Control Systems Internal Stability - LTI systems L. Lanari
outline LTI systems: definitions conditions South stability criterion equilibrium points Nonlinear systems: equilibrium points examples stable equilibrium state (see slides StabilityTheory by Prof. G. Oriolo) indirect method of Lyapunov (see slides StabilityTheory by Prof. G. Oriolo) Lanari: CS - Internal stability 2
definitions (LTI systems) (AS) - A system S is said to be asymptotically stable if its state zero-input response converges to the origin for any initial condition (MS) - A system S is said to be marginally stable if its state zero-input response remains bounded for any initial condition (U) - A system S is said to be unstable if its state zero-input response diverges for some initial condition note: only interested in the free state evolution (ZIR) note: use of any/some state transition matrix (t) =e At (t, t ) LTI (Linear Time Invariant) LTV (Linear Time Variant) Lanari: CS - Internal stability 3
possible behaviors we saw that the x ZIR (t) =e At x is a linear combination of A diagonalizable mg( i) = ma( i) for all i real i complex i = i + j!i aperiodic modes e i t pseudoperiodic modes e it [sin( i t + R )u re + cos( i t + R )u im ] A not diagonalizable (defective matrix A) mg( i) < ma( i) real i complex i = i + j!i...,..., t n k (n k )! e i t t n k (n k )! e it sin i t max dimension of Jordan block Jk Lanari: CS - Dynamic response - Time 4
stability and eigenvalues (stability criterion) A LTI system is asymptotically stable if and only if all the eigenvalues have strictly negative real part A LTI system is marginally stable if and only if all the eigenvalues have non positive real part and those which have zero real part have scalar Jordan blocks equivalent to mg( i) = ma( i) for all i (only for those with real part) A LTI system is unstable if and only if there exists at least one eigenvalue with positive real part or a Jordan block corresponding to an eigenvalue with zero real part of dimension greater than equivalent to mg( i) < ma( i) (only for those with real part) Lanari: CS - Internal stability 5
stability and eigenvalues (stability criterion) it all depends upon the positioning of the eigenvalues of matrix A in the complex plane Im Re asymptotic stability all eigenvalues in the open left half-plane Im Re distinct eigenvalues case marginal stability some eigenvalues may be on the Im axis (ma( i) = case) Im Re instability at least one eigenvalue with positive real part (the case Re( i) = and Jordan block dim > is not shown) Lanari: CS - Internal stability 6
remarks stability is an intrinsic characteristic of the system, depends only on A stability does not depend upon the applied input nor from B, C or D example t u(t) ( AS system ẋ = x + u t x(t) t u(t) y = x t x(t) but system is asymptotically stable diverging response due to the input not to the system Lanari: CS - Internal stability 7
remarks unstable systems can have bounded or converging solutions for some specific initial conditions x(t) =e t u v T x + e 2t u 2 v2 T x aperiodic > unstable modes system 2 < x 5 has no component along the unstable eigenspace the time evolution is a decaying exponential along the eigenspace stable eigenspace 2 < x 5 x 6 u 2 u x7 > unstable eigenspace x 4 has a component c 4 along the unstable eigenspace c 4 u = v T x 4 x 4 c 4 as long as the initial condition does not have a component along the unstable eigenspace the zero input response does not diverge Lanari: CS - Internal stability 8 x 3 x 2 x
remarks if the system is asymptotically stable then the output ZIR also converges to (the converse is not true) if the system is stable then the output ZIR is bounded (the converse is not true) if the system is unstable it does not necessarily imply that the output will diverge for some initial condition (it may never diverge) y = Ce At x = nx i= e it Cu i v T i x this term may be zero for some ui example A = 2, B =, C = (compute yzir(t)) Lanari: CS - Internal stability 9
examples MSD with no friction and no spring eigenspace V is generated by m s = f A = = ma( ) = 2 and therefore mg( ) = < ma( ) system in unstable (with a non-zero initial velocity, the mass will move with constant velocity and the position will grow linearly with time) MSD with no spring m s + µṡ = f A = µ/m = 2 = -µ/m < since ma( ) = = mg( ) for the zero eigenvalue =, the system is marginally stable (from a generic initial condition, the ZIR velocity will go to zero while the ZIR position will asymptotically stop at a constant value which depends upon the initial conditions) Lanari: CS - Internal stability
LTI stability criterion: Routh criterion In order to establish if a LTI system is asymptotically stable we do not need to compute the eigenvalues but just the sign of their real parts generic polynomial of order n p( )=a n n + a n n + a n 2 n 2 + + a + a A necessary condition in order for the roots of p( ) = to have all negative real part is that the coefficients need to have all the same sign if all the roots of p( ) = have negative real part then the coefficients have the same sign if a coefficient a i is null then the coefficients do not have the same sign and therefore the necessary condition is not satisfied Lanari: CS - Internal stability
Routh-Hurwitz stability criterion In order to state a necessary and sufficient condition we need to build a table Routh table row n row n- row n-2 row row p( )=a n n + a n n + a n 2 n 2 + + a + a a n a n 2 a n 4 a n a n 3 a n 5 b b 2 c c 2 computed d as. missing terms can be set to directly from p( ) b = b 2 = a n a n 2 a n a n a n 3 a n a n 4 a n a n a n 5 c = b a n a n 3 b b 2 the Routh table has a finite number of elements and has n+ rows an entire row can be multiplied by a positive number without altering the result c 2 = b a n a n 5 b b 3 Lanari: CS - Internal stability 2
