Stability of Feedback Control Systems

Transcription

1 Stability of Feedback Control Systems

2 Equilibrium point Concept of Stability stable: system does not go far from the equilibrium point with small disturbance. (a). stable equilibrium point (b). conditional stable equilibrium point (c). unstable equilibrium point N UE SE CS

3 Stability Analysis (a). Free motion type stability small disturbance large disturbance (b). Force motion type stability bounded-input bounded-output (BIBO) stability A stable system is a dynamic system with a bounded response to a bounded input stability in the sense of Lyapunov asymptotic stability

4 Stability of Linear Systems Zero-input stability Equilibrium state x e : x(t) = x e for all t t 0 (1) x e = 0 () x e is an eigenvector corresponding to a zero eigenvalue, i.e. Atx () = λi = 0 e Consider xt &( ) = Atxt () () with xt ( ) = x 0 0

5 Stability in the sense of Lyapunov Equilibrium state x e = 0 of the system xt &( ) = Atxt () () is stable if for any ε x t ) < δ ( t, ). ( 0 0 ε ε implies x( t) < ε t t 0 0 x(0) δ

6 Asymptotically stable The system is asymptotically stable if it is stable in the sense of Lyapunov, and δ ( t0) > 0, x( t0) < δ ( t0) with lim t x( t) = 0. x(0) 0 δ ε

7 Remark 1 Consider xt &( ) = Axt () with xt ( ) = x 0 0 (1). The system is stable in the sense of Lyapunov iff Re{ λ i } 0, and the eigenvalue with zero real parts are distinct. (). The system is asymptotically stable iff Re{λ i } < 0, for all t. (3). For time invariant linear systems, asymptotical stability implies exponential stability. *** For time varying systems, the system can be unstable even if Re{λ i } < 0, for all t.

8 Zero-State Stability xt &( ) = Atxt () () + Btut () () yt () = Ctxt () () + Dtut () () with x(t 0 ) = 0, where A(t), B(t), C(t) and D(t) are continuous and bounded.

9 Bounded-input Bounded-Output (BIBO) Stable For all bounded input u(t) t t 0, the output y(t), is bounded for t t 0. Theorem: The system is BIBO stable iff there exist a finite constant c such that t t 0 Gt (, τ) dτ c t t 1 0 where G(t,τ) = C(τ)φ(t,τ)B(τ) and Gt (, τ) = max g (, t τ) 1 j n i = 1 ij

10 Bounded-input Bounded-Output (BIBO) Stable From the convolution integral relating u(t), y(t) and G(t) y( t and ) = y( t ) 0 or If u(t) is bounded, i.e., ( t ) M, then Therefore, if 0 u( t τ )g( τ )dτ 0 u( t τ ) g( τ )dτ Q < g( τ )dτ or Then y(t) will be bounded. y( t ) u y( t ) M M 0 g( τ )dτ N = 0 < u( t τ )g( τ )dτ 0 g( τ )dτ

11 Remark BIBO stability of the linear time invariant system (A, B, C, D) does not always guarantee asymptotic stability. e.g., If any canceled pole has a positive real part but all the remaining poles have negative real parts the system is BIBO stable but not asymptotically stable.

12 Asymptotically stable det( si A) = 0 BIBO stable Gs () = Cadj( si A) B + D det( si A) det( si A)

13 Routh-Hurwitz Stability Criterion The closed loop linear time invariant system is stable The roots of the closed loop characteristic equation are all in the open left-half plane (have negative real parts). Hurwitz polynomial: A polynomial with real coefficients and all its roots have negative real parts.

14 Routh-Hurwitz Stability Criterion The number of roots of the characteristic polynomial with positive real parts is equal to the number of changes in sign of the first column of the Routh array. For a given characteristic equation a n s n + a n n 1 1 s a1s + a0 = 1 0 then, the Routh Array is defined as follows

15 where b n 1 s n a n a n- a n-4... s n-1 a n-1 a n-3 a n-5... s n- b n-1 b n-3 b n-5... s n-3 c n-1 c n-3 c n s 0 h n-1 = a a a a a n 1 n n n 3 = 1 a n 1 n 1 a a n a a n n 1 n 3 b n 3 = 1 a n 1 a a n a a n 4 n 1 n

16 Constructing the Routh Array 1. The result will not be altered if a certain row is divided/multiplied by a positive real.. No element in the first column is zero: The Routh array can be constructed by definition shown above. 3. Zeros in the first column while some other elements of the row containing a zero in the first column are non-zero: The zero can be replaced with a small positive number, say ε, which is allowed to approach zero after completing the array.

