Nonlinear Control Sy.. - Free

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A.5 Chapter 8
APPENDIX A. PROOFS
Theorem 8.3: Let H : Xe -> Xe , and assume that (I + H) is invertible in Xe , i.e., assume
that (I + H)-1 : Xe --> Xe . Define the function S : Xe --> Xe :
We have
S = (H - I)(I + H)-1 (A.43)
(a) H is passive if and only if the gain of S is at most 1, that is, S is such that
II(Sx)TIIx s IIxTIIx Vx E XefVT E Xe (A.44)
(b) H is strictly passive and has finite gain if and only if the gain of S is less than 1.
(a) : By assumption, (I + H) is invertible, so we can define
xr (I + H)-1y (A.45)
(I + H)x = y (A.46)
Hx = y - x (A.47)
= Sy = (H - I)(I + H)-1y = (H - I)x (A.48)
IISyIIT == ((H - I)x, (H - I)x >T
(Hx-x,Hx-x>T
= IIHxliT + IIxIIT - 2(x, Hx >T (A.49)
IIyIIT = (y, y >T
((I + H)x, (I + H)x >T
IIHxIIT+IIxIIT+2(x,Hx>T
Thus, subtracting (A.50) from (A.49), we obtain
IISyIIT = IIyIIT - 4(x, Hx >T
(A.50)
(A.51)
Now assume that H is passive. In this case, (x, Hx >T> 0, and (A.51) implies that
IISyIIT - IIyIIT = IISyIIT T= IIyIIT - IISyIIT - 0
and thus H is passive. This completes the proof of part (a).