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11-1 Lyapunov Based Redesign Motivation But the real system is is unknown but not necessarily small. We assume it has a known bound. Consider.

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Presentation on theme: "11-1 Lyapunov Based Redesign Motivation But the real system is is unknown but not necessarily small. We assume it has a known bound. Consider."— Presentation transcript:

1 11-1 Lyapunov Based Redesign Motivation But the real system is is unknown but not necessarily small. We assume it has a known bound. Consider

2 11-2 Problem Approach : (i) chosen so that nominal closed loop system is asymptotically stable. (ii) chosen so as to cancel the effect of uncertainty. Find a state feedback controller so that the closed loop system is stable in a sufficiently strong sense.

3 11-3 Suppose also that is a Lyapunov function that proves the following. is strictly increasing.where, i.e. Assume that results in the uniformly asymptotically stable nominal closed loop system, Solution

4 11-4 Solution (Continued)

5 11-5 Let Due to the matching condition, can wipe out There are two ways at least : when is bounded in or in Solution (Continued)

6 11-6 Solution (Continued)

7 11-7 Smooth Control Smooth Control ( case)

8 11-8

9 11-9 Smooth Control (Continued) Then take large, so that

10 11-10 - When is chosen small, we can arrive at a sharper result. Assume that such that where is positive definite if Thus choosing we have Then, when where Also when We conclude which shows that the origin is uniformly asymptotically stable.

11 11-11 Example Ex: is Hurwitz. Choose are chosen so that where

12 11-12 Example (Continued)

13 11-13 Example (Continued)

14 11-14 Backstepping Consider a system 1 2 +

15 11-15 Backstepping (Continued) (1) (1),(2) ++ A

16 11-16 Backstepping (Continued) ++

17 11-17 Backstepping (Continued) 3 4 which is similar to the original system but  has an asymptotically stable origin when the input is 0.

18 11-18 Lemma & Example Lemma: (1), (2). (1) A (1), (2) Ex:

19 11-19 Example (Continued) Let’s consider

20 11-20 Recursive Backstepping Consider the following strict feedback system

21 11-21 Recursive procedure p.d.f   Consider Then using the previous result, obtain

22 11-22 Recursive procedure (Continued) Next consider Then we recognize that Thus, similarly, obtain the state feedback control and

23 11-23  Motivation  Plant – nonlinear  Controller – linear  Design method – classical linearization  Assumption – no single linear controller satisfies the performance specification  Idea – design a set of controllers, each good at a particular operating point, and switch (schedule) the gains of the controllers accordingly  Problem – now we have a nonlinear (piecewise linear) system with time dependent jump  Solution – no good tool but some theory is being developed mostly simulation in the past Extended Linearization (Gain scheduling method)

24 11-24 -Examples  Tank system Structure & Examples Structure ControllerPlant Gain Scheduler gains y Operating point

25 11-25 Control Goal + -

26 11-26  A different angle – nonlinear actuator Nonlinear Actuator actuator Large gain u f(u)

27 11-27 Step Responses

28 11-28 Approximation + - Domain f(u) Domain 21.3 3 16

29 11-29 Results

30 11-30 Classification + - Operating point gain + - + -

31 11-31 Issues Controller :

32 11-32 Example Ex: Theorem: Proof: See Ch 5 in Nonlinear System Analysis

33 11-33  A version of scheduling on the output Formalization

34 11-34 Block Diagram

35 11-35 Conditions  


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