ODE system with 2 degrees of freedom

Question

Nader Mohamed 2021-10-27

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评论： Nader Mohamed 2021-10-28

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

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Answer 1

James Tursa 2021-10-27

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编辑：James Tursa 2021-10-27

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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

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Nader Mohamed 2021-10-28

Thank you so much!

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Answer 2

David Goodmanson 2021-10-27

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https://ww2.mathworks.cn/matlabcentral/answers/1573303-ode-system-with-2-degrees-of-freedom#answer_818273

编辑：David Goodmanson 2021-10-28

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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));