Error: Array indices must be positive integers or logical values.
显示 更早的评论
I been using ode45 to solve an coupled equation. I keep getting the error Array indices must be positive integers or logical values for this part of my code " -2*zeta2*(z(4)-z(2))-w2^2*(z(3)-z(1))-2*zeta2(z(4)-z(2))-w2^2*(z(3)-z(1))+(ki*(z(1)-z(3)-d1)+ci*(z(2)-z(4)))/m2-F1*cos(w*t);" I have tried changing the values, changing the variables and I still get the error.
%% Clear Memory
clc;
clear all;
close all
global w F1 F2 F3 F4 d1 d2
%% Excitation Amplitude
g = 9.81;
F1 = 0.05*g;
F2 = 0.1*g;
F3 = 0.5*g;
F4 = 1.0*g;
%%
Omega =[250:3:400];
%% Distance to determine which case
x1 = 2;
x2 = 12;
x = x1-x2;
%% F1
% if x<-d1
i=1; % Initilize the loop counter
for w=Omega
%Tint is the time interval for the time integration
%(start and end).
Tstep=60; % Time step, the maximum Tstep the lower time interval
Per =2*pi/w; % Period of Excitations
Tint=[0:Per/Tstep:500*Per]; % Tint is the time interval
x0v0=[0;0;0;0;0]; % Define IC's
[t,x]=ode45(@LIFun,Tint,x0v0); % Use the ode45 solver to call the Function Q1Fun get time respons
[rT,cT]=size(t); % Use the ode45 solver to get time respons
X=[x(rT-Tstep+1:rT,:)]; % Find the steady state amplitudes for the last 60 periods
XMAX1(i)=max(X(:,2))*1e3; % Find the Max amplitude from the steady state amplitude of the last 60 periods
%VMAX1(i)=max(X(:,4)); % Find the Max amplitude from the steady state amplitude of the last 60 periods
i=i+1; % Update the loop counter
end
%% Left Impact Function F1
function dzdt = LIFun(t,z)
global w F1 d1 d2
m1 = 0.056; %kg
m2 = 0.0084; %kg
k1 = 1500; %N/m
k2 = 144; %N/m
c1 = 0.1833; %Ns/m
c2 = 0.0132; %Ns/m
w1 = sqrt(k1/m1);
w2 = sqrt(k2/m2);
zeta1 = c1/(2*sqrt(k1*m1));
zeta2 = c2/(2*sqrt(k2*m2));
ki = 2000; %N/m
ci = 80; %Ns/m
d1 = 5e-3; %m
d2 = 5e-3; %m
T1 = 10e-6;
T2 = 10e-6;
S = 1;
epsilon = 8.854e-12;
epsilon_r = 1;
sigma = 5;
A = 1;
R = 10e6;
dzdt=[z(2);
-2*zeta1*w1*z(2)-w1^2*z(1)+2*zeta1*w1*(z(4)-z(2))+w1^2*(z(2)-z(1))+2*zeta1*(z(4)-z(2))+w1^2*(z(3)-z(1))+(ki*(z(3)-z(1)-d1)+ci*(z(2)-z(4)))/m1-F1*cos(w*t);
z(4);
-2*zeta2*(z(4)-z(2))-w2^2*(z(3)-z(1))-2*zeta2(z(4)-z(2))-w2^2*(z(3)-z(1))+(ki*(z(1)-z(3)-d1)+ci*(z(2)-z(4)))/m2-F1*cos(w*t);
-z(5)/(S*epsilon*R)*(T1/epsilon_r+d1+(z(1)-z(3)))+(sigma*A*(d1+(z(1)-z(3)))/(epsilon*S*R))];
采纳的回答
Steven Lord
2021-4-21
-2*zeta2(z(4)-z(2))
Is z(4)-z(2) guaranteed to be a positive integer value suitable for use as an index into the variable zeta2? Or did you mean to multiply zeta2 by the quantity (z(4)-z(2)) and accidentally omitted the multiplication sign?
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Programming 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
