Error in dynamic system integration
显示 更早的评论
Hello everyone, I am writing a control synthesis program for a dynamic system. First I discretize it, then I find a control and apply it. When I integrate the system, a strange error occurs, which says that I have symbolic variables somewhere in the integration. But I checked, they are not there. Here is my code:
clear
set(0,'DefaultTextInterpreter', 'latex');
m = 0.127;
M = 1.206;
I = 0.001;
L = 0.178;
Bc = 5.4;
Bp = 0.002;
g = 9.8;
h=0.1;
A_0 = [m+M -m*L; -m*L I+m*L^2];
A_1 = [Bc 0; 0 Bp];
A_2 = [0 0; 0 -m*g*L];
A = [zeros(2) eye(2); -inv(A_0)*A_2 -inv(A_0)*A_1];
B = [0;0; inv(A_0)*[1;0]];
A_d=expm(A*h);
syms s;
A_exps=eye(4);
for k=1:40
A_exps=A_exps+(A*s)^k;
end
B_d=int(A_exps*B, s, 0, h);
evs = eig(A_d)
C=[B_d A_d*B_d A_d^2*B_d A_d^3*B_d];
evs(2) = 0.19343;
Xi_A1 = (A_d-evs(1)*eye(4))*(A_d-evs(2)*eye(4))*(A_d-evs(3)*eye(4))*(A_d-evs(4)*eye(4));
Theta1 = -[0 0 0 1]*inv(C)*Xi_A1;
X0 = [0.0;-0.6; 0.0; 0.0];
t_k2=0.0;
count = 10;
t_all=[];x_all=[];
for k=1:count
t_k1=t_k2
t_k2 = t_k1+h
[t_tmp,X_tmp]=ode45(@(t,X)pendulum(t,X,Theta1),[t_k1 t_k2],X0) %This is line 71
t_all=[t_all;t_tmp];
x_all=[x_all;X_tmp];
X0=X_tmp(end,:)';
end
figure;
subplot(1,1,1);
plot(t_all,x_all(:,1));
function dXdt = pendulum(~,X,Theta)
m = 0.127; M = 1.206; I = 0.1e-2; L = .178; Bc = 5.4; Bp = 0.2e-2;g=9.8;
x = X(1);
phi = X(2);
dx = X(3);
dphi = X(4);
A0=[m+M -m*L*cos(phi);-m*L*cos(phi) I+m*L^2];
b=[Theta*X-Bc*dx-m*L*dphi^2*sin(phi); -Bp*dphi+m*g*L*sin(phi)];
dXdt = [dx;dphi;inv(A0)*b];
end
And here's the error:
Error using odearguments (line 113)
Inputs must be floats, namely single or double.
Error in ode45 (line 115)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in LW21 (line 71)
[t_tmp,X_tmp]=ode45(@(t,X)pendulum(t,X,Theta1),[t_k1 t_k2],X0)
Can you please tell me what could be causing this and how can I fix it?
回答（1 个）
set(0,'DefaultTextInterpreter', 'latex');
m = 0.127;
M = 1.206;
I = 0.001;
L = 0.178;
Bc = 5.4;
Bp = 0.002;
g = 9.8;
h=0.1;
A_0 = [m+M -m*L; -m*L I+m*L^2];
A_1 = [Bc 0; 0 Bp];
A_2 = [0 0; 0 -m*g*L];
A = [zeros(2) eye(2); -inv(A_0)*A_2 -inv(A_0)*A_1];
B = [0;0; inv(A_0)*[1;0]];
A_d=expm(A*h);
syms s;
A_exps=eye(4);
for k=1:40
A_exps=A_exps+(A*s)^k;
end
B_d=int(A_exps*B, s, 0, h);
evs = eig(A_d)
C=[B_d A_d*B_d A_d^2*B_d A_d^3*B_d];
evs(2) = 0.19343;
Xi_A1 = (A_d-evs(1)*eye(4))*(A_d-evs(2)*eye(4))*(A_d-evs(3)*eye(4))*(A_d-evs(4)*eye(4));
Theta1 = -[0 0 0 1]*inv(C)*Xi_A1;
X0 = [0.0;-0.6; 0.0; 0.0];
t_k2=0.0;
count = 10;
t_all=[];x_all=[];
for k=1:count
t_k1=t_k2;
t_k2 = t_k1+h;
[t_tmp,X_tmp]=ode45(@(t,X)pendulum(t,X,Theta1),[t_k1 t_k2],X0); %This is line 71
t_all=[t_all;t_tmp];
x_all=[x_all;X_tmp];
X0=X_tmp(end,:)';
end
figure;
subplot(1,1,1);
plot(t_all,x_all(:,1));
function dXdt = pendulum(~,X,Theta)
m = 0.127; M = 1.206; I = 0.1e-2; L = .178; Bc = 5.4; Bp = 0.2e-2;g=9.8;
x = X(1);
phi = X(2);
dx = X(3);
dphi = X(4);
A0=[m+M -m*L*cos(phi);-m*L*cos(phi) I+m*L^2];
b=[Theta*X-Bc*dx-m*L*dphi^2*sin(phi); -Bp*dphi+m*g*L*sin(phi)];
dXdt = double([dx;dphi;inv(A0)*b]);
end
2 个评论
Egor Semenuta
2022-4-15
Walter Roberson
2022-4-15
I = 0.1e-2
That is only one significant digit. g=9.8 is only two significant digits. Why are you concerned about whether double precision (15 to 16 significant digits) is accurate enough when your equations have high error built in to them?
类别
在 帮助中心 和 File Exchange 中查找有关 Numeric Solvers 的更多信息
2022-4-14
2022-4-15
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
