error: matrix dimensions must agree

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sxh
sxh el 23 de Jun. de 2023
Comentada: sxh el 23 de Jun. de 2023
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 comentarios
KSSV
KSSV el 23 de Jun. de 2023
ONly one csv file is uploaded....error line is not mentioned.
sxh
sxh el 23 de Jun. de 2023
Sorry about that. I just edited the code..

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Respuesta aceptada

Torsten
Torsten el 23 de Jun. de 2023
Editada: Torsten el 23 de Jun. de 2023
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 comentario
sxh
sxh el 23 de Jun. de 2023
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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Más respuestas (1)

James Tursa
James Tursa el 23 de Jun. de 2023
Editada: James Tursa el 23 de Jun. de 2023
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 comentarios
James Tursa
James Tursa el 23 de Jun. de 2023
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 el 23 de Jun. de 2023
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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