# Fit NonLinearLeastSquares to data and constrain curve to pass through point (0,1)

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Zachary Nunn 2019 年 8 月 15 日

I have data I need to fit to an equation, which I can do, but I want the equation to pass through point (0,1).
The equation is an exponential y = a*exp(-x/b)+ c*exp(-x/d)+ e
Currently, this is my code:
f3 = fitoptions('Method','NonlinearLeastSquares','Startpoint',[1,100,1,100,1])
newrelax = fittype('q*exp(-x/r)+ s*exp(-x/t)+u','options',f3);
[h,gof] = fit(timeR,stressR,newrelax)
plot(h,timeR,stressR)
Thank you,
-Zach

#### 1 件のコメント

Torsten 2019 年 8 月 19 日
Use
y = a*exp(-x/b) + c*exp(-x/d) + (1-(a+c))
as equation.

### 採用された回答

Jyotsna Talluri 2019 年 8 月 19 日
You can use a “lsqlin” function from optimization toolbox.
t=coeffvalues(h);
C=[exp(-x/t(2)) exp(-x/t(4)) ones(size(x))];
D=y;
A = []; % No inequality constraint
B= []; % No inequality constraint
%Set the specified points in C and D to find equality constraints Aeq, Beq..By substituting (0,1) they turned out to be
Aeq =[1 1 1];Beq=;
l=lsqlin(C,D,A,B,Aeq,Beq); %generates coefficients of the curve fitted passing through a specified point
Plot(timeR,C*l);

### その他の回答 (1 件)

Matt J 2019 年 8 月 19 日

This answer incorporates Torsten's advice, but I also think you should re-formulate the model to make the exponential terms asymmetric. Otherwise, the solver cannot decide which exponential term belongs to q and which belongs to s.
f3 = fitoptions('Method','NonlinearLeastSquares',...
'Startpoint',[1,100,1,100,1],'Lower',[-inf,-inf,-inf,0]);
newrelax = fittype('q*exp(-x/r)+ s*exp(-x/(r+d))+(1-(q+s))','options',f3);