Nonlinear Control Sy.. - Free

210 CHAPTER 8. PASSIVITY
Thus, H f (H1 + + Hn) is passive.
Proof of (ii) Assume that k out of the n systems H= are strictly passive, 1 < k < n. By relabeling
the systems, if necessary, we can assume that these are the systems H1, H2,- , Hk.
It follows that
and the result follows.
(x, Hx)T = (x, H1x + ... + HnX)T
= (x Hlx)T+...+(x,Hkx)T+...+(x,Hnx)T
61(x,x)T +... +6k(x,x)T+Q1 +... +/3n
Proof of (iii): Consider the following inner product:
This completes the proof.
(u, Y)T = (e + H2Y, Y)T
= (e, y)T + (H2y, Y)T
= (e, Hle)T + (y, H2y)T ? (31+02).
Remarks: In general, the number of systems in parts (i) and (ii) of Theorem 8.2 cannot be
assumed to be infinite. The validity of these results in case of an infinite sequence of systems
depends on the properties of the inner product. It can be shown, however, that if the inner
product is the standard inner product in £2, then this extension is indeed valid. The proof
is omitted since it requires some relatively advanced results on Lebesgue integration.
8.3.1 Passivity and Small Gain
The purpose of this section is to show that, in an inner product space, the concept of
passivity is closely related to the norm of a certain operator to be defined.
In the following theorem Xe is an inner product space, and the gain of a system
H : Xe -+ Xe is the gain induced by the norm IIxII2 = (x, x).
Theorem 8.3 Let H : Xe -4 Xe , and assume that (I + H) is invertible in X, , that is,
assume that (I + H)-1 : Xe -+ Xe . Define the function S : Xe -- Xe :
We have:
S = (H - I)(I + H)-1. (8.16)