exponential fit not working

i am trying to fit V*(1-exp(-x*(R/L))) to the voltage(over time) of a RL circuit using the cftool custom equation

this is my data

0.140530000000000
0.462380000000000
0.672240000000000
0.807240000000000
0.893120000000000
0.947740000000000
0.982350000000000
1.00491000000000
1.02096000000000
1.02863000000000
1.03526000000000
1.03904000000000
1.04153000000000
1.04400000000000
1.04347000000000
1.04463000000000
1.04481000000000
1.04540000000000
1.04500000000000
1.04573000000000
1.04578000000000
1.04658000000000
1.04688000000000
1.04589000000000
1.04639000000000
1.04630000000000
1.04536000000000
1.04569000000000
1.04389000000000
1.04594000000000
1.04538000000000
1.04644000000000
1.04532000000000
1.04586000000000
1.04569000000000
1.04523000000000
1.04588000000000
1.04485000000000
1.04579000000000
1.04577000000000
1.04507000000000
1.04593000000000
1.04629000000000
1.04449000000000
1.04569000000000
1.04572000000000
1.04623000000000
1.04544000000000
1.04389000000000
1.04679000000000
1.04556000000000
1.04596000000000
1.04442000000000
1.04601000000000
1.04491000000000
1.04638000000000
1.04462000000000
1.04451000000000
1.04550000000000
1.04536000000000
1.04535000000000
1.04594000000000
1.04435000000000
1.04519000000000
1.04523000000000
1.04621000000000
1.04599000000000
1.04577000000000
1.04516000000000
1.04613000000000
1.04636000000000
1.04555000000000
1.04544000000000
1.04543000000000
1.04523000000000
1.04522000000000
1.04667000000000
1.04574000000000
1.04425000000000
1.04563000000000
1.04627000000000
1.04467000000000
1.04594000000000
1.04645000000000
1.04613000000000
1.04522000000000
1.04593000000000
1.04610000000000
1.04582000000000
1.04540000000000
1.04569000000000
1.04572000000000
1.04546000000000
1.04521000000000
1.04589000000000
1.04644000000000
1.04477000000000
1.04472000000000
1.04597000000000
1.04579000000000

x data

1.21000000000000
1.21500000000000
1.22000000000000
1.22500000000000
1.23000000000000
1.23500000000000
1.24000000000000
1.24500000000000
1.25000000000000
1.25500000000000
1.26000000000000
1.26500000000000
1.27000000000000
1.27500000000000
1.28000000000000
1.28500000000000
1.29000000000000
1.29500000000000
1.30000000000000
1.30500000000000
1.31000000000000
1.31500000000000
1.32000000000000
1.32500000000000
1.33000000000000
1.33500000000000
1.34000000000000
1.34500000000000
1.35000000000000
1.35500000000000
1.36000000000000
1.36500000000000
1.37000000000000
1.37500000000000
1.38000000000000
1.38500000000000
1.39000000000000
1.39500000000000
1.40000000000000
1.40500000000000
1.41000000000000
1.41500000000000
1.42000000000000
1.42500000000000
1.43000000000000
1.43500000000000
1.44000000000000
1.44500000000000
1.45000000000000
1.45500000000000
1.46000000000000
1.46500000000000
1.47000000000000
1.47500000000000
1.48000000000000
1.48500000000000
1.49000000000000
1.49500000000000
1.50000000000000
1.50500000000000
1.51000000000000
1.51500000000000
1.52000000000000
1.52500000000000
1.53000000000000
1.53500000000000
1.54000000000000
1.54500000000000
1.55000000000000
1.55500000000000
1.56000000000000
1.56500000000000
1.57000000000000
1.57500000000000
1.58000000000000
1.58500000000000
1.59000000000000
1.59500000000000
1.60000000000000
1.60500000000000
1.61000000000000
1.61500000000000
1.62000000000000
1.62500000000000
1.63000000000000
1.63500000000000
1.64000000000000
1.64500000000000
1.65000000000000
1.65500000000000
1.66000000000000
1.66500000000000
1.67000000000000
1.67500000000000
1.68000000000000
1.68500000000000
1.69000000000000
1.69500000000000
1.70000000000000
1.70500000000000

[Walter Roberson]

When you see R or L passing through zero like that, and the range being pretty much symmetric, then the values should be considered garbage.

