least square fit "lsqcurvefit" not good enough?

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sxh 2023-7-5

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https://ww2.mathworks.cn/matlabcentral/answers/1991823-least-square-fit-lsqcurvefit-not-good-enough

评论： sxh 2023-7-6

采纳的回答： Mathieu NOE

0SM.CSV

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Hi All,

Trying to solve "K_a" by least square fitting an equation "func" to a dataset "backgroundImpIm". Bu when I plot them together, the fit doesn't seem good. I have tried to tweak FunctionTolerance and StepTolerance but the fit doesn't improve. Anyone have answers?

Thanks,

clc

close all

clear

background = csvread('0SM.CSV',3,0,[3,0,1603,2]);

backgroundImpIm = background(:,3);

f = background(:,1);

freq = f(1:100);

x0 = [1];

h = 5E-3;

a = 5E-5;

m_e = 9.1E-31;

q = 1.6E-19;

c = 3E8;

omega = 2*pi*freq;

k_0 = omega/c;

beta = k_0;

epsilon_real = 1;

epsilon_im = 0;

x = epsilon_im./epsilon_real;

func = @(K_a,freq)(-K_a./(tan(2*pi*freq*h.*(1 + 0.19./((K_a/60 +1) - 0.81))/c)));

%options = optimoptions(@lsqcurvefit,'FunctionTolerance',1e-100)

K_a = lsqcurvefit(func,x0,freq,backgroundImpIm(1:100))

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.

K_a = 5.6540

plot(freq,backgroundImpIm(1:100))

hold on

plot(freq,func(K_a,freq))

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Answer 1

Mathieu NOE 2023-7-5

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hello

you can get better results if you start the fit not at index 1 , but a bit above, as your first sample seems a bit off and your data lack some resolution in the lower freq range

this appears more clearly also if you plot the results with a x log spacing

also I am limited to the use of fminsearch as I don't have the optimization toolbox , but you can easily swith back to your original code

Result :

K_a = 50.0846

background = csvread('0SM.CSV',3,0,[3,0,1603,2]);
backgroundImpIm = background(:,3);
f = background(:,1);
ind = (2:100); % <= HERE , start at 2 at lowest , discard 1st sample
freq = f(ind);
x0 = [1];
h = 5E-3;
a = 5E-5;
m_e = 9.1E-31;
q = 1.6E-19;
c = 3E8;
omega = 2*pi*freq;
k_0 = omega/c;
beta = k_0;
epsilon_real = 1;
epsilon_im = 0;
x = epsilon_im./epsilon_real;
func = @(K_a,freq)(-K_a./(tan(2*pi*freq*h.*(1 + 0.19./((K_a/60 +1) - 0.81))/c)));
%options = optimoptions(@lsqcurvefit,'FunctionTolerance',1e-100)
% K_a = lsqcurvefit(func,x0,freq,backgroundImpIm(1:100))
K_a = fminsearch(@(x) norm(func(x,freq)-backgroundImpIm(ind)),x0)
semilogx(freq,backgroundImpIm(ind),'*-r',freq,func(K_a,freq),'b')
legend('raw data,','fit');

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Mathieu NOE 2023-7-5

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Another way to demonstrate that your fit shoud not include the first data point

background = csvread('0SM.CSV',3,0,[3,0,1603,2]);
backgroundImpIm = background(:,3);
f = background(:,1);
ind = (2:1600); % <= HERE , start at 2 at lowest , discard 1st sample
freq = f(ind);
x0 = [1];
h = 5E-3;
a = 5E-5;
m_e = 9.1E-31;
q = 1.6E-19;
c = 3E8;
omega = 2*pi*freq;
k_0 = omega/c;
beta = k_0;
epsilon_real = 1;
epsilon_im = 0;
x = epsilon_im./epsilon_real;
func = @(K_a,freq)(-K_a./(tan(2*pi*freq*h.*(1 + 0.19./((K_a/60 +1) - 0.81))/c)));
%options = optimoptions(@lsqcurvefit,'FunctionTolerance',1e-100)
% K_a = lsqcurvefit(func,x0,freq,backgroundImpIm(1:100))
K_a = fminsearch(@(x) norm(func(x,freq)-backgroundImpIm(ind)),x0)
% semilogx(freq,backgroundImpIm(ind),'*-r',freq,func(K_a,freq),'b')
semilogx(f,backgroundImpIm,'*-r',freq,func(K_a,freq),'b')
legend('raw data,','fit');