On external semi-global stochastic stabilization of linear systems with ...
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On external semi-global stochastic stabilization of linear systems with input saturation Anton A. Stoorvogel1 , Ali Saberi2 and Siep Weiland3
Abstract This paper introduces the notion of external semi-global stochastic stabilization for linear plants with saturating actuators, driven by a stochastic external disturbance, and having random Gaussiandistributed initial conditions. The aim of this stabilization is to control such plants by a possibly nonlinear static state feedback law that achieves global asymptotic stability in the absence of disturbances, while guaranteeing a bounded variance of the state vector for all time in the presence of disturbances and Gaussian distributed initial conditions. We report complete results for both continuous- and discrete-time open-loop critically stable plants.
I. I NTRODUCTION Internal and external stabilization of linear plants with actuators subject to saturation has been the subject of intense renewed interest among the control research community for the past two decades. The number of recent books and special issues of control journals devoted to this subject matter evidence this intense research focus, see for instance [1], [11], [4], [15] and the references therein. It is now considered a classical fact that both continuous- and discrete-time linear plants with saturating actuators can be globally internally stabilized if and only if all of the open-loop poles are located in the closed left half plane (in the continuous-time case), and are in inside 1
Department of Mathematics, and Computing Science, Eindhoven Univ. of Technology, P.O. Box 513, 5600 MB Eindhoven,
The Netherlands. Department of Electrical Engineering, Mathematics and Computer Science, Delft Univ. of Technology, P.O. Box 5031, 2600 GA Delft, The Netherlands. E-mail: [email protected]. 2
School of Electrical Engineering and Computer Science, Washington State University, Pullman, WA 99164-2752, U.S.A.
E-mail: [email protected]. 3
Department of Electrical Engineering, Eindhoven University of Technology, P.O. Box 513, 5600 MB Eindhoven, The
Netherlands, E-mail: [email protected].
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and/or on the unit circle (in the discrete-time case). These conditions on open-loop plants can be equivalently stated as a requirement that the plant be asymptotically null controllable with bounded control (ANCBC). It is also a classical fact that, in general, global internal stabilization of ANCBC plants requires nonlinear feedback laws. See for instance. [10], [2], [16]. By weakening the notion of global internal stabilization to a semiglobal framework, Saberi et al. showed that ANCBC plants with saturating actuators can be internally stabilized in the semiglobal sense using only linear feedback laws. Such a relaxation from a global to semiglobal framework is, from an engineering standpoint, both sensible and attractive. A body of work exists proposing design methodologies (low-gain, low-high gain, and variations) for both continuousand discrete-time ANCBC plants with saturating actiators. See for instance [3], [17], [6], [5] With regard to external stabilization, the picture is complicated. Unlike linear plants, internal stability does not necessarily guarantee external stability when saturation is present. Hence, stabilization must be done simultaneously in both the internal and external sense. There is a body of work that examines the standard notion of Lp stability for ANCBC plants with saturating actuators when the external input (disturbance) is input-additive and at the same time requiring either global or semi global internal stability in the absence of disturbnce, see [7]. For the more general case in which the disturbance is not necessarily input-additive, there are surprising results in the literature which point to the complexity of the notion of external stability for ANCBC plants with saturating actuators. Notable among these is the result from [13] of Stoorvogel et.al., that for a double integrator with saturating actuators, “any linear internally stabilizing control law can achieve Lp stability without finite gain for p ∈ [1, 2], but no linear internally stabilizing control law can achieve Lp stability for any p ∈ (2, ∞]”. Moreover, considering the recently developed notion of input-to-state stability (ISS) as a framework for simutaneous external and internal stability, such a double-integrator system cannot achieve ISS stability with a linear feedback law, see [8]. All these results point toward the delicacy of the external stability concept for linear plants with saturating actuators. Moreover, this literature also points out that, for sustained disturbances, the induced L∞ norm does not yield a suitable problem formulation. It contains mathematical functions that are not reasonable models for disturbances, and it is impossible to get a good “stable” response if we use these functions as models for disturbance. In fact one would need to consider a class of sensible sustained disturbances from an engineering point of view. All these considerations lead us to believe that a suitable notion of external December 18, 2006
