The Behavioral Approach to Systems Theory Paolo Rapisarda, Un. of - PowerPoint PPT Presentation

The Behavioral Approach to Systems Theory Paolo Rapisarda, Un. of Southampton, U.K. & Jan C. Willems, K.U.Leuven, Belgium MTNS 2006 Kyoto, Japan, July 24–28, 2006

Lecture 3: State and state construction Lecturer: Paolo Rapisarda

Outline The axiom of state Discrete-time systems First-order representations State maps The shift-and-cut map Algebraic characterization Continuous-time systems Computation of state-space representations

Questions • Are state representations “natural"?

Questions • Are state representations “natural"? • Mechanics ❀ 2nd order; SYSID, transfer functions ❀ high-order;

Questions • Are state representations “natural"? • Mechanics ❀ 2nd order; SYSID, transfer functions ❀ high-order; • First principles and “tearing and zooming" modelling ❀ systems of high-order differential equations

Questions • Are state representations “natural"? • Mechanics ❀ 2nd order; SYSID, transfer functions ❀ high-order; • First principles and “tearing and zooming" modelling ❀ systems of high-order differential equations • Algebraic constraints among variables

Questions • Are state representations “natural"? • Mechanics ❀ 2nd order; SYSID, transfer functions ❀ high-order; • First principles and “tearing and zooming" modelling ❀ systems of high-order differential equations • Algebraic constraints among variables • What makes a latent variable a “state"?

Questions • Are state representations “natural"? • Mechanics ❀ 2nd order; SYSID, transfer functions ❀ high-order; • First principles and “tearing and zooming" modelling ❀ systems of high-order differential equations • Algebraic constraints among variables • What makes a latent variable a “state"? • What does that imply for the equations?

Questions • Are state representations “natural"? • Mechanics ❀ 2nd order; SYSID, transfer functions ❀ high-order; • First principles and “tearing and zooming" modelling ❀ systems of high-order differential equations • Algebraic constraints among variables • What makes a latent variable a “state"? • What does that imply for the equations? • How to construct a state from the equations?

Questions • Are state representations “natural"? • Mechanics ❀ 2nd order; SYSID, transfer functions ❀ high-order; • First principles and “tearing and zooming" modelling ❀ systems of high-order differential equations • Algebraic constraints among variables • What makes a latent variable a “state"? • What does that imply for the equations? • How to construct a state from the equations? • How to construct a state representation from the equations?

The basic idea It’s the Mariners’ final game in the World Series. You’re late... The current score is what matters...

The basic idea • The state contains all the relevant information about the future behavior of the system • The state is the memory of the system • Independence of past and future given the state

The axiom of state Σ = ( T , W , X , B full ) is a state system if ( w 1 , x 1 ) , ( w 2 , x 2 ) ∈ B full and x 1 ( T ) = x 2 ( T ) ⇓ ( w 1 , x 1 ) ∧ T ( w 2 , x 2 ) ∈ B full ∧ T is concatenation at T : � f 1 ( t ) for t < T ( f 1 ∧ T f 2 )( t ) := f 2 ( t ) for t ≥ T

Graphically... ( w 1 , x 1 ) , ( w 2 , x 2 ) ∈ B full and x 1 ( T ) = x 2 ( T ) ⇓ ( w 1 , x 1 ) ∧ T ( w 2 , x 2 ) ∈ B full

Graphically... ( w 1 , x 1 ) , ( w 2 , x 2 ) ∈ B full and x 1 ( T ) = x 2 ( T ) ⇓ ( w 1 , x 1 ) ∧ T ( w 2 , x 2 ) ∈ B full X (w ,x ) 2 2 (w ,x ) 1 1 W time

Graphically... ( w 1 , x 1 ) , ( w 2 , x 2 ) ∈ B full and x 1 ( T ) = x 2 ( T ) ⇓ ( w 1 , x 1 ) ∧ T ( w 2 , x 2 ) ∈ B full (w ,x ) (w ,x ) X ^ X 0 2 2 1 1 (w ,x ) 2 2 (w ,x ) 1 1 W W time time

Example 1: discrete-time system Σ = ( Z , R w , R l , B full ) , with B full := { ( w , ℓ ) | F ◦ ( σℓ, ℓ, w ) = 0 } where σ : ( R l ) Z → ( R l ) Z ( σ ( ℓ ))( k ) := ℓ ( k + 1 )

Example 1: discrete-time system Σ = ( Z , R w , R l , B full ) , with B full := { ( w , ℓ ) | F ◦ ( σℓ, ℓ, w ) = 0 } Special case: input-state-output equations σ x = f ( x , u ) y = h ( x , u ) w = ( u , y )

Example 2: continuous-time system Σ = ( R , R w , R l , B full ) , with B full := { ( w , ℓ ) | F ◦ ( d dt ℓ, ℓ, w ) = 0 } Special case: input-state-output equations d = f ( x , u ) dt x y = h ( x , u ) = ( u , y ) w

