Let ek be the standard unit vector of dimension q and let zk(.) denote ..... design a full information state-feedback controller which tracks Ï. This results in the ...
H2 -Optimal Control • The H2 -norm: Computation and Interpretation • Response of systems to white noise and coloring filters • Spectral factorization and the Positiv Real Lemma • Optimal H2 -synthesis by static state-feedback • Optimal H2 -estimation (Kalman filtering) • Optimal H2 -synthesis by output feedback (Separation Principle) Related Reading [KK]: 8, 9, 11 (stationary version) Doyle, Glover, Khargonekar and Francis, IEEE TAC 34 (1989) 831-847. 1/82
The H2 -Norm Consider the LTI system with state-space description x˙ = Ax + Bw, z = Cx and with transfer matrix T (s) = C(sI − A)−1 B of dimension p × q. Here w is a disturbance input and z is an output of interest that is desired to be small. A quantification of the effect of the input w onto the output z is the so-called H2 -norm of the transfer matrix. Definition 1 Let T have all its poles in the open left half-plane. The H2 -norm of T is defined as s Z ∞ 1 kT k2 := kT (iω)k2F dω 2π −∞ with k.kF denoting the Frobenius matrix norm. 2/82
Relation to Hardy-Spaces The Hardy-space H2p×q consists of all matrices S of dimension p × q whose elements are analytic functions in the open right half-plane s.th. Z ∞ 1 2 kS(r + iω)k2F dω is finite. kSk2 := sup r>0 2π −∞ For all such functions one can show that the limit Tˆ(iω) := lim S(r + iω) r&0
exists for almost all ω ∈ R, that ω → Tˆ(iω) is square integrable over R, and that kSk2 actually equals kTˆk2 as defined on the previous slide. One can as well show that H2p×q is a Hilbert-space. The subspace of all strictly proper and stable real rational matrices is denoted as RH2p×q . The subspace RH2p×q is dense in H2p×q . This links our discussion to the fascinating area of Hardy-spaces in Math. 3/82
Parseval and Payley-Wiener The space Lp×q 2 [0, ∞) denotes all real-matrix-valued functions on the positive half-line that are square integrable. For any F ∈ Lp×q 2 [0, ∞) the Fourier-transform is defined, in the sense of a limit in the mean, as Z ∞ ˆ F (iω) = e−iωt F (t) dt. 0
Parseval’s theorem states that Z ∞ Z ∞ 1 2 kF (t)kF dt = kFˆ (iω)k2F dω. 2π −∞ 0 One can show that Fˆ is in H2p×q . Parseval’s theorem just means that the p×q Fourier transform is a linear isometry Lp×q . A version of 2 [0, ∞) → H2 the Payley-Wiener theorem establishes that this map is even surjective. p×q Therefore Lp×q are actually isometrically isomorphic. 2 [0, ∞) and H2 See: B.A. Francis, A course in H∞ -Control, Springer LNCIS 88, 1987. 4/82
Computation It is a beautiful fact that the H2 -norm of a stable transfer matrix can be computed algebraically on the basis of a state-space realization. Theorem 2 Let A be Hurwitz and T (s) = C(sI − A)−1 B. Then: 1. kT k22 = tr CPc C T where APc + Pc AT + BB T = 0. 2. kT k22 = tr B T Po B where AT Po + Po A + C T C = 0. R∞ T Proof of 2.: Recall that Po = 0 eA t C T CeAt dt. Using Parseval’s theorem we infer Z ∞ 1 2 tr T (iω)∗ T (iω) dω = kT k2 = 2π −∞ Z ∞ = tr [CeAt B]T [CeAt B] dt = tr B T Po B . 0
The proof of 1. proceeds in an analogous fashion. 5/82
Inequality Characterization The following characterizations of a bound on the H2 -norm will be useful for controller synthesis proofs. Lemma 3 A is Hurwitz and kT k22 < γ iff there exists X 0 with AT X + XA + C T C ≺ 0 and tr(B T XB) ≺ γ. Proof of if. Since AT X + XA ≺ 0 and X 0, A is Hurwitz. By subtracting the Lyapunov equation for Po we infer AT (X − Po ) + (X − Po )A ≺ 0 and thus Po ≺ X since A is Hurwitz. This implies tr(B T Po B) ≤ tr(B T XB) < γ and thus kT k22 < γ. Proof of only if. Consider the solution P of the perturbed Lyapunov equation AT P + P A + C T C + I = 0 with > 0. Clearly P → Po for → 0. Since tr(B T Po B) ≺ γ, we can hence fix some 0 with tr(B T P0 B) ≺ γ. Then X = P0 does the job. 6/82
Deterministic Interpretation Let ek be the standard unit vector of dimension q and let zk (.) denote the response of x˙ = Ax, z = Cx, x(0) = Bek Recall that this is just the response to an impulse in the k-th input. Since zk (t) = CeAt Bek we infer Z ∞ Z ∞ AT t T At T T T e C Ce dt Bek = eTk B T Po Bek . zk (t) zk (t) dt = ek B 0
0
After summing over k the right-hand side is tr(B T Po B) = kT k22 . The squared H2 -norm is the sum of the energies of the transients of “the” impulse responses: q Z ∞ X kzk (t)k2 dt = kT k22 . k=1
0 7/82
Random Vectors Uncertain outcomes of experiments are modeled by random vectors x = (x1 · · · xn )T . Here x is a vector of n random variables x1 , . . . , xn and is characterized by its distribution function Fx : Rn → R which admits the following interpretation: If (ξ1 · · · ξn )T ∈ Rn , the probability for the event x1 ≤ ξ1 , ..., xn ≤ ξn to happen is given by Fx (ξ1 , ..., ξn ). Fx (ξ1 , ..., ξn ) has the density fx (τ1 , ..., τn ) in case that Z ξ1 Z ξn Fx (ξ1 , ..., ξn ) = ··· fx (τ1 , ..., τn ) dτn · · · dτ1 for all ξ ∈ Rn . −∞
−∞
The rv x is Gaussian if its distribution function has the density 1 1 T −1 exp (τ − m) R (τ − m) fx (τ ) = p 2 (2π)n det(R) where m ∈ Rn and R ∈ Rn×n is symmetric and positive definite. 8/82
Expectation and Covariance Suppose g : Rn → Rk×l is Borel measurable. If the random vector x = (x1 · · · xn )T has the density fx (τ1 , ..., τn ), the expectation of g(x1 , ..., xn ) is a matrix in Rk×l and defined as Z +∞ Z +∞ E[g(x1 , ..., xn )] = ··· g(τ1 , ..., τn )fx (τ1 , ..., τn ) dτn · · · dτ1 . −∞
−∞
Examples. With g(τ ) = τ we obtain the expectation E[x] of x itself. With g(τ, σ) = (τ −E[x])(σ−E[y])T we obtain the covariance matrix cov(x, y) := E[(x − E[x])(y − E[y])T ] of the two random vectors x and y. Then cov(x, x) = E[(x − E[x])(x − E[x])T ] = E[xxT ] − E[x]E[x]T < 0 is the auto-covariance matrix x. Its trace is called variance of x. Moreover, E[xxT ] is the second order moment matrix of x and its trace E[xT x] is the second order moment of x.
