Linear Sytem Fundamental MEM 355 Performance Enhancement of Dynamical Sytem Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity
Content Sytem Repreentation Stability Concept Routh-Hurwitz Stability Criterion
Sytem Repreentation
State Space & Tranfer Function Repreentation A linear ytem can be repreented by ytem of firt order differential equation: x = Ax + Bu y = Cx + Du n m p where x R i the tate, u R i the (control) input y R i the output Take the Laplace tranform to obtain and olve for State Space or time domain model X = I A x + I A BU [ ] [ ] Y [ ], [ ] in term of Y = C I A x + G U G = C I A B + D U Tranfer Function
Characteritic Equation Recall, the characteritic equation of A i n φ λ = λi A = λ +a λ + +a λ+a The root of n n φ λ = are the eigenvalue of A The tranfer function can be written ( A) CAdj Adj A B + D I A G( ) = C[ I A] B+ D= C B+ D= I A I A So we ee that in the SISO cae n G( ) =, n = CAdj A B + D I A, d =φ d The pole of G are the eigenvalue of A!
Reolvant Matrix and State Tranition Matrix The matrix [ I A] i called the "reolvent" matrix. It' invere Laplace Tranform i the i the "tate tranition matrix": ( ) Φ t I A = e = L [ ] At More about thi later In term of Φ we can write x( t) =Φ t x + Φ( t τ) Bu( τ) dτ yt () = CΦ t x t t + C Φ( t τ) Bu( τ) dτ + Du() t
A Stability Lemma Lemma: Aume the ytem pole, i.e., the eigenvalue of the A matrix are all in the trictly left complex plane, then we have the following: () The repone due to the initial tate i, lim and Φ ( tx ) = t (2) The repone to the input, t ut () x = y() t = C Φ( t τ) Bu( τ) dτ + Du() t ut () i bounded for every bounded input.
Stability Definition Definition: A linear time-invariant ytem i BIBO (Bounded-Input Bounded-Output) table if and only if every bounded input reult in a bounded output. Definition: A linear time-invariant ytem i internally table if the olution xt of xt = Axt, x = x tend toward zero a t for arbitrary x.
Stability Theorem Theorem: An LTI ytem with tranfer function G( ) i BIBO table if and only if the pole of G( ) are trictly in the left half plane. Theorem: An LTI ytem with tate pace parameter ABCD,,, i internally table if and only if the all of the eigenvalue of A are trictly in the left half plane. Note: Internal tability i a tronger condition than BIBO tability. BIBO tability only reflect the attribute of the ytem that are obervable from the output and controllable from the input. There may be hidden mode that are untable. Much more about thi later
Routh-Hurwitz Stability Criterian Given a polynomial that repreent the pole polynomial of a tranfer function or the characteritic polynomial of quare matrix it i very eay to determine the root (and therefore ae tability) uing numerical computation providing the coefficient are all pecified! But uppoe one or more of the coefficient are not pecified. Rather, we want to determine an admiible range of value uch that the ytem i table. That i the control deigner problem and where the Routh-Hurwitz criterion i ueful.
Routh-Hurwitz --2 Theorem: Conider the polynomial n n + an + + a + a = A neceary condition that all root are trictly in the left half plane i that all coefficient are trictly poitive. Note that thi provide only a neceary condition. To determine a neceary and ufficient condition we aemble the Routh array.
Routh-Hurwitz--3 Theorem: Conider the polynomial n n + an + + a + a = The aociated ytem i (BIBO or internally) table if and only if all element of the firt column of the Routh array are trictly poitive.
Routh Array n n + an + + a + a = Contruct a matrix a hown by with the firt two row uing the uing the above polynomial coefficient. Then contruct the remaining row b c an 2 an 4 a a a a =, b =,... n n 3 n n 5 2 an an an an 3 an an 5 b b b b =, c =,... 2 2 2 b b n 2 n even # column = n+ 2 n odd a n n 2 n an an 3 n 2 b b2 n 3 c c2
Routh-Hurwitz Example 5 4 3 2 p = + 5 + 74.25 + 2 + 2K + 2K 5 4 3 2 K K 74 2K 5 2 2K 65.9 9.86K 2 4.52 2 227K 89.76K 2 4.52K 2K 2 K > K K > K > 2 2 4.52, 227 89.76, 2 227 K < = 26.769 K < = 25.38 4.52 89.76
Routh-Hurwitz -- Example 2 4 3 2 p K K K K 4 K K 3 + K K 2 = + + + + 6 + + 8 + 6 8 3K + 6K + K 2 ( 2 + ) 2K 23K 26K 8K 3K + 6K 2 8K + K > K + K > K + K > 2 2, 3 6, 23 26