Curve fitting of a portion of a plot and linear equilibrium of both plots

How do I curve fit of after equilbrium points in electron temperature profile so that they both equilibriate in a single line ?? MY code:

data1 = readmatrix('lat vs time.xlsx');
x1 = data1(:,1);
y1 = data1(:,2);
data2 = readmatrix('elec vs time.xlsx');
x2 = data2(:,1);
y2 = data2(:,2);
[k, yInf, y0, yFit] = fitExponential(x1, y1);
figure(1);
plot(x1,y1,'g',x2,y2,'r','linewidth',1.5);
hold on
plot(x1,yFit,'k','linewidth',2.5);
hold off

% apply corrective factor on fitted curve to math the other curve asymptote
y2_asymp = mean(y2(round(end/2):end));
correction_factor = y2_asymp/yFit(end);
yFit = yFit*correction_factor;
figure(2);
plot(x1,yFit,'g','linewidth',4);
hold on;
plot(x2,y2,'r','linewidth',2);
hold off;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [k, yInf, y0, yFit] = fitExponential(x, y)
% FITEXPONENTIAL fits a time series to a single exponential curve. 
% [k, yInf, y0] = fitExponential(x, y)
%
% The fitted curve reads: yFit = yInf + (y0-yInf) * exp(-k*(x-x0)).
% Here yInf is the fitted steady state value, y0 is the fitted initial
% value, and k is the fitted rate constant for the decay. Least mean square
% fit is used in the estimation of the parameters. 
% 
% Outputs:
% * k: Relaxation rate
% * yInf: Final steady state
% * y0: Initial state
% * yFit: Fitted time series
% 

% improve accuracy by subtracting large baseline
yBase           = y(1);

y               = y - y(1); 

shafin    = @(param) norm(param(2)+(param(3)-param(2))*exp(-param(1)*(x-x(1))) - y, 2);

initGuess(1)    = -(y(2)-y(1))/(x(2)-x(1))/(y(1)-y(end));
initGuess(2)    = y(end);
initGuess(3)    = y(1);
param           = fminsearch(shafin,initGuess);
k               = param(1);
yInf            = param(2) + yBase;
y0              = param(3) + yBase;
yFit            = yInf + (y0-yInf) * exp(-k*(x-x(1)));
end

[Shahriar]

electron and lattice curves

Answer 1

[Torsten]

How do I curve fit of after equilbrium points in electron temperature profile so that they both equilibriate in a single line ??

What are "both" ?

If you want to prescribe yInf, just remove param(2) from the parameters to be fitted and fix it as "equilibrium value - y(1)".

[Shahriar]

@Torsten i mean electron's scatterred part should be curve fitted to a straight line such that it can match with the lattice part(Green line graph 2) like this.Blue line is the electron curve fitted line.

[William Rose]

@Shahriar,

@Torsten's suggestion is a good one.

I see that you fit an exponential curve to the lattice data (noisy green trace on your plot 1). Then you apply a multiplier to the entire fitted curve so that its asymptotic value will match the asymptotic value of the electron data (red trace in your plot 1). This accomplishes your goal of having a model of the lattice, whose asymptotic value matches the experimental electron asymptote. What more do you want?

@Torsten suggests that you fix the asymptote of the latice model initially: set it equal to the electron asymptote. Adjust y0 and k when you fit the model. That works.

A third approach is to fit y0, yInf, and k, then adjust the fitted data by a vertical offset, instead of using a multiplicative factor.

The choice among these options should be guided by your understanding of the electron and lattice data, and why their measured asymptotes differ, and why you think the asymptotes should be equal.

Some code, with model S=Shahriar and model T=Torsten:

data1 = readmatrix('lat vs time.xlsx');     % x1,y1=lattice data
x1 = data1(:,1); y1 = data1(:,2);
data2 = readmatrix('elec vs time.xlsx');    % x2,y2=electron data
x2 = data2(:,1); y2 = data2(:,2);
%% Fit Model T
pT0=[1,100];                                % initial guess, model T
yInfT=mean(y2(round(end-2):end));           % yInf, model T
funT = @(p,x)yInfT+(p(2)-yInfT)*exp(-p(1)*(x-x(1)));    % model T function
fprintf('Model T initial guess: pT0(1)=%.2f, pT0(2)=%.1f.  yInf=%.1f\n',pT0,yInfT)

Model T initial guess: pT0(1)=1.00, pT0(2)=100.0.  yInf=1236.7

[pT,SSErrT] = lsqcurvefit(funT,pT0,x1,y1);             % fit pT=[k,y0]

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

RMSErrT=(SSErrT/length(y1)).^0.5;              % Model T root mean square error
yFitT=funT(pT,x1);

%% Fit Model S
pS0=[1,100,200];                            % initial guess, model S
funS = @(p,x)p(3)+(p(2)-p(3))*exp(-p(1)*(x-x(1)));    % model T function
fprintf('Model S initial guess: pS0(1)=%.2f, pS0(2)=%.1f, pS0(3)=%.1f.\n',pS0)

Model S initial guess: pS0(1)=1.00, pS0(2)=100.0, pS0(3)=200.0.

[pS,SSErrS] = lsqcurvefit(funS,pS0,x1,y1);             % fit pT=[k,y0]

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

yFitS=funS(pS,x1);
yFitSadj=yFitS*mean(y2(round(end/2):end))/yFitS(end);   % Model S fit, adjusted
RMSErrS=(sum((yFitSadj-y1).^2)/length(y1)).^0.5;        % Model S root mean square error

%% Display results on console
fprintf('Model T: k=%.3f, y0=%.1f, yInf=%.1f, RMS Err=%.1f\n',pT,yInfT,RMSErrT)

Model T: k=0.001, y0=1085.8, yInf=1236.7, RMS Err=291.3

fprintf('Model S: k=%.3f, y0=%.1f, yInf=%.1f, RMS Err=%.1f\n',pS,RMSErrS)

Model S: k=1.929, y0=306.6, yInf=1095.3, RMS Err=318.4

%% plot results
figure;
plot(x1,y1,'.g',x2,y2,'.r',x1,yFitT,'-m',x1,yFitSadj,'-b');
legend('Lattice','Electron','Model T','Model S'); grid on; 

In the fits above, I also tried using "intelligent" initial guesses, like you did. And I tried shifting the data by yBase before fitting, like you did. But these efforts did not alter the fits significantly.

The latice data does not allow reliable identification of the rate constant. I would sample faster, and for a shorter duration, to get a better estimate for the rate constant. And I would try to reduce the noisiness of the lattice data. But I assume you have already done that as much as possible.

Good luck @Shahriar.

[Shahriar]

@Torsten@William Rose thanks ,Got it.I was all along thinking about the electron profile .

[Torsten]

@William Rose

Sorry, it's your work that should have been accepted as answer here.

[William Rose]

@Torsten, no issue! I knew when I posted it that I was making a comment on your good answer.