Notes of robust control

Notes of robust control • • • • •

Zeros (continuous-time systems) Parametrization (continuous-time systems) H2 control (continuous-time systems) H∞ control (continuous-time systems) H∞ filtering in discrete-time (discrete-time systems)

ZEROS AND POLES OF MULTIVARIABLE SYSTEMS

SUMMARY

• Transmission zeros Rank property and time-domain characterization

• Invariant zeros Rank property and time-domain characterization

• Decoupling zeros • Infinite zero structure

.

(Transmission) Zeros and poles of a rational matrix G( s) = C ( sI − A) −1 B + D,

p outputs, m inputs

Basic assumption:

normal rank [G ( s)] = min( p, m) = r Smith-McMillan form of G(s)

 f1 ( s)  0  M (s ) =  M   0  0 f i ( s) =

0 f 2 ( s) M 0 0

L L O L L

0 0 M f r ( s) 0

0 0  M  0 0 

εi ( s) , i = 1,2,L , r ψi ( s)

εi divides εi+1 ψi+1 divides ψi • The transmission zeros are the roots of ε i(s), i=1,2,…,r • The (transmission) poles are the roots of ψ i(s), i=1,2,…,r

.

Example 1  0  − 10 7 A=  0 0  −1  1

0 0 5 1

0 0  1  0  − 13 5 0 0 1   ,B= ,C =   , D = 1 2  0 0 0 1 0       0 0 

Hence

 ( s − 3)( s + 1)  1 G( s) =   ( s − 2)( s − 5) ( s − 3 )( s − 2 )  

 s − 3 1 r = 1, M ( s) =    0  ( s − 2)( s − 5) There is one transmission zero at s=3 and two poles in s=2 and s=5.

.

Rank property of transmission zeros Theorem 1 The complex number λ is a transmission zero of G(s) iff there exists polynomial vector z(s) such that z(λ)≠0 and  lim G ( s ) z ( s ) = 0  s → λ  lim G ( s )' z ( s ) = 0   s → λ

if p ≥ m if p ≤ m

Remark In the square case (p=m), it follows r

det [G ( s ) ] =

∏ε i =1 r

∏ψ

i

(s )

i

(s)

i =1

so that it is possible to catch all zeros and poles if and only if there are not cancellations. Proof of the theorem Assume that p≥ m and let M(s) be the Smith-Mc Millan form of G(s), i.e. G(s)=L(s)M(s)R(s) for suitable polynomial and unimodular matrices L(s) (of dimension p) and R(s) (of dimensions m). It is possible to write: E ( s )Ψ −1 ( s ) M (s ) =   0   where E(s)=diag{εi (s)} and Ψ(s)=diag{ψi (s)}. Now assume that λ is a zero of the polynomial eκ (s). Recall that R(s) is unimodular and take the polynomial vector z(s)=[R-1(s)]k , i.e. the k-th column of the inverse of R(s). Since ψk (λ)≠0, it follows that G(s)z(s)=L(s)E(s)Ψ −1(s)R(s)z(s)=L(s)E(s) [Ψ −1(s)]k = L(s)[E(s)]k /ψ k (s)=[L(s)]k ε k (s) /ψ k (s). This quantity goes to zero as s tends to λ. Vice-versa, if z(s) is such that G(s)z(s) goes to zero as s tends to λ, then the limit as s tends to λ of E(λ)Y-1(s)R(λ)z(λ) being zero means that at least one polynomial εk (s) should vanish at s= λ. The proof in the opposite case (p≤m) follows the same lines and therefore is omitted.

.

Example 1 ( s − 1)( s + 2)  1 G( s) =  ( s − 1) 2  ( s − 1)( s + 2) 0 Hence,

1  1 0  ( s − 1)( s + 2) G( s ) =   0 1   0 

 0  1  s − 1  ( s − 1)( s + 2 ) s + 2 

so that, with

 − ( s − 1)( s + 2)  z( s) =   1   it follows

lim G ( s) z ( s) = 0 s →1

Notice that it is not true that G(s)z(1) tends to zero as s tends to one. .

.

Time-domain characterization of transmission zeros and poles Let Σ=(Α,Β, C ,D) be the state-space system with transfer function G(s)=C(sI-A) -1B+D. Let Σ′=(A′,C′,B′,D′) the “transponse” system whose transfer function is G(s) ′=B′(sI-A′)-1C′+D′. Now, let Σ Σˆ =  Σ '

p ≥ m p ≤ m

p ≥ m p ≤ m

 G (s) Gˆ ( s ) =   G ( s )'

Finally, let δi(t) denote the i-th derivative of the impulsive “function”. Theorem 2 (a) The complex number λ is a pole of Σ iff the exists an impulsive input u (t ) =

ν

∑

α iδ i ( t )

i=0

such that the forced output yf(t) of Σ is y

f

( t ) = y 0 e λt ,

∀t ≥ 0

(b) The complex number λ is a zero of Σ iff the exists an exponential/impulsive input u (t ) = u 0 e

λt

+

ν

∑

α iδ i ( t ) ,

u

0

≠ 0

i=0

such that the forced output yf(t) of Σ is y

f

(t) = 0,

∀ t ≥ 0

Zeros blocking property. The proof of the theorem is not reported here for space limitations.

.

Invariant zeros Given a system (A,B,C,D) with transfer function G(s)=C(sI-A) -1B+D, the polynomial matrix  sI − A P (s) =   C

− B D 

is called the system matrix. Consider the Smith form S(s) of P(s):

0 α1 ( s)  0 α2 ( s)  S (s ) =  M M  0  0  0 0

L 0 L 0 O M L αν ( s) L 0

0 0 M  0 0

αi divides αi+1 . The invariant zeros are the roots of α i(s), i=1,2,…,ν

−−−−−−−−−−−−−−−−− Notice that  sI − A P (s) =   C

0   ( sI − A ) − 1 I   0

0   sI − A  G ( s )   0

so that the rank of P(s) is ϖ=n+rank(G(s))

.

− B I 

Rank property of invariant zeros Theorem 3 The complex number λ is an invariant zero of (A,B,C,D) iff P(s) looses rank in s=λ, i.e. iff there exists a nonzero vector z such that  P (λ )z = 0   P ( λ )' z = 0  

if

p ≥ m

if

p ≤ m

Notice that if T is the transformation for a change of basis in the state-space, then

 sI − A P (s ) =   C

− B T  =  D  0

0   sI − A I   C

− B  T −1  D   0

0  I

Therefore, the invariant zeros are invariant with respect to a change of basis. Proof of Theorem 3 Consider the case p≥m and let P(s)=L(s)S(s)R(s), where R(s) and R(s) are suitable polynomial unimodular matrices. Hence, if P(λ)z=0 for some z≠0, then S(λ )R(λ)z=0 so that S(λ)y=0, with y≠0. Hence, λ must be the root of at least one polynomial α i(s). Vice-versa, if α k (λ) for some k, then z=[R-1(λ)]k is such that P(λ)z=L(λ)S(λ)R(λ)z= L(λ)S(λ)R(λ)[R-1(λ)]k =L(λ)[S(λ)]k =0.

.

Time-domain characterization of invariant zeros Let Σ=(Α,Β, C ,D) be the state-space system with transfer function G(s)=C(sI-A) -1B+D. Let Σ′=(A′,C′,B′,D′) the “transponse” system whose transfer function is G(s) ′=B′(sI-A′)-1C′+D′. Now, let Σ Σˆ =  Σ '

p≥m p≤m

 G ( s) Gˆ ( s ) =   G ( s )'

p≥m p≤m

Theorem 4 The complex number λ is an invariant zero Σ iff at leat one of the two following conditions holds: (i) (ii)

λ is an eigenvalue of the unobservable part of Σ there exist two vectors x0 and u 0≠0 such that the forced output of Σ corresponding to the input u(t)=u0eλt, t≥0 and the initial state x(0)=x0 is identically zero for t≥0.

Zeros blocking property. The proof of the theorem is not reported here for space limitations.

.

