How to solve ode113 with a time independent load vector?

I'm trying to solve ode113 for a wind turbine which is charged with a wind load. This wind load is calculated from another program and is a vector of random numbers which discribes the load over time in a step of 0.01 s. The problem is that ode113 usually is subjected to a function which is dependent on time. How can I manage to solve ode113 but at the same time take into consideration the time course of my load vector?

Answer 1

[Torsten]

The example "ODE with Time-Dependent Terms" under

http://de.mathworks.com/help/matlab/ref/ode45.html

should solve your problem.

Best wishes

Torsten.

[Lisanne Meinerzhagen]

Thank you, but now I have another problem. The ode output t has always the same size, a vector of size 11x1, even if I change my tspan... my load vector is changed automatically every time, since it is a random vector. How is that possible? Should I send the code??

[Torsten]

Please include the relevant part of your code.

Best wishes

Torsten.

[Lisanne Meinerzhagen]

This is the main code:

    ta=0;
    te=10;
    deltat=0.001;
    elem=30;
    ld=3;
    nodes=elem+1; 
    u0=zeros(nodes*ld,1); 

P=zeros(nodes*ld,1);                       %Load vector per node
Pt = linspace(ta, te, (te+deltat)/deltat);   %time vector with te*100+1 Inputs 
m=length(Pt);
F=random('Normal',0,10,length(Pt),1);
 for i=1:m
    P(end-2:end,1)=0;
    P(end-2:end,1) = P(end-2:end,1)+F(i);   
    A(:,i)=P;
        end 
tspan = [ta te];
ic = 1;
opts = odeset('RelTol',1e-2,'AbsTol',1e-4);
[t,u]=ode45(@(t,u) fu1(t,u,u0,Pt,A),[ta te],u0,opts);

[Lisanne Meinerzhagen]

And than for fu1:

function u=fu1(t,u,u0,Pt,A)
A=A.';
A = interp1(Pt,A,t); % Interpolate the data set (P,A) at time t
A=A.';
n=size(A);
m=n(2);

for i=1:m
u=zeros(nodes*ld*2,1);
u(1:nodes*ld,1)=u0;  % 1:nodes*ld Displacement, nodes*ld+1:nodes*ld*2 Velocity
du=zeros(nodes*ld*2,1); % 1:nodes*ld Velocity, nodes*ld+1:nodes*ld*2 Acceleration  
du(1:nodes*ld,1)=u(nodes*ld+1:nodes*ld*2,1);   % du(1)=u(2)
du(nodes*ld+1:end,1)=M\(A(:,i)-K*u(1:nodes*ld,1)-C*u(nodes*ld+1:nodes*ld*2,1));   %Acceleration %du(2)=M\(P-K*u(1)-C*u(2)) 
u(1:nodes*ld,1)=K\(A(:,i)-M*du(nodes*ld+1:nodes*ld*2,1)-C*u(nodes*ld+1:nodes*ld*2,1));   %Displacement
u(nodes*ld+1:nodes*ld*2,1)=C\(A(:,i)-M*du(nodes*ld+1:nodes*ld*2,1)-K*u(1:nodes*ld,1));   %Velocity
%ddu=du(nodes*ld+1:nodes*ld*2,1);
%du=u(nodes*ld+1:nodes*ld*2,1);
u=u(1:nodes*ld,1);
end
end

[Lisanne Meinerzhagen]

M, K and C are inputs from again another function... they have the size nodes*ldxnodes*ld, where ld stands for degrees of freedom... Does that help in some way or is it to complex?? Thanks a lot!!

[Lisanne Meinerzhagen]

 I can change tspan, the load or my degrees of freedom, t will always remain a 11x1 vector...

[Lisanne Meinerzhagen]

Hey, there is a problem with the code I just send you... since t is zero at first it interpolates my load array A, so that there isn't any load left... A becomes a nodes*ldx1 vector, so it discribes only ONE load at a specific time! Do you have an idea how I can solve this problem? How can I manage, that t is solved for the ODE but at the same time is using my load vector as an input?? Thanks again!

[Torsten]

The ODE-integrator needs the load at the specific time t (t is a scalar, one value). That's why I gave you the link in my first response: you have to interpolate the load vector A at the time required by the ODE-integrator.

Best wishes

Torsten.

[Lisanne Meinerzhagen]

Hey Torsten, sorry for writing you again, but I don't understand, what time the ODE-integrator requires... Doesn't it calculate solutions for specific times, which I don't know it advance depending on the load?? Thanks for your answers and in advance!!

[Lisanne Meinerzhagen]

Hey, I think I solved the problem! Thanks :)