Ode45 for solution of 4 d.o.f. problem

  • #1
719
1

Main Question or Discussion Point

Hi Folks,

I have made up this code in an attempt to find the response time a 4 dof problem. However, I do not get an oscillatory response as I expect, instead I get a straight line which drops off towards the end. Not sure what is happening..any ideas..?

function f = Simple_Planetary(t,y)
f = zeros(8, 1);
f(1) = y(2);
f(2) = -y(7)+2*y(1);
f(3) = y(4);
f(4) = -y(7)+2*y(3);
f(5) = y(6);
f(6) = -y(7)+2*y(5);
f(7) = y(8);
f(8) = 3*y(7)-y(1)-y(2)-y(3)+10*cos(t);

y0 = [1;1;1;1;1;1;1;1];
[t,y] = ode45('Simple_Planetary', tspan, y0);
subplot (211)
plot (t,y (: , 1));
xlabel ('t');
ylabel ('x1 (t)');
subplot (212)

plot (t,y (: , 3));
xlabel ('t');
ylabel ('x2 (t)');

Regards
 

Answers and Replies

  • #2
11,501
5,052
Two things that may help are enclosing your source code in code tags. This will eliminate the smiley faces.

The second is it seems this is not the whole program as I can't see what span is defined as.

Lastly, what does d.o.f mean?
 
  • #3
719
1
Hi, sorry I forgot a line for the span. That is all the code.....
4 dof is a 4 degree of freedom system, ie for independent variables that we are solving for


Code:
tspan = [0: 0.01: 20];
y0 = [1;1;1;1;1;1;1;1];
[t,y] = ode45('Simple_Planetary', tspan, y0);
subplot (211)
plot (t,y (:, 1));
xlabel ('t');
ylabel ('x1 (t)');
subplot (212)

plot (t,y (:, 3));
xlabel ('t');
ylabel ('x2 (t)');
Code:
function f = Simple_Planetary(t,y)
f = zeros(8, 1);
f(1) = y(2);
f(2) = -y(7)+2*y(1);
f(3) = y(4);
f(4) = -y(7)+2*y(3);
f(5) = y(6);
f(6) = -y(7)+2*y(5);
f(7) = y(8);
f(8) = 3*y(7)-y(1)-y(2)-y(3)+10*cos(t);
 
  • #4
DrClaude
Mentor
7,167
3,310
Could you write down the system of equations you are solving?
 
  • #5
719
1
Hi

Sorry for the delay. It turns out the my K matrix was incorrect from which I derived the system of equations and hence the reason for the large numbers because my matrix was nearly singular etc.
 

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