Trasmission vs invariant zeros Theorem 5 (i) A transmission zero is also an invariant zero. (ii) Invariant and transmission zeros of a system in minimal form do coincide. Proof of point (i). Let p≥m and let λ be a transmission zero. Then, there exists an exponential/impulsive input

u (t ) = u 0 e

λt

ν

+ ∑ αi δ (i ) (t ) such that the forced output is zero, i.e. i=0

ν   y f (t ) = C ∫ e A(t −τ) B u 0 e λτ + ∑ αi δ (i ) (τ)dτ + Du 0e λt = 0, t > 0 . Letting  i =0  0 t

α0  α  1 ν x0 = B AB L A B   , it follows M   αν 

[

]

t

Ce x0 + C ∫ e A ( t −τ ) Bu 0 e λτ dτ + Du0 e λt = 0, t > 0 At

0

so that λ is an invariant zero. The proof in the case p≤m is the same, considering the transponse system. Proof of point (ii). Let p≥m. We have already proven that a trasmission zero is an invariant zero. Then, consider a minimum system and an invariant zero λ. Hence Av+Bw=λw, Cv+Dw=0. Notice that w≠0 since the system is observable. Moreover, thanks to the reachability condition there exists a vector

α 0  α  1 ν AB L A B   M   αν 

[

]

v= B

Since λ is an invariant zero there exists an initial state x(0)=v and an input u(t)=we λt , such that t

Ce v + C ∫ e A ( t −τ ) Bwe λτ dτ + Dweλt = 0, t > 0 At

0

By noticing that Ce v = Ce At

t

C∫ e 0

A( t −τ)

ν

t

ν

0

i =0

∑ A Bαi = ∫ Ce A(t −τ) B ∑αi δ (i ) (τ )dτ , it follows, i= 0

i

 λτ ν  B u 0 e + ∑ αiδ (i) (τ )dτ + Du0 e λt = 0, t > 0 ,  i =0 

trransmission zero. .

At

which

means

that

λ

is

a

Decoupling zeros  sI − A  PC ( s ) =    C 

PB ( s ) = [sI − A

− B]

Associated Smith forms

SC

 diag (s) =  

{α 0

C i

}

(s)   

[

S B ( s ) = diag

{α

B i

( s)

} 0]

The output decoupling zeros are the roots of α 1C(s) α 2C(s)… α nC(s), the input decoupling zeros are the roots of α iB(s) α 2 B(s)…α n B(s) and the inputoutput decoupling zeros are the common roots of both. Lemma (i) A complex number λ is an output decoupling zero iff there exists z≠0 such that P C(λ)z=0. (ii) A complex number λ is an input decoupling zero iff there exists w≠0 such that P B(λ)′w=0. (iii) A complex number λ is an input-output decoupling zero iff there exist z≠0 and w≠0 such that P C(λ)z=0, P B(λ)′w=0. (iv) For square systems the set of decoupling zeros is a subset of the set of invariant zeros. (v) For square invertible systems the transmission zeros of the inverse system coincide with the poles of the system and the invariant zeros of the inverse system coincide with the eigenvalues of the system.

.

Infinite zero structure Mc Millan form over the ring of causal rational matrices

Given the transfer function G(s) there exists two unimodular matrices in that ring (bi-proper rational functions) such that  M ( s ) 0 G ( s) = V ( s )  W ( s ), M ( s ) = diag s −δ1 , s −δ 2 , L, s −δ r  0  0

{

}

where δ1, δ2, …, δr are integers such that δ1≤ δ2≤…≤ δr. They describes the structure of the infinity zeros.

• Inversion • Model matching • ……

Example: compute the finite and infinite zeros of: s +1 0  s −1 1 G( s ) =  1  s  1 0   s

.

 s+1  s−1  s+1   s 3 2 s − 3s + s + 5   s 3 ( s − 3) 

PARAMETRIZATION OF STABILIZING CONTROLLERS

Summary

• BIBO stability and internal stability of feedback systems • Stabilization of SISO systems. Parameterization and characterization of stabilizing controllers • Coprime factorization • Stabilization of MIMO systems. Parameterization and characterization of stabilizing controllers • Strong stabilization

Stability of feedback systems d r

e

u C(s)

y P(s)

The problem consists in the analysis of the asymptotic stability (internal stability) of closed-loop system from some characteristic transfer functions. This result is useful also for the design.

THEOREM 1 The closed loop system is asymptotically stable if and only if the transfer matrix G(s) from the input to the output r  d   

e u   

is stable.  ( I + P ( s)C ( s)) −1 − ( I + P( s )C ( s ))−1 P( s)  G( s) =   −1 −1 ( I + C ( s ) P ( s )) C ( s ) ( I + C ( s ) P ( s ))  

Stability of interconnected SISO systems The closed loop system is asymptotically stable if and only if the three transfer functions S ( s) =

1 C ( s) P ( s) , R( s ) = , H (s ) = 1 + C ( s) P ( s) 1 + C ( s ) P ( s) 1 + C ( s ) P ( s)

are stable. • Notice that if S(s), R(s), H(s) are stable, then also the complementary sensitivity T ( s) =

C ( s) P ( s) = 1 − S ( s) 1 + C ( s ) P( s )

is stable. • The stability of the three transfer functions are made to avoid prohibited (unstable) cancellations between C(s) and P(s).

Example Let C(s)=(s-1)/(s+1), P(s)=1/(s-1) . Then S(s)=(s+1)/(s+2), R(s)=(s-1)/(s+2) e H(s)=(s+1)/(s2+s-2)

Comments The proof of Theorem 1 can be carried out pursuing different paths. A simple way is as follows. Let assume that C(s) and P(s) are described by means of minimal realizations C(s)=(A,B,C,D), P(s)=(F,G,H,E). It is possible to write a realization of the closedloop system as Σ c l=(Ac l,Bc l,Cc l,Dc l). Obviously, if Ac l is asymptotically stable, then G(s) is stable. Vice-versa, if G(s) is stable, asymptotic stability of Ac l follows form being Σ cl=(Acl,Bc l,Cc l,Dc l) a minimal realization. This can be seen through the well known PBH test. In our example, it follows − 2  0 1 1  − 1 0  Acl =  , B = , C = cl cl     ,  − 1 − 1 1 0   − 1 − 2

G( s)

 s+1 s + 2 =   s −1  s + 2

−

( s + 1)  ( s − 1)( s + 2)   s+1   s+2

SMM

>

1   ( s − 1)( s +  0 

2)

1 0  Dcl =   1 1      ( s + 1)( s − 1)

0

Hence, Acl is unstable and the closed-loop system is reachable and observable. This means that G(s) is unstable as well.

Parametrization of stabilizing controllers 1° case: SISO systems and P(s) stable

d r

e

C(s)

u

P(s)

THEOREM 2 The family of all controllers C(s) such that the closed-loop system is asymptotically stable is:

C (s ) =

Q (s ) 1 − P (s )Q (s )

where Q(s) is proper and stable and Q(∞)P(∞) ≠1.

Proof of Theorem 2 Let C(s) be a stabilizing controller and define Q(s)=C(s)/(1+C(s)P(s)). Notice that Q(s) is stable since it is the transfer function from r to u. Hence C(s) can be written as =Q(s)/(1-P(s)Q(s)), with Q(s) stable and, obviously, the stability of both Q(s) and P(s) implies that Q(∞)P(∞)=C(∞)/(1+C(∞)P(∞)) ≠1. Vice-versa, assume that Q(s) is stable and Q(∞)P(∞)≠1. Define C(s)=Q(s)/(1-P(s)Q(s)). It results that S(s)=1/(1+C(s)P(s))=1-P(s)Q(s), R(s)=C(s)/(1+C(s)P(s))=Q(s), H(s)=P(s)/(1+C(s)P(s))=P(s)(1-P(s)Q(s)) are stable. This means that the closed-loop system is asymptotically stable.

Parametrization of stabilizing controllers 2° case: SISO systems and generic P(s)

It is always possible to write (coprime factorization) P (s ) =

N (s) M (s )

where N(s) and M(s) are stable rational coprime functions, i.e. such that there exist two stable rational functions X(s) e Y(s) satisfying (equation of Diofanto, Aryabatta, Bezout) N ( s) X ( s) + M (s )Y ( s) = 1

THEOREM 3 The family of all controllers C(s) such that the closed-loop system is well-posed and asymptotically stable is:

C( s) =

X (s ) + M (s )Q (s ) Y ( s) − N ( s)Q( s)

where Q(s) is proper, stable and such that Q(∞)N(∞) ≠ Y(∞).

