Data fitting weibull and integration

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Nuno
Nuno on 18 Apr 2011
Hi all, first of all let me say that i'm a bit inexperienced regrading programming and Matlab. I have made a m file with a bit of code that tries to fit a weibull distribution function to some data from a excel file. this part is done by the code but i also want to do a definite integral on that function multiplied by x but i always get the same error, says that cannot find explicit integral... i don't know if the problem is because of the function hand le or other thing... any suggestions?
CODE:
clc
% Vectors (x, fi) to create histogram with the experimental data for length
%%#1
% Go to excel spreadsheet and retrieve the data for x, fi, vectors
fi = xlsread('C:\Users\Nuno\Documents\MATLAB\dados_raiz_working.xlsx','Dados raiz','f4:f42');
x = xlsread('C:\Users\Nuno\Documents\MATLAB\dados_raiz_working.xlsx','Dados raiz','b4:b42');
Lmean=xlsread('C:\Users\Nuno\Documents\MATLAB\dados_raiz_working.xlsx','Weibull','o7')
%bar graph
bar(x,fi)
fi=fi';
x=x';
%%#2
% Weibull function
modelFun = @(p,x) p(3) .* (x ./ p(1)).^(p(2)-1) .* exp(-(x ./ p(1)).^p(2));
startingVals = [2.6057 0.67657 1];
coefEsts = nlinfit(x', fi', modelFun, startingVals)
xgrid = x;
%%#3
%Writes on excel spreadsheet the data obtained from weibull distribuiton
write_column_excel=xlswrite('C:\Users\Nuno\Documents\MATLAB\dados_raiz_working.xlsx',modelFun(coefEsts, xgrid)', 'Weibull', 'D3');
%%#4
% line graph
line(xgrid, modelFun(coefEsts, xgrid), 'Color','red','LineWidth',2.5)
title('\fontsize{14}Histograma com distribuição de Weibull')
xlabel('\fontsize{12}Comprimentos(mm)');
ylabel('\fontsize{12}Frequência (%)')
h = legend('Dados Raiz','Distribuição Weibull',1);
set(h,'Interpreter','none')
text(1.28,6,[{'\bf \fontsize{12}Goodness of fit'}])
text(1.3,5,['SSE=19.65'])
text(1.3,4.4,['R^2=0.94'])
text(1.3,3.8,['RMSE=0.7387'])
hold on
%%#5
%In here i try to determine a specific coeficient that is obtained by definite integration between 0 and inf of the weibull function multiplied by "x" and (1/Lmean) (PROBLEM!!).
a=coefEsts(1); %Coeficientes dados pelo método de regressão linear
b=coefEsts(2); %na célula 3
c=coefEsts(3);
syms x
modelFun = @(x) c .* (x ./a).^(b-1) .* exp(-(x ./ a).^b);
Coef_ajust_length=eval(int(modelFun(x)*x*(1/Lmean), x, 0, inf));
  1 Comment
Nuno
Nuno on 19 Apr 2011
I've changed the code as proecsm suggested but it gives the same error, "explicit integral could not be found".... the error occurs on the #5 cell when i try to do the integration of "modelFun(x)*x" if i do it without the multiplication by "x" it doesn't give me errors...

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Accepted Answer

bym
bym on 19 Apr 2011
I had no problem getting an answer using assumed values for a,b,c. Upon closer inspection, you are using x as both a numeric variable as in :
x = xlsread('C:\Users\Nuno\Documents\MATLAB\dados_raiz_working.xlsx','Dados raiz','b4:b42');
and as symbolic
syms x
I would suggest using different variables names and see if the warning goes away. As for you comment about the fractions, symbolic computations return answers as exact rational expressions instead of floating point approximations (i.e. 1/3 instead of .33333...)

More Answers (2)

bym
bym on 18 Apr 2011
you have not defined the variable p in your function. just use
ModelFun = @(x)c .* (x ./a).^(b-1) .* exp(-(x ./ a).^b)
  2 Comments
bym
bym on 18 Apr 2011
also, drop the eval() from your integration
Nuno
Nuno on 19 Apr 2011
If i dont use eval() the result doesn't appear, but instead a fraction... like 1234567890/0987654321...! don't know why this happens...

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Nuno
Nuno on 19 Apr 2011
Hi, just to let you know that i think i solve my problem, in the integration part of my code, i jut add the double() part:
Coef_ajust_length=double(int(modelFun(x)*x*(1/Lmean), x, 0, inf));
the warning still appears, but i now get a result.
Don't know if this is the best way to solve the problem, but it seems to get the job done.
Thank you

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