Curve fitting for multiple variable

I am trying to fit the curves as shown below. Equivalent resistance R is varying with SoC. But I have third variable Vi (0.8 -1p.u) which shifts the R downwards. I am trying to find a general fit for these curves such that F ( SoC, Vi).

I have attached data as well if anyone can help me out or any idea to approach this problem. I have tried with Cftool but the data dimensions are not same .

[Matt J]

It would be easiest if you showed us the model function you want applied to the different curves.

[Haroon Zafar]

I have no specific requirement of the function type. Using Cftool I found first order power function to be satisfactory for individual curves. But how can I include the effect of Vi , which is shifting curves downwards and varies from 0.8-1.0.

Answer 1

[Star Strider]

I have no clear idea what you want to do.

This approach fits all the parameters at once —

T1 = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1201063/ReqFittingData.xlsx', 'VariableNamingRule','preserve')

T1 = 26×6 table
    SoC    0.8Vi     0.85Vi    0.9Vi     0.95Vi     1Vi  
    ___    ______    ______    ______    ______    ______

     5     60.989    57.342    54.322    51.526    50.204
    10     60.212    56.618    53.619    50.862     49.53
    15     59.832    56.268    53.278     50.53    49.205
    20     59.334    55.776    52.813    50.061    48.783
    25     58.914    55.404    52.422    49.712    48.434
    30     58.567    55.057    52.131    49.431    48.144
    35     58.187    54.701    51.772    49.087    47.813
    40     58.012    54.567    51.637    48.956    47.708
    45      57.91    54.439    51.523     48.86     47.59
    50      57.79    54.318    51.407    48.758    47.474
    55     57.644    54.191    51.278    48.613    47.346
    60     57.532    54.066    51.181    48.509    47.242
    65     57.384    53.956     51.07    48.427    47.143
    70      57.25     53.79    50.946     48.31    47.033
    75     57.157    53.707    50.825    48.184    46.918
    80     57.075     53.64    50.785    48.141    46.864

VN = T1.Properties.VariableNames

VN = 1×6 cell array
    {'SoC'}    {'0.8Vi'}    {'0.85Vi'}    {'0.9Vi'}    {'0.95Vi'}    {'1Vi'}

Viv = regexp([VN{:}], '(\d*\.\d*)|\d*', 'match')

Viv = 1×5 cell array
    {'0.8'}    {'0.85'}    {'0.9'}    {'0.95'}    {'1'}

Viv = str2double(Viv)

Viv = 1×5

0.8000 0.8500 0.9000 0.9500 1.0000

Viym = T1{:,2:6};
Viyv = Viym(:);                                                 % 'Vi' Vector

Vivm = repmat(Viv, size(T1,1), 1);
Vivv = Vivm(:);                                                 % 'Viv' Vector

y = Viyv;
SoCv = repmat(T1.SoC, size(Viym,2), 1)                           % 'SoC' Vector

SoCv = 130×1

5 10 15 20 25 30 35 40 45 50

x = [SoCv Vivv];                                                % Concatenate Vectors

objfcn = @(b,x) exp(b(1)).*x(:,1).^b(2) + b(3).*x(:,2);

% [B,resn] = lsqcurvefit(objfcn, rand(3,1), x, y)

mdl = fitnlm(x, y, objfcn, rand(3,1))

mdl = 
Nonlinear regression model:
    y ~ exp(b1)*x1^b2 + b3*x2

Estimated Coefficients:
          Estimate         SE         tStat       pValue   
          _________    __________    _______    ___________

    b1       4.6505     0.0067979     684.11    2.3549e-228
    b2    -0.014195    0.00067264    -21.103     2.4122e-43
    b3      -52.331       0.72897    -71.788    1.1374e-104


Number of observations: 130, Error degrees of freedom: 127
Root Mean Squared Error: 0.588
R-Squared: 0.978,  Adjusted R-Squared 0.977
F-statistic vs. zero model: 3.4e+05, p-value = 1.03e-247