Comments The proof of the previous theorem can be carried out following different ways. However, it requires a preliminary discussion on the concept of coprime factorization and on the stability of factorized interconnected systems.

d r

e

Mc (s)-1

z1

Nc (s)

u

-1

z2

M(s)

y N(s)

LEMMA 1 Let P(s)=N(s)/M(s) e C(s)=Nc(s)/Mc(s) stable coprime factorizations. Then the closed-loop system is asymptotically stable if and only if the transfer matrix K(s) from [r′ d′]′ to [z1′ z2′]′ is stable. r   d 

K ( s) =

 z1     z2 

N ( s)   M c (s )   − N ( s ) M ( s )  c 

−1

Proof of Lemma 1 Define four stable functions X(s),Y(s),Xc(s),Yc(s) such that X ( s) N ( s) + Y ( s ) M ( s) = 1 X c ( s) Nc (s ) + Yc ( s) M c (s ) = 1

To proof the Lemma it is enough to resort to Theorem 1, by noticing that the transfer functions K(s) and G(s) (from [r′ d′]′ to [e′ u′]′) are related as follows:

K ( s) =

G ( s)

 Yc ( s )   − X (s )

I =  0

X c (s ) Y (s )

 0 +  I   Nc

0

 G ( s) − 

−

N 0

 0  − X

 K (s ) 

Xc 0

 

Proof of Theorem 3 Assume that Q(s) is stable and Q(∞)N(∞) ≠ Y(∞). Moreover, define C(s)=(X(s)+M(s)Q(s))/(Y(s)-N(s)Q(s)). It follows that 1=N(s)X(s)+M(s)Y(s)=N(s)(X(s)+M(s)Q(s))+M(s)(Y(s)N(s)Q(s)) so that the functions X(s)+M(s)Q(s) e Y(s)-N(s)Q(s) defining C(s) are coprime (besides being both stable). Hence, the three characterist transfer functions are: S(s)=1/(1+C(s)P(s))=M(s)(Y(s)-N(s)Q(s)), R(s)=C(s)/(1+C(s)P(s))=M(s)(X(s)+M(s)Q(s)), H(s)=P(s)/(1+C(s)P(s))=N(s) (Y(s)-N(s)Q(s)) Since they are all stable, the closed-loop system is asymptotically stable as well (Theorem 1). Vice-versa, assume that C(s)=Nc(s)/Mc(s) (stable coprime factorization) is such that the closed-loop system is well-posed and asymptotically stable. Define Q(s)=(Y(s)Nc(s)X(s)Mc(s))/(M(s)Mc(s)+N(s)Nc(s)). Since the closed-loop system is asymptotically stable and (Nc,Mc) are coprime, then, in view of Lemma 1 it follows that Q(s) = (Y(s)Nc(s)-X(s)Mc(s))/(M(s)Mc(s)+N(s)Nc(s)) = = [0 I]K(s)[Y(s)′-X(s)′]′ is stable. This leads to C(s)=(X(s)+M(s)Q(s))/(Y(s)-N(s)Q(s)). Finally, Q(∞)N(∞) ≠ Y(∞), as can be easily be verified.

Coprime Factorization 1° case: SISO systems

Lemma 1 makes reference to a factorized description of a transfer function. Indeed, it is easy to see that it is always possible to write a transfer function as the ratio of two stable transfer functions without common divisors (coprime). For example

G ( s) =

s+1 ( s − 1)(s + 10)( s − 2)

N (s ) =

1 ( s − 1)(s − 2) , M ( s) = (s + 10)( s + 1) (s + 1)2

=

N ( s ) M ( s) − 1

with

It results

N (s ) X ( s ) + M (s ) X (s ) = 1 with

X ( s) =

4(16 s − 5) (s + 15) , Y (s ) = (s + 1) (s + 10)

Euclide’s Algorithm

Coprime Factorization 1° case: MIMO systems In the MIMO case, we need to distinguish between right and left factorizations.

Right and left factorization Given P(s), find four proper and stable transfer matrices such that:

P(s ) = N d (s )M d (s )− 1

=

M s ( s) − 1 N s ( s)

N d (s ), M d (s ) right coprime X d (s ) N d (s ) + Yd ( s) M d (s ) = I N s ( s), M s (s ) left

coprime N s ( s) X s (s ) + M s (s )Ys (s ) = I

Double coprime factorization Given P(s), find eight proper and stable transfer functions such that:

( i)

P ( s ) = N d ( s ) M d ( s) − 1 = M s ( s) − 1 N s ( s )

(ii)

 Yd ( s )   − N s (s )

X d ( s )   M d ( s) M s ( s)   N d (s )

Hermite’s algorithm Observer and control law

−

X s (s ) Ys (s ) 

=

I

Construction of a stable double coprime factorization Let (A,B,C,D) a stabilizable and detectable realization of P(s) and conventionally write P=(A,B,C,D) .

THEOREM 4 Let F e L two matrices such that A+BF e A+LC are stable. Then there exists a stable double coprime factorization, given by:

Md=(A+BF,B, F,I),

Nd=(A+BF,B,C+DF,D)

Ys=(A+BF,L,-C-DF,I),

Xs=(A+BF,L,F,0)

Ms=(A+ LC,L,C,I),

Ns=(A+LC,B+LD,C,D)

Yd=(A+LC,B+LD,- F,I),

Xd=(A+LC,L,F,0)

Proof of Theorem 4 The existence of stabilizing F e L is ensured by the assumptions of stabilizability and detectability of the system. To check that the 8 transfer functions form a stable double coprime factorization notice first that they are all stable and that Md e Ms are biproper systems. Finally one can use matrix calculus to verify the theorem. Alternatively, suitable state variables can be introduced. For example, from

x& = Ax + Bu

=

( A + BF ) x + Bv

y = Cx + Du = (C + DF ) x + Dv u

=

Fx + v control law

one obtains y=P(s)u=Nd(s)v, u=Md(s)v, so that P(s)Md(s)=Nd(s). Similarly,

x& = Ax + Bu y = Cx + Du q& = A q + Bu + L h   h =

Cq

+

Du − y

observer

implies η=Ns(s)u-Ms(s)y=Ns(s)u-Ms(s)P(s)y=0 (stable autonomous dynamic) so that Ns(s)=Ms(s)P(s). Analoguously one can proceed for the rest of the proof.

Parametrization of stabilizing controllers 3° case: MIMO systems and generic P(s)

(i) P( s) = N d ( s) M d (s )−1 = M s ( s) −1 N s ( s) X d (s )  M d (s ) − X s (s )  Yd (s ) (ii)    N ( s) =I − N ( s ) M ( s ) Y ( s )  s s  d s 

C(s)

P(s)

THEOREM 5 The family of all proper transfer matrices C(s) such that the closed-loop system is well-posed and asymptotically stable is:

C (s ) = ( X s (s ) + M d ( s )Qd ( s))(Ys ( s ) − Nd (s )Qd ( s ))−1 = (Yd (s ) − Qs ( s ) N s (s ))−1 ( X d ( s ) + Qs (s )M s (s )) where Qd(s) [Qs(s)] is stable and such that Nd(∞)Q d(∞)≠Ys(∞) [Qs(∞)Ns(∞)≠Yd(∞)].

Comments The proof of Theorem 5 follows the same lines as that of Theorem 3. Thowever, the generalization of Lemma 1 to MIMO is required.

d r

e

-1

Mc (s)

z1

Nc (s)

u

-1

Md(s)

z2

y Nd(s)

LEMMA 2 Let P(s)=Nd(s)Md(s)-1 and C(s)=Nc(s)Mc(s)-1 stable right coprime factorizations. Then the closed-loop system is asymptotically stable if and only if the transfer matrix K(s) from [r′ d′]′ to [z1′ z2′]′ is stable.

 M c ( s) N d ( s)  K ( s) =   − N ( s ) M ( s )  c d 

−1

Proof of Lemma 2 It is completely similar to the proof of Lemma 1. Proof of Theorem 5 Assume that Q(s) is stable and define C(s)=(Xs(s)+Md(s)Q(s))(Ys(s)-Nd(s)Q(s))-1 It results that I=Ns(s)Xs(s)+Ms(s)Ys(s) =Ns(s)(Xs(s)+Md(s)Q(s))+Ms(s)(Y(s)s-Nd(s)Q(s)) so that the functions Xs(s)+Md(s)Q(s) e Ys(s)-Nd(s)Q(s) defining C(s) are right coprime (besides being both stable). The four transfer matrices characterizing the closed-loop are: (1+PC)-1= (Ys-NdQ)Ms -(1+PC)-1P= (Ys-NdQ)Ns (1+CP)-1C=C(I+PC)-1= (Xs+MdQ)Ms (1+CP)-1=I-C(I+PC)-1C= I-(Xs+MdQ)Ns They are all stable so that the closed-loop system is asymptotically stable. Vice-versa, assume that C(s)=Nc(s)Mc(s)-1 (right coprime factorization) is stabilizing. Notice that the matrices

X d (s )   Yd ( s)  − N ( s) M ( s )   s s 

and

M d ( s ) N c ( s )   N ( s) M ( s )  ≈ K  d c 

have stable inverse. Hence,  Yd ( s )  − N s ( s )

X d (s)  M d (s)  M s (s)  N d (s)

Nc (s)  I = M c ( s )  0

(*) has stable inverse as well. Then,

Yd ( s ) N c ( s ) X d ( s ) N c ( s )   N s (s) N c (s) + M s (s)M c (s)

Q(s)=(Yd(s)Nc(s)-Xd(s)Mc(s))(Ms(s)Mc(s)+Ns(s)Nc(s))-1 is well–posed and stable. Premultiplying equation (*) by [Md(s) –Xs(s);Nd(s) Ys(s)] it follows Nc(s)Mc(s)-1=(Xs(s)+Md(s)Qd(s))(Ys(s)-Nd(s)Qd(s))-1.

