How to solve exceed number of array element (1) problem?

2 vues (au cours des 30 derniers jours)
shahin sharafi
shahin sharafi le 5 Août 2021
Commenté : darova le 5 Août 2021
Hi
As I try to run the bellow code, I will encounter the Index exceeds the number of array elements (1). As I tried the Time_Delay_Bra as a constant, the problem will be solved. However as I use it as "Time_Delay_Bra=0.1:0.05:0.2", it causes problem for me.
Could you please help me with this mattter.
Thank youo in advance
%%
tic
clear all
clc
close all
%%
Mass=60;Lenght=1;Ja=60;g=9.81;beta2=411.67;
Kp_Bra=2.15;Kd_Bra=0.75;
alpha=(Mass*Lenght*g)/Ja;Zeta3=beta2/Ja;
N=7;
Time_Delay_Bra=0.1:0.05:0.2;
Time_Delay_Exo=0.1:0.05:0.2;
AreaCount=1;
AreaVector=zeros(1,AreaCount);
AreaMatrix=zeros(length(Time_Delay_Exo),length(Time_Delay_Bra));
for j=1:length(Time_Delay_Bra)
for jj=1:length(Time_Delay_Exo)
%% Legendre Polynomial
s1=0;
s2=-Time_Delay_Bra(j);
s3=-Time_Delay_Exo(jj);
[Phi_0_s,Phi_BTD_s,Phi_ETD_s]=Shape_Function(N,s1,s2,s3);
%%
Time_Delay_Bra=Time_Delay_Bra(j);
[M,C]=M_C(N,Time_Delay_Bra);
%%
count=1;
KP_Exo=0:20:800;
KD_Exo=0:20:800;
x=zeros(1,count);
y=zeros(1,count);
MaximunEigenValuesMatrix=zeros(length(KP_Exo),length(KD_Exo));
WantedEigenValuesforDefinedGaines=zeros(1,1);
for ii=1:length(KP_Exo)
KP_Exo(ii)
for i=1:length(KD_Exo)
KP1=KP_Exo(ii)/Ja;
KD1=KD_Exo(i)/Ja
Final_Matirx=Evaluation(KP1,KD1,alpha,Phi_0_s,Phi_BTD_s,Phi_ETD_s,M,C,Zeta3,Kp_Bra,Kd_Bra);
Max_Real_EignValues=max(real(eig(Final_Matirx)));
if Max_Real_EignValues<0
MaximunEigenValuesMatrix(ii,i)=max(real(eig(Final_Matirx)));
x(1,count)=KP1*Ja;
y(1,count)=KD1*Ja;
count=count+1;
end
hold on
end
end
k = boundary(x',y');% generate boundary of data points
plot(x(k),y(k))
Area= polyarea(x,y);
AreaMatrix(jj,j)=Area;
end
end
toc
%%
function [Phi_0_s,Phi_BTD_s,Phi_ETD_s]=Shape_Function(N,s1,s2,s3)
% % Shape Function for the Zero PArt
Phi_0_s(1)=1;
Phi_0_s(2)=1+2*s1/(-s2);
for k=3:N
Phi_0_s(k)=((2*k-3)*Phi_0_s(2)*Phi_0_s(k-1)-(k-2)*Phi_0_s(k-2))/(k-1);
end
Phi_0_s=Phi_0_s';
%% Shape Function for the Brain TimeDelay Part
Phi_BTD_s(1)=1;
Phi_BTD_s(2)=1+2*s2/(-s2);
for k=3:N
Phi_BTD_s(k)=((2*k-3)*Phi_BTD_s(2)*Phi_BTD_s(k-1)-(k-2)*Phi_BTD_s(k-2))/(k-1);
end
Phi_BTD_s=Phi_BTD_s';
%% Shape Function for the Exo TimeDelay Part
Phi_ETD_s(1)=1;
Phi_ETD_s(2)=1+2*s3/(-s2);
for k=3:N
Phi_ETD_s(k)=((2*k-3)*Phi_ETD_s(2)*Phi_ETD_s(k-1)-(k-2)*Phi_ETD_s(k-2))/(k-1);
end
Phi_ETD_s=Phi_ETD_s';
end
function [M,C]=M_C(N,Time_Delay_Bra)
Delta=zeros(N,N);
M=zeros(N,N);
C=zeros(N,N);
for i=1:N
for j=1:N
if i==j
Delta(i,j)=1;
else
Delta(i,j)=0;
end
end
end
for i=1:N
for j=1:N
M(i,j)=(Time_Delay_Bra*Delta(i,j))/(2*i-1);
if i<j
if rem(i+j, 2) == 1
C(i,j)=2;
else
C(i,j)=0;
end
end
end
end
end
function [L]=Evaluation(KP1,KD1,alpha,Phi_0_s,Phi_BTD_s,Phi_ETD_s,M,C,Zeta3,Kp_Bra,Kd_Bra)
kp_exo=KP1;
kd_exo=KD1;
kp_bra=Zeta3*Kp_Bra;
kd_bra=Zeta3*Kd_Bra;
%% G1&G2
G1=alpha*Phi_0_s'-kp_exo*Phi_ETD_s'-kp_bra*Phi_BTD_s';
G2=-kd_bra*Phi_BTD_s'-kd_exo*Phi_ETD_s';
%% C3
C3=Phi_0_s'*inv(M)*Phi_0_s;
%% X Matrix
X1=(Phi_0_s*Phi_0_s')/C3;
X2=-((Phi_0_s*Phi_0_s'/M)*C)/C3;
X3=(Phi_0_s*G1)/C3;
X4=(Phi_0_s*G2-(Phi_0_s*Phi_0_s'/M)*C)/C3;
%% L Matrix
L1=C/M+X2/M;
L2=X1/M;
L3=X3/M;
L4=C/M+X4/M;
L=[L1 L2;L3 L4];
end

Réponse acceptée

darova
darova le 5 Août 2021
Here is the problem. You are redefining the variable inside for loop and it has (1) element again
  2 commentaires
shahin sharafi
shahin sharafi le 5 Août 2021
Thank you very much. It does work! You saved me
darova
darova le 5 Août 2021

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Creating and Concatenating Matrices dans Help Center et File Exchange

Produits


Version

R2019a

Community Treasure Hunt

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

Start Hunting!

Translated by