9/82
Wiener-Processes In control the disturbance w on slide 2 is often considered as white noise, which is associated with irregular signals having a flat spectrum. Loosely speaking, this boils down to viewing w as the derivative of the (normalized) Wiener-process or Brownian motion. It would take us too far astray and it is not required for our purposes to develop the whole theory of stochastic differential equations (based on ˆIto calculus). Instead let us just collect some basic facts that are required in the sequel. 1) There exists a Wiener-process W (t) for t ≥ 0 with intensity 1: • Initialized at zero: W (0) = 0 with probability one. • Independent increments: For all 0 ≤ t1 ≤ t2 ≤ t3 ≤ t4 , the random variables W (t2 ) − W (t1 ) and W (t4 ) − W (t3 ) are independent. • Gaussian increments: For all 0 ≤ t1 ≤ t2 , the increment W (t2 ) − W (t1 ) is Gaussian with expectation 0 and variance t2 −t1 = 1|t2 −t1 |. 10/82
Simulation of Standard Wiener Process Realizations of Wiener Process 8 6 4 2 0 −2 −4
0
1
2
3
4
5
6
7
8
9
10
• Sample paths are continuous with probability one. • W (t) for t > 0 is Gaussian with density fW (t) (τ ) =
τ2
√ 1 e− 2t 2πt
.
• The process W itself is Gaussian: For all k ∈ N and all pairwise different time instances t1 , . . . , tk > 0 the random vector T W (t1 ) · · · W (tk ) is Gaussian. 11/82
Wiener-Processes 2) For a square integrable real-valued function f on [a, b] (0 ≤ a ≤ b), Z b the random variable f (t) dW (t) a
is defined analogously to standard Lebesgue-Stieltjes integrals as follows: • For a step-function s(.) which takes the values s1 , . . . , sN on the partition [tk , tk+1 ], k = 1, . . . , N (a = t1 < · · · < tN +1 = b), define Z b N X s(t) dW (t) := sk [W (tk+1 ) − W (tk )]. a
k=1
• With a sequence sν of step-functions that converges to f in L2 [a, b], Z b Z b sν (t) dW (t), f (t) dW (t) := lim ν→∞ a | {z } {z } |a I
Iν
2
in the sense of E[(I − Iν ) ] → 0 for ν → ∞. 12/82
Wiener-Processes 3) The integral is a Gaussian random variable. Just by direct calculation for step-functions and taking limits, one proves for x, y ∈ L2 [a, b] that Z b E x(t) dW (t) = 0, a
Z E
b
Z x(t) dW (t)
a
b
Z b y(τ ) dW (τ ) = x(t)y(t) dt.
a
a
ˆ is a Wiener-process that is independent from W then If W Z b Z b ˆ E x(t) dW (t) y(τ ) dW (τ ) = 0. a
a
All these properties about expectation and correlation of integrals of deterministic functions with respect to Wiener-processes are the only facts from stochastics that remain unproven here. L. Arnold, Stochastic Differential Equations: Theory and Applications, Wiley, 1974 13/82
Wiener-Processes The q-dimensional Wiener-process W = col(W1 , . . . , Wq ) is a vector of q Wiener processes W1 , . . . , Wq (with intensity 1) that are pairwise independent. If X and Y are matrix-valued functions of dimension p × q with square integrable elements on [a, b] (0 ≤ a ≤ b), then the random vectors Z b Z b x= X(t) dW (t), y = Y (t) dW (t) a
a
of dimension p are defined elementwise. They both have mean zero and and their correlation matrix is given by Z b T E[xy ] = X(t)Y (t)T dt. a
Proof. Verify these properties elementwise. 14/82
White Noise and System Response Let us come back to slide 2 with w being interpreted as the “derivative ˙ . In this sense “W of a Wiener process”; we (formally) denote it by W can be obtained by integrating white noise”: Z t Z t ˙ W (t) = “ W (τ ) dτ “ = dW (τ ) for t ≥ 0. 0
0
The middle expression is NOT sensible mathematically. But we can now just define precisely what we mean by the state-response of a linear system to a white noise input and a random initial condition ξ. Tacitly, we assume that ξ is Gaussian and independent from W (t) for all t ≥ 0. ˙ Definition 4 The response of the linear system x˙ = Ax + B W with x(0) = ξ (and ξ independent from W ) is defined as Z t At x(t) := e ξ + eA(t−τ ) B dW (τ ) for t ≥ 0. 0
15/82
White Noise and System Response According to our preparatory remarks x(.) is a Gaussian process. If just applying the rules given above, the expectation of x(t) and the covariance matrix of x(t1 ) and x(t2 ) are easily determined. ˙ , x(0) = ξ. Theorem 5 Let x(.) be the response of x˙ = Ax + B W Then E[x(t)] = eAt E[ξ] for t ≥ 0 and cov(x(t1 ), x(t2 )) = e
At1
cov(ξ, ξ)e
AT t2
Z +
t1
eA(t1 −τ ) BB T eA
T (t −τ ) 2
dτ
0
for 0 ≤ t1 ≤ t2 . The specialization to A = 0 and ξ = 0 implies x(t) = BW (t) and thus E[BW (t)] = 0 as well as E[BW (t)W (t)T B T ] = tBB T . 16/82
Stochastic Interpretation of H2 -Norm Corollary 6 Let eig(A) ⊂ C− and x(.), z(.) be the state- and ˙ , z = Cx, x(0) = ξ. Then output-responses of x˙ = Ax + B W E[x(t)] → 0 and E[z(t)] → 0 for t → ∞. Moreover: • The state’s auto-covariance satisfies lim cov(x(t), x(t)) = Pc . t→∞
• The output-variance satisfies lim tr cov(z(t), z(t)) = kT k22 . t→∞
kT k22 is the asymptotic variance limt→∞ E[z(t)T z(t)] − E[z(t)]T E[z(t)] of the output of a stable linear system driven by white noise. Proof. E[x(t)] = eAt E[ξ] → 0 and E[z(t)] → 0 for t → 0 are clear. T Since lim eAt cov(ξ, ξ)eA t = 0, we infer that lim cov(x(t), x(t)) is t→∞ t→∞ Z t Z t T T lim eA(t−τ ) BB T eA (t−τ ) dτ = lim eAσ BB T eA σ dσ = Pc . t→∞
t→∞
0
0
This also implies lim tr cov(z(t), z(t)) = lim tr C[cov(x(t), x(t))]C T = tr(CPc C T ).
t→∞
t→∞
17/82
Example √ ˙ , z = x for p > 0. Simulate the system and Consider x˙ = −px + 2p W plot the output process and a histogram thereof for p = 10, 100: Squared H2−norm: 1
Estimated variance 1.0747
4
300
2
200
0 100
−2 −4 0
5
10
0 −5
Squared H2−norm: 1 400
2
300
0
200
−2
100 5
5
Estimated variance 1.0272
4
−4 0
0
10
0 −5
0
5 18/82
Colored Noise Definition 7 We say that w˜ is colored noise if there exists ˜ B, ˜ C) ˜ with eig(A) ˜ ⊂ C− such that w˜ is the output of (A, ˜x + B ˜W ˙ , w˜ = C˜ x˜, x˜(0) = 0. x˜˙ = A˜ This is also expressed as w˜ emerging through filtering white noise with ˜ ˜ −1 B. ˜ the coloring filter T˜(s) = C(sI − A) The response of the linear system x˙ = Ax + B w, ˜ z = Cx + Dw, ˜ x(0) = ξ driven by colored noise w˜ is defined as the output of 0 x˙ A B C˜ x x(0) ξ ˙ = + ˜ W, = ˙x˜ ˜ x ˜ x ˜ (0) 0 B 0 A x z = C DC˜ . x˜ 19/82
Remarks R t A(t−τ ˜ dW (τ ) for t ≥ 0, we can assume w.l.o.g. ˜ ˜ )B • Since w(t) ˜ = 0 Ce ˜ B, ˜ C) ˜ is minimal, and that the coloring filter T˜ (and not that (A, its realization) captures all properties of w. ˜ This raises the question about how to determine such coloring filters in practice. • The response of a linear system to colored noise is reduced to that for white noise and for the series interconnection of the system and the coloring filter. • In the theory of stochastic differential equations it is shown that our definitions are consistent. More precisely, the response of x˙ = Ax + B w, ˜ z = Cx + Dw, ˜ x(0) = ξ can indeed be represented as it should, namely as Z t At x(t) = Ce ξ + CeA(t−τ ) B w(τ ˜ ) dτ + Dw(t) ˜ for t ≥ 0, 0
if defining the integral appropriately.
