Modelling and Control of Dynamic Systems Controllability and - PowerPoint PPT Presentation

Modelling and Control of Dynamic Systems Controllability and Observability Sven Laur University of Tartu

Closed-loop Controllers: Basic Structure

Closed-loop control with state estimation System Controller r [ k ] u [ k ] y [ k ] ˆ G [ z ] State Estimator The instability of a open loop controller is caused by gradual accumulation of disturbances. The control signal becomes unsynchronised with the system. ⊲ If we can estimate the system state from the output, we can circumvent such synchronisation errors. The system must be observable for that. ⊲ Even if we know the system state, it might be impossible to track the reference signal. In brief, the system must be controllable . 1

Controllability The state equation is controllable if for any input state x 0 and for any final state x 1 there exists an input u that transfers x 0 to x 1 in a finite time. ⊲ T6. The n dimensional pair ( A , B ) is controllable iff the controllability � � A 2 B A n − 1 B matrix has the maximal rank n . B AB The desired input can be computed from the system of linear equations x 1 = A n x 0 + A n − 1 Bu [0] + A n − 2 Bu [1] + · · · + Bu [ n − 1] . Although the theorem T6 gives an explicit method for controlling the system, it is an off-line algorithm with a time lag n . A good controller design should yield a faster and more robust method. 2

Observability The state equation is observable if for any input state x 0 and for any input signal u , finite the output y sequence determines uniquely x 0 . ⊲ T7. The pair ( A , C ) is observable iff the observability matrix   C CA     . . .   CA n − 1 has the maximal rank n . 3

Offline state estimation algorithm Again, the input state x 0 can be computed from a system of linear equations  y [0] = Cx 0 + Du [0]      y [1] = CAx 0 + CBu [0] + Du [1]  · · ·      y [ n − 1] = CA n − 1 x 0 + · · · + CBu [ n − 2] + Du [ n − 1]  Although this equation allows us to find out the state of the system, it is offline algorithm, which provides a state estimation with a big time lag. A good state space estimator must be fast and robust. 4

General structure of state estimators System u [ k ] y [ k ] x [ k ] = Ax [ k − 1] + Bu [ k − 1] y [ k ] = Cx [ k ] + Du [ k ] x [ k ] ˆ State Estimator A state estimate ˆ x [ k ] is updated according to u [ k ] and y [ k ] : ⊲ Update rules are based on linear operations. ⊲ The state estimator must converge quickly to the true value x [ k ] ⊲ The state estimator must tolerate noise in the inputs u [ k ] and y [ k ] . 5

Example g [ z ] = 0 . 5 z +0 . 5 Consider a canonical realisation of the transfer function ˆ z 2 − 0 . 25 � � � � 0 0 . 25 1 A = B = 1 0 0 C = [0 . 5 0 . 5] D = 0 Then a possible stable state estimator is following x [ k + 1] = A ˆ ˆ x [ k ] + Bu [ k ] + 1 ( y [ k ] − C ˆ x [ k ] ) . � �� � y [ k ] ˆ In general, the feedback vector 1 can be replaced with any other vector to increase the stability of a state estimator. 6

Kalman Decomposition

Equivalent state equations Two different state descriptions of linear systems � � x [ k + 1] = A 1 x [ k ] + B 1 u [ k ] x [ k + 1] = A 2 x [ k ] + B 2 u [ k ] y [ k ] = C 1 x [ k ] + D 1 u [ k ] y [ k ] = C 2 x [ k ] + D 2 u [ k ] are equivalent if for any initial state x 0 there exists an initial state x 0 such that for any input u both systems yield the same output and vice versa. ⊲ T8. Two state descriptions are (algebraically) equivalent if there exists an invertible matrix P such that A 2 = P A 1 P − 1 B 2 = P B 1 C 2 = C 1 P − 1 D 2 = D 1 . 7

Linear state space transformations The basis e 1 = (0 , 1) and e 2 = (1 , 0) is a canonical base in R 2 . However a basis a 1 = (1 , 1) and a 2 = (1 , − 1) is also a basis. Now any state x = x 1 e 1 + x 2 e 2 can be represented as x = x 1 a 1 + x 2 a 2 and vice versa. The latter is known as a basis transformation:  x 1 = x 1 + x 2 �  x 1 = x 1 + x 2  2 x 2 = x 1 − x 2 x 2 = x 1 − x 2   2 For obvious reasons, we can do all computations wrt the basis { a 1 , a 2 } so that the underlying behaviour does not change. The same equivalence of state descriptions hold for other bases and larger state spaces, as well. 8

Kalman decomposition ⊲ T9. Every state space equation can be transformed into an equivalent description to a canonical form         x co [ k + 1] x co [ k ] A co A 13 B co 0 0 x co [ k + 1] x co [ k ] A 21 A co A 23 A 24 B co          =  +  u [ k ]         x co [ k + 1] A co x co [ k ] 0 0 0 0      x co [ k + 1] x co [ k ] A 43 A co 0 0 0 y [ k ] = [ C co 0 ] x [ k ] + Du [ k ] C co 0 where ⋄ x co is controllable and observable ⋄ x co is controllable but not observable ⋄ x co is observable but not controllable ⋄ x co is neither controllable nor observable 9

