fitting a function of three output

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Nadav Wetzler
Nadav Wetzler 2021-10-21
回答： MULI 2024-4-25
Hi everyone!
I need you advise on fitting coefficientts of a function that output 3 valus:
E = calcE(N, Ds, Dd, k), where E is a vector: E(1), E(2), E(3)
Ds, Dd and k are know. I need to find the set of N, where N is a vector N(1),N(2),N(3), that result E1,2,3 that is equal to a given e: e(1), e(2), e(3).
I tied:
function N1 = calcN(Ds, Dd, k, e)
b0 = [1, 1, 1];
fun = @(N)calcE(N,Ds, Dd,k)./e;
N1 = lsqcurvefit(fun,b0, N, e);
end
tryig to force E ./ e =1
But did't work...
with calcE
function E = calcE(N,Ds, Dd,k)
S(1) = -sqrt((1-N(1)^2-N(2)^2) / (1-N(2)^2+(N(1)^2)*((((k^2) * (1-N(1)^2))/N(2)^2)+2*k)));
S(2) = (S(1)*N(1)*k/N(2));
S(3) = -sqrt(1-S(1)^2 - S(2)^2);
E(1) = sqrt((N(1)*Dd(1))^2 + (S(1)*Ds(1))^2);
E(2) = sign(k)*sqrt((N(2)*Dd(2))^2 + (S(2)*Ds(2))^2);
E(3) = sqrt((N(3)*Ds(3))^2 + (S(3)*Dd(3))^2);
end
Thanks!
Nadav

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回答（1 个）

MULI
MULI 2024-4-25
Hi Nadav,
I understand you are trying to get optimized values of (N) which are used to compute vector (E) that should match with target vector (e).
The function "lsqcurvefit” is designed to minimize the difference between function's outputs and target values. It is typically more effective to minimize the absolute or squared differences than trying to make each ratio equal to 1.
Below is the modified code which you can try:
function N_opt = calcN(Ds, Dd, k, e)
b0 = [1, 1, 1]; % Initial guess for the N vector
% Objective function: modeled E based on N, Ds, Dd, k
fun = @(N, xdata) calcE(N, Ds, Dd, k); % x is not used in calcE but is required by lsqcurvefit syntax
% Use lsqcurvefit to find the N that minimizes the difference between calcE's output and e
N_opt = lsqcurvefit(fun, b0, [], e); % xdata is set to [] since it's not used
end
You may refer to this documentation for more information and examples related to “lsqcurvefit”.
https://www.mathworks.com/help/releases/R2020a/optim/ug/lsqcurvefit.html
Hope this answers your query!

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