Model reduction for linear systems by balancing

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1 Model reduction for linear systems by balancing Bart Besselink Jan C. Willems Center for Systems and Control Johann Bernoulli Institute for Mathematics and Computer Science University of Groningen, Groningen, the Netherlands 30 November 2017

2 Outline Main topics Problem formulation Observability Gramian Controllability Gramian Balanced truncation Balanced singular perturbation Example Conclusions More advanced topics Hankel operator Fundamental lower bound Positive real balancing Conclusions 2/36

3 Linear systems u Σ y State-space representation with { ẋ = Ax + Bu Σ : y = Cx + Du x R n, u R m, y R p, m, p n 3/36

4 Linear systems u Σ y State-space representation with { ẋ = Ax + Bu Σ : y = Cx + Du x R n, u R m, y R p, m, p n Transfer function (input-output representation) G(s) = C(sI A) 1 B + D 3/36

5 Problem statement Problem. Given a dynamical system Σ, nd a reduced-order system ˆΣ that approximates its input-output behavior u Σ y u ˆΣ ŷ Σ : { ẋ = Ax + Bu y = Cx + Du with x R n, n large G(s) = C(sI A) 1 B + D ˆΣ k : { ξ = Âξ + ˆBu ŷ = Ĉξ + ˆDu with ξ R k, k < n small Ĝ k (s) = Ĉ(sI Â) 1 ˆB + ˆD 4/36

6 Problem statement Problem. Given a dynamical system Σ, nd a reduced-order system ˆΣ that approximates its input-output behavior u Σ y u ˆΣ ŷ Σ : { ẋ = Ax + Bu y = Cx + Du with x R n, n large ˆΣ k : { ξ = Âξ + ˆBu ŷ = Ĉξ + ˆDu with ξ R k, k < n small Objectives 1. Preservation of stability 4/36

7 Problem statement Problem. Given a dynamical system Σ, nd a reduced-order system ˆΣ that approximates its input-output behavior u Σ ˆΣ + y ŷ e Σ : { ẋ = Ax + Bu y = Cx + Du with x R n, n large ˆΣ k : { ξ = Âξ + ˆBu ŷ = Ĉξ + ˆDu with ξ R k, k < n small Objectives 1. Preservation of stability 2. Error e = y ŷ "small" 4/36

8 Signal and system norms u Σ y Signal norm: L 2 signal norm for x : [0, ) R n x L2 = x(t) 2 dt, with x(t) 2 = x T (t)x(t) 0 5/36

9 Signal and system norms u Σ y Signal norm: L 2 signal norm for x : [0, ) R n x L2 = x(t) 2 dt, with x(t) 2 = x T (t)x(t) 0 System norm: L 2 gain Σ L2 = y L2 sup (for x 0 = 0) u L 2, u 0 u L2 5/36

10 Signal and system norms u Σ y System norm: L 2 gain Σ L2 = y L2 sup (for x 0 = 0) u L 2, u 0 u L2 5/36

11 Signal and system norms u Σ y System norm: L 2 gain Properties Σ L2 = y L2 sup (for x 0 = 0) u L 2, u 0 u L2 Let γ Σ L2. For any input function u, it holds that y L2 γ u L2 For stable Σ, the L 2 gain equals the H norm of G, i.e., ( ) Σ L2 = G H = sup σ max G(jω) ω R 5/36

12 Objectives u Σ ˆΣ + y ŷ e Objectives 1. Preservation of stability 2. Bound ε on the L 2 gain of the reduction error e = y ŷ, i.e., y ŷ L2 ε u L2 Equivalently, G Ĝ H ε 6/36

13 Observability energy function x 0 Σ y Observability energy function L o (x 0 ) = 0 y(t) 2 dt, x(0) = x 0, u(t) = 0 t 0 The observability energy function gives the energy associated by observing the output of Σ for initial condition x 0 7/36

14 Observability energy function x 0 Σ y Observability energy function L o (x 0 ) = 0 y(t) 2 dt, x(0) = x 0, u(t) = 0 t 0 The observability energy function gives the energy associated by observing the output of Σ for initial condition x 0 Theorem. If A is Hurwitz, then L o (x 0 ) = x T 0 Qx 0, Q = 0 e ATt C T Ce At dt 7/36

