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Unformatted text preview: once the gain matrix
K is chosen, there is no further freedom in the choice of L and H .
The specific form of the new matrix L in (12) suggests another option for
implementation of the dynamics of the reduced-order observer, namely: ˆˆ
z = Ax2 + Ly + Hu (14) where L = A 21 − KA11 (15) A block-diagram representation of this option is given in Figure 1 (b) The selection of the gain matrix K of the reduced-order observer may be accomplished
by any of the methods that can be used to select the gains of the full-order observer as
discussed in the previous article. In particular, pole-placement, using any convenient
algorithm is feasible.
The gain matrix can also be obtained as the solution of a reduced-order Kalman filtering
problem, taking into account the cross-correlation between the observation noise and the
process noise. Suppose the dynamic process is governed by ©Encyclopedia of Life Support Systems (EOLSS) CONTROL SYSTEMS, ROBOTICS AND AUTOMATION - Vol. VIII - Reduced-Order State Observers - Bernard Friedland x1 = A11x 1 + A 12 x 2 + B1 u + F1 v (16) x2 = A 21 x1 + A 22 x 2 + B 2 u + F2 v (17) with the observation being noise-free: y = x1 (18) In this case the gain matrix is given by ′
K = ( PA12 + F2QF1 ) R −1 (19) U
S where ′
R = F1QF1 and P is the ( n − m) × ( n − m) covariance matrix of the estimation error e 2 , as given by ′
P = AP + PA′ − PA12 R −1A12P + Q (20) where ′
A = A 22 − F2QF1 R −1A12 (21) ′
Q = F2QF2 − F2QF1 R −1F1QF2 (22) The initial condition on (20) is () P t 0 = P0 the covariance matrix of the initial uncertainty of the substate x2 . Note that (20) becomes homogeneous when Q=0 (23) In this case it is possible that lim P (t ) := P ( ∞ ) = 0 t →∞ ©Encyclopedia of Life Support Systems (EOLSS) (24) CONTROL SYSTEMS, ROBOTICS AND AUTOMATION - Vol. VIII - Reduced-Order State Observers - Bernard Friedland which means that the steady-state error in estimating x2 converges to zero! We can’t
expect to achieve anything better than this. Unfortunately, P ( ∞ ) = 0 is not the only
possible steady-state solution to (23). To test whether it is, it is necessary to check
whether the eigenvalues of the resulting observer dynamics matrix ˆ
A = A 22 − F2F1−1A12 (25) lie in the open left half-plane. If not, (24) is not the correct steady-state solution to (20). U
S The eigenvalues of the “zero steady-state variance” observer dynamics matrix (25) have
an interesting interpretation: as shown in by...
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- Spring '14