A nemlineáris rendszer- és irányításelmélet alapjai Relative degree and Zero dynamics (Lie-derivatives)

Transcription

1 A nemlineáris rendszer- és irányításelmélet alapjai Relative degree and Zero dynamics (Lie-derivatives) Hangos Katalin BME Analízis Tanszék Rendszer- és Irányításelméleti Kutató Laboratórium MTA Számítástechnikai és Automatizálási Kutató Intézete Budapest NONL-3 p. 1/21

2 Lie-derivative NONL-3 p. 2/21

3 Lie-derivative Consider λ R n R, f R n R n with U =dom(λ) =dom(f) R n open, λ is C 1 on U Derivative of λ along f: L f λ(x) = λ(x) x f(x) = n i=1 λ(x) x i f i (x) =< dλ(x),f(x) > Repeated use: L g L f λ(x) = (L fλ(x)) g(x) x L k fλ(x) = (Lk 1 f λ(x)) x f(x) NONL-3 p. 3/21

4 Lie-derivative: examples 1. One-dimensional example, i.e. x R λ(x) = x 2, f(x) = x Both L f λ and L λ f can be computed: 2. Two-dimensional example, x R 2 L f λ(x) = 2x x, L λ f(x) = 1 2 x x2 λ(x 1, x 2 ) = x 2 1, f(x 1, x 2 ) = f(x) = x2 1 + x2 2 x x3 2 and L f λ(x 1, x 2 ) = 2x 1 (x x 2 2) NONL-3 p. 4/21

5 Zero dynamics NONL-3 p. 5/21

6 The relative degree 1 The SISO nonlinear system ẋ = f(x) + g(x)u y = h(x) has relative degree r at a point x 0 if 1. L g L k f h(x) = 0 for all x in a neighborhood of x0 and all k < r 1 2. L g L r 1 f h(x 0 ) 0. Remark: there may be points where the relative degree is not defined! If L g L k fh(x) = 0 for all k 0 then the output of the system is not affected by the input for all t near t 0. NONL-3 p. 6/21

7 The relative degree 2 Example: Van der Pool oscillator ẋ = x u 2ωζ(1 µx 2 1)x 2 ω 2 x 1 for y = x 1 : relative degree is 2 L g h(x) = h g(x) = [1 0] 0 x 1 = 0 L f h(x) = x 2 L g L f h(x) = L fh g(x) = [0 1] x 0 1 = 1 for y = sin(x 2 ): relative degree is 1 (where defined) NONL-3 p. 7/21

8 The relative degree 3 Interpretation: the number of times y has to be differentiated at t = t 0 in order to have the value u(t 0 ) of the input explicitly appearing. g(x) is vector-valued, that is g(x) = g 1 (x).. g n (x) k = 0: y = h(x) dy dt = h xẋ = h x (f(x) + g(x)u) = L fh + L g h u k = 1: assume Lg h = 0 (condition 1) then ẏ = L f h, and d 2 y dt 2 = (L fh) (f(x) + g(x)u) = L 2 f x h + L gl f h u If L g L f h 0 then r = 2 (see condition 2), otherwise we continue the procedure, and so on. NONL-3 p. 8/21

9 Special case: LTI systems ẋ = Ax + Bu y = Cx L k f h(x) = CAk x and L g L k f h(x) = CAk B (i.e. the Markov-parameters) Relative degree is r if CA k B = 0 k < r 1 CA r 1 B 0 which is equal to the difference between the degree of the denominator and the numerator of the transfer function H(s) = C(sI A) 1 B. NONL-3 p. 9/21

10 Coordinates transformation Suppose the system has relative degree r at x 0 (r n). Set φ 1 (x) = h(x) φ 2 (x) = L f h(x)... φ r (x) = L r 1 f h(x). If r < n then it is always possible to find n r more functions φ r+1,...,φ n such that the mapping (local coordinates transformation) Φ(x) = φ 1 (x)... φ n (x) has a Jacobian matrix which is nonsingular at x 0. NONL-3 p. 10/21