15 Observability Gramian Q = Properties 0 e ATt C T Ce At dt If A is Hurwitz, Q exists and is the unique solution of A T Q + QA + C T C = 0 Q is symmetric and positive (semi-)denite, i.e., Q = Q T 0 If (A, C) is observable, then Q 0 8/36

16 Observability Gramian: interpretation Singular value decomposition Q = U o Σ o Uo T n = σ o,i u o,i uo,i T 1 σo,2 u o,2 x 2 1 σo,1 u o,1 x 1 i=1 with σ o,i σ o,i+1 The ellipse gives all initial states x 0 for which x T 0 Qx 0 = c, c > 0, i.e., all initial states that give the same output energy u o,1 : easily observable direction (σ o,1 large) u o,2 : poorly observable direction (σ o,2 small) 9/36

17 Observability Gramian: interpretation Singular value decomposition Q = U o Σ o Uo T n = σ o,i u o,i uo,i T 1 σo,2 u o,2 x 2 1 σo,1 u o,1 x 1 i=1 with σ o,i σ o,i+1 Unobservable subspace. If σ o,k+1 =... = σ o,n = 0 and σ o,k > 0, C CA N = span{u o,k+1,..., u o,n } = ker. CA n 1 9/36

18 Controllability energy function x 0 u Σ Controllability energy function { 0 } L c (x 0 ) = min u(t) 2 dt u L 2, x( ) = 0, x(0) = x 0 The controllability energy function gives the least input energy needed to steer the system state from 0 to x 0 in innite time 10/36

19 Controllability energy function x 0 u Σ Controllability energy function { 0 } L c (x 0 ) = min u(t) 2 dt u L 2, x( ) = 0, x(0) = x 0 The controllability energy function gives the least input energy needed to steer the system state from 0 to x 0 in innite time Theorem. If A is Hurwitz and (A, B) is controllable, then L c (x 0 ) = x T 0 P 1 x 0, P = 0 e At BB T e ATt dt 10/36

20 Controllability Gramian P = Properties 0 e At BB T e ATt dt If A is Hurwitz, P exists and is the unique solution of AP + PA T + BB T = 0 P is symmetric and positive (semi-)denite, i.e., P = P T 0 If (A, B) is controllable, then P 0 11/36

21 Controllability Gramian: interpretation Singular value decomposition x 2 σc,1 u c,1 P = U c Σ c Uc T n = σ c,i u c,i uc,i T i=1 with σ c,i σ c,i+1, i.e., n P 1 = σ 1 c,i u c,iuc,i T i=1 σc,2 u c,2 x 1 The ellipse gives all states x 0 for which x T 0 P 1 x 0 = c, c > 0, i.e., all initial states that can be reached with bounded energy c u c,1 : easily reachable direction (σ c,1 large) u c,2 : poorly reachable direction (σ c,2 small) 12/36

22 Controllability Gramian: interpretation Singular value decomposition x 2 σc,1 u c,1 P = U c Σ c Uc T n = σ c,i u c,i uc,i T i=1 with σ c,i σ c,i+1, i.e., n P 1 = σ 1 c,i u c,iuc,i T i=1 σc,2 u c,2 x 1 Reachability subspace. If σ c,k+1 =... = σ c,n = 0 and σ c,k > 0, W = span{u c,1,..., u c,k } = im [ B AB... A n 1 B ] 12/36

23 Example ẋ = [ ] 1 0 x [ ] 100 u, y = [ ] x 1 x 2 constant input energy u c,1 x 1 From a controllability perspective, x 1 is important 13/36

24 Example ẋ = [ ] 1 0 x [ ] 100 u, y = [ ] x 1 x 2 constant output energy u o,1 u c,1 x 1 From an observability perspective, x 2 is important 13/36

25 Example ẋ = [ ] 1 0 x [ ] 100 u, y = [ ] x 1 x 2 u o,1 u c,1 x 1 States that are easy to control (σ c,1 large) are not necessarily easy to observe (σ o,1 large) or vice versa 13/36