Routh-Hurwitz stability criterion If the Routh table can be completed then we have the following N&S condition All the roots of p( ) = have negative real part iff there are no sign changes in the first column of the Routh table applied to the characteristic polynomial we have the following stability criterion A LTI system is asymptotically stable iff the Routh table built from the characteristic polynomial has no changes in the first column if the table cannot be completed (due to some in the first column) then not all the roots have negative part the number of sign changes in the first column of the Routh table is equal to the number of roots with positive real part Lanari: CS - Internal stability 3
Routh table example p( )= 5 + 4 +2 3 + 2 +3 + added to compute the next elements + 5 2 3 5 2 3 4 3 5 2 3 4 3 2 5 2 3 4 3 2 5 2 3 4 3 2 + + 4 3 2 2 2 2-2 - 3 - + 2-3 + 2 3 2 2 3 3 3 3 the table has been completed, 2 sign changes in the first column (from row 3 to row 2 and from row 2 to row ) so 2 roots with positive real part Lanari: CS - Internal stability 4
Routh table example second order polynomial p( )=a 2 + b + c Routh table a c b a for a second order polynomial, the necessary condition is also sufficient (for the 2 roots to have negative real part) if c has different sign than a and b, then root has positive real part if b has different sign than a and c, then both roots have positive real part Lanari: CS - Internal stability 5
Routh table example we want to use the Routh criterion in order to state N&S condition for the roots of a polynomial to have real part less than a given since Re[ ] < Re[ - ] < setting - = Re[ ] < for p( ) Re[ ] < for p( ) = p( ) = + in order to check if the roots of p( ) = all have real part smaller than, we can apply the Routh criterion to the polynomial p( ) Im Re this corresponds to a translation of the Im axis Lanari: CS - Internal stability 6
linear systems - equilibrium states the origin is a particular state: at the origin the state velocity is if no inputs are applied therefore if we start from the origin, the state will stay there in the ZIR mathematically = A. we can look for any state x e with such a property i.e. a state x e such that these are defined as equilibrium states Ax e = all the equilibrium states belong to the nullspace of A if A nonsingular then only one equilibrium state (the origin) if A singular then infinite equilibrium states (subspace) note that A singular means that is i = is an eigenvalue of A det (A ) = det (A -.I) = Lanari: CS - Internal stability 7
linear systems - equilibrium states therefore if A has no eigenvalue i = then the system has a unique equilibrium point which is necessarily the origin (physical example: MSD system) if A has at least one eigenvalue i = then the system has infinite equilibrium points (physical example: point mass with friction) example A = = 2 = stable eigenspace unique equilibrium point unstable eigenspace x 2 5 4 3 2 - -2-3 -4 initial state Phase portrait (lambda = -, lambda 2 = ) initial state initial state initial state -5-5 -4-3 -2-2 3 4 5 x Lanari: CS - Internal stability 8
linear systems - equilibrium states example 2.25.25 A =.25.25 det (A ) = =.5 2 = 2 = u 2 = stable eigenspace every equilibrium state is of the form 4 3 2 eigenspace relative to i = initial state x e = infinite equilibrium points Phase portrait (lambda = -.5, lambda 2 = ) xe x 2e = x e initial state initial state the ZIR for arbitrary initial conditions will not always tend to the origin: following the velocity directions, we end in an equilibrium point ( ) different from the origin x 2 - -2-3 initial state -4-5 -4-3 -2-2 3 x Lanari: CS - Internal stability 9
from linear to nonlinear Nonlinear systems (see slides StabilityTheory by Prof. G. Oriolo): equilibrium points examples stable equilibrium state indirect method of Lyapunov the next slides are supplementary Lanari: CS - Internal stability 2
nonlinear systems - equilibrium states pendulum example damping coefficient µ m `2 # + mg` sin # + µ # = # ` m x = x x 2 ẋ = x 2 = # # ẋ 2 = g` sin x in the general form ẋ = f(x) µ m `2 x 2 we are going to look for those states x e (equilibrium states) for which that is for which f(x e ) = ẋ = 2 equilibrium states x e = down rest position x e2 = upright rest position Lanari: CS - Internal stability 2
solution on phase plane: pendulum no damping case (µ = ) m `2 # + mg` sin # = phase portrait from this initial state the pendulum will never stop rotating #(t) x 2 6 4 2 Phase portrait undamped pendulum #() particular initial condition from which the pendulum reaches autonomously (asymptotically) the upright rest position x e -2 x e2 (stable) equilibrium state -4-6 -4-2 2 4 6 8 x (unstable) equilibrium state #(t) Lanari: CS - Time response 22
solution on phase plane: pendulum no damping case (µ = ) m `2 # + mg` sin # = x e stable equilibrium state for a given neighbourhood of radius " of x e we can find a neighbourhood of radius ± such that the stability condition is verified 5 4 3 2 Phase portrait undamped pendulum x e (stable) equilibrium state x 2 - -2-3 -4 ± " x e2 (unstable) equilibrium state -5-4 -3-2 - 2 3 4 x Lanari: CS - Time response 23
solution on phase plane: pendulum solutions with non-zero damping from these initial states the pendulum will go over the upright position to finally asymptotically stop in the down rest position 8 6 Phase portrait damped pendulum 4 this equilibrium state x e is now asymptotically stable 2 x 2-2 -4-6 -8-5 5 Lanari: CS - Internal stability 24 x
solution on phase plane: Van der Pol oscillator ẍ b( x 2 )ẋ + x = for a given neighbourhood of radius " of x e there is no neighbourhood of radius ± such that the stability condition is verified Phase portrait Van der Pol equation b =.4 3 2 x 2 " - limit cycle -2-3 -3-2 - 2 3 x Lanari: CS - Internal stability 25