optimization problem for exponential coefficients

1 Ansicht (letzte 30 Tage)
fima v
fima v am 24 Mär. 2024
Kommentiert: fima v am 29 Mär. 2024
Hello , i have the attached code which produces certain radiation pattern shown below.
basicly in matlab view its AF,theta plot.
In in code the coeeficients of real_voltage phases are known, however i want to create a code which given AF,theta it whould return me the best optimized real_voltage phases coefficients.
Thanks.
theta=linspace(-pi/2,pi/2,1000);
f=5.6;
N=6;
lambda=300/f;
d=0.8*lambda;
theta_0=pi*((0)/180);
amplitude_norm_V=[0.05,0.3,0.75,1,0.75,0.3,0.05];
k=(2*pi)/lambda;
delta=k*d*cos(theta_0);
x=sqrt(1/(sum(amplitude_norm_V.^2)));
real_voltage=amplitude_norm_V.*x;
real_power=(real_voltage.^2);
sum(real_power);
phases=[0,0,0,0,0,0,0];
total=0;
tot=0;
for z=1:6
AF=real_voltage(z)*exp(1i*(phases(z)/180*pi))*exp(-1i*(z-1)*k*d*cos(theta)+1i*(z-1)*delta);
total=total+AF;
end
plot((theta/pi)*180,20*log10(total)), grid on
Warning: Imaginary parts of complex X and/or Y arguments ignored.

Antworten (1)

Morgan
Morgan am 25 Mär. 2024
This sounds like a non-linear regression problem if I'm understanding your question correctly.
The function you're fitting to looks like:
Use fitnlm() and how to use fitnlm() to solve for the best fit coefficients in your case.
  3 Kommentare
Morgan
Morgan am 26 Mär. 2024
Bearbeitet: Morgan am 26 Mär. 2024
% NUMBER OF FIT PARAMETERS
N = 6;
% THETA ARRAY
theta = linspace(-pi/2,+pi/2,1000);
% V DATA
V = ... % Your data for V that has length of N
% AF DATA
AF = ... % Your data for AF the same size as theta
% COEFFICIENT GUESS
beta0 = zeros(1,N);
% SOLVE NONLINEAR REGRESSION MODEL
mdl = fitnlm(theta,AF,@(b,theta) modelfun(b,theta,V),beta0)
% FUNCTION MODEL
function y = modelfun(b,theta,V)
y = 0;
for n = 1 : length(V)
y = y + V(n)*exp(1i*b(n))*exp(1i*(n-1)*(delta - k*d*cos(theta)));
end
end
fima v
fima v am 29 Mär. 2024
Hello, what is my expression has not jus V but also phases.How does the code will change then?
Thanks.

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by