26 Coordinate transformations { ẋ = Ax + Bu Σ : y = Cx + Du Coordinate transformation T (T R n n nonsingular) x = Tx In new coordinates, { x = Ā x + Bu Σ : y = C x + Du with Ā = TAT 1, B = TB, C = CT 1, D = D Properties Stability properties and input-output behavior are unchanged 14/36

27 Hankel singular values Gramians in new coordinates x = Tx P = TPT T, Q = T T QT 1 15/36

28 Hankel singular values Gramians in new coordinates x = Tx P = TPT T, Q = T T QT 1 Transformation of PQ P Q = TPQT 1 The eigenvalues of PQ are invariant under transformation The eigenvalues of PQ equal the Hankel singular values σ i σ i = λ i (PQ), i {1,..., n}, σ 1 σ 2... σ n 0 15/36

29 Hankel singular values Gramians in new coordinates x = Tx P = TPT T, Q = T T QT 1 Transformation of PQ P Q = TPQT 1 The eigenvalues of PQ are invariant under transformation The eigenvalues of PQ equal the Hankel singular values σ i σ i = λ i (PQ), i {1,..., n}, σ 1 σ 2... σ n 0 Theorem. There exists a transformation T such that P = Q = Σ := diag{σ 1, σ 2,..., σ n } A realization for which P = Q = Σ is called balanced. Then, σ i = σ c,i = σ o,i 15/36

30 Balanced realization: interpretation (1) σ σ 2 0 P = Q = Σ = σ n Interpretation. P and Q are Equal. States that are easy to control are also easy to observe and vice versa (σ c,i = σ o,i = σ i and u c,i = u o,i ) Diagonal. Controllability and observability can be interpreted per state component (u c,i = u o,i = e i ) Ordered. The rst state is the easiest to control and easiest to observe (σ i σ i+1 ) 16/36

31 Balanced realization: interpretation (1) σ σ 2 0 P = Q = Σ = σ n x 2 x 2 x = Tx x 1 x 1 controllability observability 16/36

32 Balanced realization: interpretation (2) Observability and controllability energy functions L o ( x 0 ) = x T 0 Σ x 0, L c ( x 0 ) = x T 0 Σ 1 x 0 Consider x 0 = e 1 = [ ] T L o (e 1 ) = σ 1, Starting at e 1 gives large output energy L c (e 1 ) = σ 1 1, Small input energy needed to reach e 1 17/36

33 Balanced realization: interpretation (2) Observability and controllability energy functions L o ( x 0 ) = x T 0 Σ x 0, L c ( x 0 ) = x T 0 Σ 1 x 0 Consider x 0 = e 1 = [ ] T L o (e 1 ) = σ 1, Starting at e 1 gives large output energy L c (e 1 ) = σ 1 1, Small input energy needed to reach e 1 Consider x 0 = e n = [ ] T L o (e n ) = σ n, Starting at e n gives small output energy L c (e n ) = σn 1, Large input energy needed to reach e n 17/36

34 Balanced realization: interpretation (2) Observability and controllability energy functions L o ( x 0 ) = x T 0 Σ x 0, L c ( x 0 ) = x T 0 Σ 1 x 0 Consider x 0 = e 1 = [ ] T L o (e 1 ) = σ 1, Starting at e 1 gives large output energy L c (e 1 ) = σ 1 1, Small input energy needed to reach e 1 Consider x 0 = e n = [ ] T L o (e n ) = σ n, Starting at e n gives small output energy L c (e n ) = σn 1, Large input energy needed to reach e n In a balanced realization, states are ordered according to their contribution to the input-output behavior 17/36

35 Balancing algorithm 1. Obtain P and Q as solutions of the Lyapunov equations AP + PA T + BB T = 0 A T Q + QA + C T C = 0 2. Determine U through a Cholesky factorization of P as P = UU T 3. Perform an eigenvalue decomposition of U T QU as U T QU = KΣ 2 K T to obtain K and Σ 4. The transformation matrix T and its inverse are given as T = Σ 1 2 K T U 1, T 1 = UKΣ 1 2 In Matlab, use lyap(), chol(), and eig() 18/36

36 Balanced realization: partitioning Consider a balanced realization { ẋ = Ax + Bu Σ : y = Cx + Du Partitioning of the energy functions and Gramian with L o (x 0 ) = x T 0 Σx 0 = x T 1,0Σ 1 x 1,0 + x T 2,0Σ 2 x 2,0 L c (x 0 ) = x T 0 Σ 1 x 0 = x T 1,0Σ 1 1 x 1,0 + x T 2,0Σ 1 2 x 2,0 x = [ x1 x 2 ] [ ] Σ1 0, Σ = 0 Σ 2 Σ 1 R k k contains large Hankel singular values Σ 2 R (n k) (n k) contains small Hankel singular values 19/36

37 Balanced realization: partitioning Consider a balanced realization { ẋ = Ax + Bu Σ : y = Cx + Du Partitioning of the energy functions and Gramian with L o (x 0 ) = x T 0 Σx 0 = x T 1,0Σ 1 x 1,0 + x T 2,0Σ 2 x 2,0 L c (x 0 ) = x T 0 Σ 1 x 0 = x T 1,0Σ 1 1 x 1,0 + x T 2,0Σ 1 2 x 2,0 x = [ x1 x 2 ] [ ] Σ1 0, Σ = 0 Σ 2 Partitioning of system matrices [ ] A11 A A = 12, B = A 21 A 22 [ B1 B 2 ], C = [ C 1 C 2 ] 19/36

38 Model reduction: truncation Reduced-order model { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du Properties Obtained by setting x 2 = 0 Approximation is exact for s, i.e., G( ) = Ĝ t ( ) 20/36

39 Model reduction: truncation Reduced-order model { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du Properties Obtained by setting x 2 = 0 Approximation is exact for s, i.e., G( ) = Ĝ t ( ) Theorem. Let ˆΣ t be obtained by balanced truncation. Then, ˆΣ t is in balanced realization with controllability Gramian Σ 1 and observability Gramian Σ 1 If σ k > σ k+1, then A 11 is Hurwitz, i.e., ˆΣ t is asympt. stable 20/36

40 Model reduction: truncation Reduced-order model { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du Properties Obtained by setting x 2 = 0 Approximation is exact for s, i.e., G( ) = Ĝ t ( ) Theorem. The following error bound holds: G(s) Ĝ t (s) n H 2 i=k+1 Note: multiplicities of σ i do not have to be counted σ i 20/36

41 Model reduction: singular perturbation Reduced-order model { ξ = ( A 11 A 12 A ˆΣ 1 22 s : A ) ( 21 ξ + B1 A 12 A 1 B 22 2) u ŷ s = ( C 1 C 2 A 1 A ) ( ξ + D C2 A 1 B 22 2) u Properties Obtained by setting ẋ 2 = 0 Approximation is exact for s = 0, i.e., G(0) = Ĝ s (0) 21/36

42 Model reduction: singular perturbation Reduced-order model { ξ = ( A 11 A 12 A ˆΣ 1 22 s : A ) ( 21 ξ + B1 A 12 A 1 B 22 2) u ŷ s = ( C 1 C 2 A 1 A ) ( ξ + D C2 A 1 B 22 2) u Properties Obtained by setting ẋ 2 = 0 Approximation is exact for s = 0, i.e., G(0) = Ĝ s (0) Theorem. Let ˆΣ t be obtained by balanced truncation. Then, ˆΣ s is in balanced realization with controllability Gramian Σ 1 and observability Gramian Σ 1 If σ k > σ k+1, then A 11 A 12 A 1 22 A 21 is Hurwitz 21/36

43 Model reduction: singular perturbation Reduced-order model { ξ = ( A 11 A 12 A ˆΣ 1 22 s : A ) ( 21 ξ + B1 A 12 A 1 B 22 2) u ŷ s = ( C 1 C 2 A 1 A ) ( ξ + D C2 A 1 B 22 2) u Properties Obtained by setting ẋ 2 = 0 Approximation is exact for s = 0, i.e., G(0) = Ĝ s (0) Theorem. The following error bound holds: G(s) Ĝ s (s) n H 2 i=k+1 Note: multiplicities of σ i do not have to be counted σ i 21/36

44 Truss frame example u, y Dynamics M q + D q + Kq = Su State-space form with x = [ q T q T ] T R 72 [ ] [ ] 0 I 0 ẋ = x + u, y = Cx M 1 K M 1 D M 1 S 1/1 22/36

45 Truss frame example: results Reduction to k = 12 using balanced truncation σi [-] 10 9 G [-] i [-] G G f [Hz] 23/36

46 Truss frame example: results Reduction to k = 6 using balanced truncation σi [-] 10 9 G [-] i [-] G G f [Hz] 23/36

47 Truss frame example: modes Balancing - truss frame example - modes 26/51 The Balancing most important "modes""mode" are dependent depends on onthe input input and andoutput u,y /w u,y 24/36

48 Conclusions (so far...) Gramians as measure of controllability and observability Balancing to identify most important states Two methods for reduction: 1. Truncation 2. Singular perturbation Preservation of stability Error bound in terms of the truncated Hankel singular values 25/36

49 Conclusions (so far...) Gramians as measure of controllability and observability Balancing to identify most important states Two methods for reduction: 1. Truncation 2. Singular perturbation Preservation of stability Error bound in terms of the truncated Hankel singular values Questions What do the parameters σ i represent? What is the most accurate reduced-order model possible? Can we preserve additional properties (for example, passivity or a bounded L 2 gain)? 25/36

50 Observability and controllability operator x 0 Σ y Observability operator Ψ o : R n L p 2 ([0, )) Ψ o x 0 = Ce At x 0, t 0 26/36

51 Observability and controllability operator x 0 Σ y Observability operator Ψ o : R n L p 2 ([0, )) Properties Ψ o x 0 = Ce At x 0, t 0 Ψ o is bounded for A Hurwitz Relation to the observability energy function for y = Ψ o x 0 L o (x 0 ) = y 2 L 2 = Ψ o x 0, Ψ o x 0 L 2 = x 0, Ψ oψ o x 0 R n = x T 0 Qx 0 with Ψ o the adjoint of Ψ o 26/36

52 Observability and controllability operator x 0 u Σ Controllability operator Ψ c : L m 2 ((, 0]) Rn Ψ c u = 0 e At Bu(t) dt 26/36

53 Observability and controllability operator x 0 u Σ Controllability operator Ψ c : L m 2 ((, 0]) Rn Properties Ψ c u = 0 e At Bu(t) dt If (A, B) is controllable, then im(ψ c ) = R n Relation to the controllability energy function L c (x 0 ) = min { u 2 L 2 Ψ c u = x 0 } = u opt 2 L 2 = P 1 x 0, Ψ c Ψ cp 1 x 0 R n = x T 0 P 1 x 0 with u opt = Ψ cp 1 x 0 26/36

54 Hankel operator Hankel operator H : L m((, 0]) 2 Lp 2 ([0, )) Hu = Ψ o Ψ c u = Ce At 0 e Aτ Bu(τ) dτ, t 0 27/36

55 Hankel operator Hankel operator H : L m((, 0]) 2 Lp 2 ([0, )) Hu = Ψ o Ψ c u = Ce At 0 e Aτ Bu(τ) dτ, t 0 u t Hu The Hankel operator maps past inputs to future outputs t 27/36

56 Hankel operator Hankel operator H : L m((, 0]) 2 Lp 2 ([0, )) Properties Hu = Ψ o Ψ c u = Ce At 0 H is a bounded operator for A Hurwitz e Aτ Bu(τ) dτ, t 0 H is a nite-dimensional operator with dim im(h) n 27/36

57 Hankel operator Hankel operator H : L m((, 0]) 2 Lp 2 ([0, )) Properties Hu = Ψ o Ψ c u = Ce At 0 H is a bounded operator for A Hurwitz e Aτ Bu(τ) dτ, t 0 H is a nite-dimensional operator with dim im(h) n Singular value decomposition of the Hankel operator n Hu = σ i u, ūi ȳ L 2 i i=1 with ū i L m((, 0]) 2 and ȳ i L p 2 ([0, )) orthonormal, i.e., ūi, ū j = L { 1, i = j ȳ 2 i, ȳ j = L 2 0, i j 27/36

58 Hankel singular values Singular value decomposition of the Hankel operator n Hu = σ i u, ūi ȳ L 2 i i=1 Theorem. The Hankel singular values σ i satisfy PQ x i = σi 2 x i, i.e., σi 2 = λ i (PQ) 28/36

59 Hankel singular values Singular value decomposition of the Hankel operator n Hu = σ i u, ūi ȳ L 2 i i=1 Theorem. The Hankel singular values σ i satisfy PQ x i = σi 2 x i, i.e., σi 2 = λ i (PQ) Hankel norm. Let σ 1 σ 2... σ n 0 and y = Hu. Then, { } y L p Σ H = sup 2 (R +) u u Lm 2 ((, 0]), u 0 = σ 1 L m 2 (R ) 28/36

60 Hankel singular values Singular value decomposition of the Hankel operator n Hu = σ i u, ūi ȳ L 2 i i=1 Theorem. The Hankel singular values σ i satisfy PQ x i = σi 2 x i, i.e., σi 2 = λ i (PQ) Hankel norm. Let σ 1 σ 2... σ n 0 and y = Hu. Then, { } y L p Σ H = sup 2 (R +) u u Lm 2 ((, 0]), u 0 = σ 1 L m 2 (R ) Properties The Hankel norm is a lower bound on the L 2 norm, i.e., Σ H Σ L2 = G H 28/36

61 A fundamental lower bound u Σ ˆΣ + y ŷ e Theorem. Let Σ be a minimal realization with A Hurwitz. For any asymptotically stable ˆΣ k of dimension k < n, Σ ˆΣ k L2 = G Ĝ k H σ k+1 with σ 1 σ 2... σ n > 0 the Hankel singular values of Σ 29/36

62 A fundamental lower bound u Σ ˆΣ + y ŷ e Theorem. Let Σ be a minimal realization with A Hurwitz. For any asymptotically stable ˆΣ k of dimension k < n, Σ ˆΣ k L2 = G Ĝ k H σ k+1 with σ 1 σ 2... σ n > 0 the Hankel singular values of Σ Properties Lower bound is independent of the reduction procedure For balanced truncation and balanced singular perturbation, σ k+1 Σ ˆΣ k L2 2 ( σ k+1 + σ k σ n ) 29/36

63 Passive systems u Σ y Passivity. Σ is passive if there exists V : R n [0, ) such that V (x(t 1 )) V (x(t 0 )) + t1 for all t 0 t 1 and for all trajectories of Σ t 0 u T (t)y(t) dt 30/36

64 Passive systems u Σ y Passivity. Σ is passive if there exists V : R n [0, ) such that V (x(t 1 )) V (x(t 0 )) + t1 for all t 0 t 1 and for all trajectories of Σ t 0 u T (t)y(t) dt Properties Examples: physical systems, e.g., mechanical or electrical Special case of dissipativity theory, with storage function V and supply rate s(u, y) = u T y 30/36

65 Passive systems u Σ y Passivity. Σ is passive if there exists V : R n [0, ) such that V (x(t 1 )) V (x(t 0 )) + t1 for all t 0 t 1 and for all trajectories of Σ t 0 u T (t)y(t) dt Properties Examples: physical systems, e.g., mechanical or electrical Special case of dissipativity theory, with storage function V and supply rate s(u, y) = u T y Question. Can we nd a reduced-order model that preserves passivity? 30/36

66 Postive real lemma Theorem. Let Σ be minimal. Then, the following are equivalent 1. Σ is passive 2. The transfer function G is positive real, i.e., G(s) + G (s) 0, Re(s) > 0 31/36

67 Postive real lemma Theorem. Let Σ be minimal. Then, the following are equivalent 1. Σ is passive 2. The transfer function G is positive real, i.e., G(s) + G (s) 0, Re(s) > 0 3. Σ is passive with V (x) = 1x T Kx for some K = K T /36

