Nonlinear Control Sy.. - Free

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A.3. CHAPTER 6
V
315
Thus,
f If
P = (DT(T,t)Q(T)a(D(T,t) dr+J
4T(T,t)Q(T)41(T,t)
dT - Q(t)
J
4pT
(T, t)Q(T)4D (T, t) dTl - Q(t)
-l A(t)-AT(t) 00
which implies that
A.3 Chapter 6
P = -P(t)A(t) - AT(t)p(t) - Q(t).
Theorem 6.1 Consider a function P(s) E R(s). Then F(s) E A if and only if (i) P(s) is
proper, and (ii) all poles of F(s) lie in the left half of the complex plane.
Proof: Assume first that P(s) satisfies (i) and (ii), and let n(s)andd(s) E R[s] respectively,
be the numerator and denominator polynomials of F(s). Dividing n(s) by d(s), we can
express F(s) as:
F(s) = k + d s) = k + G(s) (A.9)
where k E 1R and d(s) is strictly proper. Expanding G(s) in partial fractions and antitransforming
(A.9), we have
n
f (t) = G-1{F(s)} = kb(t) + E g2(t)
where n is the degree of d(s) and each g2 has one of the following forms [u(t) represents
the unit step function : (i) ktii-1e)tu(t), if A < 0 is a real pole of multiplicity m; and
(ii) ke(°+.3.+)tu(t), a < 0, if A = (v +,yw) is a complex pole. It follows that f (t) E A, and
so F(s) E A. To prove the converse, notice that if (i) does not hold, then f (-) contains
derivatives of impulses and so does not belong to A._Also, if (ii) does not hold, then, clearly,
for some i, g2 ¢ G1 and then f (t) V A and F(s) il A, which completes the proof.
Theorem 6.2 Consider a linear time-invariant system H, and let represent its impulse
response. Then H is L stable if and only if hoo(t) + ha,(t) E A and moreover, if H
is Gp stable, then JjHxjjr < jh11A11x11p.
2=1
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