MATLAB Answers

lsqcurvefit not adjusting some parameters as expected

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Leor Greenberger
Leor Greenberger on 26 Jul 2019
Answered: Alex Sha on 22 Feb 2020
I am trying to fit some data to a 3 parameter curve expressed as .
What I am finding when using lsqcurvefit is that it converges on what appears to be good values for and but not for c. In fact, c remains unchanged from the guessed value, regardless of its initial value.
First I try with c = p(3) = 1000
x = 0:2047;
y = load('y.mat'); % see attachment
fun = @(p,x)abs(p(1)*(x-p(3))+p(2)*(x-p(3)).^3);
options = optimoptions(@lsqcurvefit,'StepTolerance',1e-10, 'Display', 'iter-detailed', 'FunctionTolerance', 1E-12);
pGuess = [2.5E-5 2.6E-12 1000];
[p,fminres] = lsqcurvefit(fun,pGuess,x,y, [], [], options)
Norm of First-order
Iteration Func-count f(x) step optimality
0 4 0.00635297 1.03e+08
1 8 0.00632173 3.02663e-13 99.9
2 12 0.00623569 9.85058e-05 1.13e+04
3 16 0.00623569 3.31307e-17 0.0114
Optimization stopped because the relative sum of squares (r) is changing
by less than options.FunctionTolerance = 1.000000e-12.
p =
2.58614068198555e-05 1.75916049599052e-12 1000.00009850203
fminres =
0.00623568562340633
Now I try with c = p(3) = 1400:
pGuess = [2.5E-5 2.6E-12 1400];
[p,fminres] = lsqcurvefit(fun,pGuess,x,y, [], [], options)
Norm of First-order
Iteration Func-count f(x) step optimality
0 4 0.168905 9.79e+09
1 8 0.10571 6.45539e-12 1.16e+06
2 12 0.105681 9.10932e-06 1.43e+08
3 16 0.105681 0.000231759 1.43e+08
4 20 0.105681 5.79397e-05 1.43e+08
5 24 0.105681 1.44849e-05 1.43e+08
6 28 0.105681 3.62123e-06 1.43e+08
7 32 0.105681 9.05308e-07 1.43e+08
8 36 0.105681 2.26327e-07 1.43e+08
9 40 0.105681 5.65817e-08 1.43e+08
10 44 0.105681 1.41454e-08 1.43e+08
11 48 0.105681 3.53636e-09 1.43e+08
12 52 0.105681 8.8409e-10 1.43e+08
13 56 0.105681 2.21022e-10 1.43e+08
14 60 0.105681 5.52556e-11 1.43e+08
Optimization stopped because the norm of the current step, 5.525560e-11,
is less than options.StepTolerance = 1.000000e-10.
p =
2.4936115641411e-05 -3.9025705054523e-12 1399.9999908909
fminres =
0.105680830464589
From the data set it is clear that the min occurs at x = 1066.
>> [m,k] = min(y)
m =
0.000292016392169768
k =
1067
>> x(k)
ans =
1066

  3 Comments

Leor Greenberger
Leor Greenberger on 26 Jul 2019
I knew I forgot something! It is attached now. Thanks for looking into this!
Walter Roberson
Walter Roberson on 27 Jul 2019
Your function would benefit from a constraint.
fun = @(p,x)abs(p(1)*(x-p(3))+p(2)*(x-p(3)).^3);
if you feed in -p(1) and -p(2) then you the result will have the same abs() as original p(1) and p(2) . Therefore you can constrain one of the values, such as p(1) to be >= 0, which will reduce searching.

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Accepted Answer

Matt J
Matt J on 27 Jul 2019
It helps to pre-normalize your x,y data and to use polyfit to generate a smart initial guess,
%data pre-normalization
y=y.'/max(y);
[~,imin]=min(y);
x=(x-x(imin))/max(x);
%generate initial guess
p0=polyfit(x(imin:end),y(imin:end),3);
pGuess = [p0(1),p0(3), 0];
fun = @(p,x)abs(p(1)*(x-p(3))+p(2)*(x-p(3)).^3);
options = optimoptions(@lsqcurvefit,'StepTolerance',1e-10, 'Display', 'iter-detailed', 'FunctionTolerance', 1E-12);
[p,fminres,~,ef] = lsqcurvefit(fun,pGuess,x,y, [], [], options)
plot(x,y,'o-',x(1:20:end),fun(p,x(1:20:end)),'x--y'); shg
untitled.png

  3 Comments

Matt J
Matt J on 27 Jul 2019
Strictly speaking, your model function is "illegal" because it has no Jacobian at
p(1)*(x-p(3))+p(2)*(x-p(3)).^3=0
However, that doesn't seem to harm the fit...
Leor Greenberger
Leor Greenberger on 27 Jul 2019
Thank you very much for helping with this! The normalization you did is interesting. Ultimately what I need to do is create an algorithm that can find the inflection point with the least number of data samples. That is, I have a digital system with an 11 bit register and what you see here is my characterization of it from code 0 to 2047. I set the register and measured the output of the system. I wonder if this will work if I have enough samples and guess min(y).
Matt J
Matt J on 27 Jul 2019
You're welcome, but please Accept-click the answer if you are satisfied that the fitting code is now working.

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More Answers (1)

Alex Sha
Alex Sha on 22 Feb 2020
The best solution seems to be:
Root of Mean Square Error (RMSE): 8.58712889537096E-5
Sum of Squared Residual: 1.50943288116718E-5
Correlation Coef. (R): 0.999947601349384
R-Square: 0.999895205444387
Adjusted R-Square: 0.999895102905683
Determination Coef. (DC): 0.99989229572485
Chi-Square: 0.00435400203642259
F-Statistic: 9515701.1004098
Parameter Best Estimate
---------- -------------
p1 -2.54821313582336E-5
p2 -2.52958254532916E-12
p3 1062.58024839597

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