors" but show no errors nor is there a description as to what the errors are.

The model and the data can only estimate R/J and not R and J separately. Evidence of this issue (beside observing the structure of the model) is that the estimate of the parameter correlation matrix is singular:

nlm = NonlinearModelFit[data, {k[t] /. \[Omega] -> 350.73, 
  {0.0001 <= R <= 0.0005,  0.0013 <= J <= 0.0018}}, {{R, 0.0003}, {J, 0.0015}}, t]
nlm["BestFitParameters"]
(* {R -> 0.000127627, J -> 0.00162485} *)
nlm["CorrelationMatrix"]
(* {{1., -1.}, {-1., 1.}} *)
POSTED BY: Jim Baldwin
Michael Helmle
Michael Helmle
Posted 7 years ago
POSTED BY: Michael Helmle
Mor Bo
Mor Bo
Posted 7 years ago

Ah Thank you for the clear explanation and the examples. I will definitely take it to heart. However, I still have the issue of trying to find approx. values for my R and J...

The original model, for a flywheel driven by electric motor is:

\[Phi][t] -> (t u \[Tau])/R - (E^(-((R t)/J)) J C[1])/R + C[2]

I took the derivative, to get the angular velocity function, as the angle of the flywheel is entirely unknown and not valuable. 

Derivative[1][\[Phi]][t] -> (u \[Tau])/R + E^(-((R t)/J)) C[1]

Here, u -> 0 so the first term falls away, C[1] is the initial velocity, R the friction, J the inertia...both values I have a good idea in the range they should be but I need something more accurate...

Do you have a suggestion on how I might find those values? As you say, my data does not give me a straight line when using ListLogPlot...Otherwise I'm at a loss on the best way to find them. Thanks again for the help, I appreciate it!
POSTED BY: Mor Bo
Mor Bo
Mor Bo
Posted 7 years ago

I have updated the code and described the 'errors' I meant...hopefully my problem is clearer now, Thank you however, for the explanation of the fitting for R/J...I will attempt to find another way to determine the values I need...Do you happen to know why the dataset and the function do not appear to be on the same scale...or atleast why my dataset is plotting in the negative Y, when using Show[] ?

Thanks again for the help!
POSTED BY: Mor Bo
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