20/82
Coloring Filters Clearly E[w(t)] ˜ = 0 for all t ≥ 0. For a fixed τ ∈ R let us consider the asymptotic covariance matrix of w(t) ˜ and w(t ˜ + τ ): R(τ ) := lim E[w(t ˜ + τ )w(t) ˜ T ]. t→∞
It turns out that R(τ ) can be easily obtained algebraically as follows: Theorem 8 If A˜ is Hurwitz and P˜ is ˜B ˜T Lyapunov equation A˜P˜ + P˜ A˜T + B ( ˜ ˜ ˜T ˜ Aτ Ce PC R(τ ) = T ˜ C˜ P˜ e−A τ C˜ T
the unique solution of the = 0 then for τ ≥ 0, for τ < 0.
˜ ⊂ C− , R(τ ) decays exponenNote that R(−τ )T = R(τ ). Since eig(A) tially for τ → ±∞ and has a well-defined Fourier transform. ˆ of R is called the spectral Definition 9 The Fourier transform R density of the process w. ˜ 21/82
Proof For t ≥ 0 and τ > 0 the filter state satisfies Z t ˜ T ˜B ˜ T eA˜T (t−σ) dσ = E[˜ x(t + τ )˜ x(t) ] = eA(t+τ −σ) B 0
=e
˜ Aτ
Z
t
˜T ˜ ˜ ˜ ˜T A B e σ dσ → eAτ P˜ for t → ∞. eAσ B
0
Similarly for t ≥ 0 and τ ∈ [−t, 0] we get Z t+τ ˜ T ˜B ˜ T eA˜T (t−σ) dσ = E[˜ x(t + τ )˜ x(t) ] = eA(t+τ −σ) B 0
Z =
t+τ
˜ ˜ ˜T A ˜T ˜T eAσ B B e (σ−τ ) dσ → P˜ e−A τ for t → ∞.
0
The proof is finished with the observation ˜ x(t + τ )˜ R(τ ) = lim CE[˜ x(t)T ]C˜ T . t→∞
22/82
Coloring Filters Theorem 10 The spectral density of w˜ is given by ˆ R(iω) = T˜(iω)T˜(iω)∗ . ˆ Hence R(iω) is Hermitian and positive semi-definite for all ω ∈ R. ˆ Proof. By definition R(iω) equals Z ∞ Z 0 ˜ ˜ ˜T ˜T τ ˜ T −iωτ ˜ ˜ −A ˜ −iωτ eAτ P C dτ = C dτ + Ce CP e e 0 −∞ Z ∞ Tτ ˜ iωτ A −1 ˜ P˜ C˜ T = e e = C˜ P˜ dτ + (iωI − A) 0 ˜ −∗ + (iωI − A) ˜ −1 P˜ C˜ T . = C˜ P˜ (iωI − A) ˜B ˜ T = 0 also satisfies the equation Now note that A˜P˜ + P˜ A˜T + B ˜ P˜ + P˜ (iωI − A) ˜ ∗=B ˜B ˜ T and thus (iωI − A) ˜ −∗ + (iωI − A) ˜ −1 P˜ = (iωI − A) ˜ −1 B ˜B ˜ T (iωI − A) ˜ −∗ . P˜ (iωI − A) Plugging this into the above relation proves the formula.
23/82
Example Consider the colored noise on slide 18. The filter has the realization √ −p − 2p ˜ T = 1 0 and hence P˜ = 1. Therefore −p|τ |
R(τ ) = e
ˆ and R(iω) =
√ √ 2p 2p 2p . = 2 p + iω p − iω ω + p2
The graphs for p = 10 (blue) and p = 100 (red) look as follows: R(τ)
|T~(iω)|
0
1
10
0.8 0.6
−2
10 0.4 0.2 0 −0.1
−4
−0.05
0
0.05
0.1
10
−2
10
0
10
2
10
4
10
24/82
Coloring Filters: Construction We are now prepared for discussing how to determine coloring filters in practice. By measurements one (statistically) estimates the spectral ˆ density R(iω) of the process under scrutiny. One then approximates the experimentally determined spectral density by G(iω) where G(s) is a strictly proper real rational function without poles in C0 and with G(iω) = G(iω)∗ and G(iω) < 0 for all ω ∈ R.
(∗)
Finally, the coloring filter is obtained by spectral factorization. Theorem 11 Suppose that the strictly proper real-rational function G(s) without poles in C0 satisfies (∗). Then there exists a strictly proper and stable transfer matrix T with G(s) = T (s)T (−s)T . This implies G(iω) = T (iω)T (iω)∗ such that T is a coloring filter for modeling noise with the spectral density G, just as we desired. Often T is also called a spectral factor of G.
25/82
Example Consider the transfer function G(s) =
1−s2 . s4 −13s2 +36
We have
1 + ω2 > 0 for ω ∈ R. ω 4 + 13ω 2 + 36 Hence, by Theorem 11, G has a spectral factorization. This can be seen directly as follows. The numerator and denominator of G(s) can be factorized as G(iω) =
(1 + s)(1 − s) and (3 + s)(2 + s)(2 − s)(3 − s) respectively. The symmetric location of the poles and zeros with respect to the imaginary axis is a consequence of G(iω) being real and positive for ω ∈ R. Obviously, both transfer functions 1+s 1−s and T+ (s) = T− (s) = (3 + s)(2 + s) (3 + s)(2 + s) are spectral factors of G. They distinguish themselves in that T− shares it stable zero(s) with G, while T+ shares its anti-stable zero(s) with G. 26/82
System Description for Design Let us now consider the following generalized plant x˙ = Ax + Bw w + Bu z = Cz x + Dzw w + Dz u y = Cx + Dw w + Du
w
z y
P
u
with an external disturbance input w (that cannot be influenced), a control input u, a performance output z (to-be-rendered small) and a measurement output y. Recall how various concrete configurations of disturbance rejection as discussed in the lecture on regulation can be subsumed to this paradigm. The goal is to find a feedback controller which stabilizes this system and which minimizes the H2 -norm of the closed-loop transfer matrix. For simplicity of the exposition let us assume Dzw = 0 and D = 0.
