Is it possible to extract the time and two output variables from ODE45?

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Tyler Bikaun 2017-1-17

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评论： Jan 2017-1-20

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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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Jan 2017-1-17

Where does the "new_output" come from? What do you expect in this variable?

Tyler Bikaun 2017-1-19

Sorry Jan, the syntax in the 3rd code above is just off of my head and not what I actually expect to be able to do. I want to be able to extract z(1) from my odescript (see comments below), I thought that that may be a way to achieve it.

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Answer 1

Torsten 2017-1-17

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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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Tyler Bikaun 2017-1-19

编辑：Tyler Bikaun 2017-1-19

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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.

So, I've added zdot(9) = z(1) as a new state space equation within the odess function and updated z0 for an additional initial condition.

As z(1) is supposed to be angular displacement of a gear, the output seems to be reasonable as it is not oscillating in value like the others outputs (but I cannot be 100% sure of this, could be a coincidence). However, I'm not sure if the inclusion of that line of code actually is integrated by ode45 when its ran, or just outputs the value of z(1), if its the former, then its not outputting what is required.

Updated ODE function:
function zdot = odess(t,z)
% defining global variables
global r_p r_g I_p I_g m_p m_g k_mb
%%state space equations are defined below:
zdot(2) = ((r_p*k_mb)/I_p)*(-r_p*z(1)+r_g*z(3)+z(5)-z(7));  % Eq 1ss
zdot(6) = (k_mb/m_p)*(r_p*z(1)-r_g*z(3)-z(5)+z(7));         % Eq 2ss
zdot(4) = ((r_g*k_mb)/I_g)*(-r_g*z(3)+r_p*z(1)+z(7)-z(5));  % Eq 3ss
zdot(8) = (k_mb/m_g)*(-r_p*z(1)+r_g*z(3)-z(7)+z(5));        % Eq 4ss
zdot(1) = z(2); % Eq 5ss
zdot(3) = z(4); % Eq 6ss
zdot(5) = z(6); % Eq 7ss
zdot(7) = z(8); % Eq 8ss
zdot(9) = z(1); % Angular displacement?.
% Outputs
zdot = zdot';

Updated main script:

% Mainscript for vibration analysis (16/01/2017).
clear all;
clc;
%%Defining global variables
global r_p r_g I_p I_g m_p m_g k_mb
k_mb = 5;  % torsional stiffness.
r_p = 10;   % pinion pitch(?) radius
r_g = 15;   % gear pitch(?) radius
I_p = 100;  % pinion inertia.
I_g = 200;  % gear inertia.
m_p = 50;   % pinion mass
m_g = 100;  % gear mass
%==========================================================================
%%initial coditions for ode solver
%==========================================================================
z0 = [1,0,0,0,0,0,0,0,0]; % zero initial conditions.
t_step = [0:1:10];        % timestep 0 to 5 seconds.
opts = odeset('Reltol', 1e-3, 'AbsTol', 1e-3);  % conservative tolerances.
[t,z] = ode45(@odess, t_step, z0, opts);
%==========================================================================
%%Screen Output and Plotting 
%==========================================================================
[t,z(:,:)]
p_velocity = t./z(:,9);
subplot(3,1,1)
plot(t,z(:,9));
subplot(3,1,2)
plot(t,p_velocity);

Torsten 2017-1-20

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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 2017-1-20

@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.

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