Linear Matrix Inequalities in Control

Transcription

1 Linear Matrix Inequalities in Control Guido Herrmann University of Leicester Linear Matrix Inequalities in Control p. 1/43

2 Presentation 1. Introduction and some simple examples 2. Fundamental Properties and Basic Structure of Linear Matrix Inequalities (LMIs) 3. LMI-problems 4. Tricks in Matrix Inequalities - Approaches to create LMIs from Matrix Inequalities (a) (b) (c) (d) (e) Congruence Transformation Change of Variables Projection Lemma S-procedure Schur Complement 5. Examples (L 2 -gain computation, non-linearities, etc) 6. Conclusions Linear Matrix Inequalities in Control p. 2/43

3 A linear system Introduction - A Simple Example ẋ = Ax is stable if and only if there is a positive definite P for and V(x) = x T Px (i.e. V(x) > 0 f or x 0) x T PAx+x T A T Px < 0 x 0 Linear Matrix Inequalities in Control p. 3/43

4 A linear system Introduction - A Simple Example ẋ = Ax is stable if and only if there is a positive definite P for V(x) = x T Px (i.e. V(x) > 0 f or x 0) and x T PAx+x T A T Px < 0 x 0 The two matrix inequalities involved here are and PA+A T P < 0 P > 0. Linear Matrix Inequalities in Control p. 3/43

5 A linear system Introduction - A Simple Example ẋ = Ax is stable if and only if there is a positive definite P for V(x) = x T Px (i.e. V(x) > 0 f or x 0) and x T PAx+x T A T Px < 0 x 0 The two matrix inequalities involved here are and PA+A T P < 0 P > 0. The matrix problem here is to find P so that these inequalities are satisfied. The inequalities are linear in P. Linear Matrix Inequalities in Control p. 3/43

6 Introduction - LQR-optimal control We would like to compute a state feedback controller u = Kx controlling with an initial condition of x(0) = x 0. The cost function J = 0 ẋ = Ax+Bu (x T Qx+u T Ru)dt is to be minimized. We know that the solution to this problem is K = R 1 B T P, A T P+PA+PBR 1 B T P+Q = 0 and J = min (x T Qx+u T Ru)dt = x T u 0 Px 0. 0 How can we express this problem in terms of an LMI? Linear Matrix Inequalities in Control p. 4/43

7 Introduction In control the requirements for controller design are usually 1. Closed Loop Stability 2. Robustness 3. Performance 4. Robust Performance Control design requirements are usually best encoded in form of an optimization criterion (e.g. robustness in terms of L 2 /small gain-requirements, performance via linear quadratic control, H -requirements etc.) Linear Matrix Inequalities in Control p. 5/43

8 Introduction We have seen that stability of a linear autonomous system can be easily expressed via a linear matrix inequality We will see that linear quadratic control problems can be expressed in terms of LMIs L 2 or H analysis/design problems can be expressed as LMI-problems Some classes of nonlinearities are easily captured via matrix inequalities This creates a synergy which allows to express a control design problem via different seemingly contradictive requirements For LMIs, very reliable numerical solution tools are available Linear Matrix Inequalities in Control p. 6/43

9 Fundamental LMI properties A matrix Q is defined to be positive definite if it is symmetric and This is signified by x T Qx > 0 x 0 Q > 0 Linear Matrix Inequalities in Control p. 7/43

10 Fundamental LMI properties A matrix Q is defined to be positive definite if it is symmetric and This is signified by x T Qx > 0 x 0 Q > 0 Likewise, Q is said to be positive semi-definite if it is symmetric and x T Qx 0 x thus Q 0 Linear Matrix Inequalities in Control p. 7/43

11 Fundamental LMI properties A matrix Q is defined to be positive definite if it is symmetric and This is signified by x T Qx > 0 x 0 Q > 0 Likewise, Q is said to be positive semi-definite if it is symmetric and x T Qx 0 x thus Q 0 A matrix Q is negative definite if it is symmetric and x T Qx < 0 x 0 thus Q < 0 or negative semi-definite if it is symmetric and x T Qx 0 x 0 thus Q 0 Linear Matrix Inequalities in Control p. 7/43

12 The Basic Structure of an LMI Any linear matrix inequality (LMI) can be easily rewritten as F(v) = F 0 + m i=1 v i F i > 0 where v R m is a variable and F 0,F i are given constant symmetric matrices. This matrix inequality is linear in the variables v i. Linear Matrix Inequalities in Control p. 8/43

13 The Basic Structure of an LMI Any linear matrix inequality (LMI) can be easily rewritten as F(v) = F 0 + m i=1 v i F i > 0 where v R m is a variable and F 0,F i are given constant symmetric matrices. This matrix inequality is linear in the variables v i. For instance for the simple linear matrix inequality in the symmetric P PA+A T P < 0 the variables v R m are defined via P R n n. Hence, m = n(n+1) 2 in this case! Linear Matrix Inequalities in Control p. 8/43