27/82
The H2 -Control Problem Open-loop system P : x˙ = Ax + Bw w + Bu z = Cz x + Dz u y = Cx + Dw w Controller K: x˙ K = AK xK + BK y u = C K xK
w
z P y
u K
Controlled system described as ξ˙ = Aξ + Bw, z = Cξ with A BCK Bw A B = BK C AK BK Dw . C D Cz Dz CK 0 Problem: Minimize the H2 -norm kC(sI −A)−1 Bk2 of the controlled closed-loop system over all controllers which render A Hurwitz. 28/82
LQG-Control One particular scenario is worth mentioning. Consider the system ˙ 1 + Bu x˙ = Ax + B1 W ˙ 1 . Suppose the measurewith control input u and process noise B1 W ˙ 2 where W1 and W2 are indements Cx are corrupted by white noise W pendent Wiener processes. The measured output hence is ˙ 2. y = Cx + D2 W As in LQ-control, we are interested in keeping the linear combinations C1 x and D1 u of the states and the controls small. Hence we choose C1 x z= D1 u as the performance output. The LQG-control goal is to find a stabilizing controller which minimizes lim tr cov(z(t), z(t)), which equals t→∞ lim E[z(t)T z(t)] = lim E x(t)T C1T C1 x(t) + u(t)T D1T D1 u(t) . t→∞
t→∞
29/82
LQG-Control Let us define the following generalized plant x˙ = Ax + B1 0 w + Bu C1 0 z = x+ u 0 D1 y = Cx + 0 D2 w. Finding a stabilizing output-feedback controller which minimizes the asymptotic variance of z for a while noise w is the classical so-called Linear-Quadratic-Gaussian (LQG) optimal control problem. In view of our preparation this is merely a special case of H2 -control! Note that the process and measurement noises are defined via the Wiener-processes B1 W1 and D2 W2 with auto-covariances E B1 W1 (t)W1 (t)T B1T = B1 B1T t, E D2 W2 (t)W2 (t)T D2T = D2 D2T t. Hence B1 B1T and D2 D2T are the intensities of these processes.
30/82
LQG-Control ˙ 1 is not white but colored noise w˜1 , one absorbs the related coloring If W ˜ ˜ −1 B ˜ into the generalized plant and solves the filter T (s) = C(sI − A) H2 -problem for the weighted generalized plant 0 0 x˙ A B1 C˜ x B = + ˜ w+ u ˜ x˜ x˜ 0 B 0 0 A C1 0 x 0 z = + u 0 0 x˜ D1 x y = C 0 + 0 D2 w. x˜ The block-diagram of the weighted generalized plant is more instructive: w˜1
z P y
w2
T
w1
u 31/82
Static State-Feedback Synthesis Let us first consider the case that the whole state is available for control. For this purpose we consider the open-loop system x˙ = Ax + Bw w + Bu z = Cz x + Dz u under the following hypotheses: • (A, B) is stabilizable. • (A, Cz ) has no unobservable modes in C0 . • DzT Cz Dz = 0 I . Controlling the system by static state-feedback as u = −F x leads to the closed-loop system x˙ = (A − BF )x + Bw w z = (Cz − Dz F )x.
32/82
Optimal H2 -Control by Static State-Feedback The goal is to infimize k(Cz − Dz F )(sI − A + BF )−1 Bw k2 over all F such that A − BF is Hurwitz. It turns out that the infimum is actually attained. Moreover, both the optimal value and an optimal state-feedback gain can be computed by solving an LQ-Riccati equation. Theorem 12 Let P denote the stabilizing solution of AT P + P A − P BB T P + CzT Cz = 0. Then the optimization problem γopt :=
min
F , A−BF Hurwitz
k(Cz − Dz F )(sI − A + BF )−1 Bw k22
has the optimal value γopt = tr(BwT P Bw ) and F = B T P is an optimal solution. 33/82
Proof Let γ > γopt . Then there exists F for which A − BF is Hurwitz and such that k(Cz − Dz F )(sI − A + BF )−1 Bw k22 < γ. Due to Lemma 3 we can choose X 0 with (A − BF )T X + X(A − BF ) + (Cz − Dz F )T (Cz − Dz F ) ≺ 0 and tr(BwT XBw ) < γ. Exploiting DzT Cz = 0 and DzT Dz = I allows to rearrange to AT X + XA + CzT Cz − XBB T X + (B T X − F )T (B T X − F ) ≺ 0. Therefore AT X + XA + CzT Cz − XBB T X ≺ 0 and thus P ≺ X by the result on slide 68. This implies that tr(BwT P Bw ) 4 tr(BwT XBw ) < γ. Since γ > γopt was arbitrary we conclude tr(BwT P Bw ) ≤ γopt . 34/82
Proof Let us now choose F = B T P . Note that A − BF = A − BB T P is Hurwitz, just because of the choice of P as the stabilizing solution of the ARE. Moreover with the ARE we trivially have AT P + P A + CzT Cz − P BB T P + (B T P − F )T (B T P − F ) = 0. As above this can be re-arranged to (A − BF )T P + P (A − BF ) + (Cz − Dz F )T (Cz − Dz F ) = 0 Viewed as a Lyapunov equation, this shows that k(Cz + Dz F )(sI − A − BF )−1 Bw k22 = tr(BwT P Bw ). This proves that tr(BwT P Bw ) is attained by the stabilizing state-feedback gain F = B T P which indeed confirms tr(BwT P Bw ) = γopt and optimality of F . 35/82
Example For the following model of a DC-motor ([F] p.20) 0 1 0 φ x˙ = x+ u, x = . 0 −1 1 ω design a full information state-feedback controller which tracks φ. This results in the following response: Control action 200 0 −200 0
2
4 6 Reference (red) and output
8
10
8
10
2 0 −2 0
2
4
6
36/82
Example Assume that the control input is affected by colored noise with filter 134.2/(s + 10) and having H2 -norm 30 in order to clearly display its effect. We get the following substantially deteriorated response: Control action 200 0 −200 0
2
4 6 Reference (red) and output
8
10
8
10
2 0 −2 0
2
4
6
37/82
Example With the coloring filter included in the system description, design an optimal H2 -state-feedback gain for the cost function 2000|φ|2 + 0.3|u|2 . The feed-forward gain is adjusted appropriately. The cost has been tuned so that the noise-free response resembles the one we obtained earlier: Control action 200 0 −200 0
2
4 6 Reference (red) and output
8
10
8
10
2 0 −2 0
2
4
6
38/82
Example The noisy closed-loop response shows that the effect of the noise onto the to-be-tracked output is visibly reduced: Control action 200 0 −200 0
2
4 6 Reference (red) and output
8
10
8
10
2 0 −2 0
2
4
6
39/82
Kalman Filtering Consider again the full generalized plant x˙ = Ax + Bw w + Bu, z = Cz x + Dz u, y = Cx + Dw w. If w = 0, an observer for this system is defined as xˆ˙ = Aˆ x + Bu + L(y − yˆ), zˆ = Cz xˆ + Dz u, yˆ = C xˆ where L is taken with eig(A − LC) ⊂ C− ; then the observer state asymptotically reconstructs the system’s state. If w does not vanish and is a white-noise disturbance, the quantity lim E (z(t) − zˆ(t))T (z(t) − zˆ(t)) (err) t→∞
can serve as a measure for how well zˆ(t) approximates z(t) for t → ∞. An observer for which the asymptotic variance (err) of z − zˆ is minimized is called a Kalman Filter for the generalized plant. 40/82
Optimal H2 -Observer Synthesis By considering the dynamics of the state-error ξ = x − xˆ, one easily checks that the transfer matrix from w to z − zˆ admits the description ξ˙ = (A − LC)ξ + (Bw − LDw )w, z − zˆ = Cz ξ. H2 -optimal observer synthesis problem Find L which renders A − LC Hurwitz and minimizes the H2 -norm of the transfer matrix from w to z − zˆ. We stress again that this formulation admits various interpretations, according to what can be subsumed to H2 -norm minimization. The problem is solved under the following simplifying assumptions: • (A, C) is detectable. • (A, Bw ) has no uncontrollable modes in C0 . • Dw BwT DwT = 0 I . 41/82
Optimal H2 -Observer Synthesis We hence need to determine L which renders A − LC Hurwitz and which minimizes kCz (sI − A + LC)−1 (Bw − LDw )k2 . The solution to this problem is dual to Theorem 12. Theorem 13 Let Q denote the stabilizing solution of AQ + QAT − QC T CQ + Bw BwT = 0. Then the optimization problem γopt :=
min
L, A−LC Hurwitz
kCz (sI − A + LC)−1 (Bw − LDw )k22
has the optimal value γopt = tr(Cz QCzT ) and L = QC T is an optimal solution. 42/82
Proof by Duality Note that we have kCz (sI − A + LC)−1 (Bw − LDw )k2 = = k(BwT − DwT LT )(sI − AT + C T LT )−1 CzT k2 . This observation reduces the problem to one of static state-feedback control for the system ˇ xˇ˙ = AT xˇ + CzT wˇ + C T uˇ, zˇ = BwT xˇ + DwT w. An application of Theorem 12 finishes the proof. (Provide the details.)