Observation Taking Q(s)=0 we have the so-called central controller C0(s) =Xs(s)Ys(s)-1 which coincides with the controller designed with the pole assignment technique. Taking r=d=0 one has: ξ& = Aξ + Bu + L (Cξ + Du − y ) u = Fξ

LEMMA 3

[

]

−1

C( s ) = C0 ( s) + Yd ( s) −1Q( s ) ( I − Ys ( s) −1 N d ( s )Q( s)) Ys ( s) −1

u

y P C0 Yd-1 Nd Q

Ys-1

Strong Stabilization

C(s)

P(s)

The problem is that of finding, if possible, a stable controller which stabilizes the closed-loop system.

Example P (s) =

s −1 ( s − 2)( s + 2)

is not stabilizable with a stable controller. Why? P (s) =

s −2 ( s − 1)( s + 2)

is stabilizable with a stable controller. Why? Stabilizazion of many plants Two-step stabilization

Interpolation Let consider a SISO control system with P(s)=N(s)/M(s) e C(s)=Nc(s)/Mc(s) (stable coprime stabilization). Then the closedloop system is asymptotically stable if and only if U(s)=N(s)Nc(s)+M(s)Mc(s) is an unity (stable with stable inverse). Since we want that C(s) be stable, we can choice Nc(s)=C(s) and Dc(s)=1. Then,

C( s) =

U (s) − M (s) N (s)

Of course, we must require that C(s) be stable. This fact depends on the role of the right zeros of P(s). Indeed, if N(b)=0, with Re(b)≥0 (b can be infinity), then the interpolation condition must hold true: U (b) = M (b)

Consider the first example and take M(s)=(s-2)(s+2)/(s+1)2, N(s)=(s-1)/(s+1)2. Then, it must be: U(1)=-0.75, U(∞)=1. Obviously this is impossible. Consider the second example and take M(s)=(s-1)/(s+2) e N(s)=(s2)/(s+2)2. It must be U(1)=0.25, U(∞)=1, which is indeed possible.

Parity interlacing property (PIP)

THEOREM 6 • P(s) è strongly stabilizable if and only if the number of poles of P(s) between any pair of real right zeros of P(s) (including infinity) is an even number. • We have seen that the PIP is equivalent to the existence of the existence of a unity which interpolates the right zeros of P(s). Hence, if the PIP holds, the problem boils down to the computation of this interpolant. • In the MIMO case, Theorem 6 holds unchanged. However it is worth pointing out that the poles must be counted accordingly with their Mc Millan degree and the zeros to be considered are the so-called blocking zeros.

REFERENECES J.G.Truxal, Control Systems Synthesis, Mc-Graw Hill, NY, 1955. J.R. Ragazzini, G.E. Franklin, Sampled-data Control Systems, Mc-Graw Hill, NY, 1958. M. Vidyasagar, Control System Synthesis: a Factorization Approach, MIT Press, Cambridge, MA, 1985. J.C. Doyle, B.A. Franklin, A.R. Tannenbaum, Feedback Control Theory, Mc Millan, NY, 1992. D.C. Youla, J.J. Bongiorno, N.N. Lu, Single-loop feedback stabilization of linear multivariable dynamical plants”, Automatica, 10, p. 159-175, 1974. G. Zames, “Feedback and optimal sensitivity in model reference transformation, multiplicative seminorms, approximate inverses, IEEE Trans. AC-26, p. 301-320, 1981. V. Kucera, Discrete Linear Control: The Polynomial Equation Approach, Wiley, NY, 1979.

H2 CONTROL

SUMMARY

• Definition and characterization of the H2 norm • State-feedback control (LQ) • Parametrization, mixed problem, duality • Disturbance rejection, Estimation (Kalman filter), Partial Information • Conclusions and generalizations

The L2 space

L2 space: The set of (rational) functions G(s) such that ∞

1 2π

∫ trace [G(− jω )'G( jω )]dω < ∞

−∞

(strictly proper – no poles on the imaginary axis!) With the inner product of two functions G(s) and F(s):

G (s ), F ( s ) =

1 2π

∞

∫ trace [G(− jω )' F ( jω )]dω

−∞

The space L2 is a pre-Hilbert space. Since it is also complete, it is indeed a Hilbert space. The norm, induced by the inner product, of a function G(s) is ∞   1 G ( s) =  trace [G (− jω )' G ( jω ) ]dω 2 2π  −∞

∫

1/ 2

   

The subspaces H2 and H2⊥ The subspace H2 is constituted by the functions of L2 which are analytic in the right half plane. (strictly proper – stable !) The subspace H2⊥ is constituted by the functions of L2 which are analytic in the left half plane. (stricty proper – untistable !)

Note: A rational function in L2 is a strictly proper function without poles on the imaginary axis. A rational function in H2 is a strictly proper function without poles in the closed right half plane. A rational function in H2⊥ is a strictly proper function without poles in the closed left half plane. The functions in H2 are related to the square integrable functions of the real variable t in (0,∞]. The functions in H2⊥ are related to the square integrable functions of the real variable t in (-∞,0].

A function in L2 can be written in an unique way as the sum of a function in H2 and a function in H2: G(s)=G1(s)+G2(s). Of course, G1(s) and G2(s) are orthogonal, i.e. =0, so that

G ( s)

2 2

= G1 (s )

2 2

+

G2 ( s )

2 2

System theoretic interpretation of the H2 norm Consider the system

x& (t ) = Ax (t ) + Bw(t ) z(t ) = Cx(t ) x(0) = 0 And let G(s) be its transfer function. Also consider the quantity to be evaluated: m ∞

J1 =

∑∫

z (i ) ' (t ) z (i ) (t )dt

i =1 0

where z(i) represents the output of the system forced by an impulse input at the i-th component of the input. m ∞

∞

1 m (i ) ( i) J1 = ∑∫ z ' (t)z (t)dt = Z ( − j ω )' Z ( jω)dω = ∑ ∫ 2π −∞ i=1 i=1 0 (i )

∞

( i)

∞

1 m 1 2 e ' G ( − j ω )' G ( j ω ) e d ω = trace [ G ( − j ω )' G ( j ω ) ] = G ( s ) ∑i i 2 2π −∫∞ i=1 2π −∫∞

Computation of the norm Lyapunov equations

“Control-type Lyapunov equation” A' Po + Po A + C ' C = 0 “Filter-type Lyapunov equation” Pr A'+ Pr A + BB ' = 0

m ∞

G ( s)

2 2

= J1 =

∑∫ z

∞ m ( i) ' (t ) z (i) (t ) dt =

i =1 0

∞

∫ ∑ trace [e ' B ' e i

0 i =1

m   A ' t At  = trace e C ' Ce Bei ei ' B ' dt     i =1  0

∑

∫

 ∞    = trace  B ' e A't C ' Ce At dt B  = trace [B ' P0 B ]  0 

∫

 ∞    = trace C e At BB ' e A't dt C ' = trace [CPr C ']  0 

∫

]

A't C ' Ce At Be dt = i

Other interpretations Assume now that w is a white noise with identity intensity and consider the quantity:

J 2 = lim E ( z ' (t ) z (t )) t→∞

It follows: t   t A(t −τ ) A'(t −σ ) C 'dσ   E trace E Ce Bw ( τ ) d τ w ( σ )' B ' e J 2 = lim ∫ ∫   0   = 0  t →∞

t t  = lim trace ∫ ∫ Ce A ( t −τ) BE [w(τ ) w(σ )']B' e A '( t −σ) C ' d τ dσ  = t →∞ 0 0  t  A ( t−τ) A'( t −τ ) = lim trace ∫ Ce BB 'e C ' dτ  = G ( s ) t →∞ 0 

2 2

Finally, consider the quantity

1 T J 3 = lim E ∫ z' ( t ) z ( t )dt T →∞ T 0 and let again w(.) be a white noise with identity intensity. It follows: E ( z ' (t ) z (t )) = trace [CP(t )C '],

t

P (t ) = ∫ e Aτ BB ' e A 'τ dτ 0

Notice that J3=(||G(s)||2)

