using gradient with ode45
6 ビュー (過去 30 日間)
古いコメントを表示
i have this code, where gradient(u,t) should be the derivative of u:
function [xdot]=pid_practica9_ejercicio3_ec_diferencial_prueba2(t,x)
Vp=5;
u=Vp*sin(2*pi*t)+5;
xdot = [
x(2, :);
1.776*0.05252*20*gradient(u,t)-10*x(1, :)-7*x(2, :);
];
%[t,x]=ode45('pid_practica9_ejercicio3_ec_diferencial_prueba2',[0,10],[0,0])
% plot(t,x)
but in the solution i get only zeros (and they shouldn't be)
is it possible to work with a derivative in the definition of a diferential equation like i do?
2 件のコメント
回答 (1 件)
Walter Roberson
2018 年 7 月 7 日
The t and x values passed into your function will be purely numeric, with t being a scalar and x being a vector the length of your initial conditions (so a vector of length 2 in this case.)
You calculate u from the scalar t value, and you pass the scalar u and scalar t into gradient -- the numeric gradient routine. The numeric gradient() with respect to scalar F and scalar H is always 0.
8 件のコメント
参考
カテゴリ
Help Center および File Exchange で Ordinary Differential Equations についてさらに検索
製品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!