error: matrix dimensions must agree

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sxh
sxh 2023-6-23
评论： sxh 2023-6-23
Hi
I am banging my head over this least square curve fitting even when following the simplest procedure I found. The error says,
"Error using /
Matrix dimensions must agree." - in line where the function 'func' is defined. Anyone have answers?
Thanks
clc
close all
clear
background = csvread(['C:\Users\shadh\Downloads\vnaSweep\ar2000mT\5.5mm length\0.1mmReCal\0w\0SM.csv'],3,0,[3,0,1603,2]);
backgroundImpIm = background(:,3);
f = background(:,1);
freq = f(100:1601);
x0 = [1,1];
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)));
K_a = lsqcurvefit(func,x0,freq,backgroundImpIm(100:1601))
  2 个评论
KSSV
KSSV 2023-6-23
ONly one csv file is uploaded....error line is not mentioned.
sxh
sxh 2023-6-23
Sorry about that. I just edited the code..

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采纳的回答

Torsten
Torsten 2023-6-23
编辑：Torsten 2023-6-23
Ran in:
K_a in your function definition is a scalar. Thus you have to change x0 to be a scalar, too.
clc
close all
clear
background = csvread(['0SM.CSV'],3,0,[3,0,1603,2]);
backgroundImpIm = background(:,3);
f = background(:,1);
freq = f(100:1601);
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)));
K_a = lsqcurvefit(func,x0,freq,backgroundImpIm(100:1601))
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 = 48.3478
  1 个评论
sxh
sxh 2023-6-23
Youve got good eyes mate. Thanks. Spent almost a full day for the error, wonder how I missed this one.
Thanks a bunch,
S

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更多回答（1 个）

James Tursa
James Tursa 2023-6-23
编辑：James Tursa 2023-6-23
I would presume you may need element-wise operators. Try this:
func = @(K_a,freq)(-K_a./(tan(2*pi*freq*h.*(1 + 0.19./((K_a/60 +1) - 0.81))/c)));
  3 个评论
显示 1更早的评论
James Tursa
James Tursa 2023-6-23
Type the following at the command line:
dbstop if error
Then run your code. When the error happens, the program will pause with all variables intact. Examine them to figure out which variables are causing the dimension problem, then backtrack in your code to figure out why the dimensions are not what you expected.
sxh
sxh 2023-6-23
The only culprit I can think of is 'K_a' but this is the one I am trying to solve for.
I replaced 'K_a' with just a scalar value 1 to see if the function value the same as the size of the YDATA. They are the same size.

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