represent differencital equation with ode45
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jose luis guillan suarez
2018 年 5 月 20 日
i got this differential equation:
function xdot=tresorden(t,x)
xdot=zeros(3,1);
Vp=5;
Vi=Vp*square(2*pi*t)+5;
xdot(1)=x(2);
xdot(2)=x(3);
xdot(3)=6*Vi-6*x(1)-11*x(2)-6*x(3);
xdot=[xdot(1);xdot(2);xdot(3)];
how can i represent x(1)?
2 件のコメント
Anderson Francisco Silva
2020 年 8 月 29 日
And if he wanted to use the last vector, to be entered in another function he could do it like this? :
xdot(3)=6*Vi-6*x(1)-11*x(2)-6*x(3);
x_dot=[xdot(1);xdot(2);xdot(3)]; (I chance the name of vector, for no replaces xdot)
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Jan
2018 年 5 月 20 日
編集済み: Jan
2018 年 5 月 29 日
This integrates the function from the start point x=[1,2,3] over the time 0 to 7:
[EDITED - bug concerning t.' fixed]
function main
[t, x] = ode45(@tresorden, [0, 7], [1,2,3]);
plot(t, x(:, 1));
xdot = tresorden(t.', x.').';
end
function xdot = tresorden(t, x)
Vp = 5;
Vi = Vp * (2*pi*t)^2 + 5; % Or what is square() ?
xdot = [x(2, :); ...
x(3, :); ...
6 * Vi - 6 * x(1, :) - 11 * x(2, :) - 6 * x(3, :)];
end
Note: Due to square you are integrating a non-smooth system. This causes numerical instabilities. See http://www.mathworks.com/matlabcentral/answers/59582#answer_72047.
0 件のコメント
jose luis guillan suarez
2018 年 5 月 21 日
1 件のコメント
Jan
2018 年 5 月 21 日
Sure? I'd expect:
xdot(1) = x(2);
xdot(2) = x(3);
xdot(3) = Vi - 6*x(3) - 11*x(2) - 6*x(1);
if you convert the 3rd order equation to a system of 1st order.
But even then: ODE45 is used to solve initial value problems numerically. If you want the values of x(1), you need to run the integration from an initial value.
Please do not post parts of the question in the section for answer. And explain, what "represent differencital equation with ode45" means exactly.
jose luis guillan suarez
2018 年 5 月 22 日
編集済み: jose luis guillan suarez
2018 年 5 月 22 日
11 件のコメント
Jan
2018 年 5 月 27 日
@jose: You have posted and removed another equation formerly. The solution of how to get the 3rd derivative has been given repeatedly and it even occurs in the original question.
Currently my best assumption is that your "numerical checking" contains a mistake.
i checked numerically and the [...] it's not the 3rd derivative.
My best assumption is that your "numerical check" contains a mistake.
After 6 days it could not be clarified, what the actual question is or why the obvious and already posted solution does not satisfy you. Therefore I will leave this thread now.
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