Control Systems Internal Stability - LTI systems L. Lanari
definitions (AS) - A system S is said to be asymptotically stable if its state zeroinput 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: state behavior note: any/some state transition matrix Φ(t) =e At Φ(t, t 0 ) LTI (Linear Time Invariant) LTV (Linear Time Variant) Lanari: CS - Internal stability 2
possible behaviors x ZIR (t) =e At x 0 a linear combination of A diagonalizable mg( i) = ma( i) for all i real i complex i = i + j!i aperiodic mode e λ it pseudoperiodic mode e αit [sin(ω i t + ϕ R )u re + cos(ω i t + ϕ R )u im ] A not diagonalizable mg( i) < ma( i) real i complex i = i + j!i...,..., t n k 1 (n k 1)! eλ it t n k 1 (n k 1)! eα it sin ω i t max dimension of Jordan block Jk Lanari: CS - Dynamic response - Time 3
stability and eigenvalues 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 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 1 equivalent to equivalent to mg( i) < ma( i) mg( i) = ma( i) for all i Lanari: CS - Internal stability 4
stability and eigenvalues it all depends upon the positioning of the eigenvalues in the complex plane Im Re all eigenvalues in the open left half-plane Im Re distinct eigenvalues case some eigenvalues may be on the Im axis Im Re at least one eigenvalue with positive real part (the case Re( i) = 0 and Jordan block dim > 1 is not shown) Lanari: CS - Internal stability 5
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) diverging response Lanari: CS - Internal stability 6
remarks unstable systems can have bounded or converging solutions for some specific initial conditions x(t) =e λ1t u 1 v1 T x 0 + e λ2t u 2 v2 T x aperiodic λ 1 > 0 unstable 0 modes system λ 2 < 0 eigenspace λ 2 < 0 x 0 λ 1 > 0 eigenspace x 0 u 2 u 1 x 0 x 0 x 0 x 0 x 0 as long as the initial condition does not have component along the unstable eigenspace Lanari: CS - Internal stability 7
remarks if the system is asymptotically stable then the output ZIR also converges to 0 (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 0 = n i=1 e λ it Cu i v T i x 0 this term may be zero for some ui example A = λ1 0 1 λ 2, B = 1 1, C = 1 0 (compute yzir(t)) Lanari: CS - Internal stability 8
examples MSD with no friction and no spring m s = f eigenspace V1 is generated by 1 0 A = 0 1 0 0 1 = 0 ma( 1) = 2 and therefore mg( 1) = 1 < ma( 1) 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 = 0 1 0 µ 1 = 0 2 = - µ < 0 since ma( 1) = 1 = mg( 1) for the zero eigenvalue 1 = 0, 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 Lanari: CS - Internal stability 9
LTI stability 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(λ) =a n λ n + a n 1 λ n 1 + a n 2 λ n 2 + + a 1 λ + a 0 A necessary condition in order for the roots of p( ) = 0 to have all negative real part is that the coefficients need to have all the same sign If all the roots of p( ) = 0 have negative real part then the coefficients have the same sign If a coefficient is null then the coefficients do not have the same sign Lanari: CS - Internal stability 10
Routh-Hurwitz stability criterion In order to state a necessary and sufficient condition we need to build a table Routh table (n+1 rows) row n row n-1 row n-2 row 1 row 0 p(λ) =a n λ n + a n 1 λ n 1 + a n 2 λ n 2 + + a 1 λ + a 0 a n a n 2 a n 4 a n 1 a n 3 a n 5 b 1 b 2 c 1 c 2 computed d 1 as. missing terms can be set to 0 N.B.: an entire row can be multiplied by a positive number without compromising the table directly from p( ) b 1 = 1 a n 1 b 2 = 1 a n 1 a n a n 1 a n a n 1 a n 3 c 1 = 1 a n 1 b 1 b 1 b 2 a n 5 c 2 = 1 a n 1 b 1 b 1 b 3 a n 2 a n 3 a n 4 a n 5 Lanari: CS - Internal stability 11
Routh-Hurwitz stability criterion If the Routh table can be completed then we have the following N&S condition All the roots of p( ) = 0 have negative real part iff we have 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 0 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 12
Routh table example p(λ) =λ 5 + λ 4 +2λ 3 + λ 2 +3λ +1 5 1 2 3 0 4 1 1 1 0 3 2 1 0 5 1 2 3 0 5 1 2 3 0 5 1 2 3 0 4 1 1 1 0 4 1 1 1 0 4 1 1 1 0 3 1 2 0 3 1 2 0 0 3 1 2 0 0 2 1 0 added to compute the next elements 2-1 1 0 2-1 1 0 1 0 1 3 0 0 0 + + + - + + 5 1 2 3 4 1 1 1 3 1 2 0 2-1 1 0 1 3 0 0 0 1 0 0 1 1 1 2 1 1 1 1 1 3 1 1 1 1 1 0 1 0 1 1 1 1 1 2 1 1 1 1 1 0 1 1 1 0 0 0 1 1 1 2 1 1 1 1 1 0 1 0 1 3 1 1 3 0 1 3 1 0 3 0 the table has been completed, 2 sign changes in the first column (from row 3 to row 2 and from row 2 to row 1) so 2 roots with positive real part Lanari: CS - Internal stability 13
Routh table example Second order polynomial p(λ) =aλ 2 + bλ + c Routh table a c b 0 c for a second order polynomial, the necessary condition is also sufficient (in order to guarantee that the 2 roots have negative real part) Lanari: CS - Internal stability 14
Routh table example we want to use 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[ - ] < 0 setting - = Re[ ] < for p( ) Re[ ] < 0 for p( ) = p( ) = + this corresponds to a translation of the Im axis Im Re Lanari: CS - Internal stability 15