Additional comments: Hoagg and Bernstein (IEEE Control Systems magazine, July 2007) say that numerical testing suggests that the statement may be true, but that a proof of this conjecture is open. First, note that the statement contains the words at least which seems to have been missed by Hoagg and Bernstein. Second, as mentioned above, the basis for the statement (conjecture) is actually a personal communication from Leon de la Barra (1998). As noted by Hoagg and Bernstein, the earlier paper by la Barra (1994) has a more restrictive stament, where it is proved that the number of zero crossings is equal to the number of positive zeros for strictly proper transfer functions with only real poles and zeros.
Yp(:,i) = C(:,diag(E)==p(i))/norm(C(:,diag(E)==p(i)));
G = [g11 0 0; g21 g22 g23; g31 0 g33]is not triangular, but RGA(G)=I for any non-zero value of the entries gij. Note that mu(E)=0 for this system (so it is clearly diagonally dominant) and stable individual loops gurantee overall stability.
However, RGA=I does not generally imply mu(E)=0 (or even diagonally dominance) as is shown by the 4x4 counterexample in the book.
Thus, this 4x4 counterexample is worse from a decentralized control point of view.
To see how to generate such "counterexamples" note that RGA(G)=I occurs when det G is the product of the diagonal elements gii (thanks to Nima Monshizadeh Naini for noting this). To see this consider the formula for RGA in (A.78): lambda_ij = (-1)^i+j gij*detG^ij / det G. To have RGA=I (lambda_ii=1) we must then require det G = gii * det G^ii (for all i), and since det G^ii is independent of gii this proves that det G is the product of the diagonal elements gii (assuming that we are looking for a "structural" property that holds for any value of gii). In particular, for a 3*3 plant the above relations yield:
For i=1: det(G) = g11(g22*g33-g23*g32) = g11*g22*g33-g11*g23*g32 For i=2: det(G) = g22(g11*g33-g13*g31) = g11*g22*g33-g22*g13*g31 For i=3: det(G) = g33(g11*g22-g12*g21) = g11*g22*g33-g33*g12*g21and to get det(G)=g11*g22*g33 (and have RGA=I for the 3x3 case) we must require g23*g32=0, g13*g31=0 and g12*g21=0. This is satisfied for the above 3x3 counterexample which has g12=0, g13=0 and g32=0.
A = [0.2 0; 0.5 -0.2]; B = [-0.2; 0]; C = [1 0; -0.96 0.784]; D = [0;0];We note from the state space realization that the system is unstable, but to understand clearer the properties of the system, note that it corresponds to the following transfer functions:
y1 = 1/(-5s+1) * u; y2 = (-4.8s+1)/(5s+1) * y1This may be viewed as an unstable system with a single state (y1=x1), but we also have a measurement y2 of y1 which has a RHP-zero close to the RHP-pole. This problem is motivated by a real problem where we want to stabilize slug flow. y1 is then the pressure at the seabottom which may be difficult to measure. y2 is the pressure at the top which is easy to measure. We now consider two cases: Case 1. Measure both states (including y1=x1). It turns out the optimal LQR- controller in terms of stabilization with minimum input usage is to use state feedback (P-controller) only from x1. This moves the unstable pole to its mirror image (from 0.2 to -=.2; which is a well-known result). Case 2. Measure only y2. The optimal is to use a state observer (Kalman filter) which estimates the states (x1 and x2), and then combine it with the above LQR-controller (this is the separation principle). However, there are problems here because of the RHP-zero (at z=1/4.8=0.2083) which is located close to the RHP-pole (at p=0.2), In fact, we know from (5.21) that the sensitivity function S from u to y2 must have a peak of at least |z+p|/|z-p| = 49 (in practice, the system is impossible to stabilize if we only have y2 as a measurement). Thus, we know that good control is not possible - even with the best state estimator in the world. This is confirmed by simulations with the LQR (see details).
Let z=Aw where the vector w is in a given set and we want to evaluate the two-norm of the vector z. We consider three sets for w: (1) ||w||_2 \le 1 (2) ||w||_infty \le 1 (largest element in w less than 1) (3) w is N(0,I) (normally distributed with zero mean and variance=1) and two cases for the evaluation of the two-norm of z: (a) worst-case (for a specific allowed w) (b) average/expected (for all allowed w) We then have the following results (Kariwala et al., Ind.Eng.Chem.Res., 2008, 1150-1158): Case (allowed set for w) | (a) worst-case ||z||_2 | (b) average ||z||_2 --------------------------------|-------------------------------|---------------------- 1 ||w||_2 \le 1 | sigma(A) | ||A||_F*sqrt(1/3nw) 2 ||w||_infty \le 1 | sigma(A)*sqrt(nw) (UPPER) | ||A||_F*sqrt(1/3) 3 w is N(O,I) | infinity | ||A||_F Notes: A uniform disturibution for w is assumed in cases 1b and 2b nw is the length of the vector w sigma(A)=||A||_2 is the maximum singular value of A The value for Case 2a is an upper bound; the lower bound is sigma(A).