6.241 Dynamic Systems and Control

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1 6.241 Dynamic Systems and Control Lecture 24: H2 Synthesis Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology May 4, 2011 E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

2 Standard setup Consider the following system, for t R 0 : where ẋ(t) = Ax(t) + B w w(t) + B u u(t), x(0) = x 0 z(t) = C z x(t) + D zw w(t) + D zu u(t) y(t) = C y x(t) + D yw w(t) + D yu u(t), w is an exogenous disturbance input (also reference, noise, etc.) u is a control input, computed by the controller K z is the performance output. This is a virtual output used only for design. y is the measured output. This is what is available to the controller K It is desired to synthesize a controller K (itself a dynamical system), with input y and output u, such that the closed loop is stabilized, and the performance output is minimized, given a class of disturbance inputs. In particular, we will look at controller synthesis with H 2 and H criteria. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

3 Interpretation of the H 2 norm deterministic Consider a stable, causal CT LTI system with state-space model (A, B, C, D), transfer function G (s), and impulse response G (t). The H 2 norm of G measures: A) The energy of the impulse response: G := g L 2 ij (t) 2 2 dt = G(t) 2 F dt i j 0 0 [ + ] [ ] = Tr G(t) + 1 G (t) dt = 2 π Tr G (jω) 2 G (jω) dω =: G H 0 B) The energy of the response to initial conditions, of the form x(0) = Bu 0, for u 0 = (1, 1,..., 1). Set u(t) = u 0δ(t) to see this. Clearly, in order for G L2 = G H2 to be finite, it is necessary that lim ω G (jω) = 0, i.e., that the system is strictly causal D = E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

4 Interpretation of the H2 norm stochastic Consider a stable, strictly causal CT LTI system with state-space model (A, B, C, 0), transfer function G (s), and impulse response G (t). Consider a hypothetical stochastic input signal u such that E[u(t)] = 0, and E[u(t)u(t + τ) ] = I δ(τ ). This is called white noise, and is just a mathematical abstraction, since it is a signal with infinite power. The H 2 norm of G measures: C) The (expected) power of the response to white noise: [ ( T )] 1 E lim Tr y (t)y(t) dt T + T 0 T t t ( [ ] ) 1 = lim Tr E G (t τ 1)u(τ 1)u(τ 2) G (t τ 2) dτ 1dτ 2 dt T + T ( T t ) 1 = lim Tr G (t τ )G(t τ ) dτ dt T + T 0 0 [ T ] = lim Tr G (T τ )G(T τ) d(t τ) T = G L2 = G H2. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

5 Computation of the H2 norm Computation of the H 2 norm is easy through state-space methods. In fact, [ + ] [ + ] 2 G H = Tr G (t) G (t) dt = Tr G (t)g(t) dt Since G (t) = Ce At B (recall that D = 0 is necessary for the H 2 norm to be finite), we get [ + ] 2 G H2 = Tr B e A t C Ce At B dt = Tr [B QB] where 0 [ + ] = Tr Ce At BB e A t C dt = Tr [CPC ], 0 Q is the observability Gramian, satisfying AQ + AQ = CC. P is the reachability Gramian, satisfying A P + AP = B B. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

6 Structure of the D block We will make the following assumptions on the structure of the D block: D yu = 0. D zw = 0 We can always make this assumption, since u is known. The H 2 norm of a system that is not strictly proper (i.e., such that lim s G (s) = G > 0) is +. Note that T zw ( ) = D zw + D zu D uy D yw. If there is no D uy such that T zw ( ) = 0, then the problem is ill-posed. If there is one such Duy 0, then define ũ = u + Duy 0 y, and rewrite the problem as follows: D 0 A = A + B u yu C y B w = B + B u D 0 w uy D uw C z = C z + D D 0 zu uy C y D zw = D zw + D zu D 0 D yw = 0. uy E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

7 The LQR problem The LQR problem is the special case of H 2 synthesis in which we assume: Full state feedback: C y = I ; No disturbance input: w = 0. Objective: find a control signal u(t, x) L 2 that minimizes 2 z 2 = 0 z 2 = + C z x + D zu u 2 2 dt, given the initial condition x(0). Note that if C z = [ Q 0 ] and Dzu = [ ] 0 R then we get (x Qx + u Ru) dt, which is the usual way the LQR problem is formulated. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