14 The Basic Structure of an LMI Another very generic way of writing down an LMI is F(V 1,V 2,...,V n ) = F 0 + G 1 V 1 H 1 + G 2 V 2 H = F 0 + n i=1 G i V i H i > 0 where the unstructured V i R q i p i are matrix variables, n i=1 q i p i = m. We seek to find V i as they are variables. The matrices F 0,G i,h i are given. From now on, we will mainly consider LMIs of this form. Linear Matrix Inequalities in Control p. 9/43

15 System of LMIs A system of LMIs is F 1 (V 1,...,V n ) > 0. > 0 F p (V 1,...,V n ) > 0 where F j (V 1,...,V n ) = F 0 j + This can be easily changed into a single LMI... n i=1 G i j V i H i j Linear Matrix Inequalities in Control p. 10/43

16 System of LMIs Let s define F 0, G i, H i,ṽ i as F 0 = F F F 0p = diag(f 01,...,F 0p ) G i = diag(g i1,...,g ip ) H i = diag(h i1,...,h ip ) Ṽ i = diag(v i,...v i ) Linear Matrix Inequalities in Control p. 11/43

17 System of LMIs Let s define F 0, G i, H i,ṽ i as F 0 = F F F 0p = diag(f 01,...,F 0p ) G i = diag(g i1,...,g ip ) H i = diag(h i1,...,h ip ) Ṽ i = diag(v i,...v i ) We then have the inequality F big (V 1,...,V n ) := F 0 + n i=1 G i Ṽ i H i > 0 which is just one single LMI. But be aware that this time the new variable Ṽ i is structured, i.e. not all elements of Ṽ i are free parameters! Linear Matrix Inequalities in Control p. 11/43

18 Different classes of LMI-problems: Feasibility Problem We seek a feasible solution {V 1,...,V n } such that F(V 1,...,V n ) > 0 We are not interested in the optimality of the solution, only in finding a solution, which satisfies the LMI and may not be unique. Linear Matrix Inequalities in Control p. 12/43

19 Different classes of LMI-problems: Feasibility Problem We seek a feasible solution {V 1,...,V n } such that F(V 1,...,V n ) > 0 We are not interested in the optimality of the solution, only in finding a solution, which satisfies the LMI and may not be unique. Example: A linear system ẋ = Ax is stable if and only if there is a matrix P satisfying and PA+A T P < 0 P > 0. Linear Matrix Inequalities in Control p. 12/43

20 LMI-problems: Linear Objective Minimization Minimization (or maximization) of a linear scalar function, α(.), of the matrix variables V i, subject to LMI constraints: minα(v 1,...,V n ) s.t. F(V }{{} 1,...,V n ) > 0 such that, sub ject to where we have used the abbreviation s.t. to mean such that. Linear Matrix Inequalities in Control p. 13/43

21 LMI-problems: Linear Objective Minimization Minimization (or maximization) of a linear scalar function, α(.), of the matrix variables V i, subject to LMI constraints: minα(v 1,...,V n ) s.t. F(V }{{} 1,...,V n ) > 0 such that, sub ject to where we have used the abbreviation s.t. to mean such that. Example: Calculating the H norm of a linear system. ẋ = Ax+Bw z = Cx+Dw the H norm of the transfer function matrix T zw from w to z is computed by: min γ s.t. A T P+PA PB C T B T P γi D T C D γi < 0, P > 0. The LMI variables are P and γ! The value of γ is unique, P is not. Linear Matrix Inequalities in Control p. 13/43

22 LMI-problems: Generalized eigenvalue problem minλ s.t. F 1 (V 1,...,V n )+λf 2 (V 1,...,V n ) < 0 F 2 (V 1,...,V n ) < 0 F 3 (V 1,...,V n ) < 0 Note that in some cases, a GEVP problem can be reduced to a linear objective minimization problem, through an appropriate change of variables. Linear Matrix Inequalities in Control p. 14/43

23 LMI-problems: Generalized eigenvalue problem minλ s.t. F 1 (V 1,...,V n )+λf 2 (V 1,...,V n ) < 0 F 2 (V 1,...,V n ) < 0 F 3 (V 1,...,V n ) < 0 Note that in some cases, a GEVP problem can be reduced to a linear objective minimization problem, through an appropriate change of variables. Example: Bounding the decay rate of a linear system. The decay rate is the largest α such that x(t) exp( αt)β x(0), β >, x(t) Let s choose the Lyapunov function V(x) = x T Px > 0 and ensure that V(x) 2αV(x). The problem of finding the decay rate could be posed as min α s.t. A T P+PA+2αP < 0, P < 0, Linear Matrix Inequalities in Control p. 14/43

24 LMI-problems: Generalized eigenvalue problem minλ s.t. F 1 (V 1,...,V n )+λf 2 (V 1,...,V n ) < 0 F 2 (V 1,...,V n ) < 0 F 3 (V 1,...,V n ) < 0 Note that in some cases, a GEVP problem can be reduced to a linear objective minimization problem, through an appropriate change of variables. Example: Bounding the decay rate of a linear system. The decay rate is the largest α such that x(t) exp( αt)β x(0), β >, x(t) Let s choose the Lyapunov function V(x) = x T Px > 0 and ensure that V(x) 2αV(x). The problem of finding the decay rate could be posed as min α s.t. A T P+PA+2αP < 0, i.e. F 1 (P) := A T P+PA P < 0, i.e. F 2 (P) := 2P i.e. F 3 (P) := I Linear Matrix Inequalities in Control p. 14/43