Outline The axiom of state Discrete-time systems First-order representations State maps The shift-and-cut map Algebraic characterization Continuous-time systems Computation of state-space representations

First-order discrete-time representations Theorem: A ‘complete’ latent variable system Σ = ( Z , R w , R x , B full ) is a state system iff B full can be described by F ◦ ( σ x , x , w ) = 0

First-order discrete-time representations Theorem: A ‘complete’ latent variable system Σ = ( Z , R w , R x , B full ) is a state system iff B full can be described by F ◦ ( σ x , x , w ) = 0 0-th order in w , 1st order in x

First-order discrete-time representations Theorem: A ‘complete’ latent variable system Σ = ( Z , R w , R x , B full ) is a state system iff B full can be described by F ◦ ( σ x , x , w ) = 0 Linear case: E σ x + Fx + Gw = 0

First-order discrete-time representations Theorem: A ‘complete’ latent variable system Σ = ( Z , R w , R x , B full ) is a state system iff B full can be described by F ◦ ( σ x , x , w ) = 0 Linear case: E σ x + Fx + Gw = 0 1st order in x is equivalent to state property!

Proof (linear case)         a x ( 1 ) a    | ∃ ( x , w ) ∈ B full s. t.  = V := b x ( 0 ) b       w ( 0 ) c c � E G � V linear ⇒ ∃ E , F , G s.t. V = ker ( F )

Proof (linear case)         a x ( 1 ) a    | ∃ ( x , w ) ∈ B full s. t.  = V := b x ( 0 ) b       w ( 0 ) c c � E G � V linear ⇒ ∃ E , F , G s.t. V = ker ( F ) ⇓ [( x , w ) ∈ B full = ⇒ E σ x + Fx + Gw = 0 ]

Proof (linear case)         a x ( 1 ) a    | ∃ ( x , w ) ∈ B full s. t.  = V := b x ( 0 ) b       w ( 0 ) c c � E G � V linear ⇒ ∃ E , F , G s.t. V = ker ( F ) Converse by induction, using axiom of state: E σ x + Fx + Gw = 0 on [ 0 , k ] = ⇒ ( w , x ) | [ 0 , k ] ∈ B full | [ 0 , k ] Then apply completeness of B

State construction: basic idea Problem: Given kernel or hybrid description, find a state representation E σ x + Fx + Gw = 0

State construction: basic idea Problem: Given kernel or hybrid description, find a state representation E σ x + Fx + Gw = 0 First compute polynomial operator in the shift acting on system variables, inducing a state variable: � � w X ( σ ) w = x X ( σ ) = x ℓ

State construction: basic idea Problem: Given kernel or hybrid description, find a state representation E σ x + Fx + Gw = 0 First compute polynomial operator in the shift acting on system variables, inducing a state variable: � � w X ( σ ) w = x X ( σ ) = x ℓ Then use original eqs. and X to obtain 1st order representation.

State maps for kernel representations X ∈ R •× w [ ξ ] induces a state map X ( σ ) for ker ( R ( σ )) if the behavior B full with latent variable x , described by R ( σ ) w = 0 X ( σ ) w = x satisfies the axiom of state.

Example B = { w | r ( σ ) w = 0 } where r ∈ R [ ξ ] , deg ( r ) = n . (Minimal) state map induced by     1 w ξ σ w     . .      ❀ . .  .  .  ξ n − 1 σ n − 1 w

The axiom of state revisited A linear system Σ = ( T , W , X , B full ) with latent variable x is a state system if ( w , x ) ∈ B full and x ( T ) = 0 ⇓ ( 0 , 0 ) ∧ T ( w , x ) ∈ B full

The axiom of state revisited A linear system Σ = ( T , W , X , B full ) with latent variable x is a state system if ( w , x ) ∈ B full and x ( T ) = 0 ⇓ ( 0 , 0 ) ∧ T ( w , x ) ∈ B full • Time-invariance = ⇒ can choose T = 0;

The axiom of state revisited A linear system Σ = ( T , W , X , B full ) with latent variable x is a state system if ( w , x ) ∈ B full and x ( T ) = 0 ⇓ ( 0 , 0 ) ∧ T ( w , x ) ∈ B full • Time-invariance = ⇒ can choose T = 0; • Concatenability with zero trajectory is key.

When is w ∈ B concatenable with zero? R 0 w + R 1 σ w + . . . + R L σ L w = 0 0 0 w ( 0 ) w ( 1 ) w ( 2 ) w ( 3 ) . . . . . . . . . . . . k = − 2 k = − 1 k = 0 k = 1 k = 2 k = 3

When is w ∈ B concatenable with zero? R 0 w + R 1 σ w + . . . + R L σ L w = 0 0 0 R 0 R 1 R 2 R 3 . . . . . . 0 0 w ( 0 ) w ( 1 ) w ( 2 ) w ( 3 ) . . . . . . . . . . . . k = − 2 k = − 1 k = 0 k = 1 k = 2 k = 3 R 0 w ( 0 ) + R 1 w ( 1 ) + . . . + R L w ( L ) = 0