8 Towards a solution of the LQR problem (intuition) Consider a stabilizing control law of the form u = Fx, and assume C z D zu = 0. By assumption, A F = A + B u F is stable. + ( ) 2 A z F t A 2 C F t A = x F t e z C z e + e F D Fe A F t x 0 dt, 0 0 zud zu i.e., z 2 = x 0 X F x 0, where X F is the observability gramian of the pair 2 (C F, A F ), with C F = C z + D zu F, and A F X F + X F A F = C F C F. Since we know that the closed-loop is stable, we can also rewrite the above equation as + 2 d z 2 = 0 dt (x(t) X F x(t)) dt. The integrand can be written as x(t) (A X F + F B u X F + X F A + X F B u F ) x(t) E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

9 Towards a solution of the LQR problem (intuition) Assume there is a matrix S such that F = SX F. Then, the integrand becomes x(t) (A X F + X F S B u X F + X F A + X F B u SX F ) x(t) In other words, it must be that = x(t) (C z C z + X F S D zu D zu SX F ) x(t) A X F + X F A + X F S B X F + X F B u SX F + C C z + X F S D D zu SX F = 0 u z zu Set S = (D zu D zu ) 1 B u. Then, X F must satisfy A X F + X F A + X F B u (D ) 1 B X F + C C z = 0 zud zu u z and F = (D zu D zu ) 1 B u X F Is this solution indeed stabilizing/optimal? E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

10 On Riccati equations We have already encountered a matrix equation that plays a major role in control, i.e., the (continuous-time) Lyapunov equation: A X + XA + Q = 0. This equation can be used, among other things, to check stability of a LTI system, and to compute reachability/observability gramians. The Lyapuov equation is linear in X, and can be easily solved. Another important equation in control theory it the (c.t.) algebraic Riccati equation: A X + XA + XRX + Q = 0. The Riccati equation is quadratic in X ; what can we say about its solutions, and how do we compute them? E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

11 Hamiltonian matrices It turns out that to each Riccati equation we can associate a Hamiltonian matrix of the form [ ] A R H :=, Q A which will be used to compute solutions to the Riccati equation. The spectrum of H is symmetric with respect to the imaginary axis. To see this, consider the similarity transformation: [ ] 1 [ ] [ ] [ ] [ ] [ ] 0 I A R 0 I 0 I R A A Q = = = H. I 0 Q A I 0 I 0 A Q R A In other words, H and H are similar, and hence if λ is an eigenvalue of H so is λ. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

12 Computing solutions to the Riccati equation Assume that H has no eigenvalues on the imaginary axis. Then H will have n eigenvalues in the open left half plane, and n in the open right half plane. Let X be the subspace spanned by the eigenvectors associated with the eigenvalues with [ negative ] real part, and find n n matrices X 1 and X 2 such X that X 1 = Ra. X2 If X 1 is nonsingular, then set X := X 2 X 1 1. Note that X is unique, since any other set of basis vectors satisfies X 1 = X 1 S, X 2 = X 2 S, for some invertible matrix S, and X := X 2X 1 1 = X 2 SS 1 X 1. Theorem Assume that (i) H has no eigenvalues on the imaginary axis, and that (ii) the matrix X 1 in the above construction is not singular. Then, 1 X is real symmetric; 2 X satisfies the Riccati equation A X + XA + XRX + Q = 0 3 All the eigenvalues of the matrix A + RX are in the open left half plane. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

13 Computing solutions to the Riccati equation proof 1 1 X = X 2 X 1 is real symmetric. Note that there exists a stable n n matrix H such that [ ] [ ] X 1 X 1 H = H X 2 X 2 [ ] [ ] X 1 0 I Premultiply by : X 2 I 0 [ ] [ ] [ ] [ ] [ ] [ ] X 1 0 I H X1 X 1 0 I X 1 = H X 2 I 0 X 2 X 2 I 0 X, 2 The left hand side is Hermitian so the right hand side is also Hermitian, and ( X 1 X 2 + X 2 X 1)H + H ( X 1 X 2 + X 2 X 1) = 0. This is a Lyapunov equation, and since H is stable, it has a unique solution X 1 X 2 = X 2 X 1. Hence the matrix X := X 2X 1 1 = (X 1 1 ) (X 1 X 2)X 1 1 is Hermitian. Since X 1 and X 2 can be chosen to be real, and X is unique, X is real and symmetric. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