25 Tricks: Congruence transformation We know that for Q R n n Q > 0 and a real W R n n such that rank(w) = n, the following inequality holds W QW T > 0 Definiteness of a matrix is invariant under pre and post-multiplication by a full rank real matrix, and its transpose, respectively. Often W is chosen to have a diagonal structure. Linear Matrix Inequalities in Control p. 15/43

26 Tricks: Change of variables By defining new variables, it is sometimes possible to linearise nonlinear MIs Example: State feedback control synthesis Find F such that the eigenvalues of A+BF are in the open left-half complex plane This is equivalent to finding a matrix F and P > for (A+BF) T P+P(A+BF) < 0 or A T P+PA+F T B T P+PBF < 0! Terms with products of F and P are nonlinear or bilinear! Linear Matrix Inequalities in Control p. 16/43

27 Tricks: Change of variables By defining new variables, it is sometimes possible to linearise nonlinear MIs Example: State feedback control synthesis Find F such that the eigenvalues of A+BF are in the open left-half complex plane This is equivalent to finding a matrix F and P > for (A+BF) T P+P(A+BF) < 0 or A T P+PA+F T B T P+PBF < 0! Terms with products of F and P are nonlinear or bilinear! Multiply with Q := P 1 > 0 (A very simple case of congruence transformation): QA T + AQ+QF T B T + BFQ < 0 This is a new matrix inequality in the variables Q > 0 and F (still non-linear). Linear Matrix Inequalities in Control p. 16/43

28 Tricks: Change of variables Define a second new variable L = FQ QA T + AQ+QF T B T + BFQ < 0 QA T + AQ+L T B T + BL < 0 We now have an LMI feasibility problem in the new variables Q > 0 and L. Recovery of F and P by F = LQ 1, P = Q 1. Linear Matrix Inequalities in Control p. 17/43

29 Tricks: Schur complement Schur s formula says that the following statements are equivalent: i. Φ = [ Φ 11 Φ 12 Φ T 12 Φ 22 < 0 ii. Φ 22 < 0 Φ 11 Φ 12 Φ 1 22 ΦT 12 < 0 The main use is to transform quadratic matrix inequalities into linear matrix inequalities. Linear Matrix Inequalities in Control p. 18/43

30 Tricks: Schur complement Schur s formula says that the following statements are equivalent: i. Φ = [ Φ 11 Φ 12 Φ T 12 Φ 22 < 0 ii. Φ 22 < 0 Φ 11 Φ 12 Φ 1 22 ΦT 12 < 0 The main use is to transform quadratic matrix inequalities into linear matrix inequalities. Example: Making a LQR-type quadratic inequality linear (Riccati inequality) A T P+PA+PBR 1 B T P+Q < 0 where P > 0 is the matrix variable and Q,R > 0 are constant. Linear Matrix Inequalities in Control p. 18/43

31 Tricks: Schur complement Schur s formula says that the following statements are equivalent: i. Φ = [ Φ 11 Φ 12 Φ T 12 Φ 22 < 0 ii. Φ 22 < 0 Φ 11 Φ 12 Φ 1 22 ΦT 12 < 0 The main use is to transform quadratic matrix inequalities into linear matrix inequalities. Example: Making a LQR-type quadratic inequality linear (Riccati inequality) A T P+PA+PBR 1 B T P+Q < 0 where P > 0 is the matrix variable and Q,R > 0 are constant. The Riccati inequality can be transformed into [ A T P+PA+Q PB R < 0 Linear Matrix Inequalities in Control p. 18/43

32 Tricks: Schur complement Example: Making a LQR-type quadratic inequality linear (Riccati inequality) A T P+PA+PBR 1 B T P+Q < 0 is equivalent to [ A T P+PA+Q PB R < 0 where P > 0 is the matrix variable and Q,R > 0 are constant. This inequality can be used to minimize the cost function J = 0 (x T Qx+u T Ru)dt for the computation of the state feedback controller u = Kx controlling ẋ = Ax+Bu with an initial condition of x(0) = x 0. Linear Matrix Inequalities in Control p. 19/43

33 Tricks: Schur complement Example: Making a LQR-type quadratic inequality linear (Riccati inequality) A T P+PA+PBR 1 B T P+Q < 0 is equivalent to [ A T P+PA+Q PB R < 0 where P > 0 is the matrix variable and Q,R > 0 are constant. This inequality can be used to minimize the cost function J = 0 (x T Qx+u T Ru)dt for the computation of the state feedback controller u = Kx controlling ẋ = Ax+Bu with an initial condition of x(0) = x 0. We know that the solution to this problem is K = R 1 B T P, A T P+ PA+ PBR 1 B T P+Q = 0 and J = min (x T Qx+u T Ru)dt = x T u 0 Px 0. 0 Linear Matrix Inequalities in Control p. 19/43

