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Download Adaptive backstepping control of uncertain systems: by Jing Zhou, Changyun Wen PDF

By Jing Zhou, Changyun Wen

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"‘The publication is useful to benefit and comprehend the elemental backstepping schemes’. it may be used as an extra textbook on adaptive keep an eye on for complicated scholars. keep an eye on researchers, specifically these operating in adaptive nonlinear keep watch over, also will broadly make the most of this book." (Jacek Kabzinski, Mathematical reports, factor 2009 b)

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Extra info for Adaptive backstepping control of uncertain systems: Nonsmooth nonlinearities, interactions or time-variations

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4), f0 is selected as the upper bound of Vρ (0)+ 0 f2 (θ˜T Γ −1 θ+ t Mρ e−f (t−τ ) dτ, g(t) = bm (t). 1, we can conclude that Vρ (t) 0 and χ(t), hence zi , (i = 1, . . , ρ), θ, q and a are bounded. Since z1 and yr are bounded, y is also bounded. 36) are bounded as A0 is Hurwitz. Since the system is minimum phase in Assumption 4 and the boundedness of y, we have λ1 , . . , λm+1 are bounded. 59) gives ˆ qˆ, y (i−1) , λ1 , . . , λm+i−1 ) vm,i = zi + αi−1 (y, ξ, Ξ, θ, r i = 2, 3, . . 101) ˆ qˆ, yr and y˙ r proves that Let i = 2, the boundedness of λ1 , .

In a conclusion, we can find a subsequence that leads to a contradiction in both g(t) > 0 and g(t) < 0. Therefore, χ(t) has upper bound. Case 2): χ has no lower bound on [0, tf ]. Define χ = −w. Accordingly, w has no upper bound. 16) Thus, there must exist a monotone increasing variable {wi = w(ti )} with w0 = |w(t0 )| > 0, limi→∞ ti = tf , and limi→∞ wi = ∞. Following the procedure as in Case 1), we can also construct a subsequence that leads to a contradiction. Accordingly, we can claim that w has upper bound on [0, tf ).

G(t) > 0. 8) where lm1 = 4m + 1 − χ0 . 9) System Model and Problem Formulation 37 where cm1 ∈ (0, 1), here we select cm1 = 12 . 10) ∗ where g0 = 2cm1 gmin e−f (t2 −t1 ) cos(πcm1 /2) > 0, and g1 = 2cm1 e−f 0. 11) Note that em grows faster than m when m → ∞. 11), we know that Vg (χ0 , χ2 ) = Vg (χ0 , 4m + 3) → −∞ as m → ∞. 2. g(t) < 0. 12) where lm2 = 4m − 1 − χ0 . 14) ∗ where g2 = 2cm1 gmin e−f (t1 −t3 ) cos(πcm1 /2) > 0, and g3 = 2cm1 e−f 0. 15), we know that Vg (χ0 , χ1 ) = Vg (χ0 , 4m + 1) → −∞ as m → ∞.

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