Solve nonlinear equation using ODE45 function with different initial conditions
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I have written a function plus a script to solve a non linear single degree of freedom pendalum system.I don't know how to solve the nonlinear equation with ODE45 function for different values of initial conditions.How could I do that?.I have used a for loop but it gives me just the state vector of the last initial condtion.I want the state vector for all initial condtions.Here are my function and the script:
function xDot = of(x,g,L,u)
xDot = zeros(2,1);
xDot(1) = x(2);
xDot(2) = ((g./L)*sin(x(1)))+u;
end
%% Main Script
clc;clear;close all;
L = 1;
g = 9.81;
h = 0.25;
t = [0:h:5];
A = [0 1;(g/L) 0];
B =[0 1]';
Ics = [pi,0;pi/2 0;pi/5 0;0.001 0;pi 0.5;pi/2 0.5;pi/5 0.5;0.001 0.5];
[Poles,~] = eig(A); %Poles Of Closed LOop System
R = 0.001;
Q = eye(2);
K = lqr(A,B,Q,R);
u = @(x)-K*(x);
for i = 1:size(Ics, 1)
[~, X] = ode45(@(t, x)of(x, g, L, u(x)), t, Ics(i, :));
end
0 Kommentare
Antworten (1)
Stephan
am 7 Dez. 2020
Bearbeitet: Stephan
am 7 Dez. 2020
Corrected:
%% Main Script
L = 1;
g = 9.81;
h = 0.25;
t = [0 5];
A = [0 1;(g/L) 0];
B =[0 1]';
Ics = [pi,0;pi/2 0;pi/5 0;0.001 0;pi 0.5;pi/2 0.5;pi/5 0.5;0.001 0.5];
[Poles,~] = eig(A); %Poles Of Closed LOop System
R = 0.001;
Q = eye(2);
K = lqr(A,B,Q,R);
u = @(x)-K*(x);
for ii = 1:size(Ics, 1)
[t, X(:,:,ii)] = ode45(@(t, x)of(t,x, g, L, u), t, Ics(ii,:));
end
function xDot = of(~,x,g,L,u)
xDot = zeros(2,1);
xDot(1) = x(2);
xDot(2) = ((g./L)*sin(x(1)))+u(x);
end
X is saved as a 3D array - To show the results for the single IC's use:
>> plot(t,X(:,:,1)) % first set of IC
>> plot(t,X(:,:,2)) % second set of IC
0 Kommentare
Siehe auch
Kategorien
Mehr zu Ordinary Differential Equations finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!