43/82
Remarks • Note that the optimal observer does not depend on Cz or Dz ! In particular it is as well an optimal H2 -observer for the full state x (with the choices Cz = I and Dz = 0) with optimal value tr(Q). • An optimal H2 -estimator is an unstructured LTI system with inputs u and y, which generates an asymptotic state-estimate xˆ such that Cz xˆ +Dz u is an optimal estimate of z in the H2 -sense. One can prove that general estimators do not offer any benefit over observers! • In view of the stochastic interpretation of the H2 -norm, optimal H2 observers minimize the asymptotic variance of z − zˆ if w is white noise. Then the optimal observer is the celebrated Kalman-Filter. R.E. Kalman, A new approach to linear filtering and prediction problems, Journal of basic Engineering, 82 (1960) 35-45. (13560 citations in Google Scholar as of February 7, 2013!) 44/82
Example Consider again the two-compartment model ([AM] pp.85): c˙ = y =
−k0 − k1 k1 k2 −k2 1 0 c + λ2 w 2
c+
0 λ1
w1 +
b0 0
u
k0 > 0, k1 > 0, k2 > 0 in which the second state and the output are corrupted by (independent) white noises w1 and w2 and λ1 and λ2 scale their intensities. For k0 = 1, k1 = 1, k2 = 2 and b0 = 1 we design a pole-placing observer with pole locations −1.5, −1.6 and compare it with the Kalman filter. Note that the poles have been chosen such that the responses resemble those for the Kalman filter if the intensities λ1 and λ2 are small (noise-free case). The simulations on the next slides illustrate a substantial reduction in noise sensitivity if using the Kalman filter (in the lower plots). 45/82
Example System and observer responses as well as errors for λ1 = λ2 = 0.01: 2
2
1
1
0
0
−1
−1
−2 0
5
10
−2 0
2
2
1
1
0
0
−1
−1
−2 0
5
10
−2 0
1 0 −1
5
10
−2 0
5
10
5
10
1 0 −1
5
10
−2 0
46/82
Example System and observer responses as well as errors for λ1 = λ2 = 0.1: 4 2 0 −2 0
5
10
2
4
1
2
0
0
−1
−2
−2
−4 0
5
10
2
1
5
10
5
10
1 0
0
0
−1 −2
−1 −2 0
−3 0
5
10
−4 0
−2 5
10
−3 0
47/82
Example System and observer responses as well as errors for λ1 = λ2 = 1: 4
4
4
2
2
2
0
0
0
−2
−2
−2
−4 0
5
10
4 2 0 −2 0
5
10
−4 0
5
10
−4 0
4
4
2
2
0
0
−2
−2
−4 0
5
10
−4 0
5
10
5
10 48/82
The Output-Feedback H2 -Control Problem Open-loop system P : x˙ = Ax + Bw w + Bu z = Cz x + Dz u y = Cx + Dw w Controller K: x˙ K = AK xK + BK y u = C K xK
w
z P y
u K
Controlled system described as ξ˙ = Aξ + Bw, z = Cξ with A BCK Bw A B = BK C AK BK Dw . C D Cz Dz CK 0 Problem: Minimize the H2 -norm kC(sI −A)−1 Bk2 of the controlled closed-loop system over all controllers which render A Hurwitz. 49/82
Hypotheses We derive a solution to this optimal synthesis problem in terms of AREs under the following assumptions: 1. (A, B) is stabilizable and (A, C) is detectable. Are required for the existence of a stabilizing controller. Bw 0 T T Dw = . 2. Dz Cz Dz = 0 I and Dw I It’s essential that Dz and Dw have full column and row rank. The other properties are introduced to simplify the formulas. 3. (A, Cz ) has no unobservable and (A, Bw ) no uncontrollable modes on the imaginary axis. Are required for the existence of stabilizing solutions of AREs. A fully general solution without hypotheses can be obtained with LMIs. 50/82
Output-Feedback Control: Riccati Equation Solution Under the given hypotheses the H2 -problem admits an optimal solution. This main result of this lecture is formulated as follows. Theorem 14 With the stabilizing solutions P and Q of the ARE’s AT P + P A − P BB T P + CzT Cz = 0, AQ + QAT − QC T CQ + Bw BwT = 0, an H2 -optimal controller is given as x˙ K = (A − BB T P − QC T C)xK + QC T y, u = −B T P xK and the corresponding optimal closed-loop H2 -norm is p tr(BwT P Bw ) + tr(B T P QP B). The proof is given in the appendix. The following remarks on the interpretation and on generalizations of this result are essential. 51/82
Output-Feedback Control: The Separation Principle The H2 -optimal state-feedback gain is F = B T P while L = QC T is the H2 -optimal observer gain. With these, the optimal H2 -controller can be written as x˙ K = (A − BF − LC)xK + Ly, u = −F xK or as x˙ K = AxK + Bu + L(y − yˆ), yˆ = CxK , u = −F xK . • Hence the controller is an optimal H2 -observer for the system’s state. • If the system state is available for control then u = −F x is optimal. Otherwise, one just has to replace x by an optimal estimate xK of the state and optimally control the system with u = −F xK . In short: Optimal state-feedback + Optimal estimation = Optimal output-feedback.
• The controller can as well be seen as to provide an H2 -optimal estimate −F xK of the unavailable but to-be-implemented signal −F x. The “extra cost” for this estimation is tr(F QF T ) if compared to the cost for optimal state-feedback controller. 52/82
Output-Feedback Control: Comments In Matlab’s robust control toolbox the command h2syn allows to solve the H2 -control problem. The algorithm is applicable for systems P x˙ = Ax + Bw w + Bu z = Cz x + Dzw w + Dz u y = Cx + Dw w + Du
w
z y
P
u
whose descriptions satisfy the following hypotheses: 1. (A, B) is stabilizable and (A, C) is detectable. There exists some N with Dzw + Dz N Dw = 0. A − iωI B 2. Dz and have full column rank for all ω ∈ R. Cz Dz A − iωI Bw 3. Dw and have full row rank for all ω ∈ R. C Dw These properties are natural generalizations of those on slide 50.
