On the assignment of eigenvalues in LTI systems - Decision and Control, 1999. Proceedings of the 38th IEEE Conference on

In this note we discuss the assignability of natural frequencies of LTI systems, given their Hankel singular values. For details we refer to [l]. Preliminaries. Consider the LTI system I=: k( t ) = Az( t ) + Bu(t), y ( t ) = Cz( t ) , A E PnX", B,CT E P", We will assume that C is controllable, observable, and stable. There are two important invariants associated with C: the natural frequencies or poles A;@), and the Hankel singular values o;(C). The former quantities are defined as the eigenvalues of A: X;(C) = X;(A), i = l , . , n . The latter are defined as the singular values of the Hankel operator Xc, associated with E: Xx: Ut+ y+ = X ( u ) , where y+(t):=J_= h(t-T)u(T)dT, t > 0, and h(t) = CeAtB , t 2 0, is the impulse response of C. It turns out that 3Cc has a finite number of non-zero singular values: u;(c) = Jm, i = 1, ,n, where a1 1-.1 an > 0, and P > 0, Q > 0 are the controllability, observability grammians of E , respectively, satisfying the Lyapunov equations: 0 A P + PA^ + B B ~ = 0, A ~ & + Q A + C ~ C = o The eigenvalues X;(C) describe the dynamics of the system. The Hankel singular values .;(E) on the other hand just as the singular values in the case of constant matrices describe how well a linear dynamical system can be approximated by a similar system of lower dimension. The problem which arises is the relationship between &(E) and u;(C). More specifically, given the former, to what extent can one influence the latter, and vice versa. We will actually address a more specific question, namely: given A;@), to what extent is it possible to reduce the condition number of the singular values, and given ai@), the condition number of A. The former problem can be solved by noting that one can always construct an all-pass system with given poles (an all-pass system is perfectly conditioned, since all its Hankel singular values are equal). Concerning the distribution of the eigenvalues of A for pre-assigned Hankel singular values, we first note that 0-7803-5250-5/99/$10.00