An alternative approach uses a loop to fit each column individually —

B0 = mdl.Coefficients.Estimate;

for k = 1:size(Viym,2)
    x = [T1.SoC Vivm(:,k)]
    y = Viym(:,k)
    mdl = fitnlm(x, y, objfcn, B0)
end

x = 26×2

5.0000 0.8000 10.0000 0.8000 15.0000 0.8000 20.0000 0.8000 25.0000 0.8000 30.0000 0.8000 35.0000 0.8000 40.0000 0.8000 45.0000 0.8000 50.0000 0.8000

y = 26×1

60.9887 60.2122 59.8316 59.3342 58.9138 58.5672 58.1869 58.0121 57.9105 57.7896

Warning: Iteration limit exceeded. Returning results from final iteration.

mdl = 
Nonlinear regression model:
    y ~ exp(b1)*x1^b2 + b3*x2

Estimated Coefficients:
          Estimate        SE        tStat      pValue 
          _________    ________    ________    _______

    b1       5.0195       3.948      1.2714    0.21629
    b2    -0.010636    0.043594    -0.24397    0.80942
    b3       -109.3      746.47    -0.14643    0.88486


Number of observations: 26, Error degrees of freedom: 23
Root Mean Squared Error: 0.117
R-Squared: 0.992,  Adjusted R-Squared 0.991
F-statistic vs. constant model: 1.37e+03, p-value = 1.25e-24

x = 26×2

5.0000 0.8500 10.0000 0.8500 15.0000 0.8500 20.0000 0.8500 25.0000 0.8500 30.0000 0.8500 35.0000 0.8500 40.0000 0.8500 45.0000 0.8500 50.0000 0.8500

y = 26×1

57.3419 56.6183 56.2681 55.7763 55.4043 55.0568 54.7007 54.5669 54.4393 54.3177

Warning: Iteration limit exceeded. Returning results from final iteration.

mdl = 
Nonlinear regression model:
    y ~ exp(b1)*x1^b2 + b3*x2

Estimated Coefficients:
          Estimate        SE        tStat      pValue 
          _________    ________    ________    _______

    b1       4.9841      4.0116      1.2424     0.2266
    b2    -0.010412    0.043331     -0.2403    0.81223
    b3      -101.15      689.07    -0.14679    0.88457


Number of observations: 26, Error degrees of freedom: 23
Root Mean Squared Error: 0.11
R-Squared: 0.992,  Adjusted R-Squared 0.991
F-statistic vs. constant model: 1.38e+03, p-value = 1.08e-24

x = 26×2

5.0000 0.9000 10.0000 0.9000 15.0000 0.9000 20.0000 0.9000 25.0000 0.9000 30.0000 0.9000 35.0000 0.9000 40.0000 0.9000 45.0000 0.9000 50.0000 0.9000

y = 26×1

54.3216 53.6187 53.2782 52.8128 52.4222 52.1308 51.7724 51.6367 51.5228 51.4068

Warning: Iteration limit exceeded. Returning results from final iteration.

mdl = 
Nonlinear regression model:
    y ~ exp(b1)*x1^b2 + b3*x2

Estimated Coefficients:
          Estimate        SE        tStat      pValue 
          _________    ________    ________    _______

    b1       4.9387      3.9118      1.2625    0.21941
    b2    -0.010408    0.042233    -0.24645    0.80752
    b3      -91.846      606.51    -0.15143    0.88095


Number of observations: 26, Error degrees of freedom: 23
Root Mean Squared Error: 0.102
R-Squared: 0.992,  Adjusted R-Squared 0.991
F-statistic vs. constant model: 1.46e+03, p-value = 6.04e-25

x = 26×2

5.0000 0.9500 10.0000 0.9500 15.0000 0.9500 20.0000 0.9500 25.0000 0.9500 30.0000 0.9500 35.0000 0.9500 40.0000 0.9500 45.0000 0.9500 50.0000 0.9500

y = 26×1

51.5259 50.8622 50.5304 50.0606 49.7115 49.4312 49.0868 48.9559 48.8601 48.7579

Warning: Iteration limit exceeded. Returning results from final iteration.