34 Tricks: Schur complement The alternative solution to the optimization problem is given by the following LMI-problem: [ minx T 0 Px 0 s.t. A T P+PA+Q PB R < 0 P < 0 for which the optimal controller is given by K = R 1 B T P. Linear Matrix Inequalities in Control p. 20/43

35 Tricks: The S-procedure We would like to guarantee that a single quadratic function of x R m is such that F 0 (x) 0 F 0 (x) := x T A 0 x+2b 0 x+c 0 whenever certain other quadratic functions are positive semi-definite F i (x) 0 F i (x) := x T A i x+2b 0 x+c 0, i {1,2,...,q} Linear Matrix Inequalities in Control p. 21/43

36 Tricks: The S-procedure We would like to guarantee that a single quadratic function of x R m is such that F 0 (x) 0 F 0 (x) := x T A 0 x+2b 0 x+c 0 whenever certain other quadratic functions are positive semi-definite F i (x) 0 F i (x) := x T A i x+2b 0 x+c 0, i {1,2,...,q} Illustration: Consider i = 1. We need to ensure F 0 (x) 0 for all x such that F 1 (x) 0. If there is a scalar, τ > 0, such that F aug (x) := F 0 (x)+τf 1 (x) 0 x s.t.f 1 (x) 0 then our goal is achieved. F aug (x) 0 implies that F 0 (x) 0 if τf 1 (x) 0 because F 0 (x) F aug (x) if F 1 (x) 0. Linear Matrix Inequalities in Control p. 21/43

37 Tricks: The S-procedure F aug (x) := F 0 (x)+τf 1 (x) 0 x s.t.f 1 (x) 0 Extending this idea to q inequality constraints: F 0 (x) 0 whenever F i (x) 0 ( ) holds if F 0 (x)+ q i=1 τ i F i (x) 0, τ i 0 ( ) Linear Matrix Inequalities in Control p. 22/43

38 Tricks: The S-procedure F aug (x) := F 0 (x)+τf 1 (x) 0 x s.t.f 1 (x) 0 Extending this idea to q inequality constraints: F 0 (x) 0 whenever F i (x) 0 ( ) holds if F 0 (x)+ q i=1 τ i F i (x) 0, τ i 0 ( ) The S-procedure is conservative; inequality ( ) implies inequality ( ) Equivalence is only guaranteed for i = 1. The τ i s are usually variables in an LMI problem. Linear Matrix Inequalities in Control p. 22/43

39 Tricks: The Projection Lemma We sometimes encounter inequalities of the form Ψ(V)+G(V)ΛH T (V)+H(V)Λ T G T (V) < 0 ( ) where V and Λ are the matrix variables, Λ is an unstructured matrix variable. Ψ(.),G(.),H(.) are (normally affine) functions of V. Linear Matrix Inequalities in Control p. 23/43

40 Tricks: The Projection Lemma We sometimes encounter inequalities of the form Ψ(V)+G(V)ΛH T (V)+H(V)Λ T G T (V) < 0 ( ) where V and Λ are the matrix variables, Λ is an unstructured matrix variable. Ψ(.),G(.),H(.) are (normally affine) functions of V. Inequality ( ) is satisfied for some V if and only if { W T G(V) Ψ(V)W G(V) < 0 W T H(V) Ψ(V)W H(V) < 0 Linear Matrix Inequalities in Control p. 23/43

41 Tricks: The Projection Lemma We sometimes encounter inequalities of the form Ψ(V)+G(V)ΛH T (V)+H(V)Λ T G T (V) < 0 ( ) where V and Λ are the matrix variables, Λ is an unstructured matrix variable. Ψ(.),G(.),H(.) are (normally affine) functions of V. Inequality ( ) is satisfied for some V if and only if { W T G(V) Ψ(V)W G(V) < 0 W T H(V) Ψ(V)W H(V) < 0 where W G(V) and W H(V) are the orthogonal complements of G(V) and H(V), i.e. W G(V) G(V) = 0 W H(V) H(V) = 0. and [WG(V) T G(V), [W H(V) T H(V) are both full rank. Linear Matrix Inequalities in Control p. 23/43

42 Tricks: The Projection Lemma The main point is that we can transform a matrix inequality which is a function of two variables, V and Λ, into two inequalities which are functions of one variable: (i) (ii) It can facilitate the derivation of an LMI. There are less variables for computation. Linear Matrix Inequalities in Control p. 24/43

43 Tricks: The Projection Lemma The main point is that we can transform a matrix inequality which is a function of two variables, V and Λ, into two inequalities which are functions of one variable: (i) (ii) It can facilitate the derivation of an LMI. There are less variables for computation. It is often the approach is to solve for V using { W T G(V) Ψ(V)W G(V) < 0 W T H(V) Ψ(V)W H(V) < 0 and then for Λ using Ψ(V)+G(V)ΛH T (V)+H(V)Λ T G T (V) < 0 Note that this can be numerically unreliable!! Linear Matrix Inequalities in Control p. 24/43