17 Constructing the Routh Array Let a modified be p ( s) = p( s) ( s + β ), β > 0, then construct the Routh array for the modified polynomial. 4. Zero in the first column, and the other elements of the row containing the zero are also zero: This case occurs when the polynomial contains singularities that are symmetrically located about the origin in the s-plane, e.g. {( s+3), (s-3)}. In this case an auxiliary polynomial, which excludes the symmetrical zeros, is used.

18 Example δ ( s) = s + 5s + 11s + 5s + 36s + 30s + 36 s s s s s 1 6 s 1 5 s s + 30s + 36 is a factor of polynomial δ(s). ' Let δ() s = δ () s ( s + 3)( s + ) and δ ' () s = s + 5s + 6. By checking the roots of δ(s) it is obtained that there are four roots on the imaginary axis and the others are in the LHP.

19 Example δ ( s) = s + s + 10s + 7s + 15s + 40 s s s s s s 0 40 Two RHP roots

20 3 Example δ( s) = s + s + 4 s + k. Find the range of k such that the polynomial is Hurwitz. s s k s 1 8 s 0 k k Conditions for polynomial to be Hurwitz: 8 > k > 0 For k = 8, there are two roots on imaginary axis. For k = 0, there are one root at the origin.

21 Example 4 3 δ ( s) = s + s + s + s + 5 (1) s s 3 1 s 0 5 s 1 s s 3 1 s s 1-5/0 + s 0 5 Two RHP roots

22 () Let δ'( s) = δ( s) ( s + 1) s s s 3 9 s s 1 11 s 0 10 Two RHP roots

23 Kharitonov's Theorem Stability testing for interval polynomial δ( s, q) = q + q s q s 0 1 with { + q Q q q q q } i = i i i All polynomial in δ are Hurwitz if and only if the following four polynomial K i (s) are Hurwitz. n n K ()= s q + q s + q s + q s K ()= s q + q s + q s + q s K ()= s q + q s + q s + q s K ()= s q + q s + q s + q s

24 t Lyapunov Function A scalar function V(x, t) is called a Lyapunov function if for all t 0 and all vectors x in the neighbourhood of the origin, it satisfies the following conditions. 1. V(0, t) = 0. V( x, t), V( x, t) V( x, t) and, i = 1,..., n, all exist and t x are continuous. i 3. V( x, t) > α( x ) > 0, for all x 0 and t t 0, where α( ) is a non-decreasing function of x and α( 0) = 0, that is, V(x, t) is bounded below by a non-zero function.

25 Lyapunov Stability Analysis Theorem: (Sufficient condition) The equilibrium state x e = 0, of the system xt &( ) = f( x, t), f(, 0 t) = 0 is asymptotically stable if a Lyapunov function V(x, t) can be found such that V & ( x, t ) < 0 for all x 0 and t t 0 Theorem: (Sufficient condition) The equilibrium state x e = 0, of the system xt &( ) = f( x, t), f(, 0 t) = 0 is unstable if a Lyapunov function V(x, t) can be found such that V & ( x, t ) > 0 for all x 0 and t t 0

26 Example 0 1 x& = 1 xt (). Determine the stability of the above 15 t system choosing V( x, t) = [ x1 + ( t + 3) x ] Solution:. & V( x, t) V ( x, t ) = + t n i= 1 V & x x i i = 1 = x 1 + x (30t 1 x& 1 + ( t + 89) x + 3) x This indicates that the origin is asymptotically stable. < 0 x&

27 Stability Theorem The linear system xt &( ) = Atxt ()() where A(t) is continuous and bounded for t t 0, is uniformly asymptotically stable, iff given a positive definite real matrix Q(t), there exists a symmetric positive definite real matrix P(t) which satisfies Pt & () A T + () tpt () + Pt () At () = Qt (), t t0

28 Proof of Theorem Let V( x, t) = x ( t) P( t) x( t). T V( & T T x,t ) = x& ( t )P( t )x( t ) + x ( t )P( t )x( & t ) + x( t ) T P( & t )x( t ) T = x ()( t A () tpt () + Pt () At () + Pt & ())() xt = x T T () t Q()() t x t V & ( x, t ) is a negative definite quadratic form and the system is asymptotically stable. Note: For linear time invariant system xt &( ) = Axt (), the corresponding equation to be used is T A P + PA + Q = 0

29 Example [P] = lyap( A, Q) Check system stability of x& x& x1 = 1 1 x Choose Q = I p p + p p 1 p p p p = 0 1 p p p p = p11 = > 0, det( P) = > 0, P is positive definite 4 Therefore, the system is asymptotically stable.

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