53/82
How to Enforce Satisfaction of Hypotheses? Condition 1 is necessary for the existence of a stabilizing controller which renders the closed-loop transfer matrix strictly proper. If condition 2 fails choose Cze , Dze such that A − iωI B Dz Dz have full column rank ∀ ω ∈ R. , Cz Dze Cze Cze If condition 3 fails choose Bwe , Dwe such that A − iωI Bw Bwe Dw Dwe , have full row column ∀ ω ∈ R. C Dw Dwe These properties can be always achieved, as the following simple but rough choices demonstrate: Bwe I 0 I 0 Cze Dze = and = . 0 I Dwe 0 I 54/82
How to Enforce Satisfaction of Hypotheses? Then define, with small > 0, the perturbed open-loop system P : x˙ z ze y
= = = =
Ax + Bw w + Bwe we + Bu Cz x + Dzw w + Dz u Cze x + Dze u Cx + Dw w + Dwe we + Du
z ze y
P
w we u
The perturbed system satisfies the required hypothesis and one can find an optimal H2 -controller K with optimal value γ . Now interconnect K with the the original system P with closed-loop transfer matrix T . One can show that K also stabilizes P and achieves an H2 -norm level of at least γ : kT k2 ≤ γ . Moreover, γ converges monotonically to the optimal achievable H2 norm level for P (although an optimal controller might not exist). 55/82
Example In the lectures we provide a demo on designing an output-feedback tracking controller for the motor on slide 36. We obtain the following responses for a pole-placement and LQG-synthesis with low noise-levels: Control action 20 0 −20 0
2
4 6 Reference (red) and output
8
10
8
10
8
10
8
10
2 0 −2 0
2
4
6
Control action 20 0 −20 0
2
4 6 Reference (red) and output
2 0 −2 0
2
4
6
56/82
Example For increased noise-levels we see the benefit of LQG-control: Control action 20 0 −20 0
2
4 6 Reference (red) and output
8
10
8
10
8
10
8
10
2 0 −2 0
2
4
6
Control action 20 0 −20 0
2
4 6 Reference (red) and output
2 0 −2 0
2
4
6
57/82
Example However, comparisons of this sort can be misleading: • Obviously the response of the pole-placement observer to non-zero initial conditions is faster than that of the LQG controller. This explains its higher sensitivity to noise. Slowing down the observer poles does not alter the tracking behavior (a lot), but it reduces the sensitivity to noise. • For this simple example, one can obtain similar designs by poleplacement and LQG-synthesis after tuning. In practice, modern synthesis tools rather serve to reduce the time required for tuning a controller, while the optimality properties are not that crucial. • LQG output-feedback controllers suffer from an essential deficiency: There are no guarantees for robustness! This lead to the development of dedicated tools for robust controller synthesis. You are now wellprepared to enter this exciting field of control. 58/82
Appendix
59/82
Proof of Theorem 11: Step 1 Choose a minimal realization G(s) = CG (sI −AG )−1 BG . Since G has no poles in C0 , the matrix AG has only eigenvalues in C− or C+ . In suitable coordinates we can assume that the realization has the structure A1 0 B1 0 A2 B2 , eig(A1 ) ⊂ C− , eig(A2 ) ⊂ C+ . C1 C2 0 Therefore G(s) = C1 (sI − A1 )−1 B1 + C2 (sI − A2 )−1 B2 . Observe that G(−s)T = B2T (−sI − AT2 )−1 C2T + B1T (−sI − AT1 )−1 C1T . Since G(iω)∗ = G(−iω)T , we have G(s) = G(−s)T for s ∈ C0 and hence for all s ∈ C (different from poles of G(s) and G(−s)T ). Thus the stable and anti-stable parts in the additive decomposition of G(s) and G(−s)T coincide.
59/82
Proof of Theorem 11: Step 1 Therefore C1 (sI −A1 )−1 B1 = B2T (−sI −AT2 )−1 C2T = (−B2T )(sI −(−AT2 ))−1 C2T . By realization minimality, there exists a non-singular T such that A1 = −T AT2 T −1 , B1 = T C2T , C1 = −B2T T −1 . Performing yet another state-coordinate change proves the following intermediate fact. G admits a minimal realization A 0 B 0 −AT −C T with eig(A) ⊂ C− . C BT 0 Observe that minimality of the realization is equivalent to (A, B) being controllable and (A, C) being observable. 60/82
Proof of Theorem 11: Step 2 If we define J(s) = C(sI − A)−1 B we infer G(iω) = J(iω) + J(iω)∗ < 0 for all ω ∈ R. Hence J is said to be positive real. This has the following consequence, the result of which is the celebrated Positive Real Lemma. There exists some X = X T such that AX + XAT < 0 and B + XC T = 0. ˜B ˜ T = AX + XAT , we infer as on slide 23: If we then just factorize B ˜B ˜ T (iωI − A)−∗ C T = C(iωI − A)−1 B = −CX(iωI − A)−∗ C T − C(iωI − A)−1 XC T for ω ∈ R. Due to XC T = −B, the right-hand side is J(iω)∗ + J(iω) = G(iω). ˜ is the filter to-be-constructed. This proves that T (s) = C(sI − A)−1 B 61/82
Proof of the Positive Real Lemma For ν > 0 let us consider the perturbed algebraic Riccati equation AX + XAT − ν(XC T + B)(XC T + B)T = 0. It suffices to show that this ARE has a solution Xν which does converges to some X for ν → ∞. Indeed, we can then conclude AXν +Xν AT < 0 which implies AX + XAT < 0; moreover from 1 (AXν + Xν AT ) = (Xν C T + B)(Xν C T + B)T ν we infer Xν C T + B → 0 and thus XC T + B = 0. In order to show existence of Xν , we observe that the ARE can be written as (A − νBC)X + X(A − νBC)T − νXC T CX − νBB T = 0.