mdl = 
Nonlinear regression model:
    y ~ exp(b1)*x1^b2 + b3*x2

Estimated Coefficients:
          Estimate        SE        tStat      pValue 
          _________    ________    ________    _______

    b1       4.9102      3.9509      1.2428    0.22647
    b2    -0.010229    0.041892    -0.24418    0.80926
    b3      -85.939      564.01    -0.15237    0.88022


Number of observations: 26, Error degrees of freedom: 23
Root Mean Squared Error: 0.0967
R-Squared: 0.992,  Adjusted R-Squared 0.992
F-statistic vs. constant model: 1.48e+03, p-value = 5.01e-25

x = 26×2

5 1 10 1 15 1 20 1 25 1 30 1 35 1 40 1 45 1 50 1

y = 26×1

50.2041 49.5297 49.2049 48.7825 48.4339 48.1437 47.8131 47.7076 47.5903 47.4742

Warning: Iteration limit exceeded. Returning results from final iteration.

mdl = 
Nonlinear regression model:
    y ~ exp(b1)*x1^b2 + b3*x2

Estimated Coefficients:
          Estimate        SE        tStat      pValue 
          _________    ________    ________    _______

    b1       4.9107      3.9557      1.2414    0.22696
    b2    -0.010039    0.041135    -0.24405    0.80936
    b3      -83.087      536.75     -0.1548    0.87833


Number of observations: 26, Error degrees of freedom: 23
Root Mean Squared Error: 0.0933
R-Squared: 0.993,  Adjusted R-Squared 0.992
F-statistic vs. constant model: 1.54e+03, p-value = 3.3e-25

The ‘objfun’ function uses a simple power relation and adds a term for the ‘Vi’ value.

I am not certain how the ‘Vi’ values relate the the ‘Eqivalent Resistance’, or for that matter what these data are. So this is essentially a prototype approach to demonstrate how to incorporate all the necessary data, even though I’m not certain how to correctly incorporate it.

I defer to you to determine the desiured result, and how to actually get it.

.

[Haroon Zafar]

@Star Strider Thanks a lot! Approach 1 exactly does what I was looking for as I wanted to implement one general equation for all the curves and include impact of Vi.

[Star Strider]

As always, my pleasure!

[Alex Sha]

try the model function below, where, x1=SoC, x2=Vi, y=Req

Sum Squared Error (SSE): 0.426970556518359
Root of Mean Square Error (RMSE): 0.0573095881706985
Correlation Coef. (R): 0.999892090395877
R-Square: 0.999784192436237

Parameter	Best Estimate       
---------	-------------       
p1       	-0.0700360286852488 
p2       	0.000891267300607825
p3       	0.0904543618535223  
p4       	248.161725009268    
p5       	2.57600192708312    
p6       	-0.799835129723361  
p7       	1.10012417863406    
p8       	-10036.5979452884   
p9       	-3.96886303014785   
p10      	-239.992195659297   
p11      	3.75637702611614    
p12      	0.183843198467239   
p13      	10069.7969832811

if want simple model function:

Sum Squared Error (SSE): 1.61466290688424
Root of Mean Square Error (RMSE): 0.111447224725919
Correlation Coef. (R): 0.999591859922242
R-Square: 0.999183886422807

Parameter	Best Estimate       
---------	-------------       
b0       	-190.708200295556   
b1       	-0.140261199618056  
b2       	0.00181016348907718 
b3       	-8.89079975979054E-6
b4       	1050.98456432426    
b5       	-1362.74435026589   
b6       	553.465126422919 

[Haroon Zafar]

Thanks a lot Alex. Is there any english version of 1stOpt avaialble? I could only find chinese versions and they just get stuck in between fitting.