44 Examples: L 2 gain- Continuous-time systems Linear systems: The H norm is equivalent to the maximum RMS (Root-Mean-Square) energy gain, the H -gain of a linear system, the L 2 gain. Linear Matrix Inequalities in Control p. 25/43

45 Examples: L 2 gain- Continuous-time systems Linear systems: The H norm is equivalent to the maximum RMS (Root-Mean-Square) energy gain, the H -gain of a linear system, the L 2 gain. A system with input w(t) and output z(t) is said to have an L 2 gain of γ if where w 2 = t=0 w (t)w(t)dt. z 2 < γ w 2 + β, β > 0 The L 2 gain is a measure of the output relative to the size of its input. Linear Matrix Inequalities in Control p. 25/43

46 Examples: L 2 gain- Continuous-time systems Linear systems: The H norm is equivalent to the maximum RMS (Root-Mean-Square) energy gain, the H -gain of a linear system, the L 2 gain. A system with input w(t) and output z(t) is said to have an L 2 gain of γ if where w 2 = t=0 w (t)w(t)dt. z 2 < γ w 2 + β, β > 0 The L 2 gain is a measure of the output relative to the size of its input. The H norm of ẋ = Ax+Bw, z = Cx+Dw is given by: min γ s.t. A T P+PA PB C T B T P γi D T C D γi < 0, P > 0. Linear Matrix Inequalities in Control p. 25/43

47 Examples: L 2 gain- Continuous-time systems A T P+PA PB C T B T P γi D T C D γi < 0, P > 0. The Schur complement gives [ A T P+PA+ 1 γ CT C PB+ 1 γ CT D B T P+ 1 γ DT C γi + 1 γ DT D Linear Matrix Inequalities in Control p. 26/43

48 Examples: L 2 gain- Continuous-time systems A T P+PA PB C T B T P γi D T C D γi < 0, P > 0. The Schur complement gives [ A T P+PA+ 1 γ CT C PB+ 1 γ CT D B T P+ 1 γ DT C γi + 1 γ DT D In terms of [ x T w T T, it follows that we need to find the minimum of γ so that [ x w T [ A T P+PA+ 1 γ CT C PB+ 1 γ CT D B T P+ 1 γ DT C γi + 1 γ DT D [ x w < 0, [ x T w T T 0 Linear Matrix Inequalities in Control p. 26/43

49 Examples: L 2 gain- Continuous-time systems A T P+PA PB C T B T P γi D T C D γi < 0, P > 0. The Schur complement gives [ A T P+PA+ 1 γ CT C PB+ 1 γ CT D B T P+ 1 γ DT C γi + 1 γ DT D In terms of [ x T w T T, it follows that we need to find the minimum of γ so that [ x w T [ A T P+PA+ 1 γ CT C PB+ 1 γ CT D B T P+ 1 γ DT C γi + 1 γ DT D [ x w < 0, [ x T w T T 0 or x T A T Px+x T PAx+ 1 γ xt C T Cx+x T (PB+ 1 γ CT D)w+w T (B T P+ 1 γ DT C)x+w T 1 γ DT Dw γw T w = x T A T Px+x T PA T x+2x T PBw+ 1 γ zt z γw T w < 0 Linear Matrix Inequalities in Control p. 26/43

50 Examples: L 2 gain- Continuous-time systems Defining V = x T Px V = x T A T Px+x T PAx+2x T PBw Thus, we require: x T A T Px+x T PA T x+2x T PBw+ 1 γ zt z γw T w = V + 1 γ zt z γw T w < 0 Linear Matrix Inequalities in Control p. 27/43

51 Examples: L 2 gain- Continuous-time systems Defining V = x T Px V = x T A T Px+x T PAx+2x T PBw Thus, we require: x T A T Px+x T PA T x+2x T PBw+ 1 γ zt z γw T w = V + 1 γ zt z γw T w < 0 and integration in the interval [0, ) implies V(t = ) V(t = 0) + t=0 t=0 t=0 z T (s)z(s)ds < t=0 t=0 1 γ zt (s)z(s)ds t=0 γw T (s)w(s)ds < 0 γ 2 w T (s)w(s)ds+γ(v(t = 0) V(t = )) z T (s)z(s)ds < γ 2 w T (s)w(s)ds+γv(t = 0) t=0 z T (s)z(s)ds < γ w T (s)w(s)ds+ γv(t = 0) t=0 Linear Matrix Inequalities in Control p. 27/43

52 Examples: L 2 gain- Continuous-time systems 1 V(t = ) V(t = 0) + t=0 γ zt (s)z(s)ds γw T (s)w(s)ds < 0 t=0 z T (s)z(s)ds < γ w T (s)w(s)ds+ γv(t = 0) t=0 t=0 z 2 < γ w 2 + β }{{} γv(t=0) Thus, the linear system (A,B,C,D) has indeed an L 2 gain γ. Linear Matrix Inequalities in Control p. 28/43