11 The transformed normal form It is always possible to choose φ r+1,...,φ n in such a way that L g φ i (x) = 0 for all r + 1 i n and all x around x 0. State space description in the new coordinates z = Φ(x): ż 1 = z 2 ż 2 = z 3... ż r 1 = z r ż r = b(z) + a(z)u ż r+1 = q r+1 (z)... ż n = q n (z) y = z 1 where q i (z) = L f φ i (Φ 1 (z)), a(z) = L g L r 1 f h(φ 1 (z)), b(z) = L r f h(φ 1 (z)) NONL-3 p. 11/21

12 Zero dynamics the problem Problem: how does the system (i.e. the state variables) behave when its output is identically zero? Zero-output constrained dynamics in the original coordinates m ẋ = f(x) + g i (x)u i for a pair (u i,y) (SISO case). i=1 0 y = h(x) (constrained!) NONL-3 p. 12/21

13 Zero dynamics example Simple example: linear output function ẋ 1 = x x u ẋ 2 = 5x x 3 2 x 2 u 0 y = x 1 with x 1 = 0 with ẋ 1 = 0 ẋ 1 = x u ẋ 2 = 3x 3 2 x 2 u u = 2x 2 2 ẋ 2 = 3x 3 2 2x 2 x 2 2 = x 3 2 NONL-3 p. 13/21

14 Zero dynamics normal form Normal form for systems with relative degree r ż 1 = z 2 ż 2 = z 3... ż r 1 ż r = z r = b(ξ,η) + a(ξ,η)u η = q(ξ, η) where ξ = [z 1... z r ] T, η = [z r+1... z n ] T, a(ξ,η) = L g L r 1 f h(φ 1 (ξ,η)) and b(ξ,η) = L r f h(φ 1 (ξ,η)). NONL-3 p. 14/21

15 Controller form realization ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) with A c = a 1 a 2... a n , B c = C c = [ b 1 b 2... b n ] with the coefficients of the polynomials a(s) = s n + a 1 s n a n 1 s + a n and b(s) = b 1 s n b n 1 s + b n that appear in the transfer function H(s) = b(s) a(s) NONL-3 p. 15/21

16 Zero dynamics Problem of Zeroing the Output is to find, pairs consisting of an initial state x and input function u, such that the corresponding output y(t) of the system is identically zero. Zero dynamics for η = [z r+1... z n ] T η(t) = q(0,η(t)), η(0) = η 0 With a(ξ,η) = L g L r 1 f h(φ 1 (ξ,η)) and b(ξ,η) = L r f h(φ 1 (ξ,η)) the input is u(t) = b(0,η(t)) a(0,η(t)) NONL-3 p. 16/21

17 Zero dynamics Example 1 ẋ 1 = 0.4x 1 x 2 1.5x 1 ẋ 2 = 0.8x 1 x 2 1.5x u y = x 2 System parameters f(x) = 0.4x 1x 2 1.5x 1 0.8x 1 x 2 1.5x 2, g(x) = h(x) = [ x 2 ] NONL-3 p. 17/21

18 Zero dynamics Example 1 Relative degree: r = 1 y = x 2 = 0 ẋ 1 = 1.5x 1 ẋ 2 = 1.5u Zero dynamics: ẏ = x 2 = 0 u 0 output zeroing input ẋ 1 = 1.5x 1 NONL-3 p. 18/21

19 Zero dynamics Example 2 ẋ 1 ẋ 2 = x x x 2 1x 2 + x u = 3x 1 x x 2 1x 2 + x 1 u y = x 2 System parameters f(x) = x x x 2 1x 2 3x 1 x x 2 1x 2 + x 1, g(x) = x h(x) = [ x 2 ] NONL-3 p. 19/21

20 Zero dynamics Example 2 Relative degree: r = 1 y = x 2 = 0 ẋ 1 ẋ 2 = x u = x 1 u Zero dynamics: ẏ = x 2 = 0 u = x 1 output zeroing input ẋ 1 = x NONL-3 p. 20/21

21 Homework 1. Compute the relative degree of your system. 2. Compute the output zeroing input and the zero dynamics of your system. 3. Is your zero dynamics stable? What does it mean? NONL-3 p. 21/21