Minimal realisation ⊲ T10 . All minimal realisations are controllable and observable. A realisation of a proper transfer function ˆ g [ z ] = N ( z ) /D ( z ) is minimal iff it state space dimension dim( x 0 ) = deg D ( z ) . 10

Closed-loop Controllers: Design Principles

General setting According to Kalman decomposition theorem, the state variables can be divided into four classes depending on controllability and observability. ⊲ We cannot do anything with non-controllable state variables. ⊲ Non-observable variables can be controlled only if they are marginally stable. We can do it with an open-loop controller . ⊲ For controllable and observable state variables, we can build effective closed-loop controllers. Simplifying assumptions ⊲ From now on, we assume that we always want to control state variables that are both controllable and directly observable: y [ k ] = x [ k ] . ⊲ If this is not the case, then we must use state estimators to get an estimate of x . The latter just complicates the analysis. 11

Unity-feedback configuration System r [ k ] Controller u [ k ] y [ k ] + p ˆ g [ z ] ˆ C [ z ] -1 Design tasks ⊲ Find a compensator ˆ c [ z ] such that system becomes stable. ⊲ Find a proper value of p such that system starts to track reference signal. It is sometimes impossible to find p such that y [ k ] ≈ r [ k ] . ⊲ The latter is impossible if ˆ g [1] = 0 , then the output y [ k ] just dies out. 12

Overall transfer function Now note that p ˆ g [ z ]ˆ c [ z ] y = ˆ ˆ g [ z ]ˆ c [ z ]( p ˆ r − ˆ y ) ⇐ ⇒ y = ˆ c [ z ]ˆ r 1 + ˆ g [ z ]ˆ and thus the configuration is stable when the new transfer function p ˆ g [ z ]ˆ c [ z ] ˆ g ◦ [ z ] = 1 + ˆ g [ z ]ˆ c [ z ] has poles lying in the unit circle. Now observe g [ z ] = N ( z ) c [ z ] = B ( z ) pB ( z ) N ( z ) ˆ D ( z ) , ˆ = ⇒ ˆ g ◦ [ z ] = A ( z ) A ( z ) D ( z ) + B ( z ) N ( z ) 13

Pole placement We can control the denominator of the new transfer function ˆ g ◦ [ z ] . Let F ( z ) = A ( z ) D ( z ) + B ( z ) N ( z ) be a desired new denominator. Then there exists polynomials A ( z ) and B ( z ) for every polynomial F ( z ) provided that D ( z ) and N ( z ) are coprime. The degrees of the polynomials satisfy deg B ( z ) ≥ deg F ( z ) − deg D ( z ) . ℑ ( s ) ℑ ( z ) BIBO stable BIBO ℜ ( s ) ℜ ( z ) stable 14

Signal tracking properties Let ˆ g ◦ [ z ] be the overall transfer function. ⊲ The system stabilises if ˆ g ◦ [ z ] is BIBO stable. ⊲ The system can track a constant signal r [ k ] ≡ a if ˆ g ◦ [1] = 1 . ⊲ The system can track a ramp signal r [ k ] ≡ ak if ˆ g ◦ [1] = 1 and ˆ g ′ ◦ [1] = 0 . The latter follows from the asymptotic convergence g ′ y [ k ] → a ˆ ◦ [1] + ka ˆ g ◦ [1] for an input signal r [ k ] = ak . Moreover, let ˆ g ◦ = N ( z ) /D ( z ) . Then g ′ N (1) = D (1) ∧ N ′ (1) = D ′ (1) . ˆ g ◦ [1] = 1 ∧ ˆ ◦ [1] = 0 ⇐ ⇒ 15

Trade-offs in pole placement A pole placement is a trade-off between three design criteria: ⊲ response time ⊲ overshoot ratio ⊲ maximal strength of the input signal There is no general recipe for pole placement. Rules of thumb are given in ⊲ C.-T. Chen. Linear System Theory and Design. page 238. 16

Illustrative example 1 Consider a feedback loop with the transfer function g [ z ] = z − 2 . Then we can try many different pole placements � z − 0 . 5 , z + 0 . 5 , z 2 − 0 . 25 , z 2 + z + 0 . 25 , z 2 − z + 0 . 25 � F ( z ) ∈ The corresponding compensators are � 3 � 2 , 5 15 24 9 c [ z ] ∈ ˆ 2 , 4 z + 8 , 4 z + 12 , , 4 z + 4 � 1 � 3 , 3 1 9 1 p ∈ 5 , 5 , , , . 25 9 They can be found systematically by solving a system of linear equations. 17

Robust signal tracking Sometimes the system changes during the operation. The latter can be modelled as an additional additive term w [ k ] in the input signal. If we know the poles of reference signal r [ k ] and w [ k ] ahead, then we can design a compensator that filters out the error signal w [ k ] . 1 For instance, if w [ k ] is a constant bias, then adding an extra pole z − 1 cancels out the effect of bias. See pages 277–283 for further examples. In our example, the robust compensator for F [ z ] = z 2 − 0 . 25 is c [ z ] = 12 z − 9 ˆ p = 1 . 4 z − 4 18