53 Examples: Discrete-time systems A linear discrete system is asymptotically stable if and only if there is x(k+1) = Ax(k) V(x) = x T Px, P > 0. and V(x(k+1)) V(x(k)) = V(x(k+1)) = x T (k)a T PAx(k) x T (k)px(k) < 0 x(k) 0 or A T PA P < 0. Linear Matrix Inequalities in Control p. 29/43

54 Examples: l 2 gain- Discrete-time systems A system with input w(t) and output z(t) is said to have an L 2 gain of γ if z 2 < γ w 2 + β, β > 0 where w 2 = k=0 wt (k)w(k). Linear Matrix Inequalities in Control p. 30/43

55 Examples: l 2 gain- Discrete-time systems A system with input w(t) and output z(t) is said to have an L 2 gain of γ if z 2 < γ w 2 + β, β > 0 where w 2 = k=0 wt (k)w(k). For linear systems x(k+1) = Ax(k)+Bw(k) y = Cx(k)+Dw(k) the value of the finite l 2 -gain, γ, (H norm; the maximum RMS energy gain) is: min γ [ A T PA P+ 1 γ CT C A T PB+ 1 γ CT D 1 γ DT C γi + B T PB+ 1 γ DT D s.t. < 0 P < 0 for P > 0. Linear Matrix Inequalities in Control p. 30/43

56 Examples: l 2 gain- Discrete-time systems The l 2 gain relationship readily follows for V = x T Px from: V(x(k+1)) + 1 γ yt (k)y(k) γw T (k)w(k) < 0 k=0 V(x(k+1)) } {{ } V(x( )) V(x(0)) + 1 γ k=0 y T (k)y(k) γ k=0 w T (k)w(k) < 0 Linear Matrix Inequalities in Control p. 31/43

57 Examples: l 2 gain- Discrete-time systems The l 2 gain relationship readily follows for V = x T Px from: V(x(k+1)) + 1 γ yt (k)y(k) γw T (k)w(k) < 0 k=0 V(x(k+1)) } {{ } V(x( )) V(x(0)) + 1 γ k=0 y T (k)y(k) γ k=0 w T (k)w(k) < 0 Problem : The matrix inequality [ A T PA P+ 1 γ CT C A T PB+ 1 γ CT D 1 γ DT C γi + B T PB+ 1 γ DT D < 0 is not linear. P > 0 and γ > 0 are variables. Linear Matrix Inequalities in Control p. 31/43

58 Examples: l 2 gain- Discrete-time systems The l 2 gain relationship readily follows for V = x T Px from: V(x(k+1)) + 1 γ yt (k)y(k) γw T (k)w(k) < 0 k=0 V(x(k+1)) } {{ } V(x( )) V(x(0)) + 1 γ k=0 y T (k)y(k) γ k=0 w T (k)w(k) < 0 Problem : The matrix inequality [ A T PA P+ 1 γ CT C A T PB+ 1 γ CT D 1 γ DT C γi + B T PB+ 1 γ DT D < 0 is not linear. P > 0 and γ > 0 are variables. The Schur Complement implies A T PA P A T PB C T B T PA γi + B T PB D T C D γi < 0 Linear Matrix Inequalities in Control p. 31/43

59 Examples: l 2 gain- Discrete-time systems [ A T PA P+ 1 γ CT C A T PB+ 1 γ CT D 1 γ DT C γi + B T PB+ 1 γ DT D Congruence transformation & Change of variable approach: < 0 [ γa T PA γp+c T C D T C γa T PB+C T D γ 2 I + γb T PB+D T D < 0 Defining Q = Pγ and µ = γ 2 : [ A T QA Q+C T C D T C min µ s.t. A T QB+C T D µi + B T QB+D T D < 0 Q < 0 Q > 0 and the scalar µ > 0 are variables. The l 2 -gain is readily computed with γ = µ. Linear Matrix Inequalities in Control p. 32/43

60 Examples: Sector boundedness The saturation function is defined as sat(u) = [sat 1 (u 1 ),...,sat 2 (u m ) T and sat i (u i ) = sign(u i ) min{ u i,ū i }, ū i > 0 i {1,...,m} ū i is the i th saturation limit Linear Matrix Inequalities in Control p. 33/43

61 Examples: Sector boundedness The saturation function is defined as sat(u) = [sat 1 (u 1 ),...,sat 2 (u m ) T and sat i (u i ) = sign(u i ) min{ u i,ū i }, ū i > 0 i {1,...,m} ū i is the i th saturation limit It is easy to verify that the saturation function, sat i (u i ) satisfies the following inequality u i sat i (u i ) sat 2 i (u i ) Linear Matrix Inequalities in Control p. 33/43

62 Examples: Sector boundedness It is easy to verify that the saturation function, sat i (u i ) satisfies the following inequality u i sat i (u i ) sat 2 i (u i ) or sat i (u i )[u i sat i (u i )w i 0 for some w i > 0. Linear Matrix Inequalities in Control p. 34/43

63 Examples: Sector boundedness It is easy to verify that the saturation function, sat i (u i ) satisfies the following inequality u i sat i (u i ) sat 2 i (u i ) or sat i (u i )[u i sat i (u i )w i 0 for some w i > 0. We can write for some diagonal W = diag(w 1,w 2,...) > 0. sat(u) T W[u sat(u) 0 This inequality can be easily used in an S procedure approach where the elements of the diagonal matrix W act as free parameters if necessary/possible. Linear Matrix Inequalities in Control p. 34/43

64 Examples: A slightly more detailed example Consider the L 2 -gain for the SISO-system with saturated input signal u: ẋ = Ax+bsat(u), x R n and a limited measurement range of the output y: y = sat(cx+dsat(u)). Linear Matrix Inequalities in Control p. 35/43

65 Examples: A slightly more detailed example Consider the L 2 -gain for the SISO-system with saturated input signal u: ẋ = Ax+bsat(u), x R n and a limited measurement range of the output y: y = sat(cx+dsat(u)). The limits at the actuator inputs u can be due to mechanical limits (e.g. valves) or due to digital-to-analogue converter voltage signal limits. Output signals can be constrained due to sensor voltage range limits or simply by analogue-to-digital converter limits. The analysis of such systems is vital to practical control systems and will be pursued in greater detail later. We may consider here an L 2 gain analysis. Linear Matrix Inequalities in Control p. 35/43

66 Examples: A slightly more detailed example We may define Hence, it follows s = sat(u). For the output signal y: sw 1 (u s) 0, w 1 > 0 yw 2 (cx+ds y) 0, w 2 > 0 Linear Matrix Inequalities in Control p. 36/43

67 Examples: A slightly more detailed example We may define Hence, it follows s = sat(u). For the output signal y: sw 1 (u s) 0, w 1 > 0 We know that from yw 2 (cx+ds y) 0, w 2 > 0 V + 1 γ y2 γu 2 0 follows that our system has the L 2 -gain γ. We have to consider the two saturation nonlinearities! Linear Matrix Inequalities in Control p. 36/43

68 Examples: A slightly more detailed example With the S-procedure V + 1 γ y2 γu 2 + 2sw 1 (u s)+2yw 2 (cx+ds y) < 0 f or [ x T u y 0 the system has also an L 2 -gain of γ. The expression V implies: x T A T Px+xPAx+x T Pbs+sb T Px + 1 γ y2 γu 2 + 2sw 1 (u s)+2yw 2 (cx+ds y) 0 Linear Matrix Inequalities in Control p. 37/43

69 Examples: A slightly more detailed example With the S-procedure V + 1 γ y2 γu 2 + 2sw 1 (u s)+2yw 2 (cx+ds y) < 0 f or [ x T u y 0 the system has also an L 2 -gain of γ. The expression V implies: x T A T Px+xPAx+x T Pbs+sb T Px + 1 γ y2 γu 2 + 2sw 1 (u s)+2yw 2 (cx+ds y) 0 Rewriting: x s u y T A T P+PA Pb 0 c T w 2 b T P 2w 1 w 1 dw 2 0 w 1 γ 0 w 2 c w 2 d 0 2w γ x s u y < 0 for [ x T s u y 0. Linear Matrix Inequalities in Control p. 37/43

70 Examples: A slightly more detailed example This is equivalent to A T P+PA Pb 0 c T w 2 b T P 2w 1 w 1 dw 2 0 w 1 γ 0 w 2 c w 2 d 0 2w γ < 0. We would like to minimize γ, while P, w 1, w 2 are variables. Not an LMI! Linear Matrix Inequalities in Control p. 38/43

71 Examples: A slightly more detailed example This is equivalent to A T P+PA Pb 0 c T w 2 b T P 2w 1 w 1 dw 2 0 w 1 γ 0 w 2 c w 2 d 0 2w γ < 0. We would like to minimize γ, while P, w 1, w 2 are variables. Not an LMI! Using Projection Lemma twice, we can derive a significantly simpler matrix inequality which delivers the L 2 -gain. First Step: A T P+PA Pb 0 c T w 2 b T P 0 0 dw γ 0 + w 2 c w 2 d 0 2w γ 0 1 [ 0 w [ w < 0. Linear Matrix Inequalities in Control p. 38/43

72 Examples: A slightly more detailed example First Step: A T P+PA Pb 0 c T w 2 b T P 0 0 dw γ 0 + w 2 c w 2 d 0 2w γ 0 1 [ 0 w [ w < 0. Defining the matrices g 1 = , h 1 = , Ψ 1 = A T P+PA Pb 0 c T w 2 b T P 0 0 dw γ 0 w 2 c w 2 d 0 2w γ, allows us to write Ψ 1 + g 1 w 1 h T 1 + h 1w T 1 gt 1 < 0. Linear Matrix Inequalities in Control p. 39/43

73 Examples: A slightly more detailed example g 1 = , h 1 = , Ψ 1 = A T P+PA Pb 0 c T w 2 b T P 0 0 dw γ 0 w 2 c w 2 d 0 2w γ, The null space matrices W g1 and W h1 satisfy [ W T g1 g 1 & [W T h 1 h 1 f ull rank; W g1 g 1 = 0, W h1 h 1 = 0 Linear Matrix Inequalities in Control p. 40/43

74 Examples: A slightly more detailed example g 1 = , h 1 = , Ψ 1 = A T P+PA Pb 0 c T w 2 b T P 0 0 dw γ 0 w 2 c w 2 d 0 2w γ, The null space matrices W g1 and W h1 satisfy [ W T g1 g 1 & [W T h 1 h 1 f ull rank; W g1 g 1 = 0, W h1 h 1 = 0 Hence, W g1 = I , W h1 = I Linear Matrix Inequalities in Control p. 40/43

75 Examples: A slightly more detailed example g 1 = , h 1 = , Ψ 1 = A T P+PA Pb 0 c T w 2 b T P 0 0 dw γ 0 w 2 c w 2 d 0 2w γ, The null space matrices W g1 and W h1 satisfy [ W T g1 g 1 & [W T h 1 h 1 f ull rank; W g1 g 1 = 0, W h1 h 1 = 0 Hence, W g1 = I , W h1 = I Hence, it follows W g1 Ψ 1 W T g 1 = A T P+PA 0 c T w 2 0 γ 0 w 2 c 0 2w γ, W h1 Ψ 1 Wh T 1 = A T P+PA Pb c T w 2 b T P γ dw 2 w 2 c w 2 d 2w γ Linear Matrix Inequalities in Control p. 40/43

76 Examples: A slightly more detailed example W g1 Ψ 1 W T g 1 = A T P+PA 0 c T w 2 0 γ 0 w 2 c 0 2w γ, W h1 Ψ 1 Wh T 1 = A T P+PA Pb c T w 2 b T P γ dw 2 w 2 c w 2 d 2w γ If W h1 Ψ 1 W T h 1 < 0 then also W g1 Ψ 1 W T g 1 < 0 (easily seen from a further analysis using the Projection lemma). We may carry on investigating W h1 Ψ 1 W T h 1 only Linear Matrix Inequalities in Control p. 41/43

77 Examples: A slightly more detailed example W g1 Ψ 1 W T g 1 = A T P+PA 0 c T w 2 0 γ 0 w 2 c 0 2w γ, W h1 Ψ 1 Wh T 1 = A T P+PA Pb c T w 2 b T P γ dw 2 w 2 c w 2 d 2w γ If W h1 Ψ 1 W T h 1 < 0 then also W g1 Ψ 1 W T g 1 < 0 (easily seen from a further analysis using the Projection lemma). We may carry on investigating W h1 Ψ 1 W T h 1 only W h1 Ψ 1 W T h 1 = Ψ 2 + g 2 w 2 h T 2 + h 2w 2 g T 2 where g 2 = 0 0 1, h 2 = c T d 1, Ψ 2 = A T P+PA Pb 0 b T P γ γ This allows us to derive the null space matrices W g2 and W h2 for g 2 and h 2 W g2 = [ I , W h2 = [ I 0 c T 0 1 d Linear Matrix Inequalities in Control p. 41/43

78 Examples: A slightly more detailed example W g2 = [ I , W h2 = [ I 0 c T 0 1 d, Ψ 2 = A T P+PA Pb 0 b T P γ γ Thus, W g2 Ψ 2 W T g 2 = [ A T P+PA b T P Pb γ, W h2 Ψ 2 W T h 2 = [ A T P+PA+ ct c γ b T P+ dc γ Pb+ ct d γ γ + d2 γ. Linear Matrix Inequalities in Control p. 42/43

79 Examples: A slightly more detailed example W g2 = [ I , W h2 = [ I 0 c T 0 1 d, Ψ 2 = A T P+PA Pb 0 b T P γ γ Thus, W g2 Ψ 2 W T g 2 = [ A T P+PA b T P Pb γ, W h2 Ψ 2 W T h 2 = [ A T P+PA+ ct c γ b T P+ dc γ Pb+ ct d γ γ + d2 γ. W g2 Ψ 2 W T g 2 < 0 is always satisfied if W h2 Ψ 2 W T h 2. Hence, the L 2 gain is computed using [ A T P+PA+ ct c γ b T P+ dc γ Pb+ ct d γ γ + d2 γ < 0, P > 0 The L 2 -gain of the linear system (A,b,c,d) is an upper bound of the non-linear operator. The L 2 -gain of the non-linear and linear operator are identical. Linear Matrix Inequalities in Control p. 42/43

80 Summary Matrix inequalities have shown to be versatile tool to 1. represent L 2, H, linear quadratic performance constraints, H 2 etc. 2. analyze linear parameter varying systems, mild non-linear systems 3. combine several analysis problems in one frame work Matrix inequalities can often be transformed into linear matrix inequalities by congruence transformation, change of variable approach, etc. Existence of a large variety of powerful tools for solving LMIs (semi-definite programming) LMIs have become a standard tool in the analysis and controller design of linear and non-linear control systems Linear Matrix Inequalities in Control p. 43/43