"Limitations" for DARE function

Hey all. I have been trying to better understand the "limitations" of matlab's dare() function but cannot seem to do so. Please help me out. This is the link to the page. Specifically how did [Q S; S.' R] > 0 come to be (call this condition 1)? I don't see how condition 1 implies (Q,A) being detectible?

I wrote a little piece of code to try and figure this out, it is shown below along with the output. I am desperate for an answer or at least some guidance. Any input will be appreciated. Thank you.

    clear; 
    clc; 
    n=2;
    E = eye(2);
    A = [-2 0; 0 -2];
    B = eye(2);
    S = [-1000 0; 0 -1000];
    Q = 1000*[1.0005 0.0005; 0.0005 1.0005];
    R = [1000 0; 0 1000];

    %%%%%%%%%%%%%%%%%%%%%%[RICCATI CONVERGENCE TESTS]%%%%%%%%%%%%%%%%%%%%%%
    %%%matlab stability check
    t1=0;
    if all(eig([Q S; S.' R]) > 0)
        t1=1;
    else
        t1=0;
    end

    %%%check if Q is PSD
    t2=0; 
    if issymmetric(Q) && all(eig(Q) >= 0)
        t2 = 1;
    else
        t2=0;
    end

    %%%check if R is PD
    t3=0; 
    if issymmetric(R) && all(eig(R) > 0)
        t3 = 1;
    else
        t3=0;
    end

    %%%PBH TEST FOR STABILIZIBILITY
    t4=0;
    e = eig(A);
    i = 0;
    while i < size(e,1) && rank([eye(n)*e(i+1)-A eye(n)]) == n
        i = i+1;
    end

    if i == size(e,1)
        t4=1;
    else
        t4=0;
    end

    %%%PBH TEST FOR DETECTABILITY
    t5=0;
    if t2
        i = 0;
        C = chol(Q);
        %%%sub test
        if rank(C)==rank(Q)
            t5=1;
        else
            t5=0;
        end

        while i < size(e,1) && rank([eye(n)*e(i+1)-A; Q]) == n
            i = i+1;
        end

        if i == size(e,1) && t5==1
            t5=1;
        else
            t5=0;
        end
    else
        t5=0;
    end
    t = [t1 t2 t3 t4 t5];
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    if all(t)
        fprintf('[*] All tests PASSED (DARE convergence)\n')
    else
        fprintf('[*] One or more tests FAILED (no DARE convergence)\n')
    end
    fprintf('[-] [Q zeros(2); zeros(2) R] > 0: %d\n', round(t1))
    fprintf('[-] Q is PSD: %d\n', round(t2))
    fprintf('[-] R is PD: %d\n', round(t3))
    fprintf('[-] Stabilizible: %d\n', round(t4))
    fprintf('[-] Detectible: %d\n', round(t5))

Results:

[*] One or more tests FAILED (no DARE convergence)
[-] [Q zeros(2); zeros(2) R] > 0: 0
[-] Q is PSD: 1
[-] R is PD: 1
[-] Stabilizible: 1
[-] Detectible: 1

I don't see how the first test fails but all the other four tests pass.

Answer 1

[Pritesh Shah]

Try edit dare.

You will able to see dare code.

Refer,

References [1] Arnold, W.F., III and A.J. Laub, "Generalized Eigenproblem Algorithms and Software for Algebraic Riccati Equations," Proc. IEEE®, 72 (1984), pp. 1746-1754

[TheElder777]

I have already done that but it does not answer my question...