1/2

since

P& ( t ) = AP (t ) + P( t ) A'+ BB ' ,

1T AY + YA'+ BB ' = 0, Y = lim ∫ P (τ ) dτ T →∞ T 0

Optimal Linear Quadratic Control (LQ) versus Full-Information H2 control

x& = Ax + Bu , x (0) = x0 ,

∞

J = ∫ ( x ' Qx + 2u ' Sx + u ' Ru )dt 0

Assume that

Q S  R > 0, H =  ≥0  S ' R  And notice that these conditions are equivalent to

R > 0, W = Q − SR− 1S ' ≥ 0 Find (if any) u(⋅) minimizing J

The above matrix result is the well known Schur Lemma. The proof is as follows. If H≥0, then

[

]

 I  − SR −1 H  −1  ≥ 0 − R S ' Vice-versa, assume that H≥0 and R>0, and suppose by contradiction that W is not positive semidefinite, then Wx=νx, for some ν 0 (Af,Bf) stabilizable (A-B2C 1) stable

Af = A-B1 D21′(D21D21′)-1 C2 B1f = B1(I-D 21′ (D21D21′)-1D21)

Solution of the Output Estimation problem Theorem 2 There exists an admissible controller (filter) minimizing ||Tzw||2. It is given by ξ& = Aξ + B u + L (C ξ − y ) 2

2

2

u = −C1ξ where L2 = −( PC2 + B2D'21 )(D 21D 21 )

-1

and P is the positive semidefinite and stabilizing solution of the Riccati equation AP+ PA'−(PC2 '+B1D21')(D21D21' )−1(C2P + D21B1' ) + B1B1' = 0 Aff = A−(PC2'+B1D21' )(D21D21' )−1 = A + L2C2 stable

Solution of the Output Estimation problem (cont.) The optimal norm is 2

Tzw 2 = C1Pf (s) = trace(C1PC1'), Pf (s) = (sI − A− L2C2 )−1(B1 + L2D21) 2

2

The set of all controllers generating all ||Tzw||2 < γ is given by u

y

Aff -B2C 1 L∞

-B2

M

C1 C2

I 0

Q

M(s)= 0 I

Where Q(s) is proper, stable and such that ||Q(s)||22 < γ2−||C1Pf(s)||22

Proof of Theorem 2 It is enough to observe that the “transpose system”

λ& = A' λ + C '1 ς + C2 'ν µ = B1 ' λ + D21 'ν ϑ = B2 ' λ + ς has the same structure as the system defining the DF problem. Hence the solution is the “transpose” solution of the DF problem. It is worth noticing that the H2 solution does not depend on the particular linear combination of the state that one wants to estimate.

Filtering Given the system x& = Ax + ς1 + B22u y = Cx + ς2 where [z′1 z′2]′ are zero mean white Gaussian noises with intensity W12  W W =  11 , W22 > 0  W12 ' W22  find an estimate of the linear combination Sx of the state such as to minimize J = lim E [( Sx(t ) − u (t ))' ( Sx(t ) − u (t ))] t →∞

Letting −1

C1 = − S , C2 = W22 C , D21 = [ 0 I ], −1

B11B11 ' = W11 − W12W22 W12 ', y = W22

−1 / 2

B1 = [ B11 W12W22

−1 / 2

]

 ς1   B11 W12W22 −1 / 2  y , z = C1 x + u ,   =  w 1/2 W22 ς2   0 

the problem is recast to find a controller that minimizes the H2 norm from w to z.

The partial information problem (LQG)

w u

P K(s)

x& = Ax + B1w + B2u z = C1 x + D12u y = C2 x + D21w Assumptions: FI + DF.

z y

Theorem 3 There exists an admissible controller minimizing ||Tzw|| 2. It is given by

ξ& = Aξ + B2 u + L2 (C2ξ − y ) u = − F2ξ The optimal norm is 2

2

Tzw 2 = Pc (s)L2 2 + C1Pf (s) 2 = Pc (s)B1 2 + F2Pf (s) 2 =γ 02 2

2

2

The set of all controllers generating all ||Tzw||2 < γ is given by u

y

A +B2F2+L2C 2 L2 -B 2

S

-F2 C2

Q

S(s)= 0 I

I 0

Where Q(s) is proper, stable and such that ||Q(s)||22 < γ2−γο2

Proof Separation principle: The problem is reformulated as an Output Estimation problem for the estimate of -F2 x.

Important topics to discuss • Robustness of the LQ regulator (FI) • Robustness of the Kalman filter (OE) • Loss of robustness of the LQG regulator (Partial Information) • LTR technique

THE H∞ CONTROL PROBLEM

SUMMARY

• Definition and characterization of the H∞ norm • Description of the uncertainty: standard problem • State-feedback control • Parametrization, mixed problem, duality • Disturbance rejection, Estimation, Partial Information • Conclusions and generalizations

.

What the H∞ norm is? Bode Diagrams From: U(1) 20 10 0 -10 -20 50 0 -50 -100 -150 0 10

1

10

2

10

Frequency (rad/sec)

G( s) rational stable proper G (s) ∞ = sup Re( s ) ≥ 0 G ( s) = max ω G ( jω) Ω = σ (Ω) = λmax (Ω * Ω) = λmax (ΩΩ*)

For the frequency-domain definition we can consider a transfer function without poles on the imaginary axis. It is easy to understand that the norm of a function in L∞ can be recast to the norm of its stable part given through the so-called inner-outer factorization. The time-domain definition, given in the next page, requires, for obvious reasons, the stability. .

Time-domain characterization G(s) = D+C(sI-A)-1B x& = Ax + Bw z = Cx + Dw A = asymptotically stable

G (s) ∞ = sup w∈L2

z

2

w2

There does not exist a direct procedure to compute the infinity norm. However, it is easy to establish whether the norm is bounded from above by a given positive number γ.

The symbol indifferently the space of square integrable or the space of the strictly proper rational function without poles on the imaginary axis. If w(.) is a white noise, the infinity norm represents the square root of the maximal intensity of the output spectrum.

.

BOUNDED REAL LEMMA Let γ be a positive number and let A be asymptotically stable. THEOREM 1 The three following conditions are equivalent each other: (i)

||G(s)||∞ < γ

(ii) ||D||0  B ' P + D' C  2 γ I − D' D 

.

Complex Stability Radius and quadratic stability

x& = ( A + L∆N ) x ,  x& =  z = w = 

Ax + Lw Nx ∆x

∆ ≤α ∆

A,L,N

The system is said to be quadratically stable if there exists a solution (Lyapunov function) to the following inequality, for every ∆ in the set

( A + L∆N )' P + P( A + L∆N ) < 0 THEOREM 3 The system is quadratically stable if and only if ||N(sI-A)-1L||∞0 satisfying XLL ' X + N'N < 0 α −2 then A+L∆N is asymptotically stable for every ∆, ||∆|| ≤α, with the same Lyapunov function (quadratic stability). This happens if ||N(sI-A)-1L||∞0, we have that there exists s=jω that violates (*), a contradiction.

.

Real stability radius A= stable (eigenvalues with strictly negative real part) rr ( A, B, C ) = inf { ∆ : ∆ ∈ R m× p and A + B∆C is unstable}

= inf inf { ∆ : ∆ ∈ R m× p and det (sI − A − B∆C ) = 0} s∈ jω

= inf inf { ∆ : ∆ ∈ R m× p and det (I − ∆C ( sI − A) −1 B ) = 0} s∈ jω

Linear algebra problem: compute

[

]

µ r (M ) = inf { ∆ : ∆ ∈ R m× p and det (I − ∆M ) = 0}

−1

Proposition   Re M µ r ( M ) = inf σ 2   −1 γ ∈( 0 ,1]  γ Im M

− γ Im M    Re M  

Taking M=G(s)=C(sI-A)-1B it follows   Re G ( jω ) − γ Im G ( jω )   rr ( A, B, C ) −1 = sup inf σ 2   −1  γ ∈ ( 0 , 1 ] Re G ( jω )   ω  γ Im G ( jω )

Example 20 − 30 − 20  79 0.2190  − 41 − 12 0.0470 17 13  A= , B=  167 40 0.6789 − 60 − 38    − 14.5 − 11 33.5 9  0.6793 0.0346 0.5297 0.0077 0.0668 C=  0.0535 0.6711 0.3834 0.4175

3

0.9347 0.3835 0.5194  0.8310

2 1.8

2.5 1.6 1.4 2 1.2 1.5

1 0.8

1 0.6 0.4

0.5

0.2 0 0

5

10

15

20

25

30

0 0

− 0.4996 0.1214 ∆ worst =    0.1214 0.4996

5

10

15

20

25

30

Entropy Consider a stable system G(s) with state space realization (A,B,C,0), and assume that ||G(s)||∞0  − XAc '− Ac X − B1B1 '+γ 2 B2 B2 ' XC1c '  >0 2 C1c X γ I  

Ac = A-B2(D 12′D12)-1 D12′C 1 C1c = (I-D12 (D 12′D12)-1 D12′)C 1

Proof of Theorem 6 Assume that there exists K such that A+B2K is stable and the norm of the system (A+B2K,B1,C 1+D12K) is less than γ. Then, from the we know that there exists a positive definite solution of the inequality

( A + B2 K )' P + P( A + B2 K ) + γ −2 PB1B1 ' P + + (C1 + D12 K )'(C1 + D12 K ) < 0

(*)

Now, defining

F = −(D12 ' D12 ) (B2 ' P + D12 ' C1 ) -1

the Riccati inequality can be equivalently rewritten as

A' P + PA − ( PB2 + C ' D12 )( D12 ' D12 ) −1 ( B2 ' P + D12 ' C1 ) + P

B1 B1 ' P + C1 ' C1 + ( K − F )' ( K − F ) < 0 2 γ

so concluding the proof. Vice-versa, assume that there exists a positive definite solution of the inequality. Then, inequality (*) is satisfied with K=F, so that with such a K, the norm of the closed-loop transfer function is less than γ (BRL).

