Van der pol equation
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Given the Van der pol equation
+ 𝝁 ( – 1) x + kx =0
Where 𝝁 = k = -1
Obtain the non-linear stage space representation of the system
Find the Jacobian matrix at the equilibrium point
Use the Lyaponov direct method to check for the stability of the system
(Hint V(x)= Px : PA + P = -1)
1 commentaire
Sam Chak
le 22 Août 2024
It is true that (- 1)*(- 1)*x + (- 1)*x = 0.
Réponses (1)
Rahul
le 22 Août 2024
I understand that you are trying to obtain the non-linear stage space representation of the Van der pol system. You require the Jacobian matrix at the equilibrium point and want to use Lyaponov direct method to check stability of the system.
Note: I am using ‘ (single quote) notation to demonstrate derivative of a variable in equations below.
The Van der Pol equation is given by: x'' + 𝝁 (x^2 - 1) x' + kx = 0
Substituting 𝝁 = -1 and k = -1, we have: x'' - (x^2 - 1) x' - x = 0 ]
To convert this into a state-space representation, define the state variables:
- x1 = x
- x2 = x'
The equilibrium point is where x1' = 0 and x2' = 0.
For this system, the equilibrium point is (x1, x2) = (0, 0).
Hence on calculating the partial derevatives we get the Jacobian Matrix "A" as: A = [0 1; 1 -1];
To check the stability using the Lyapunov direct method, the function "lyap" can be used.
Further, the eigen values of the matrix obtained from the "lyap" function can be used to check the stability of the system.
So you can follow this code to achieve the desired result:
% Jacobian matrix
A = [0 1; 1 -1];
I = eye(2);
% Lyapunov equation
P = lyap(A', I);
disp('Matrix P:');
disp(P);
% Check for stability
isPositiveDefinite = all(eig(P) > 0);
if isPositiveDefinite
disp('The system is stable.');
else
disp('The system is not stable.');
end
You can refer to the following documentations to know more about these functions:
Hope this helps! Thanks.
6 commentaires
Hi @Rahul
Why is the function lyap() necessary when the eigenvalues of the state matrix A can be obtained directly? Additionally, is the Van der Pol system truly "unstable" with this approach to stability analysis? Your answer will guide @Chike to understand system dynamics using MATLAB.
A = [0 1; 1 -1];
eig(A)
isNegativeDefinite = all(real(eig(A)) < 0);
if isNegativeDefinite
disp('The system is stable.');
else
disp('The system is not stable.');
end
Chike
le 23 Août 2024
John D'Errico
le 23 Août 2024
Modifié(e) : John D'Errico
le 23 Août 2024
Please do not do obvious homework questions for students. You do not help the student, who has now learned there is someone out there who will do their work for them, with no effort made.
It does not help the forum, as it teaches this student to come back, and ask every homework question, with no effort made. Worse, it actively hurts the forum, as it teaches other students to do the same.
And it does not help the teacher who assigned this problem, as now they need to decide if this student has learned something, or merely plagiarized your work. It forces the teacher to in the future, to find other questions to assign for their students to do.
Chike
le 26 Août 2024
i have been trying to resolve for a week now but i wasnt sure if i was right or wrong.
@Chike, could you please clarify which specific part of your attempted MATLAB-based solution you were uncertain about? We'd be happy to review the relevant section(s) and provide any feedback.
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