Is it possible to extract the time and two output variables from ODE45?
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Hello,
I have written a function which holds state space equations for ODE45 to solve, being in the nature of:
function zdot = odess(t,z)
In my mainscript which calls upon this function and has the initial conditions for the ode, I am recalling the outputs by:
[t,z] = ode45(@odess, t_step, z0, opts);
However, is it possible to get a second variable from the 2nd code listed above? For example:
[t,z,new_output] = ode45(@odess, t_step, z0, opts);
Thanks for your time.
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답변 (2개)
Torsten
2017년 1월 17일
The easiest way is to reculculate new_output from t and z after the call
[t,z] = ode45(@odess, t_step, z0, opts);
Best wishes
Torsten.
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Torsten
2017년 1월 20일
If you set up your code as above, the variable z(9) is z(1), integrated over time, i.e.
z_9(t)=integral_{0}^{t} z_1(t) dt.
If you want to output z(1), just use your previous code without any changes and use the command
plot (t,z(:,1));
after the call to ode45, as I already suggested.
Your assertion
z(1) isn't included in the output matrix "z" from [t,z] = ode45(@odess,t_step,z0, opts), i think what is included are only zdots.
is simply wrong.
Best wishes
Torsten.
Jan
2017년 1월 20일
@Tyler: An integrator as ODE45 gets the start position (or "state vector") as input, than calls the function to be integrated, which replies the derivatives and accumulates them. For each successful time step, the new position is replied. Therefore the output of the integrator contains the positions (state varaibales) and the first element is its first component.
You can simply check this manually: Computer your odess(t,z) for the inital time and start vector and check the first row of the output of the integrator.
Walter Roberson
2017년 1월 17일
편집: Walter Roberson
2017년 1월 17일
No, that cannot be done. See though
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