Parametrization of all algebraic “state-feedback” controllers

Define:

AR = [ A B2 ]

CR = [C1 D12 ] W = [W1 W2 ]

THEOREM 7 The set of all controllers u=Kx such that the H∞ norm of the closed-loop system is less than γ is given by: −1

K = W2 'W1 W1 > 0

 − ARW − WAR '− B1 B1 ' WC R '  2 >0 C W ' γ I  R 

Proof of Theorem 7 We prove the theorem in the simple case in which C1′D12=0 and D12′D12=I. The LMI is equivalent to −1

K = W2 'W1 W1 > 0

AW1 + W1 A'+ B1 B1 '+W2 B2 '+ B2W2 '+γ − 2W1C1 'C1W1 + γ − 2W2W2 ' < 0 If there exists W satisfying such an inequality, it follows that the inequality

P( A + B2 K ) + ( A + B2 K )' P + γ −2 PB1 B1 ' P + C1 ' C1 + KK ' < 0 Is satisfied with K=W2′W 1-1 and P=γ2W1-1. The result follows from the BRL. Vice-versa, assume that there exists P>0 satisfying this last inequality. The conclusion directly follows by letting W1==γ2P-1 e and W2=W1K′.

Example

0 1 0 x& =  x + (w + u )    − 1 − 1 1  0 0 robustness ∆A =  1 0 0  Ω 0 z=    x + 1 u 0 0     γinf = 0.75, γ = 0.76

||T(z1,w)||∞

2.5

2

1.5 H2

1 H

0.5

0 -1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Ω

Mixed Problem H2/H∞ x& = Ax + B1 w + B 2 u z = C1 x + D12 u ξ = Lx + Mu The problem consists in finding u=Kx in such a way that

min K T (ξ , w, s , K ) 2 : T ( z , w, s , K ) ∞ ≤ γ • For instance, take the case z=ξ. The problem makes sense since a blind minimization of the H∞ infinity norm may bring to a serious deterioration of the H2 (mean squares) performances. • Obviously, the problem is non trivial only if the value of γ is included in the interval (γ inf γ2), where γing is the infimun achievable H∞ norm for T(z,w,s,K) as a function of K, whereas γ2 is the H∞ norm of T(z,w,s,KOTT) where KOTT is the optimal unconstrained H2 controller. • This problem has been tackled in different ways, but an analytic solution is not available yet. In the literature, many sub-optimal solutions can be found (Nash-game approach, convex optimization, inverse problem, …)

Post-Optimization procedure x& = Ax + B1 w + B 2 u z = C1 x + D12 u ξ = Lx + Mu

simplifying assumptions L′M=0, M′M=I

Let Ksub a controller such that ||T(z,w,s,Ksub)||∞0) q must be such that (0) (1) (2) (3)

q minimum norm q stable q(pi)=0 (1+gq)(pi)=0

Letting

q w = qa = q∏i

s + pi pi − s

it follows that qw must be such that (0) (1) (2)

qW minimim norm qW stable qW(pi)=-ag-1(pi)

A simple example δ

z u

w y

g

k

where

g=

s+2 ( s + 1)( s − 1)

Find k in such a way that ||d||∞ is maximized. Take

a=

s +1 1− s

and find qw stable of minimum norm such that such that qw (1)=-ag-1(1)=4/3. It follows qw =4/3 so that ||d||∞max =3/4 and

qw 4( s + 1) k= = − (3s + 5) 1+ qwg

In state-space A=[0 1;1 0], B1=[0;0], B2=[0;1], C1=[0 0], D12=1, C2=[2 1], D21=1. Stabilizing solution of: A′P+PA-PB2B2′P+PB1B1′P/γ2+C1′C1=0 Stabilizing solution of: AQ+QA′-PC2′C2Q+PC1′C1Q/ γ2+B1B1′=0 rs(PQ) 0

Theorem 3 N

N

t =0

t =0

J = ∑ y( t )' y( t ) − γ 2 ∑ w( t )' w( t ) − γ 2 x( 0 )' Q0−1x ( 0 )

< 0 , ∀( w( . ), x( 0 ) ) ≠ 0

iff there exists Q(t)≥0 in [0,N] such that γ 2 I − DD' − CQ ( t )C' > 0

Proof of Theorem 3 Consider the cost. N N ~ J ( N ) = ∑ z ( t )' z (t ) − γ 2 ∑ w( t )' w(t ) + x ( N + 1)' PN +1 x( N +1) < 0, x( 0) = 0,∀ w(.) ≠ 0 t =0

t =0

We proof the dual result:

~ J ( N ) < 0, x( 0) = 0,∀ w(.) ≠ 0 iff there exists the solution of the equation

P ( t ) = A' P ( t + 1 ) A + + ( A' P ( t + 1 ) B + C' D )( γ2 I − D ' D − B' P ( t + 1 ) B )−1 ( A' P ( t + 1 ) B + C' D )' + C' C , P ( N + 1 ) = PN +1 > 0 with γ 2I − D' D − B' P( t + 1)B > 0 , in [0,N+1]. Sufficient condition: if there exists the solution with the indicated properties, then pre-premultiplying the equation by x(t)' and post-multiplying by x(t), and summing all terms, it results N

N

∑ z (t )' z (t) − γ ∑ w(t )' w(t) + x( N + 1)' P 2

t= 0

N +1

t =0

N

~(t ))' ( w( t ) − w ~ (t )) x( N + 1)= x( 0)' P (0) x( 0) − ∑ ( w(t ) − w t =0

(*)

where

~( t ) = (γ 2 I − D' D − B' P(t + 1) B) −1 ( A' P(t + 1) B + C ' D)' x( t ) . w

Hence , being x(0)=0 we have Vice-versa, assume that

~ J ( N ) < 0, ∀w(.) ≠ 0 .

~ J ( N ) < 0, ∀w(.) ≠ 0 and take w(.)=0 in [0,N-1]. Being x(0)=0, with such a choice we have

0 > w ( N )' − γ 2I + D' D + BPN +1B' w ( N ), ∀w ( N ) ≠ 0 so that − γ 2I + D' D + BPN+1B' > 0 . Furthermore, from

~ ~ ~ ( N ))' ( w( N ) − w ~( N )) < 0, ∀w(.) ∈ [0, N ] J ( N ) = J ( N − 1) − ( w( N ) − w it turns out taht

~ J ( N − 1) > 0,∀w(.) ∈ [0, N − 1] . By iterating this reasoning we have γ 2I − D' D − B' P( t + 1)B > 0 in [0,N]. Theorem 3 is proven from (*) by noticing that P( t ) = Q( t ) −1γ 2 .

