Download e-book for iPad: Introduction to Mathematical Systems - Linear Systems, by Christiaan Heij, André C.M. Ran, F. van Schagen

By Christiaan Heij, André C.M. Ran, F. van Schagen

This publication offers an advent to the idea of linear structures and regulate for college kids in company arithmetic, econometrics, computing device technology, and engineering; the point of interest is on discrete time platforms. the themes taken care of are one of the imperative issues of deterministic linear procedure thought: controllability, observability, recognition thought, balance and stabilization by way of suggestions, LQ-optimal keep an eye on concept. Kalman filtering and LQC-control of stochastic platforms also are mentioned, as are modeling, time sequence research and version specification, in addition to version validation.

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34 Chapter 3. State Space Models As Θ1 is controllable this implies that SA1 = A2 S. 10) that k−1 k−1 C2 Aj2 B2 uj = C2 Sx = j=0 C1 Aj1 B1 uj = C1 x, j=0 so that C2 S = C1 . We have proved that A2 = SA1 S −1 , B2 = SB1 , C2 = C1 S −1 , and hence Θ1 and Θ2 are similar. To prove the uniqueness of S, let S : Rn → Rn be invertible with A2 = SA1 S −1 , B2 = SB1 , C2 = C1 S −1 . Then C2 Aj2 S = C1 Aj1 = C2 Aj2 S, j ≥ 0, and as (A2 , C2 ) is observable this implies that S = S. We now come to the two central results of realization theory.

Here: x(t) is the unknown state of the system, y(t) is the measured output of the system, z(t) is the known estimated state of the system, u(t) is the computed input of the system, y¯(t) is the known estimate of the output based on the estimated state. By eliminating u(t), y(t) and y¯(t) from Σcl one finds for the state (x(t)T z(t)T )T the following for t = 0, 1, . : Σcl x(t + 1) = Ax(t) + BF z(t), z(t + 1) = RCx(t) + (A − RC + BF )z(t). Thus the state transition matrix is A RC BF A − RC + BF . 4.

M2j , so that (AT )j C T CAj ≤ c0 m2j for some constant c0 . 3) converges. 2), and as (AT )j C T CAj is positive semidefinite for all j it follows that also P is positive semidefinite. It remains to show that P is non-singular. 3) implies that (AT )j C T CAj x = 0, and hence (AT )j C T CAj x, x = CAj x, CAj x = CAj x 2 = 0. So CAj x = 0 for j ≥ 0, and as (A, C) is observable this implies x = 0. 2) is also unique in this case. 1. 3). Then PQ −1 ∞ = j=0 ∞ = j=0 ∞ = (AT )j C T CAj Q−1 (AT )j QAj Q−1 − (AT )j QAj Q−1 − j=0 ∞ j=0 ∞ (AT )j AT QAAj Q−1 (AT )j QAj Q−1 = QQ−1 = I, j=1 so that P = Q.

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