[Math] How to make a minimal realization of a transfer function

control theorylinear-controlMATLABoptimal control

Let's say that I have this transfer function.

$$G (s) = \frac {s^4 + 4s^3 + 7s^2 + 10s + 8}{s^5 + 6s^4 + 15s^3 + 25s^2 + 32s + 20}$$

And with MATLAB function minreal, I can get this transfer function

$$G_{min} (s) = \frac {s^3 + 2s^2 + 3s + 4}{s^4 + 4s^3 + 7s^2 + 11s +10}$$

If I have the numerators and denomerators from $G (s) $ in two vectors $num, den $. Which MATLAB command should I use to cansle out poles against zeros so I can get the minimal realization?

I have the poles and zeros in two separate vectors. How can I (easy) use MATLAB 's smart functions to check if I have equation zeros and poles and I have, remove them.

Best Answer

The command you are looking for is minreal:

num = [1, 4, 7, 10, 8];
den = [1, 6, 15, 25, 32, 20];

G = tf(num, den);

tol = 1.0e-9;

G_min = minreal(G, tol);

Gives you the desired minimal realization. Here, tol is the tolerance which is used for comparing poles and zeros.

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[Math] MIMO state space <--> MIMO transfer function – How

For any continuous time state space model, so SISO, MISO, SIMO or MIMO you can always use the following formula to convert the state space model into a transfer function matrix

$$ G(s) = C (s\,I - A)^{-1} B + D. $$

[Math] Minimal realization of a state space model

The easiest way of finding the minimal state space model I think would be using the Kalman decomposition. This allows you to find a similarity transformation that makes it easy to split the state space model into a minimal (controllable and observable) and non-minimal (not controllable or not observable) form.

When calculating this decomposition you could use the Hautus test, but you could also use the controllability and observability matrix. I made a quick implementation of both in MATLAB, whose code can be seen below.

Hautus test:

function [M, Ak, Bk, Ck, N] = kalman_decomp_hautus(A, B, C)

[V,D] = eig(A);
n = length(A);

M1 = []; M2 = []; M3 = []; M4 = [];
N = zeros(1, 4);
for k = 1 : n
    tempC = rank([A - D(k,k) * eye(n) B]) - n;
    tempO = rank([A - D(k,k) * eye(n); C]) - n;
    if tempC == 0
        if tempO == 0
            M2 = [M2 V(:,k)];
            N(2) = N(2) + 1;
        else
            M1 = [M1 V(:,k)];
            N(1) = N(1) + 1;
        end
    else
        if tempO == 0
            M4 = [M4 V(:,k)];
            N(4) = N(4) + 1;
        else
            M3 = [M3 V(:,k)];
            N(3) = N(3) + 1;
        end
    end
end
M = [M1 M2 M3 M4];

Ak = M \ A * M;
Bk = M \ B;
Ck = C * M;

Using this implementation does have the downside that the decomposed state space model matrices can have complex numbers, because if the original $A$ matrix (assuming it is real valued) has complex conjugate eigenvalues then its eigenvectors will be complex conjugate as well, and therefore the similarity transformation also. You could of course try to detect this, but I just wanted a working example.

Controllability and observability matrix:

function [M, Ak, Bk, Ck, N] = kalman_decomp_matrix(A, B, C, tol)

cc = ctrb(A, B);
oo = obsv(A, C);

Sc = rref(cc')'; Sc = Sc(:,1:rank(cc));
Nc = null(cc');
So = rref(oo)'; So = So(:,1:rank(oo));
No = null(oo);

M1 = rref(round(Sc * (Sc \ No),tol)')';
N(1) = rank(M1);
M1 = M1(:,1:N(1));
M2 = rref(round(Sc * (Sc \ So),tol)')';
N(2) = rank(M2);
M2 = M2(:,1:N(2));
M3 = rref(round(Nc * (Nc \ No),tol)')';
N(3) = rank(M3);
M3 = M3(:,1:N(3));
M4 = rref(round(Nc * (Nc \ So),tol)')';
N(4) = rank(M4);
M4 = M4(:,1:N(4)); 
M = [M1 M2 M3 M4];

Ak = M \ A * M;
Bk = M \ B;
Ck = C * M;

I have not done many tests, but with one test I noticed that finding the intersect of two spans of vectors could give wrong results due to limited numerical accuracy, so I added rounding (using tol=13 seems to omit errors in my case).

Once you have your Kalman decomposition then the minimal realisation of the state space model can be constructed using:

list = N(1) + 1:N(1) + N(2);
Am = Ak(list,list);
Bm = Bk(list,:);
Cm = Ck(:,list);
Dm = D;

If you just want to minimal realisation then you might be able to calculate it faster. But you do need to ensure that the total $M$ matrix is square and full rank.

Best Answer

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