STOCHASTIC FILTERING x (t + 1) = Ax (t ) + v1 (t )   z (t ) = Lx(t )  y (t ) = C x (t ) + v (t ) 2  where v1 ( t ), v2 ( t ) are zero-mean white gaussian noises . Moreover,

 v (t )  E 1  = v2 (t ) 

 V1 V12  V ' V , V2 > 0  12 2 

The problem is to find an estimate zˆ(t ) in H ∞ of a linear combination z (t ) = Lx (t ) of the state, i.e. one wants to find an estimator based on the observations of {y (i), i ≤ t − N } ) in such a way that the maximum of the output spectrum of the error ε = z − zˆ due to [v1 ' v2 ']' is minimized (or bounded from above by a certain positive scalar γ). Letting u(t ) = − zˆ(t ), ε (t ) = Lx(t ) + u(t ),

[ ], B = [(V − V V D = 0 V2 1

C =C

1/ 2

12

2

−1

V12 ' )1 / 2

V12V2

−1 / 2

y (t ) = V 2

− 1/ 2

y (t )

]

and defining with w ( t ) a zero mean white gaussian noise with identity covariance, it follows:

BASIC FILTERING PROBLEM

ε S

x(t + 1) = Ax (t ) + Bw(t)  S : ε (t) = Lx(t ) + u (t )  y (t ) = Cx(t) + Dw(t) 

 xˆ(t + 1) = AF xˆ(t) + BF y(t − N ) F :  u (t) = CF xˆ (t ) + DF y (t − N ) Basic problem: Find AF , BF , C F , D F such that • •

(S,F) is asymptotically stable Tεw ∞ < γ

________________________________________________________________

zˆ (t ) the estimated variable, it follows u (t ) = − zˆ (t ) . Hence ε( t ) represents the filtering error z (t ) = z (t ) − zˆ (t ) = ε(t ) . If w(t) is a white noise, the problem is that of minimizing If z(t)=Lx(t) is the variable to be estimated and

the peak value of the error spectrum If w(t) is a generic signal in l2 , the problem is that of minimizing the l2 “gain” between the error and the disturbance.

SOLUTION OF THE FILTERING PROBLEM

Ipotesi:

DD' > 0 A = stable ( A, B, C , D ) does not have unstable invariant zeros

Theorem 4 There exists a filter F solving the basic problem iff ∃ Q≥0 solution of −1

CQL'  CQA'+ DB' CQA'+ DB'  DD'+CQC' i) Q = AQA'+ BB'−   − γ 2 I + LQL'  LQA'   LQA'   LQC ' '

ii ) V = γ 2 I − LQL '+ LQC ' ( DD'+CQC ' ) −1 CQL '> 0

filter feasibility

−1

CQL' CQA'+ DB '  DD '+CQC '  C  iii) Aˆ = A −   − γ 2 I + LQL'  L   LQA'   LQC ' '

stable

CENTRAL FILTER ξ (t + 1) = Aξ (t ) + F1 ( y(t ) − C ξ (t )), F1 = ( AQC '+ BD ' )(CQC '+ DD' ) −1 zˆ(t ) = Lξ (t ) + F2 ( y(t ) − Cξ (t )), F2 = LQC ' (CQC '+ DD' ) −1 _________________________________________________________________________ The assumption on the zeros is equivalent to the stabilizability of the pair

~ ~ ~ ~ ( A, B ), A = A − BD' ( DD ' ) −1 C ,B = B( I − D' ( DD' ) − 1 D) . The assumption of stability of A can be weaked to

the assumption of detectability of (A,C) if one is only interested in the stability of the filter.

NOTES •

If G≠0, one obtains the central filter ξ (t + 1) = Aξ (t) + F1 ( y(t ) − Cξ (t )) + Gu (t ) zˆ(t ) = Lξ (t ) + F2 ( y(t ) − Cξ (t ))

•

As γ → ∞ the H2 filtering is obtained, where F1 e F2 depend on the stabilizing solution of the standard Riccati equation: Q = AQA ' + BB ' − ( CQA' + DB ' )' ( DD ' +CQC ) −1 ( CQA ' + DB ' )

•

Notice that in the H∞, filter the solution depends on L, i.e. on the particular linear combination of the state to be estimated. This is a difference between H2 , and H ∞, design.

•

The “gains” F1 and F2 are related by: F2 = LK

[

]

F1 = A − BD ' ( DD ' ) −1 C K − BD ' ( DD ' ) −1 with

K = QC ' (CQC '+ DD' ) −1 In the uncorrelated case (DB'=0) these relations become F2 = LK , Hence

F1 = AK

ξ (t + 1) = Aξ (t ) + AK [ y (t ) − Cξ (t ) ] rˆ (t ) = Lξ (t ) + LK [y (t ) − Cξ (t )]

BASIC ONE-STEP PREDICTION PROBLEM Assume that we want to estimate z (t ) = Lx (t ) on the basis of the data y (i ) , i≤ t.

S

P

 x(t + 1) = Ax (t ) + Bw(t ) + Gu (t )  S : ε (t ) = Lx(t ) + u (t )  y (t ) = Cx (t ) + Dw(t ) 

 ~x (t + 1) = AP ~x (t ) + BP y (t ) P : ~  u (t ) = CP x (t ) Basic problem: Find AP , B P , C P such that: • •

(S,F) asymptotically stable Trw ∞ 0

predictor feasibility

 CQA ' + DB '  iii ) Aˆ = A −    LQA ' 

'

 DD ' + CQC '   LQC '

−1

  C     − γ 2 I + LQL '  − L  CQL '

stable

CENTRAL PREDICTOR η (t + 1) = Aη (t ) + F3 ( y (t ) − Cη (t )) ~z (t) = Lη (t) F3 = ( AQC' + BD' )( CQC ' + DD' )−1 + AX QL' V −1 LQC' ( CQC' + DD' )−1 A X = A − ( AQC' + BD' )( CQC' + DD' )−1 C V = γ 2 I − LQL' + LQC' ( DD' + CQC' )−1 CQL' _________________________________________________________________________ The assumption on the zeros is equivalent to the stabilizability

of

the

pair

~ ~ ~ ~ ( A, B ), A = A − BD' ( DD ' ) −1 C ,B = B( I − D' ( DD' ) − 1 D) . The assumption of stability of A can be weakened

to that of detectability of (A,C), if one is only interested to the stability of the predictor.

NOTES

•

If G≠0, the so-called central predictor is obtained µ (t + 1) = Aη (t ) + F3 ( y (t ) − Cη (t )) + Gu (t ) ~z (t ) = Lη (t)

•

When γ → ∞ the H2 predictor is recovered. Moreover, F3 → F1 , i.e. the gain of the predictor and the gain of the filter coincide and depend on the stabilizing solution of the standard Riccati equation: Q = AQA ' + BB ' − ( CQA' + DB ' )' ( DD ' +CQC ) −1 ( CQA ' + DB ' )

•

Notice that the H ∞ predictor depends on L, i.e. on the particular linear combination of the state to be estimated. This is a difference between H2, and H ∞, design.

•

The “gain” F3 of the predictor and the “gains” F1 and F2 of the filter are related by: F3 = F1 + A − F1C QL'V −1 F2

________________________________________________________________ In the uncorrelated case (DB'=0) and under the assumption of reachability of (A,B), the Ricacti equation of Theorem 3 can be rewritten in the following way: −1

  1 Q = AQ −1 + C' ( DD' ) −1 C − 2 L ' L A'+ BB ' γ   Feasibility condition for the predictor Feasibility condition for the filter Filter/Predictor

γ 2 I − LQL' > 0

Q −1 + C' C − L ' Lγ −2 > 0

F1 = A(Q −1 + C ' C ) −1 C ' , F2 = L(Q −1 + C ' C) −1 C' , F3 = A(Q −1 + C ' C −

1 L' L)C ' γ2

PREDICTOR VS FILTER The Riccati equation is the same, but the feasibility conditions are different System for the prediction error µ ( t + 1) = ( A − F3 C ) µ ( t ) + ( F3 C − B ) w ( t ) e( t ) = L µ ( t ) feasibility (analysis) :

γ 2 I − LQL ' > 0

System for the filtering error

η ( t + 1) = ( A − F1C )η (t ) + ( F1 D − B) w(t ) e(t ) = ( L − F2 C )η ( t ) + F2 Dw(t ) feasibility (analysis) :

γ 2 I − LQL ' + LQC ' ( DD' + CQC ' ) −1 CQL ' − ( F2 − F2 )( DD' + CQC ' )( F2 − F2 )' > 0

F2 = LQC ' ( CQC ' + DD ' ) −1 ________________________________________________________________ 1) 2)

At γ fixed, the fesibility condition of the predictor is more stringent than that of the filter. They coincide as γ tends to infinity, to witness the fact that the Kalman predictor and filter share the same existence conditions. Assuming the existence of the stabilizing solution of the Riccati equation satisfying the feasibility condition, the sufficient part of Theorem 3 is proved from the analysis result. Indeed, the analysis Riccati equation coincides with the synthesis Riccati equation once the gains F1 , F2 (for the filter) or F3 (for the predictor) are substituted there.

