Restricting solutions using lsqnonlin with complex unknowns
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
José David Castillo Blanco
le 25 Juin 2022
Commenté : José David Castillo Blanco
le 26 Juin 2022
So, I'm trying to solve a system of equations that has complex numbers, the thing is that it is supossed to be a 12x12, however I write it as a 9 unknowns with 12 equations ever since three of my unknowns
l1,L2,L3
Are complex numbers, where the real and imaginary part are two sepparate unknowns, i'm trying to solve my system with the following code:
fun = @(x)[x(1)*(exp(1i*alfa(1))-1)+x(2)*(exp(1i*x(4))-1)-x(3)*(exp(1i*rho(1))-1),x(1)*(exp(1i*alfa(2))-1)+x(2)*(exp(1i*x(5))-1)-x(3)*(exp(1i*rho(2))-1),x(1)*(exp(1i*alfa(3))-1)+x(2)*(exp(1i*x(6))-1)-x(3)*(exp(1i*rho(3))-1),x(1)*(exp(1i*alfa(4))-1)+x(2)*(exp(1i*x(7))-1)-x(3)*(exp(1i*rho(4))-1),x(1)*(exp(1i*alfa(5))-1)+x(2)*(exp(1i*x(8))-1)-x(3)*(exp(1i*rho(5))-1),x(1)*(exp(1i*alfa(6))-1)+x(2)*(exp(1i*x(9))-1)-x(3)*(exp(1i*rho(6))-1)];
[x,fval] = lsqnonlin(fun,[40+20i,40+20i,40+40i,1,1,1,1,1,1])
L1=x(1)
l2=x(2)
L3=x(3)
phi1=x(4)
phi2=x(5)
phi3=x(6)
phi4=x(7)
phi5=x(8)
phi6=x(9)
I need to restrict the solutions to only those who have phi1, phi2, phi3, phi4, phi5, phi6 stricly as real numbers, without any complex part
Full code below
%Introducción de puntos y cambio de eje de referencia
%Los puntos se escogieron de acuerdo al trabajo del mecanismo de 10 barras
clc
clear all
L=82.4;
L2=42.7;
l1=39.7;
Px=[6.56044 -4.12471 -19.6017 -18.2934 -1.8923 21.1196 17.6706];
Py=[0.995643 0.00147957 1.14159 2.07962 6.30745 11.3302 4.55826];
Theta1=[1.8073522328820310935623781790315, 1.6640119075452171534709629667332, 1.6070200513956322755772098198635, 1.6722276935006778576202574721229, 2.0499843737905114408996166747857, 2.5269236792605256431572972964069, 2.104035221140493127739748316949];
Theta2=[1.5061633838536617569715490239474, 1.3818662717816540137480189935275, 1.0670342257290320137842410556993, 1.039607130354159250776488333576, 1.0892436455770986051727755147433, 1.206386181298577448316370700124, 1.5032866660028009663082593137064];
Theta1_Grados=round(Theta1,3)*(180/pi)
Theta2_Grados=round(Theta2,3)*(180/pi)
for j=1:6
alfa(j)=(Theta1(j+1)-Theta1(j)); %Alfa representa el cambio de una posición en grados a otra
end
AlfaGrados=alfa*(180/pi)
%Angulos de entrada de la manivela para generar una unica rotacion de 360
%grados, que se distribuye de forma uniforme para la generación de función
gamma=[40 70 100 180 240 360 40];
for j=1:6
rho(j)=(gamma(j+1)-gamma(j))*(pi/180);
end
%Sistemas de ecuaciones
fun = @(x)[x(1)*(exp(1i*alfa(1))-1)+x(2)*(exp(1i*x(4))-1)-x(3)*(exp(1i*rho(1))-1),x(1)*(exp(1i*alfa(2))-1)+x(2)*(exp(1i*x(5))-1)-x(3)*(exp(1i*rho(2))-1),x(1)*(exp(1i*alfa(3))-1)+x(2)*(exp(1i*x(6))-1)-x(3)*(exp(1i*rho(3))-1),x(1)*(exp(1i*alfa(4))-1)+x(2)*(exp(1i*x(7))-1)-x(3)*(exp(1i*rho(4))-1),x(1)*(exp(1i*alfa(5))-1)+x(2)*(exp(1i*x(8))-1)-x(3)*(exp(1i*rho(5))-1),x(1)*(exp(1i*alfa(6))-1)+x(2)*(exp(1i*x(9))-1)-x(3)*(exp(1i*rho(6))-1)];
[x,fval] = lsqnonlin(fun,[40+20i,40+20i,40+40i,1,1,1,1,1,1])
L1=x(1)
l2=x(2)
L3=x(3)
phi1=x(4)
phi2=x(5)
phi3=x(6)
phi4=x(7)
phi5=x(8)
phi6=x(9)
0 commentaires
Réponse acceptée
Matt J
le 25 Juin 2022
Modifié(e) : Matt J
le 25 Juin 2022
l1,L2,L3 are complex numbers, where the real and imaginary part are two sepparate unknowns,
If so, then you have 12 unknowns, not 9. Write your function in terms of 12 unknowns so that lsqnonlin knows that:
fun = @(P) residualFunction(P,rho,alfa);
[P,fval] = lsqnonlin(fun,[40,20,40,20,40,40,1,1,1,1,1,1]);
L=complex(P(1:3), P(4:6));
phi=P(7:end);
function F=residualFunction(P,rho,alfa)
x(1:3)=complex(P(1:2:6), P(2:2:6));
x(4:9)=P(7:end);
expr= [x(1)*(exp(1i*alfa(1))-1)+x(2)*(exp(1i*x(4))-1)-x(3)*(exp(1i*rho(1))-1),x(1)*(exp(1i*alfa(2))-1)+x(2)*(exp(1i*x(5))-1)-x(3)*(exp(1i*rho(2))-1),x(1)*(exp(1i*alfa(3))-1)+x(2)*(exp(1i*x(6))-1)-x(3)*(exp(1i*rho(3))-1),x(1)*(exp(1i*alfa(4))-1)+x(2)*(exp(1i*x(7))-1)-x(3)*(exp(1i*rho(4))-1),x(1)*(exp(1i*alfa(5))-1)+x(2)*(exp(1i*x(8))-1)-x(3)*(exp(1i*rho(5))-1),x(1)*(exp(1i*alfa(6))-1)+x(2)*(exp(1i*x(9))-1)-x(3)*(exp(1i*rho(6))-1)];
F=[ real(expr(:)),imag(expr(:)) ];
end
11 commentaires
Torsten
le 26 Juin 2022
Matt's solution does not work with complex numbers directly, but with their real and imaginary parts which are "glued together" in "residualFunction" by the command
x(1:3)=complex(P(1:2:6), P(2:2:6));
So you can set constraints on the real and imaginary parts of x(1),x(2) and x(3).
To set a constraint on the absolute value of x(1),x(2) and/or x(3), you will have to use "fmincon" instead of "lsqnonlin", I guess.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Operating on Diagonal Matrices dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!