14 Computing solutions to the Riccati equation proof 2 3 X satisfies the Riccati equation A X + XA + XRX + Q = 0: [ ] [ ] X 1 X 1 Start with H = H, and left-/right-multiply as follows: X 2 X 2 ( [ ]) ([ ] ) [ ] X 1 1 [ ] X X I 1 1 H X 1 = X I H X 1, X 2 X 2 [ ] [ ] [ ] A R I [ ] X I Q A 1 = X X 2 X 2 H X 1 A + RX is stable: XA + Q + XRX + A X = 0. Similarly, ( [ ]) ([ ] ) [ ] X1 1 [ ] X1 1 I 0 H X X 1 = I 0 H X X 1, 2 A + RX = X 1H X 1, 1 i.e., A + RX is similar to a stable matrix, and hence stable. 2 E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

15 Technical conditions A1) (A, B u ) stabilizable. A2) (C z, A) detectable. [ ] A jωi B A3) u has full column rank for all ω R C z D zu A4) D = R, invertible, i.e., D zu has full column rank. zud zu A1-A3) ensure that the Riccati equation admits a solution X that is positive semi-definite. In particular, A1) is obviously necessary, A2) ensures that any unstable mode of A will be detected by the performance output, and A3) ensures that the control effort is penalized at all frequencies (this is an additional technical condition ensuring that the Hamiltonian does not have purely imaginary eigenvalues). A4) is just for convenience. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

16 LQR: optimal control It turns out that the optimal controller can be obtained from the unique, symmetric, positive-definite solution X of the (algebraic) Riccati Equation (A B R 1 D C z ) X + X (A B u R 1 D u uu zu uu zu C z ) XB u Ruu 1 B u X + C z (I D R 1 D zu uu zu )C z = 0 by setting F = R 1 (B X + D C z ). uu u zu Define A F := A + B u F, and C F := C z + D zu F. Recall that X can be interpreted as the observability Gramian of (A F, C F ), describing the energy of the impulse response of the closed-loop system (A F, I, C F, 0). Hence z 2 = x Xx E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

17 Note on the detectability of (C z, A) Claim: Since (C z, A) is detectable, if u, z L 2, then x L 2, and x 0. Proof: Design a hypothetical observer using z to compute an estimate ˆx of the state x. Then xˆ = Axˆ + B u u + L(C z xˆ z + D zu u) = (A + LC z )ˆx + (B u + LD zu )u Lz, and hence ˆx L 2, ˆx 0. Moreover, since the observer is stable, x xˆ 0. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

18 Optimality of the proposed control law Assume that u = Fx + v. Then one can write [ ] [ ] [ ] ẋ A F B u x =, x(0) = x z C 0 F D zu v Note that v L 2 x, z, u L 2 (stability of A F ), and u, z L 2 v, x L 2 (detectability of the state in the performance output). So minimizing over u L 2 is equivalent to minimizing over v L 2. Differentiate x(t) Xx(t) along system trajectories, noting that C F D zu = XB u : d x Xx = x (A F X + XA F )x + 2x XB u v dt = x C F C F x 2x C F D zu v v D zu D zu v + v D zu D zu v = z 2 + v Rv. Integrating from 0 to +, we get 2 2 x 0 Xx 0 = z 2 + Rv 2. Hence the minimum is attained for v = 0. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

19 LQE problem Kalman filter The LQE problem is the special case of H 2 synthesis addressing the design of an observer (i.e., u takes the role of the observer update), assuming Full state updates: B u = I. Zero initial conditions: x(0) = x(0) xˆ(0) = 0. Objective: find an update signal u(t, x ) L 2 that minimizes the power in the error signal due to white noise disturbance w. Note that the disturbance enters the system in two places: As process noise: x = Ax + B w w. As sensor noise: y = C y x + D yw w. If C y = [ Q 0 ] and Dyw = [ 0 noise are not correlated, R ], then the process noise and sensor E [w B w D yw w ] = 0, which is the usual way the LQE problem is formulated. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