68 Postive real lemma Theorem. Let Σ be minimal. Then, the following are equivalent 1. Σ is passive 2. The transfer function G is positive real, i.e., G(s) + G (s) 0, Re(s) > 0 3. Σ is passive with V (x) = 1x T Kx for some K = K T There exist K = K T, L, and W such that A T K + KA = LL T PB C T = LW T D + D T = WW T where 0 K K K + 31/36

69 Postive real lemma Theorem. Let Σ be minimal. Then, the following are equivalent 1. Σ is passive 2. The transfer function G is positive real, i.e., G(s) + G (s) 0, Re(s) > 0 3. Σ is passive with V (x) = 1x T Kx for some K = K T There exist K = K T, L, and W such that A T K + KA = LL T PB C T = LW T D + D T = WW T where 0 K K K + Note. The equivalence is known as the positive real lemma or Kalman-Yakubovich-Popov (KYP) lemma 31/36

70 Extremal storage functions Available storage function V a : R n [0, ) V a (x 0 ) = 1x T 2 0 K x 0 { = sup 0 } u T (t)y(t) dt x(0) = x 0, x( ) = 0, u L 2 32/36

71 Extremal storage functions Available storage function V a : R n [0, ) V a (x 0 ) = 1x T 2 0 K x 0 { = sup 0 } u T (t)y(t) dt x(0) = x 0, x( ) = 0, u L 2 Required supply function V r : R n [0, ) V r (x 0 ) = 1x T 2 0 K + x 0 { 0 } = inf u T (t)y(t) dt x( ) = x 0, x(0) = x 0, u L 2 32/36

72 Extremal storage functions Available storage function V a : R n [0, ) V a (x 0 ) = 1x T 2 0 K x 0 { = sup 0 } u T (t)y(t) dt x(0) = x 0, x( ) = 0, u L 2 Required supply function V r : R n [0, ) V r (x 0 ) = 1x T 2 0 K + x 0 { 0 } = inf u T (t)y(t) dt x( ) = x 0, x(0) = x 0, u L 2 Interpretation V a (x 0 ) gives the maximum energy one can extract from x 0 V r (x 0 ) gives the minimal energy needed to reach x 0 32/36

73 Extremal storage functions Available storage function V a : R n [0, ) V a (x 0 ) = 1x T 2 0 K x 0 { = sup 0 } u T (t)y(t) dt x(0) = x 0, x( ) = 0, u L 2 Required supply function V r : R n [0, ) V r (x 0 ) = 1x T 2 0 K + x 0 { 0 } = inf u T (t)y(t) dt x( ) = x 0, x(0) = x 0, u L 2 Interpretation V a (x 0 ) gives the maximum energy one can extract from x 0 V r (x 0 ) gives the minimal energy needed to reach x 0 Idea: perform balancing using K and K /36

74 Positive real balancing Theorem. The eigenvalues of K K+ 1 denoted as π i = λ i (K K+ 1 ) are system invariants and satisfy 0 < π i 1 33/36

75 Positive real balancing Theorem. The eigenvalues of K K+ 1 denoted as π i = λ i (K K+ 1 ) are system invariants and satisfy 0 < π i 1 Theorem. There exists a set of coordinates in which with π 1 π 2... π n > 0 K = K 1 + = Π = diag{π 1,..., π n } 33/36

76 Positive real balancing Theorem. The eigenvalues of K K+ 1 denoted as π i = λ i (K K+ 1 ) are system invariants and satisfy 0 < π i 1 Theorem. There exists a set of coordinates in which with π 1 π 2... π n > 0 K = K 1 + = Π = diag{π 1,..., π n } Notes π i are called positive real singular values A realization with K = K 1 + = Π is positively real balanced 33/36

