Prof. Dr. Eleni Chatzi System Stability
Fundamentals Overview System Stability
Assume given a dynamic system with input u(t) and output x(t). The stability property of a dynamic system can be defined from the system s impulse response x δ (t) as follows: Asymptotically stable system: The steady state impulse response tends to zero lim x δ(t) = 0 t Marginally stable system: The steady state impulse response is non-zero, but bounded (does not grow infinite): lim x δ(t) < t Unstable system: The steady state impulse response is unbounded: lim x δ(t) = t
Stable System Example - Assume the following SDOF system: ẍ(t) + 3ẋ(t) + 8x(t) = u(t) L s 2 X (s) + 3sX (S) + 8X (s) = U(s) 1 Therefore, the associated Transfer Function is: H(s) = s 2 + 3s + 8 0.2 Impulse Response of a Stable System 0.15 0.1 x(t) 0.05 0 0.05 0 2 4 6 8 10 time (s) Observe that the roots of s 2 + 3s + 8 are: s 1 = 1.5 + 2.4i, s 2 = 1.5 2.4i
Marginally System Example - Assume the following SDOF system: ẍ(t) + 9x(t) = u(t) L s 2 X (s) + 9X (s) = U(s) Therefore, the associated Transfer Function is: H(s) = 1 s 2 + 9 0.4 Impulse Response of a Marginally Stable System 0.3 0.2 0.1 x(t) 0 0.1 0.2 0.3 0.4 0 2 4 6 8 10 time (s) Observe that the roots of s 2 + 9 are: s 1 = 3i, s 2 = 3i
Unstable System Example - Assume the following SDOF system: ẍ(t) + 3ẋ(t) 3/4x(t) = u(t) L s 2 X (s) + 3sX (S) 3/4X (s) = U(s) 1 Therefore, the associated Transfer Function is: H(s) = s 2 + 3s 3/4 3 Impulse Response of an Unstable System 2.5 2 x(t) 1.5 1 0.5 0 0 2 4 6 8 10 time (s) Observe that the roots of s 2 + 9 are: s 1 = 3.23, s 2 = 0.23
Transfer Function - Poles & Zeros For the general case of linear systems examined herein. The TF will generally have the following form: H(s) = K(s + z 1)(s z 2 )... (s z m ) (s p 1 )(s p 2 )... (s p n ) m < n *Note: If m > n, then the division can be carried out and the system can eventually be rewritten using a form analogous to the above. The constants z i are called the zeros of the transfer function or signal, and p i are the poles. Viewed in the complex plane, it is clear that the magnitude of H(s) will be equal to zero at the zeros (s = z i ), and to infinity at the poles (s = p i ).
Reminder - By using partial fraction expansion we can rewrite the TF in the following form: H(s) = b 1 s + b 2 (s p 1 )(s p 2 ) + b 3 s p 3 +... Where p 1 = p 2 (complex conjugate roots) and p 3 is a real root. In order to derive b 1, b 2 multiply both sides by (s p 1 )(s p 2 ), and then evaluate at s = p 1. In order to derive b 3 multiply both sides by (s p 3 ), and then evaluate at s = p 3. It is then easy to obtain the system s response x(t) to an impulse input u(t) = δ(t) by simply applying the Inverse Laplace transform on H(s): x(t) = L 1 {H(s) U(s)} U(s)=L{δ(t)}=1 x(t) = L 1 {H(s)}
Stability Rules For systems that are linear and time-invariant (i.e their TF does not depend on time) their stability is defined by the roots of the characteristic polynomial, i.e., the poles of the TF. Specifically: Asymptotically Stable All the roots of the characteristic polynomial lie in the left half plane (Re(s) < 0) Unstable At least one root of the characteristic polynomial lies in the right half plane (Re(s) > 0) Marginally Stable No solution grows unbounded but some do not decay (Re(s) 0) Asymptotically Stable Im Unstable Marginal stability (if not repeated) Re
Interpretation The poles of the TF are essentially the roots of the characteristic polynomial that corresponds to the original ODE of the system. Therefore they govern the system s homogeneous response. A root s = σ ± ωi signifies that the homogeneous response will be of the type: x = e st = e σt (C 1 sinωt + C 2 cosωt) It is now apparent that a positive real part for s would be linked to an exponentially increasing response which is the cause for instability.
