how to solve the program?

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prajith samraj
prajith samraj 2022 年 7 月 18 日
編集済み: prajith samraj 2022 年 7 月 26 日
Unrecognized function or variable 'e1'.
s1=x3.^2+x1*x3+x1.^2;
s2=x1.^4+x1.^3*x3+x1.^2*x3.^2 +x1.^2*x3.^2+x1*x3.^3 + x3.^4;
%s1=x2.^2+x1*x2+x1.^2;
%s2=x1.^4+x1.^3*x2+x1.^2*x2.^2 +x1.^2*x2.^3 + x2.^4;
%u1 = -e1-e2;
%u2 = beta*e1*k1 +gamma*e1*k2+omegaf*e1-ff2*cos(W2*t)+ff1*cos(W1*t);
u1 = -e2-e1;
u2 = beta*e1*s1 + gamma*e1*s2+omegaf*e1-ff2*cos(W2*t)+ff1*cos(W1*t);
%sys-I
dx1=x2;
dx2=-alpha*x2-omegaf*x1-beta*x1.^3- gamma*x1.^5+ff1*cos(W1*t);
%sys -II
dx3=x4+u1;
dx4=-alpha*x4-omegaf*x3-beta*x3.^3- gamma*x3.^5+ff2*cos(W2*t)+u2;
de1=dx3-dx1;
de2=dx4-dx2;
e1 = x3-x1;
e2= x4-x2;
de1=e2+u1;
de2=-alpha*e2-omegaf*e1-beta*e1*s1- gamma*e1*s2+ff2*cos(W2*t)-ff1*cos(W1*t)+u2;
dy = [dx1; dx2; dx3; dx4];
end
  2 件のコメント
prajith samraj
prajith samraj 2022 年 7 月 18 日
i need e1 and e2 graph. what can i do?
Lateef Adewale Kareem
Lateef Adewale Kareem 2022 年 7 月 19 日
e1 = y(:,3)-y(:,1);
e2= y(:,4)-y(:,2);

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採用された回答

Sam Chak
Sam Chak 2022 年 7 月 19 日
Try manipulating the parameters to get different results.
% Solver
tspan = [0 20];
x10 = 0.11;
x20 = 0.1;
x30 = 0.21;
x40 = 0.2;
y0 = [x10; x20; x30; x40];
[t, y] = ode45(@(t, y) f(t, y), tspan, y0);
% Plots
e1 = y(:, 3) - y(:, 1);
e2 = y(:, 4) - y(:, 2);
subplot(2,1,1)
plot(t, e1), grid on, xlabel('t'), ylabel('e_{1}')
subplot(2,1,2)
plot(t, e2), grid on, xlabel('t'), ylabel('e_{2}')
function dy = f(t, y)
% parameters
alpha = 0.5;
omegaf = -1;
beta = 0.8790;
gamma = 0.3000;
W1 = 1.4450;
ff1 = 0.5381;
W2 = 1.4450;
ff2 = 0.5381;
% assignment
x1 = y(1);
x2 = y(2);
x3 = y(3);
x4 = y(4);
e1 = x3 - x1;
e2 = x4 - x2;
s1 = x3^2 + x1*x3 + x1^2;
s2 = x1^4 + (x1^3)*x3 + (x1^2)*(x3^2) + (x1^2)*(x3^2) + x1*(x3^3) + x3^4;
u1 = - e2 - e1;
u2 = beta*e1*s1 + gamma*e1*s2 + omegaf*e1 - ff2*cos(W2*t) + ff1*cos(W1*t);
% sys I
dx1 = x2;
dx2 = - alpha*x2 - omegaf*x1 - beta*x1^3 - gamma*x1^5 + ff1*cos(W1*t);
% sys II
dx3 = x4 + u1;
dx4 = - alpha*x4 - omegaf*x3 - beta*x3^3 - gamma*x3^5 + ff2*cos(W2*t) + u2;
de1 = dx3 - dx1;
de2 = dx4 - dx2;
dy = [dx1; dx2; dx3; dx4];
end
  2 件のコメント
prajith samraj
prajith samraj 2022 年 7 月 19 日
i need the value of e1dot and e2 dot graph
Sam Chak
Sam Chak 2022 年 7 月 19 日
@prajith samraj, I'm a little confused now. In your comment, you clearly mentioned that you need e1 and e2 graph. Perhaps, you unintentionally confused yourself with e1dot and e2dot?
Please edit the title of your question for clarity...

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