Proof of Theorem 4/5 For simplicity we study the case where DB'=0, DD'=I, (A,B) reachable. Moreover, we limit the proof to filters (predictors) in observer form. This is a restriction, but the analysis of all filters (predictors) ensuring the norm bounded is far from being trivial and is not within our scope. Notice that if F1 =AK e F2 =LK are the characteristic gains of a generic filter, the transfer function from the disturbance to the error is Tεw=(L-LKC)(zI-A+AKC)-1 (AKD-B)+LKD So that the analysis Riccati equation is Q = A ( X −1 − L ' Lγ −2 )−1 A' + BB' ,

X = ( I − KC ) Q ( I − KC )' + KK '

with the feasibility condition γ2 I − LXL ' > 0 and asymptotic stability of A( I − KC ) + AXL ' ( γ2 I − LXL ' )−1 ( I − KC )' L ' On the other hand, if F3 is the gain of a generic one-step predictor, the transfer function is

(1) (2) (3)

Tεw=L(zI-A+F3 C)-1 (F3 D-B), and then the analysis Riccati equation is:

Q = A(Q −1 + C' C − L ' Lγ −2 ) −1 + BB'+( F3 − AYC' ( I + CYC ' ) −1 )( I + CYC' )( F3 − AYC' ( I + CYC ' ) −1 )' Y = (Q −1 − L ' Lγ − 2 ) (4) with the feasibility condition γ2 I − LQL ' > 0

(5)

and asymptotic stability of A − F3 C + ( A + F3C ) QL ' ( γ2 I − LQL ' )−1 L'

(6)

Now, assume that there exists Q≥0 satisfying Q=A[Q-1 +C'C-L'Lγ-2 ]-1 +BB', and such that Q-1 +C'C-L'Lγ-2 >0 and A(Q-1 +C'C-L'Lγ-2 )-1 Q-1 stable. Take the filter characterized by F1 =AK, F2 =LK, with K=(Q-1 +C'C)-1 C'. From (1) it results X=(Q-1 +C'C)-1 and hence Q = AXA' + BB' + AXL' ( γ2I − LXL' )−1 LXA' = A( X − 1 − L' Lγ− 2 )−1 A' + BB' = A( Q− 1 + C' C − L' Lγ− 2 )−1 A' + BB' Therefore, the analysis Riccati equation admits a positive semidefinite γ2 I − LXL ' = γ2 I − LQL ' LQC ' ( I + CQC ' )−1 CQL ' > 0 , so that the feasibility condition is satisfied. Also the stability condition (3) is satisfied. Hence the norm is less than γ.

solution

with

Assume now that there exists Q≥0 satisfying Q=A[Q-1 +C'C-L'Lγ-2 ]-1 +BB', and such that Q-1 -L'Lγ-2 >0 and A(Q1 +C'C-L'Lγ-2 )-1 Q-1 stable. Take a predictor defined by F =A(Q-1 +C'C-L'Lγ-2 )-1 C'. From (4) it results that 3 F3=AYC'(I+CYC')-1 and hence Q=AYA'+BB'-AYC'(I+CYC')-1 CYA'=A(Q-1 +C'C-L'Lγ-2 )-1 A'+BB'. The analysis Riccati equation admits a positive semidefinite solution and Q-1 -L'Lγ-2 >0 coincides with the feasibility condition (5) for the predictor. Also the stability condition (6) is satisfied. Hence the norm is less than γ. Vice-versa, assume that there exists a filter such that the norm is less than γ. Then, there exists Q solving (1)-(3). On the other hand, X=(Q-1 +C'C)-1 +(K-(Q-1 +C'C)-1 C')(I+CQC')(K-(Q-1 +C'C)-1 C')'≥(Q-1 +C'C)-1 and hence X-1≤Q-1 +C'C implies that Q solves the inequality Q≥A(Q-1 '+C'C-L'Lγ-2 )-1 A’+BB' whereas γ2 I-LXL'>0 with X>0 implies that Q-1 +C'C-L'Lγ-2 >0. The existence of Q solving Q=A(Q-1 +C'C-L'Lγ-2 )-1 A'+BB' with Q-1 A'+C'C-L'Lγ-2 >0 and satisfying the stability condition follows form standard monotonicity results of the solutions of the Riccati equation.

Finally, assume that there exists a predictor such that the norm is less than Then, there exists Q solving (4)-(6). Hence Q≥A(Q-1 A'+C'C-L'Lγ-2 )-1 A+BB' with γ2 I-LQL'>0 The existence of Q solving Q=A(Q-1 A'+C'C-L'Lγ-2 )1 A'+BB' with γ2 I-LQL'>0 and satisfying the stability condition follows form standard monotonicity results of the solutions of the Riccati equation

J-SPECTRAL FACTORIZATION The Riccati equation underlying the problem solution can be compactly written as:

[

][

][

~ ~ ~ ~ −1 ~ ~ Q = AQA '+ BB '− AQ C '+ BD ' R + C QC ' C QA'+ DB '

]

0  ~ C  ~ D   DD' C =  , D =  , R =  − γ 2 I   L 0   0 Let define: H 1 ( z ) = D + C ( zI − A) −1 B H 2 ( z ) = L( zI − A) −1 B

 H ( z ) 0 H ( z) =  1  H 2 (z) I 

0  I J = 2  0 − γ I 

Theorem 6 Assume that A is stable. There exists a stable filter F(z) such that H2 (z)-F(z)H1 (z) is stable and H 2 ( z ) − F ( z )H 1 ( z)

stable

H 2 ( z ) − F ( z ) H1 ( z ) ∞ < γ if and only if there exists a square Ω(z), stable with stable inverse, with stable [Ω(z)-1]11, solving the factorization problem H ( z ) JH ( z −1 )' = Ω ( z ) JΩ ( z −1 )' All feasible filters can be parametrized in the following way: F ( z ) = (Ω 21 ( z ) − Ω 22 ( z )U ( z ))(Ω 11 ( z ) − Ω12 ( z )U ( z )) −1 , U ( z ) ∈ H ∞ , U ( z )

∞

0, if, for some ε :

{[ [

]]

~ ~ ~ ~ 0 < Q(1) < Qs + Y − C ' ( R + CQs C ' ) −1 C + + + εI

}

−1

(ii) The solution Q(t) of the difference Riccati equation is feasible and converges to the stabilizing solution Qs if 0 < Q( 1 ) < Qs + Y + εI

−1

____________________________________________________________ It is possible to see that the conditions introduced in the theorem are also necessary in two significant cases: L=0 (γ→∞) and C=0. In the first case Y=0 and therefore one recovers the condition of convergence for the Kalman filter (Q(1)>0), whereas, in the second case the condition of convergence is equivalent to S(1) = Q(1)-1+C'C- L'Lγ-2>Sa, where Sa is the antistabilizing solution of S=(AS-1A'+BB')-1+C'C-L'Lγ-2

γ-SWITCHING If Q(1) is "too large", nothing can be said on the convergence of Q(t,γs) towards the stabilizing solution Qs(γ) associated with a given value of γ=γs. The convergence condition is: 0 < Q( 1 ) < Qs ( γ ) + Y ( γ ) + εI

−1

.

Theorem 11 (i) The solution Y(γ) of the Lyapunov equation goes to zero for γ→∞ (ii) The stabilizingsolution Qs(γ) of the Riccati equation increases as γ increases Thanks to point (i) of the theorem, if Q(1) does not satisfy the convergence condition, one can find a value of γ=γ1>γs such that the convergence condition is satisfied for γ=γ1. The solution of the Riccati equation with γ=γ1 converges to Qs(γ1). On the other hand, thanks to point (ii) ot the theorem, one can say that Qs(γ1)≤ Qs(γs)< Qs(γs)+(Y(γs)-1+εI)-1 So that ε>0 and tε>0 are such that Q(tε,γ1)< Qs(γs)+(Y(γs)-1+εI)-1.. Finally, imposing γ , 1 ≤ t < tε γ (t ) =  1 γ s , t ≥ tε

the solution Q(t,γ(t)) tends a Qs(γs) as t→∞. _______________________________________________________ The γ-switching strategy modifies the cost: T r (t ) − rˆ(t ) ~ JF = ∑ γ 2 (t ) t =1

2

T  2 − ∑ w(t ) + ( x(0) − xˆ 0 )' Ψ ( x(0) − xˆ 0 )  t =0 

EXAMPLE A=0.5, B=[1 0], C=3, D=[0 1], L=2, T=1/12 Q( t + 1 ) = 0 .25

1 + 1 , Q( 1 ) = 4 Q( t )−1 + 9 − 4 γ −2

Feasibility condition γ2 Q( t ) < 4 − 9γ 2 Algebraic Riccati equation Q = 0 .25

1 +1 Q −1 + 9 − 4 γ −2

→

there existse Qs (γ) for γ> 0.6576

Choice: γs=0.66 Hence, Qs (γs)=1.5318 is the stabilizing solution Q(1) < 5.4724 feasibility condition Q(1)