20 Towards a solution of the LQE problem (intuition) Consider a stabilizing update law of the form u = L(C y x + D yw w), and assume B w D = 0. By assumption, A L = A + LC y is stabilizing. yw The power of the error, under white noise disturbance, is [ + ] P z = Tr (B w + LD yw ) e A L t e A Lt (B w + LD yw ) dt, 0 i.e., P z = Tr[Y L ], where Y L is the controllability gramian of the pair (A L, B L ), with B L = B w + LD yw, and A L Y L + Y L A L = B L B L. In other words, AY L + LC y Y L + Y L A + Y L C y L + B w B w + LD yw D yw L = 0 E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

21 Towards a solution of the LQE problem (intuition) Assume there is a matrix S such that L = Y L S; then, Y L must satisfy AY L + Y L SC y Y L + Y L A + Y L C y S Y L + B w B w + Y L SD yw D yw S Y L = 0 Set S = C D ) 1 y (D yw yw. Then, Y L must satisfy AY L + Y L A Y L C y (D yw D yw ) 1 C y Y L + B w B w = 0, and L = Y L C y (D yw D yw ) 1. Duality to the LQR problem is more and more apparent... E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

22 Technical conditions B1) (C y, A) detectable. B2) (A, B w ) stabilizable. [ ] A jωi B B3) w has full row rank for all ω R. C y D yw B4) Assume D yw D = R ww invertible, i.e., D yw has full row rank. yw B1-B3) ensure that the Riccati equation admits a solution Y that is positive semi-definite. In particular, B1) is obviously necessary, B2) ensures that any unstable mode of A can be excited by the disturbance, and B3) ensures that errors are penalized at all frequencies (this is an additional technical condition ensuring that the Hamiltonian does not have purely imaginary eigenvalues). B4) is just for convenience. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

23 LQE: optimal observer It turns out that the optimal observer can be obtained from the unique, symmetric, positive-definite solution Y of the (algebraic) Riccati Equation (A B D R 1 D R 1 w yw ww C y ) Y + Y (A B w yw ww C y ) YC y Rww 1 C y Y + B w (I D yw Rww 1 D )B yw w = 0 by setting L = (YC y + B w D yw )Rww 1. Define A L := A + LC y, and B L := B w + LD yw. Recall that Y can be interpreted as the reachability Gramian of (A L, B L ), describing the power of the response to a white noise input of the closed-loop system (A L, B L, I, 0). Hence P z = Y. Optimality is proven in a similar way as that of LQR. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

24 H 2 Synthesis LQG The general version of the problem can be seen as a combination of the LQR problem and of the LQE problem. This is also called the LQG problem. By the separation principle, we can design the optimal controller for LQR, and independently design the optimal observer for LQE. Can we claim that the model-based output feedback controller is indeed optimal? E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

25 Technical conditions A1, B1) (A, B u ) stabilizable, (C y, A) detectable. [ ] A jωi B A3) u has full column rank for all ω R. C z D zu [ ] A jωi B B3) w has full row rank for all ω R. C y D yw A4, B4) D zu D zu = R uu > 0, D yw D yw = R ww > 0. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

26 H 2 optimal controller Controller gain: F = R 1 (B X + D C z ), where X is the stabilizing uu u zu solution to the ARE: (A B R 1 D C z ) X + X (A B u R 1 D u 1 zu uu zuc z ) XB u Ruu 1 B u X + C z (I D zu Ruu 1 D zu )C z = 0. Observer gain: L = (YC y + B w D to the ARE: (A B D R 1 C y )Y + Y (A B w D R 1 C y ) w yw ww yw ww yw )R 1, where Y is the stabilizing solution Controller/Observer models: [ ] [ ] A + B u F I A + LC y B w + LD G yw c :=, G f :=. C z + D zu F 0 I 0 ww YC y R 1 C y Y + B w (I D R 1 D yw )B = 0. ww yw ww w E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

27 H 2 optimal controller The state-space model of the optimal controller is then given by K = A + B u F + LC y L B u F 0 I, C y I 0 where the second input and second output of K are connected through an arbitrary stable system Q. Lengthy calculations show that C zw H 2 = Tr[B XB w ] + Tr[D zu FYF D ] + Tr[R uu QR ww ]. 2 w zu Clearly, the minimum amplification is attained when Q = 0, i.e., the conjectured model-based output feedback controller is indeed optimal. E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May 4, / 27

28 MIT OpenCourseWare J / J Dynamic Systems and Control Spring 2011 For information about citing these materials or our Terms of Use, visit: