非线性系统理论2

Lemma 11 Nonlinear Damping: Ω(x)xy − kn Ω2 (x)x2 ≤ why? y2 kn
µ ¶2 y y2 y2 Ω(x)xy − kn Ω (x)x ≤ −kn Ω(x)x − + ≤ 2kn 4kn kn
2 2
Lemma 12 Barbalat’s Lemma:If f (t) ∈ L∞ ∩ L2 , and fú(t) ∈ L∞ , then
10 1 −2 4 −1 P = 1 −2 −1 1
Since all the sucessive principal minors of P are positive, then V is PD.
Definition 6 Uniformly Continuous (UC): A function f (t) is Uniformly Continuous (UC) if for any positive constant ε, there exists a positive number δ (ε) satisfying that for any t0 and t ∈ (t0 − δ , t0 + δ ) |f (t) − f (t0 )| ≤ ε Note: In the deÞnition of UC, δ is independent of t0 . Example: The function is continuous but not UC. Why? Continuous: First note that ¯ ¯ ¯ ¯ ¯1 ¯ ¯ t − t0 ¯ 1 ¯ − ¯=¯ ¯ ¯ t t0 ¯ ¯ t0 t ¯ δ = min then t >
Chapter 2. Mathematical Background
1
Fundamental Definitions
kf (t)k2 = sZ
+∞
2-Norm
f 2 (τ ) d τ
∞-norm
−∞
kf (t)k∞ =sup |f (t)|
t
If kf (t)k2 ≤ ∞, then we say that f (t) belongs to L2 , i.e., f (t) ∈ L2 . If kf (t)k∞ ≤ ∞, then we say that f (t) belongs to L∞ , i.e., f (t) ∈ L∞ . Definition 1 Positive DeÞnite (PD): V (x) > 0 for ∀x 6= 0. Definition 2 Locally Positive DeÞnite (LPD): There exists a ball containing 0, for ∀x 6= 0 in the ball, V (x) > 0. Definition 3 Negative DeÞnite (ND): if −V (x) is PD. Definition 4 Locally Negative DeÞnite (LND): if −V (x) is LPD. Definition 5 fdsa;gfasdj;j dfsa Example:
t→∞
lim g2 (t) = 0
then
t→∞
lim g1 (t) = 0, lim fú(t) = 0
t→∞
Lemma 15 If and
V (t) > 0 ú (t) ≤ −f (t), V
t→∞
f (t) > 0
and f (t) is UC (or fú(t) ∈ L∞ ), then lim f (t) = 0.
Exact Model Knowledge (EMK) Control: Choose u to cancle out the nonlinearities in the dynamics £ ¤ u= x ú d + ax3 + b sin(t) + ke then e ú = −ke and e(t) ≤ e(0)e−kt Note: Need to insure that all the signals in the closed-loop system remain bounded. e, xd ∈ L∞ =⇒ x ∈ L∞ and for the control ¯ ¯ |u| ≤ |x ú d | + |a| ¯x3 ¯ + |b| + k |e|
We can then show that Utilizing the above fact to the 2nd equation and applying the extended Barbalat’s Lemma yields x3 → 0 therefore, Asymptotic tracking. Example: For the same system but a, b unknown, how to construct the controller? Similarly, we choose h i u= x úd + a ˆ x3 + ˆ b sin(t) + ke
4
1.1
Solution
1 1 2 1 2 V = x2 1 + x2 + x3 2 2 2 ú V
Choose then
= x1 x ú 1 + x2 x ú 2 + x3 x ú3 2 = −x1 + x2 [−x1 − x2 − x3 − x1 x3 ] + x3 (x1 + 1) x2 2 = −x2 1 − x2 − x1 x2 ¸ · 2 x1 + x2 2 2 2 ≤ −x1 − x2 + 2 2 2 x + x2 = − 1 2 x1 → 0, x2 → 0
t→∞
lim f (t) = 0.
Lemma 13 Integral form of Barbalat’s Lemma: If f (t) is UC and the intergral Z t lim | f (τ )| d τ
t→∞ 0
exists and is Þnite, then
t→∞
lim |f (t)| = 0.
. .
a ˜= − a ˆ
. .
˜ b= − ˆ b 5
ú (t) can be rewritten as follows (why?, a, b are constants). Therefore, V . i h . 1 −1˜ ˆ ú = e −ke + a b sin(t) − Γ− a ˜ a ˆ − Γ b b V ˜x3 + ˜ 1 2 µ .¶ ³ ´ . 2 3 −1 − 1 = −ke + a ˜ ex − Γ1 a ˆ +˜ b e sin(t) − Γ2 ˆ b further, make the update laws as and a ˆ= Γ1 ex3
2 2 V (x1 , x2 , x3 ) = 10x2 1 + 4x2 + x3 + 2x1 x2 − 2x2 x3 − 4x1 x3
The state is x= and then
£
x1 x2 x3
Hale Waihona Puke V = xT P x 1¤T
with
Since the matrix P is positive definite (why?), then V is PD. ¯ ¯ ¯ ¯ ¯ 10 1 −2 ¯ ¯ ¯ 10 1 ¯ ¯ ¯ > 0, ¯ 1 ¯>0 4 − 1 |10| > 0, ¯ ¯ 1 4 ¯ ¯ ¯ ¯ −2 −1 1 ¯
Lemma 10 Let A be a real, symmetric, positive-deÞnite(PD) matrix, λmin (A) and λmax (A) denote the minimum and maximum eigenvalues of A, respectively, then λmin (A) kxk2 ≤ xT Ax ≤ λmax (A) kxk2 . 3
where a ˆ, ˆ b denote the on-line estimates for a, b, respectively. Substitute the controller into the dynamics to obtain i ¤ h £ e ú = x ú d + ax3 + b sin(t) − x úd + a ˆx3 + ˆ b sin(t) − ke = −ke + a ˜x3 + ˜ b sin(t) with a ˜, ˜ b denote the estimation error. Choose 1 1 1 2 1 −1˜2 a ˜ + Γ2 b ≥ 0 V (t) = e2 + Γ− 2 2 1 2 The Lyapunov function is designed from the control requirement. Objective: make e → 0, and a ˜, ˜ b ∈ L∞, thus we put it into V (t). Γ1 and Γ2 are the subsequently defined update gains for the parameter estimation. Then . . 1 −1˜ ˜ ú = ee V ú + Γ− a ˜ a ˜ + Γ b b 1 2 Note and
t0 , 2
1 f (t ) = , t > 0 t
If we choose
½