How to input a force vector in ode45 function ?

clc;
t=0:1:18;
R=[5 8 4 3 6 -7 8 -2 7 -3 8 1 6 4 8 -5 7 6 -4];
%% Calculate response without noise
y0=[0;0];
[t, y]=ode45(testode2,t,y0,R);
figure(1)
plot(y(:,1),y(:,2))
figure(2)
plot(t,y(:,1))
function f=testode2(t,y,R)
f=zeros(2,1);
a=-30; b=1000; 
f(1)= y(2);
f(2)= R-a*(1-y(1)^2)*y(2)-b*y(1);
end

I am not able to put 'R' vector in function , it's showing 'NOT ENOUGH INPUT ARGUMENT'. So please help me ,what should I do ?

Thanks in advance

Answer 1

[Walter Roberson]

[t, y]=ode45(@(t,y)testode2(t,y,R),t,y0);

[Walter Roberson]

R=[5 8 4 3 6 -7 8 -2 7 -3 8 1 6 4 8 -5 7 6 -4];

That is a vector of length 19

f(2)= R-a*(1-y(1)^2)*y(2)-b*y(1);

You use the whole vector R there, so the right hand side is a vector of length 19, but f(2) designates a scalar output location. That is a problem. You have a problem.

If your R values are intended to represent "impulses" that are being given instantaneously at each of 0:18 then you will need to only run the ode45 for one second at a time, then adjust the boundary conditions, and then start the next cycle.

If your R values are intended to represent forces that are intended to be constant for the entire second, then you have a discontinuity in the derivatives every second, and the mathematics of ode45 would require that you only run the ode45 for one second at a time (that is, the time the value stays constant.)

If your R values are intended to represent continuous values that you want to do linear interpolation for (that is, in the time between t=0 and t=1 it rises smoothly from 5 to 8) then you also have problems with the derivatives, same situation as above, have to stop processing each time you change.

If your R values can be interpreted as sampling of a smooth curve of forces, then mathematically you potentially use interp1(0:18, R, t, 'spline') -- but do not use interp1(0:18, R, t) as it has the wrong mathematics for ode45.