(∗)
Clearly ((A − νBC)T , C T ) is controllable. 62/82
Proof of the Positive Real Lemma According to slide 66, we need to show for any ω ∈ R that (A − νBC)T − iωI −νC T C det 6= 0. νBB T −(A − νBC) − iωI
(∗ ∗)
For this purpose note that G(−iω) + ν1 I 0 and hence 1 J(−iω)∗ + J(−iω) + I 0. ν In particular the left-hand side is non-singular; it is also clearly (check!) the Schur complement of T A − iωI 0 CT 0 −A − iωI −B (∗ ∗ ∗) 1 T I B C ν with respect to the left-upper block. Since A has no eigenvalues in C0 we infer that (∗ ∗ ∗) is non-singular. Hence the Schur-complement with respect to the right-lower block is also non-singular. This is (∗ ∗). 63/82
Proof of the Positive Real Lemma Let Xν− and Xν+ denote the stabilizing and anti-stabilizing solution of (ARE). Now observe for ν ≥ µ that AXν− + Xν− AT − µ(Xν− C T + B)(Xν− C T + B)T = = (ν − µ)(Xν− C T + B)(Xν− C T + B)T < 0. Again by slide 66 we infer Xν− 4 Xµ− . Combined with a similar argument for Xν+ we conclude Xµ+ 4 Xν+ 4 Xν− 4 Xµ− for ν ≥ µ. This implies that Xν− is non-increasing and bounded from below for increasing ν. Hence it converges for ν → ∞ as we desired to show. Remark: The same argument applies to Xν+ . We get two limits which lead to different spectral factors that are distinct in the properties of their zeros. A detailed exposition goes beyond this course. 64/82
Example for Theorem 11 2
1−s Consider G(s) = s4 −13s 2 +36 . Determine a minimal realization of G and block-diagonalize the state-matrix. One then gets −0.06 −2 1.73 . 0.13 0 −3 J = −0.87 0.5 0
By solving the perturbed ARE we obtain the two approximate solutions as in the Positive Real Lemma: −1.02 0.16 −1.12 0.94 , X+ = . X− = 0.16 −0.19 0.94 −6.12 These lead to the two spectral factors −s − 1 −s + 1 T− (s) = 2 , T+ (s) = 2 . s + 5s + 6 s + 5s + 6 The first shares the stable zero with G, while the second one shares the anti-stable zero with G. This is not a coincidence but a general property! 65/82
Riccati Theory: Addendum I With controllable (A, B), positive definite R and just a symmetric Q, let us consider the algebraic Riccati equation AT P + P A − P BR−1 B T P + Q = 0. Under these hypotheses the following statements are equivalent: A −BR−1 B T • H= does not have an eigenvalue in C0 . −Q −AT • The ARE has a unique solution P− for which A − BR−1 B T P− is Hurwitz. P− is called the stabilizing solution. • The ARE has a unique solution P+ for which A − BR−1 B T P+ is anti-Hurwitz. P+ is called the anti-stabilizing solution. If P satisfies AT P + P A − P BR−1 B T P + Q < 0 (or 0) then P+ 4 P 4 P− (or P+ ≺ P ≺ P− ). 66/82
Riccati Theory: Addendum I Equivalence of the first two items is Theorem V-7. Equivalence to the third item is a direct consequence as discussed in the exercises. The relations among the various solutions is proved as in Theorem V-10, by exploiting AT P2 +P2 A−P2 BR−1 B T P2 +Q−(AT P1 +P1 A−P1 BR−1 B T P1 +Q) = = (A − BR−1 B T P1 )T ∆ + ∆(A − BR−1 B T P1 ) − ∆BR−1 B T ∆ for ∆ = P2 − P1 . For example for P2 = P and P1 = P+ we infer (A − BR−1 B T P+ )T ∆ + ∆(A − BR−1 B T P+ ) < ∆BR−1 B T ∆ < 0. Since A − BR−1 B T P+ is anti-stable, we infer ∆ < 0 and thus P < P+ . (If P satisfies the strict Riccati inequality, all derived inequalities are strict as well.) 67/82
Riccati Theory: Addendum II Suppose that (A, B) is stabilizable and (A, C) has no unobservable modes on the imaginary axis. Then the ARE AT P + P A − P BB T P + C T C = 0
(ARE)
has a unique stabilizing solution P which also satisfies P < 0. Let us reveal a useful relation to the set of all solutions X 0 of the so-called strict algebraic Riccati inequality (ARI) AT X + XA − XBB T X + C T C ≺ 0.
(ARI)
If X 0 satisfies (ARI) then P ≺ X. Remark. Note that all solutions of (ARI) must be non-singular. Hence any positive semi-definite solution is actually positive definite. 68/82
Proof: Step 1 The kernel N (P ) of P is A-invariant and contained in N (C). Indeed, with x satisfying P x = 0, we infer from xT (ARE)x = 0 that kCxk2 = 0 and thus Cx = 0. Then (ARE)x = 0 implies P Ax = 0. This proves the claim. Let the columns of T2 form a basis of N (P ) and expand to a non-singular matrix T = (T1 T2 ). Then P1 0 T T PT = and P1 0. 0 0 Due to AN (P ) ⊂ N (P ) and N (P ) ⊂ N (C) we also infer A1 0 B1 −1 −1 T AT = , T B= , CT = C1 0 . A21 A2 B2 W.l.o.g. (coordinate-change - check!) we can assume that the matrices are already given in this form while X has no particular structure. 69/82
Proof: Step 2 By inspection, the ARE reads as AT1 P1 +P1 A1 −P1 B1 B1T P1 +C1T C1 = 0. This implies for Q1 = P1−1 (recall invertibility!) that (A1 + Q1 C1T C1 )T = −P1 (A1 − B1 B1T P1 )P1−1 . Therefore A + Q1 C1T C1 is anti-stable. (All eigenvalues are in the open right half-plane.) We also have A1 Q1 + Q1 AT1 − B1 B1T + Q1 C1T C1 Q1 = 0. Similarly, Y = X −1 satisfies AY + Y AT − BB T + Y C T CY ≺ 0. Just by considering the left-upper block, we infer for Y1 Y12 Y = Y21 Y2 that Y1 satisfies the strict algebraic Riccati inequality A1 Y1 + Y1 AT1 − B1 B1T + Y1 C1T C1 Y1 ≺ 0. 70/82
Proof: Step 3 - The Coup de Grˆ ace By last property on slide 66 we conclude Q1 Y1 . (This requires some work; please check!) Due to −1 Y1 Y12 X1 X12 = Y21 Y2 X21 X2 and the block-inversion formula we have −1 = X1 − X12 X2−1 X21 . P1 = Q−1 1 ≺ Y1
With X2 0 we infer P1 0 X1 − X12 X2−1 X21 0 ≺ . 0 0 0 X2 I X12 X2−1 A congruence transformation with finally leads to 0 I P1 0 X1 X12 ≺ 0 0 X21 X2 which finishes the proof. 71/82
Proof of Theorem 14 If γopt denotes the optimal value of the H2 -problem, the first step is 2 to show tr(BwT P Bw + B T P QP B) ≤ γopt . Choose γ > γopt . We can then find AK , BK , CK which render A Hurwitz and such that kC(sI − A)−1 Bk22 < γ 2 . By simply expanding the controller realization with stable uncontrollable or unobservable modes, we can assume that AK has at least the same dimension as A. Moreover, by slide 6 there exists some X 0 with AT X + X A + C T C ≺ 0 and tr(B T X B) < γ 2 . With Z := B T X B + I and suitably small > 0 we then infer X XB T T A X + X A + C C ≺ 0, 0, tr(Z) < γ 2 . (1) BT X Z A key step of the proof is to block-factorize X in the partition of A. More precisely, let (X U ) denote the upper block row of X . We can assume that U has full row rank and that the right-lower block of X is non-singular (after perturbation if necessary). This assures: 72/82
Proof of Theorem 14 There exist X, U and Y , V such that Y 0 I VT X = and S has full row rank. X U I 0 | {z } | {z } S
(2)
R
Let us first prepare some relations for the proof. In view of (1) we need SX S T SX B T T T T T SA X S + SX AS + SC CS , BT X S T Z which are equal to T
T
T
T
T
SA R + RAS + SC CS ,
RS T RB B T RT Z
.
