Linearization problem. The simplest example

Transcription

1 Linear Systems Lecture 3 1 problem Consider a non-linear time-invariant system of the form ( ẋ(t f x(t u(t y(t g ( x(t u(t (1 such that x R n u R m y R p and Slide 1 A: f(xu f(xu g(xu and g(xu exist and are continuous. The question is: How to approximate system (1 by a linear model? The simplest example Consider the static time-invariant system y(t ku(t + η. (2 Slide 2 This system is nonlinear but intuitively it seems to be almost linear. Indeed this system can easily be made linear. To this end notice that for any constant u(t u e the output y(t is also constant and given by. kue + η. Let s call the pair (y e u e the equilibrium point. y e Define now new signals: y δ (t. y(t y e and u δ (t. u(t u e which are deviations of y and u from the equilibrium point. Then y δ (t ku δ (t (2 δ which is linear. Thus nonlinear system (2 can be made linear by an appropriate (cf. equilibrium shift of its input and output signals.

2 Linear Systems Lecture 3 2 A little bit more complicated example Consider now the system ẋ(t ax(t + bu(t + ηx y(t x(t + η y. (3 Assume again that the input u(t u e is constant. Then Slide 3 x(t t e a(t s (bu e + η x dt 1 a (1 e at (bu e + η x. As lim t e at system (3 approaches it equilibrium point a (x e u e where x e 1 a (bu e + η x. a This is called the steady-state. A little bit more complicated example (contd Let s write now system (3 in terms of the deviations x δ (t. x(t x e and u δ (t. u(t u e. Slide 4 We get: ẋδ (t ax δ (t + bu δ (t y δ (t x δ (t (3 δ where y δ (t. y(t x e η y. Equation (3 δ is thus a linear equivalent of the nonlinear system (3.

3 Linear Systems Lecture 3 3 Equilibrium The pair (x e u e is called equilibrium if f ( x e u e. Slide 5 Let (x e u e be an equilibrium of (1. Then x(t y(t and u(t can be presented as follows: x(t x e + x δ (t y(t y e + y δ (t and u(t u e + u δ (t. ( where y e g xe u e. Eqn. (1 can then be rewritten as follows: ẋδ (t f ( x e + x δ (t u e + u δ (t y δ (t g ( x e + x δ (t u e + u δ (t (1 δ y e Tailor series expansion One can expand f( and g( as follows a : f ( x u f(x e u e + f xeu e x δ + f xeu e u δ + o ( x δ u δ g ( x u g(x e u e + g xeu x e δ + g xeu u e δ + o ( x δ u δ Slide 6 where o(η is any function of a variable η such that lim η o(η η. a Given an n-dimensional function φ(η of an m-dimensional argument η the partial derivative of φ with respect to η (the Jacobian matrix is defined as the following n m matrix: φ(η η. φ 1 (η... η 1. φ n(η... η 1 φ 1 (η η m. φ n(η η m.

4 Linear Systems Lecture 3 4 around (x e u e If the deviations x δ and u δ are small enough then Eqn. (1 δ can be approximated by the following linear model: ẋδ (t A δ x δ (t + B δ u δ (t y δ (t C δ x δ (t + D δ u δ (t (1 linearized Slide 7 where A δ C δ. f. xeu B δ f e xeu e. g. xeu e D δ g xeu e. Inverted pendulum Slide 8 m θ M u y Nonlinear model: (M + mÿ + ml θ cos θ ml( θ 2 sin θ + µẏ u M : mass of the cart m : mass of the pendulum l : length of the pendulum y : position of the cart θ : angular rotation u : force on the cart µ : friction coefficient g : acceleration of gravity l θ g sin θ + ÿ cos θ or equivalently [ [ [ [ M + m ml cos θ ÿ ml( θ 2 sin θ µẏ 1 + u. cos θ l θ g sin θ

5 Linear Systems Lecture 3 5 Inverted pendulum: the state-space realization The state vector: x 1 y x x 2 θ x 3 ẏ x 4 θ Slide 9 The state equation: x 1 x 3 d x 2 x 4 [ dt x 3 1 ([ [. (4 M+m ml cos x 2 mlx 2 4 sin x 2 µx u x 4 cos x 2 l g sin x 2 }} f(xu where Λ(x 2. [ M+m ml cos x 2 is nonsingular (det Λ(x 2 l(m + m sin 2 x 2. cos x 2 l Inverted pendulum: equilibriums Equilibrium points should correspond to ẋ. Then from Eq. (4 one can get that x 3 x 4 and [ Λ(x 2 1 u. g sin x 2 Slide 1 Therefore the equilibrium points are: x 1 kπ (x e u e for every x 1 R and k ±1 ±2...

6 Linear Systems Lecture 3 6 Inverted pendulum: Jacobian matrices Jacobian matrices a : [ f [ f 34 Λ(x 2 1 mlx 2 4 cos x 2 + φ 1 (x 2 x 3 x 4 sin x 2 µ 2mlx 4 sin x 2 g cos x 2 + φ 2 (x 2 x 3 x 4 sin x 2 Slide 11 where φ 1 and φ 2 are some algebraic functions of x 2 x 3 and x 4 f 1 l(m+m sin 2 x 2 l cos x 2. a Given two matrix functions M 1 (η and M 2 (η then ( M1 (η 1 M η 2 (η M 1 (η 1( M 2 (η M 1(η M η η 1 (η 1 M 2 (η. Inverted pendulum: linearization around ( Slide 12 It s readily verified that 1 f 1 m M g µ M m+m Ml g µ Ml and f 1 M 1 Ml which yield the A and B matrices of the linearized realization.

7 Linear Systems Lecture 3 7 of discrete-time systems The linearization problem for discrete-time system xk+1 f ( x k u k y k g ( x k u k Slide 13 where x R n u R m y R p has exactly the same solution as in the continuous time except that the equilibrium is obtained from f ( x e u e xe.