ODE system with 2 degrees of freedom

29 views (last 30 days)
I'm trying to write this system of odes and then solve it with ode45, but I'm having trouble writing the function for the system:
k,J1,J2 and b are known. I tried to write this code but it doesnt work:
function dydt = ODEsystem(t,y)
k = 100;
b = 10;
J1 = 0.2;
J2 = 0.1;
dydt = zeros(6,1);
dydt(1) = y(2);
dydt(2) = y(3);
dydt(3) = k/J1 * dydt(1) - k/J1 * dydt(4);
dydt(4) = y(5);
dydt(5) = y(6);
dydt(6) = b/J2 * dydt(5) + k/J2 * dydt(4) - k/J2 * dydt(1);
end

Accepted Answer

James Tursa
James Tursa on 27 Oct 2021
Edited: James Tursa on 27 Oct 2021
I only see two variables, θ1 and θ2, and each is part of a 2nd order equation. So that would mean a 2 x 2 = 4 element state vector, not 6 element state vector as you are using. E.g., I would have the states defined as
y(1) =
y(2) =
y(3) =
y(4) =
which would mean the derivative function would be:
function dydt = ODEsystem(t,y)
k = 100;
b = 10;
J1 = 0.2;
J2 = 0.1;
theta1ddot = k/J1 * y(1) - k/J1 * y(2);
theta2ddot = b/J2 * y(4) + k/J2 * y(2) - k/J2 * y(1);
dydt = [y(3);y(4);theta1ddot;theta2ddot];
end

More Answers (1)

David Goodmanson
David Goodmanson on 27 Oct 2021
Edited: David Goodmanson on 28 Oct 2021
Hi Nader,
you only need four variables, theta1, theta1dot, theta2, theta2dot (not six). Try
function dydt = ODEsystem(t,y)
k = 100;
b = 10;
J1 = 0.2;
J2 = 0.1;
% y = [theta1; theta1dot; theta2; theta2dot]
dydt = zeros(4,1);
dydt(1) = y(2);
dydt(3) = y(4);
dydt(2) = (k/J1)*(y(1)-y(3));
dydt(4) = (b/J2)*y(4) +(k/J2)*(y(3)-y(1));

Products


Release

R2018b

Community Treasure Hunt

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

Start Hunting!