EE451/551: Digital Control. Chapter 8: Properties of State Space Models
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1 EE451/551: Digital Control Chapter 8: Properties of State Space Models
2 Equilibrium State Definition 8.1: An equilibrium point or state is an initial state from which the system nevers departs unless perturbed For the general state eqn.: xk ( + 1) = f xk ( ) ( ) equilibrium points x satisfy the condition: e ( ) ( ) x e xk ( + 1) = f xk ( ) = f x = x For LTI systems, equilibrium points x satisfy: + [ ] x ( k+ 1) = Ax ( k ) = Ax = x A I x = 0 the soln. for invertible [ A I ] x e = e e asymptotic stability is: e e e n e 0 n
3 Asymptotic Stability Definition82:AnLTIsystemisasymptotically 8.2: is stable (AS) if all of its trajectories converge to zero for any initial state xk ( ) lim xk ( ) 0 = k Theorem 8.1: A discrete-time ti LTI system is AS if and only if all eigenvalues of its state matrix A d lie inside the unit [ ] [ ] x e ( ) 1 d n d n circle A I exits or rank A I = n, called full rank 0
4 Bounded Input Bounded Output Stability Definition 8.3: An LTI system is BIBO stable if its output is bounded for any bounded input, i.e., uk ( ) < b < yk ( ) < b < u Theorem 8.2: A discrete-time time LTI system is BIBO stable if and only if the norm of its impulse response is absolutely summable, i.e., hk ( ) < (see matrix norm in Appendix III) k = 0 Theorem 8.3: If a discrete-time LTI system is AS, then it is BIBO stable; furthermore, in the absence of unstable polezero cancellation, the system is AS if it is BIBO stable y
5 Stability Example Given the discrete LTI system example of Chapter 7 with TF and IR given by: H( z) ( ) ( )( ) Y ( z ) z = = U( z) z 1 z k hk ( ) = ( k) δ ( k) What can be said about the system stability Ad = >> eig(ad) ans = >> rank(ad eye(2)) ans = 1 y y and equilibrium pt.? Since the TF has a system pole (eigenvalue of Ad ) at z = 1, i.e., on the unit circle, the system is not AS; furthermore, since the IR contains a unit step fcn., it is also not absolutely summable, and therefore the system is not BIBO stable (as expected) Also, since the system is only MS there is no EQ pt. at zero [ A I ] (note is not full rank not invertible) d n
6 Questions?
7 Controllability and Stabilizability Definition 8.4: An LTI system is said to be controllable if for any initial state xk ( =, + 1,, 1, 0 ) there exists a control sequence uk ( ) for k k0 k0 kf such hthat tan arbitrary final state t x( k f ) can be reached in finite time k f In other words, the input to the system can reach all of the system states to di drive them, and therefore the system time response, in a manner that meets design specifications (as shown in Chapter 9)
8 Controllability and Stabilizability Theorem 8.5: An LTI system is completely controllable if and only if the C n mn controllability matrix n 1 = Bd, AdBd,, Ad Bd has rank n (typ. (yp computed using a CAD tool, e.g., Matlab) Definition 8.5: A system is said to be stabilizable if its uncontrollable o modes (poles) are AS Physical system are often stabilizable rather than controllable, which is not a problem provided that the uncontrollable modes decay to zero sufficiently fast, so as not to overly slow down the controllable part of the system reponse
9 Controllability Example Given the discrete LTI system example of Chapter 7 with A d =Φ c( T ) = and T = T Bd = Φc( τ) dτ Bc = = T = 0.1 The controllability matrix is given as: C = [ B A B ] = d d d having a rank of 2, meaning two linearly independent rows or columns; therefore, the system is completely controllable
10 Controllability Example Thus the system is completely controllable, meaning that it is possible to specify an input uk ( ) to achieve a desired system response using state feedback techniques shown in Chapter 9 We compute the controllability matrix and its rank using Matlab: lb >> Ad >> C=[Bd,Ad*Bd] % Alternately C=ctrb(Ad,Bd) Ad = C = >> Bd Bd = >> rank(c) ans = 2
11 Questions?
12 Observability and Detectability Definition 8.6: An LTI system is said to be observable if any initial state xk ( ) can be estimated from the meaurements y( k) 0 = 0, 0 + 1,, f f for k k k k in finite time k To control a system it is often desirable to use the sytem states to determine an appropriate control action, i.e., using state feedback Often, the system state must be estimated from the time history of the system measurements (i.e., outputs) in order to develop controls (i.e., inputs) that can drive the system time response in a manner that meets design specifications (as shown in Chapter 9)
13 Observability and Detectability Theorem 8.8: 8: An LTI system is completely observable if and only if the ln n observability matrix Cd CA d d n 1 T O = = C d CdAd CdAd n 1 CA d d has rank n (typ. computed using a CAD tool, e.g., Matlab) Definition 8.5: A system is said to be detectable if all its unobervable modes (poles) are AS; if a system is not observable, it is desirable to be at least detectable
14 Observability Example Given the discrete LTI system example of Chapter 7 with =Φ ( ) = and = 1 0 A d c T T = 0.1 C d [ ] The observability matrix is given as: 1 0 T O = [ C C A ] = having a rank of 2, d d d meaning two linearly independent rows or columns; therefore, the system is completely observable
15 Observability Example Thus the system is completely observable, meaning that it is possible to estimate the states from the output measurements to design an input uk ( ) to achieve a desired system response using state feedback techniques shown in Chapter 9 We compute the observability matrix and its rank using Matlab: >> Ad Ad = >> Cd >> rank(o) Cd = ans = >> O=[Cd;Cd*Ad] Cd*Ad] %Alt. Use O=obsv(Ad,Cd) O =
16 Questions?
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