error: matrix dimensions must agree
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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 commentaires
Réponse acceptée
Torsten
le 23 Juin 2023
Modifié(e) : Torsten
le 23 Juin 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))
Plus de réponses (1)
James Tursa
le 23 Juin 2023
Modifié(e) : James Tursa
le 23 Juin 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 commentaires
James Tursa
le 23 Juin 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.
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