Solve Riccati (DARE) with MATLAB

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Discussion Overview

The discussion revolves around the formulation and implementation of the discrete-time algebraic Riccati equation (DARE) using MATLAB's dare() function. Participants are exploring the correct definition of arguments for the function and addressing issues related to the results obtained from running their code.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the correct definitions of variables for the dare() function, presenting a specific DARE formulation and variable definitions.
  • Another participant suggests that Mathworks provides tutorials for both continuous and discrete cases, linking to relevant resources.
  • There is a repetition of the tutorial links, indicating some confusion or oversight regarding previous contributions.
  • One participant questions whether the original poster has run their code and what issues they encountered with the results, suggesting that testing the output could help clarify the situation.
  • The original poster confirms they have run the code but expresses confusion about the results, indicating a need for further clarification on their DARE formulation.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the correctness of the DARE formulation or the results obtained from the MATLAB code. There are multiple viewpoints regarding the definitions and potential issues with the implementation.

Contextual Notes

Participants have not fully explored the assumptions underlying their variable definitions or the implications of their MATLAB code results. There is also a lack of clarity on the specific nature of the confusion regarding the output.

perplexabot
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Hey all, I was wondering if you can help me define the arguments for the dare() function of matlab. Here is the DARE form I have:
<br /> \begin{equation}<br /> \begin{split}<br /> \Phi(p_\infty) &amp;= \Phi_\infty = \Phi_{33}-\Phi_{32}(\Phi_\infty+\Phi_{R})^{-1}\Phi_{23}\\<br /> &amp;=\Phi_{33}+\Phi_{32}\Phi_{R}^{-1}\Phi_\infty\Phi_{R}^{-1}\Phi_{23}-\Phi_{32}\Phi_{R}^{-1}\Phi_\infty(\Phi_\infty+\Phi_{R})^{-1}\Phi_\infty\Phi_{R}^{-1}\Phi_{23}-\Phi_{32}\Phi_{R}^{-1}\Phi_{23}<br /> \end{split}<br /> \end{equation}<br />

Here is how MATLAB's dare() function works.

Here is how I defined my variables: E = I, B = I, R = \Phi_R, Q = \Phi_{33}, S=\Phi_{32}, , A = 2\Phi_R^{-1}S^T.

Is this correct?
Thank you : )

PS: Please let me know if this thread belongs elsewhere, as I would like to maximize feedback from the PF community.
 
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perplexabot said:
That's the same link I provided in my original post : (

Sorry I didn't see as it was hidden in the "Here" text.
 
jedishrfu said:
Sorry I didn't see as it was hidden in the "Here" text.
No problem! I still appreciate the fact that you are trying to help. Thank you.
 
Have you tried to run your code? If so, what was wrong with the answer?

You could try taking the answer and testing to see if it is the solution to gain confidence in what Matlab returned.
 
jedishrfu said:
Have you tried to run your code? If so, what was wrong with the answer?

You could try taking the answer and testing to see if it is the solution to gain confidence in what Matlab returned.
I have run my code. I got an answer but I am confused with the result as it is not as I expected. I am trying to figure out why the answer is what it is. I was wondering whether my DARE formulation was correct (hence, why I have this post).
 

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