(3)
First note that T
RS =
Y 0 X U
I I V 0
=
Y Y Y X
where we exploited the fact that this matrix is symmetric. 73/82
Proof of Theorem 14 Moreover, all other blocks in (3) can be explicitly written as Y 0 Bw Y Bw RB = = , X U BK Dw XBw + U BK Dw I I T CS = Cz Dz CK = Cz + Dz CK V Cz V 0 and
T
RAS =
=
Y 0 A BCK I I = X U V 0 BK C AK Y 0 A + BCK V A = = X U BK C + AK V BK C
Y (A + BCK V ) YA X(A + BCK V ) + U BK C + U AK V XA + U BK C
.
74/82
Proof of Theorem 14 Therefore the matrices in (3) read as Y Y Y Bw T R1 R21 X XBw + U BK Dw , Y R21 R2 T Bw Y (∗)T Z
(4)
with blocks R1 = (A + BCK V )T Y +Y (A + BCK V )+(Cz + Dz CK V )T (Cz + Dz CK V ), R2 = AT X + XA + U BK C + (U BK C)T + CzT Cz , R21 = AT Y + X(A + BCK V ) + U BK C + U AK V + CzT Cz . With the equation for P and with ∆ = X − P let us note that R2 and R21 can be written as R2 = AT ∆ + ∆A + U BK C + (U BK C)T + P BB T P, R21 = AT (Y − P ) + ∆A + (P + ∆)BCK V + U BK C + U AK V + P BB T P. 75/82
Proof of Theorem 14 After these preparations we continue the proof. Since S has full row rank, (1) clearly implies that R1 ≺ 0, R2 ≺ 0 and that the second matrix in (4) is positive definite. By slide 34, R1 ≺ 0 implies P ≺ Y . On the other hand, by taking the Schur-complement we infer X XBw + U BK Dw I − Y I Bw 0. T T (XBw + U BK Dw ) Z Bw Combining the last two inequalities implies (recalling X − P = ∆) that ∆ ∆Bw + U BK Dw 0. (∆Bw + U BK Dw )T Z − BwT P Bw With L = −∆−1 U BK we have ∆Bw + U BK Dw = ∆(Bw − LDw ) and therefore Z − BwT P Bw − (Bw − LDw )T ∆(Bw − LDw ) 0. This shows tr(Bw − LDw )T ∆(Bw − LDw ) < γ 2 − tr(BwT P Bw )
(∗)
due to (1). Moreover R2 ≺ 0 as at the bottom of slide 75 reveals (A − LC)T ∆ + ∆(A − LC) + P BB T P ≺ 0.
(∗∗) 76/82
Proof of Theorem 14 By slide 6 the inequalities (∗), (∗∗) show that A − LC is Hurwitz and
A − LC Bw − LDw 2
< γ 2 − tr(BwT P Bw ).
BT P 0 2 This allows to apply our result on H2 -estimation on slide 41. Due the choice of Q we can hence conclude tr(B T P QP B) + tr(BwT P Bw ) < γ 2 . Since γ > γopt was arbitrary we finally get 2 . tr(B T P QP B) + tr(BwT P Bw ) ≤ γopt
This concludes the first part of the proof. Equality is shown by constructing an optimal controller. For this purpose we follow the above steps as much as possible in the reverse order, which includes a motivation for the structure of the to-be-constructed controller. 77/82
Proof of Theorem 14: Controller Construction Due to our result on H2 -estimation on slide 41 (applied for Cz replaced with B T P ) we know that A − QC T C is Hurwitz and leads to
A − QC T C Bw − QC T Dw 2
= tr(B T P QP B).
BT P 0 2 Therefore the solution ∆ of (A − QC T C)T ∆ + ∆(A − QC T C) + P BB T P = 0
(5)
tr(Bw − QC T Dw )T ∆(Bw − QC T Dw ) = tr(B T P QP B).
(6)
satisfies
Next we try to achieve R1 = 0, R2 = 0 and R21 = 0 on slide 75. If recalling the state-feedback problem, we clearly have R1 = 0 for Y = P , V = I and CK = −B T P . (We stress that we could take ANY nonsingular V at this point - the constructed controller would then just have a different state-space realization!) 78/82
Proof of Theorem 14: Controller Construction Next enforce R2 = 0 and R21 = 0 if represented as at the bottom of slide 75. Indeed, (5) implies R2 = 0 with U = −∆ and BK = QC T . Moreover, with all the choices so far we have R21 = ∆A − (P + ∆)BB T P − ∆QC T C − ∆AK + P BB T P = ∆(A − BB T P − QC T C − AK ) = 0 with AK = A − BB T P − QC T C. Further take X = ∆ + P such that the two representations of R2 and R21 on slide 75 are indeed correct. Let us finally define R and S as in (2). Note that S is non-singular and P + ∆ −∆ X = −∆ ∆ is easily seen to be the unique solution of (2). This renders all equations on slides 73-75 satisfied. 79/82
Proof of Theorem 14: Controller Construction Now we can conclude the proof. The constructed controller is exactly the one on slides 51 and 52. Since A − BB T P and A − QC T C are Hurwitz and due to the observer structure, it is certainly stabilizing. Moreover, by construction we assured R1 = 0, R2 = 0, R21 = 0 which implies SAT RT + RAS T + SC T CS T and thus AT X + X A + C T C = 0 by using (2) and exploiting that S is now square and non-singular. With the formula for X on the previous slide we finally infer tr(B T X B) = tr(BwT P Bw + BwT ∆Bw + DwT CQ∆QC T Dw ) = = tr(BwT P Bw + (Bw − QC T Dw )T ∆(Bw − C T QDw )) = = tr(BwT P Bw ) + tr(B T P QP B) due to (6). This finishes the proof. 80/82
References • [KK] H.W. Knobloch, H. Kwakernaak, Lineare Kontrolltheorie, Springer-Verlag Berlin 1985 • [AM] K.J. Astr¨ om, R.M. Murray, Feedback Systems: An Introduction for Scientists and Engineers, Princeton University Press, Princeton and Oxford, 2009 (available online for free with Wiki, just google.) • J.P. Hespanha, Linear Systems Theory, Princeton University Press, 2009 • [S] E.D. Sontag, Mathematical Control Theory, Springer, New York 1998 • [K] T. Kailath, Linear Systems, Prentice Hall, Englewood Cliffs, 1980 • [F] B. Friedland, Control System Design: An Introduction to State-space Methods. Dover Publications, 2005 • W.J. Rugh, Linear System Theory, Prentice-Hall, 2 1998 • R. Brockett, Finite dimensional linear systems, Wiley, 1970 • W.M. Wonham, Linear multivariable control, a geometric approach, SpringerVerlag, 3 1985
81/82
References • J. Zabczyk, Mathematical Control Theory: An Introduction, Birkh¨auser, 2007 • B.A. Francis, A course in H∞ -control theory, Springer, 1987 • H.K. Khalil, Nonlinear Systems, Prentice Hall, 3 2002 • L. Arnold, Stochastic Differential Equations: Theory and Applications, Wiley, 1974 • A. Packard, State-space Youla Notes, Course in Multivariable Control Systems, UC Berkeley, 2008 • J.M. Coron, Control and nonlinearity, Mathematical Surveys and Monographs, 2007 • Y. Yuan, G.-B. Stan, S. Warnick, and J.M. Goncalves, Minimal realization of the dynamical structure function and its application to network reconstruction, IEEE Transactions on Automatic Control, http://www.bg.ic.ac.uk/research/g.stan/, 2012 • J.S. Freudenberg with C.V. Hollot and D.P. Looze, A first graduate course in feedback control, 2003
82/82