77 Positive real balancing Theorem. The eigenvalues of K K+ 1 denoted as π i = λ i (K K+ 1 ) are system invariants and satisfy 0 < π i 1 Theorem. There exists a set of coordinates in which with π 1 π 2... π n > 0 K = K 1 + = Π = diag{π 1,..., π n } In positively real balanced coordinates, partition [ ] [ ] [ Π1 0 A11 A Π =, A = 12 B1, B = 0 Π 2 A 21 A 22 B 2 ], C = [ C 1 C 2 ] 33/36

78 Positive real balanced truncation Reduced-order model obtained by truncation with ξ R k, k < n { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du 34/36

79 Positive real balanced truncation Reduced-order model obtained by truncation with ξ R k, k < n { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du Theorem. Let Σ be minimal and passive. Then, ˆΣ t obtained by positive real balanced truncation is passive 34/36

80 Positive real balanced truncation Reduced-order model obtained by truncation with ξ R k, k < n { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du Theorem. Let Σ be minimal and passive. Then, ˆΣ t obtained by positive real balanced truncation is passive Properties There exists an extension to strictly positive real systems that preserves, in addition, minimality and asymptotic stability 34/36

81 Positive real balanced truncation Reduced-order model obtained by truncation with ξ R k, k < n { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du Theorem. Let Σ be minimal and passive. Then, ˆΣ t obtained by positive real balanced truncation is passive Properties There exists an extension to strictly positive real systems that preserves, in addition, minimality and asymptotic stability If D + D T 0, then K and K + are also extremal solutions of A T K + KA + (KB C T )(D + D T ) 1 (KB C T ) T = 0 34/36

82 Positive real balanced truncation Reduced-order model obtained by truncation with ξ R k, k < n { ξ = A 11 ξ + B 1 u ˆΣ t : ŷ t = C 1 ξ + Du Theorem. Let Σ be minimal and passive. Then, ˆΣ t obtained by positive real balanced truncation is passive Properties There exists an extension to strictly positive real systems that preserves, in addition, minimality and asymptotic stability If D + D T 0, then K and K + are also extremal solutions of A T K + KA + (KB C T )(D + D T ) 1 (KB C T ) T = 0 A similar method known as bounded real balancing preserves an L 2 gain bounded by 1, i.e., dissipativity with respect to the supply rate s(u, y) = u 2 2 y /36

83 Conclusions Summary Gramians as measure of controllability and observability Balanced truncation or singular perturbation for reduction Preservation of asymptotic stability and minimality Error bound in terms of Hankel singular values Positive real balanced truncation for preservation of passivity Use extremal storage functions for balancing 35/36

84 Conclusions Summary Gramians as measure of controllability and observability Balanced truncation or singular perturbation for reduction Preservation of asymptotic stability and minimality Error bound in terms of Hankel singular values Positive real balanced truncation for preservation of passivity Use extremal storage functions for balancing Extensions of balancing Optimal Hankel norm approximation Frequency-weighted balancing Balancing for controllers, structured systems, networks, nonlinear systems,... 35/36

85 References B.C. Moore, Principal component analysis in linear systems - controllability, observability, and model reduction. IEEE Transactions on Automatic Control, AC-26(1): 17-32, D.F. Enns. Model reduction with balanced realizations: an error bound and a frequency weighted generalization. In Proceedings of the 23rd IEEE Conference on Decision and Control, Las Vegas, USA, , K. Glover, All optimal Hankel-norm approximations of linear multivariable systems and their L -error bounds. International Journal of Control, 39(6): , K. Fernando and H. Nicholson. Singular perturbational model reduction of balanced systems. IEEE Transactions on Automatic Control, AC-27(2): , Y. Liu and B.D.O. Anderson. Singular perturbation approximation of balanced systems. International Journal of Control, 50(4): , U. Desai and D. Pal. A transformation approach to stochastic model reduction. IEEE Transactions on Automatic Control, AC-29(12): , P. Harshavardhana, E. Jonckheere, and L. Silverman. Stochastic balancing and approximation - stability and minimality. IEEE Transactions on Automatic Control, AC-29(8): , R. Ober. Balanced parametrization of classes of linear systems. SIAM Journal on Control and Optimization, SIAM, 29(6): , A.C. Antoulas. Approximation of large-scale dynamical systems. SIAM, Philadelphia, USA, /36

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