Pole Zero Plots Examining the Pole Zero plots of the previous examples, helps illustrate the previous point (MATLAB: pzmap) 2.5 Pole Zero plot Stable 3 Pole Zero plot Marginally Stable 1 Pole Zero plot Unstable 2 0.8 1.5 2 0.6 1 1 0.4 Imaginary Part 0.5 0 0.5 Imaginary Part 0 Imaginary Part 0.2 0 0.2 1 1 0.4 1.5 2 0.6 2 0.8 2.5 2.5 2 1.5 1 0.5 0 Real Part 3 1 0.5 0 0.5 1 Real Part 1 4 3 2 1 0 1 Real Part
Continuous MDOF Systems Stability for MDOF Systems Let us remember the Laplace Transform for the state-space form of MDOF systems: ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) sx (s) X (0) = AX (s) + BU(s) Y (s) = CX (s) + DU(s) The input (X (s)) output (Y (s)) TF is obtained as: Y (s) = {C(sI A) 1 B}U(s) H F (s) = C(sI A) 1 B adj(si A) H F (s) = C det(si A) B Therefore, the poles of H(s) for the MODF case, correspond to the eigenvalues of the state matrix A.
Stability Rules for Continuous MDOF Systems Asymptotically Stable A time-invariant system is asymptotically stable if all the eigenvalues of the system matrix A have negative real parts. Unstable At least one root of the eigenvalues of the system matrix A has a positive real part. Marginally Stable A time-invariant system is marginally stable if and only if all the eigenvalues of the system matrix A are zero or have negative real parts, and those with zero real parts are simple roots of the minimal polynomial of A.
Discrete MDOF Systems Stability for Discrete MDOF Systems Let us remember the relationship between Continuous & Discrete State-Space Systems ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) (cont.) x (k+1) = A d x k + B d u k y k = Cx k + Du k (disc.) where the relationship between the continuous and discrete system matrices is: A d = e A t B d = ( e A t I ) BA 1 and the relationship between the eigenvalues of the continuous and discrete matrices is: eig(b d ) = e eig(a) t This implies that the stability of the discrete system is linked to the stability of the continuous system, with the requirement that now stable eigenvalues lie within the unit circle.
Stability Rules for Continuous MDOF Systems Let λ 1,..., λ m, m n be the eigenvalues of A R n n. The system x k+1 = A d x k + B d u k is Asymptotically Stable iff λ i < 1, i = 1,..., m Unstable iff i such that λ i > 1 Marginally Stable iff λ i 1, i = 1,..., m, and the eigenvalues with unit modulus have equal algebraic and geometric multiplicity a ( ) (*) Algebraic multiplicity = number of coincident roots λ i of det(λi A d ). Geometric multiplicity = number of linearly independent eigenvectors v i, A d v i = λv i
Appendix Considerations What happens in the case of a repeated root? Assume we have a repeated root at s = p k with multiplicity l. Then the Transfer Function is written as follows: H(s) = K(s z 1)(s z 2 )... (s z m ) (s p 1 )... (s p k ) l... (s p n ) Then the partial fraction expansion, focusing on that term, will be of the form: H(s) = + b 1 b 2 + s p k (s p k ) 2 +... + b l (s p k ) l + Hence, the impulse response will be of the form x(t) = h(t) = + b 1 e p kt + b 2 te p kt +... + b l (l 1)! tl 1 e p kt +
Appendix Considerations - Repeated Root However for p k = σ + iω with R(p k ) = σ < 0 we have that: o t l 1 e p kt < This signifies that these terms are bounded, i.e., they do not grow to infinity and therefore our previously derived criteria for stability hold. 0.4 Impulse Response of a System with Repeated Root Example: Impulse Response of 1 H(s) = (s + 1) 2 Repeated root s = 1 x(t) 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0 2 4 6 8 10 time (s)