You get the large range of R and L values like that because it is only possible to obtain the ratio R/L for the reasons I explained below: that any constant multiple of both of them gives the same ratio.

[edo mor]

but the fit line dosen't agree with the data and i dont know how to make it fit

Answer 1

[Star Strider]

If you provide a delay offset for ‘x’ (that is, ‘t’) it will provide an acceptable fit:

vfcn = @(b,x) b(1).*(1-exp(-(x+b(4)).*(b(2)./b(3))));
vftns = @(b) norm(v - vfcn(b,t));
opts = optimoptions('ga', 'PopulationSize',5000, 'InitialPopulationMatrix', randi(1000, 5000, 4))
[B,fval] = ga(vftns, 4, [], [], [], [], [1 1 1 -Inf], [1 1 1 1]*Inf, [], [], opts)

figure(1)
plot(t,v, 'p')
hold on
plot(t, vfcn(B,t), '-r')
grid

B =
     1.0479e+000   640.7085e+000     9.3901e+000    -1.2070e+000

fval =
   153.3155e-003

I cannot get ga (link) to provide an acceptable fit without adding the time-delay parameter, ‘b(4)’.

Answer 2

[Walter Roberson]

To fit

 YourData = V*(1-exp(-x*(R/L)))

then we need the vector of independent variables (presumably x).

Unless you have a second independent variable:

• if x is your independent variable you can find R/L but not either alone
• if R is your independent variable you can find x/L but not either alone
• if L is your independent variable you can find x*R but not either alone

g = fittype('V*(1-exp(-x*RoL))');
result = fit((1:length(YourData)).',YourData,g)

result = 
   General model:
   result(x) = V*(1-exp(-x*RoL))
   Coefficients (with 95% confidence bounds):
     RoL =      0.3373  (0.3227, 0.352)
     V =       1.048  (1.044, 1.052)

Here, RoL is the ratio R/L . You cannot distinguish the two because for any non-zero constaint A, (A*R)/(A*L) = R/L, so increasing both by the same factor would give you the same fit.

[Walter Roberson]

g = fittype('V*(1-exp(-x*RoL))');
result = fit(t,v,g)

result = 
   General model:
   result(x) = V*(1-exp(-x*RoL))
   Coefficients (with 95% confidence bounds):
     RoL =       1.094  (0.5239, 1.664)
     V =       1.284  (1.012, 1.557)

This V agrees with the one you already found (1.285), and gives specific values for the ratio R/L (1.094). However as indicated, you cannot tell what R or L is individually without more information.

[edo mor]

this is exactly what i got in cftool but the fit line dosent agree with the data

[Walter Roberson]

None the less, that is the closest fit of the data to the model.

I ran a quite different fitting routine against that model, and found that to within a relative small margin, the parameters are 1.28450098237226129, 1.09411994131928014 . I had my routine search a pretty large parameter space.

The implication is that that equation does not fit that data.

If you examine the graph of your data, you can see that it must approach 0 somewhere close to t = 1.2. But that model approaches 0 at t = 0 and climbs rapidly from there.

Now, if you change the model to

v = V*(1-exp(-(t-t0)*(R/L)))

and search for V, R/L, and t0, then you can get excellent agreement

V = 1.04564508272744927; RoL = 88.8009466750188494; t0 = 1.20838792927323047;
f = @(t) V*(1-exp(-(t-t0)*RoL));
plot(t, v, 'k--', t, f(t), 'b.')