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Lsqnonlin to determine coefficient

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Anand Rathnam
Anand Rathnam on 7 Jun 2021
Edited: Matt J on 7 Jun 2021
I am looking to fit data to a complex equation using lsqnonlin solver. I am not sure where I am going wrong. I keep getting a response "not enough input arguments"
% Approach
% 1. Creating the following function named fit_simp.m which uses the time and Ar data(Ar = At/Ainf).
% 2. time and Ar are passed into lsqnonlin as input arguments.
% 3. Use the time and n data to calculate values values (Ar) the diffusion equation, and subtract the original Ar data from this.
% 4. The result will be the difference between the experimental data and the calculated values.
% 5. The lsqnonlin function will minimize the sum of the squares of the differences.
% 6. Condensation of the diffusion equation:
% a=0.0008;
% n1 = 2.43;
% n2= 1.4;
% Lambda = 950;
% theta = 45;
% gama = (2*n1*pi*sqrt ((sin(theta))^2-(n2/n1)^2))/(Lambda)
% Above entered in: 1-(8*gama/pi*(1-exp(-2*gama*a)))*((exp(-D*(2*n+1)^2*pi^2*t/4*a^2))*((-1)^n*2*gama +(((2*n+1)*pi)/(2*a))*exp(-2*gama*a))/((2*n+1)*((4*gama^2)+((2*n+1)*pi/2*a)^2)));
function diff = fit_simp(x,~,~) % This function is called by lsqnonlin.
% x is a vector which contains the coefficient of the equation.
% time and Ar are the option data sets that are passed to lsqnonlin
% Defining the data sets that you are trying to fit the function to.
Ar = [0.3 0.2 0.28 0.318 0.421 0.492 0.572 0.55 0.63 0.61 0.73 0.8 0.81 0.84 0.93 0.91]';
l = length(Ar);
t = [0:l-1]';
plot(t,Ar,'ro')
title('Data points')
% D is the coefficient we are looking to determine
%*******************************************************************************
a=0.0008;
n1 = 2.43;
n2=1.4;
Lambda = 950;
theta = 45;
d= x(1);
gama = (2*n1*3.14*sqrt ((sin(theta))^2-(n2/n1)^2))/(Lambda);
time=12;
for t = 1:time
for n=0:time
r = 1-((8*gama/pi)*(1-exp(-2*gama*a)))*((exp(-d.*(2*n+1)^2*pi^2*t/4*a^2))*((-1)^n*2*gama +(((2*n+1)*pi)/(2*a))*exp(-2*gama*a))/((2*n+1)*((4*gama^2)+((2*n+1)*pi/2*a)^2)));
end
result(t) = r;
end
%*******************************
diff = result - Ar;
% Initialize the coefficients of the function.
X0=[1];
% Calculate the new coefficients using LSQNONLIN.
x=lsqnonlin(@fit_simp,X0,t,Ar);
% Plot the original and experimental data.
for t = 1:time
for n=0:time
Ar_new = 1-(8*gama/pi*(1-exp(-2*gama*a)))*((exp(-x(1).*(2*n+1)^2*pi^2*t/4*a^2))*((-1)^n*2*gama +(((2*n+1)*pi)/(2*a))*exp(-2*gama*a))/((2*n+1)*((4*gama^2)+((2*n+1)*pi/2*a)^2)))
end
Ar_newv(t)=Ar_new;
end
plot(time,Ar,'+r',time,Ar_newv,'b')
  4 Comments
Matt J
Matt J on 7 Jun 2021
@Anand Rathnam Note that your loop
for n=0:time
r = 1-((8*gama/pi)*...
end
is not doing anything except repeatedly over-writing r. It is not updating r in any way. You have a similar loop later in your posted code with the same problem.

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