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stability for linear plants with saturating actuators, and indeed for general nonlinear systems in the presence of disturbances and nonzero initial conditions, is yet to be developed. In this paper we look at the simultaneous external and internal stabilization of ANCBC plants with saturating actuators when the external input is a stochastic disturbance. Specifically, we consider a linear time-invariant system subject to input saturation, stochastic external disturbances and random Gaussian distributed initial conditions, independent of the external disturbances. The aim will be to control this system by a possibly nonlinear static state feedback law that achieves global asymptotic stability in the absence of disturbances, while guaranteeing a bounded variance of the state vector for all time. This problem seems a natural extension of the results in [9], [12]. In [14], a result was claimed. In this paper we give a complete proof for the case that the system is neutrally stable, i.e., the eigenvalues of the system matrix of the continuous-time (discrete-time) system are in the closed left half plane (open unit disc) and the eigenvalues on the imaginary axis (unit circle) have equal algebraic and geometric multiplicity. The paper is organized as follows. For both discrete- and continuous-time systems, a formal problem formulation and the main results are stated in Section II. A formal proof of the discretetime case for neutrally stable systems is given in Section III. A discussion on the implications of the main results, some extensions and conclusions are collected in Section V. II. P ROBLEM
FORMULATION AND MAIN RESULTS
A. The discrete-time case In discrete time, we consider systems of the form x(k + 1) = Ax(k) + Bσ(u(k)) + Ew(k)
(1)
where the state x, the control u and the disturbance w are vector-valued signals of dimension n, m and , respectively. Here, k ∈ Z+ , w is a white noise stochastic process with variance Q and mean 0, the initial condition x0 of (1) is a Gaussian random vector independent of w(k) for all k ≥ 0. Moreover σ is the standard saturation ⎧ ⎪ ⎪ −1 ⎪ ⎪ ⎨ σ(u) = u ⎪ ⎪ ⎪ ⎪ ⎩1 December 18, 2006
function given by: u < −1 −1 ≤ u ≤ 1 u>1 DRAFT
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An admissible feedback is a nonlinear feedback of the form u(k) = f (x(k))
(2)
where f : Rn → Rm is a continuous map with f (0) = 0. We therefore consider nonlinear static state feedbacks. We will be interested in the following problem. Problem II.1 Given the system (1), the semiglobal external stochastic stabilization problem is to find an admissible feedback (2) such that the following properties hold: 1) in the absence of the external input w, the equilibrium point x = 0 of the controlled system (1)-(2) is globally asymptotically stable. 2) the variance Var(x(k)) of the state of the controlled system (1)-(2) is bounded over k ≥ 0. The fact that controllers exist that achieve global asymptotic stability in the absence of disturbances as described in condition (1) is well-known. The main objective of this paper is to look at the additional requirement on the variance. The following is the main result of this paper for discrete time systems: Theorem II.2 Consider the system (1) and suppose that (A, B) is stabilizable while A is neutrally stable, i.e. the eigenvalues of A are in the closed unit disc and the eigenvalues on the unit circle have equal geometric and algebraic multliplicity. In the above case, there exists a linear feedback which solves the semiglobal external stochastic stabilization problem as defined in Problem II.1. We claim that the condition above is only sufficient and that we can weaken the condition of A neutrally stable to the requirement that all eigenvalues of A are in the closed unit disc. However, this does, in general, require a nonlinear feedback. Conjecture II.1 Consider the system (1). Then for any given variance Q of the white noise stochastic process w there exists a feedback (2) which solves the semiglobal external stochastic stabilization problem as defined in Problem II.1 if and only if (A, B) is stabilizable while the eigenvalues of A are in the closed unit disc.
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B. The continuous-time case In continuous time we consider the stochastic differential equation dx(t) = Ax(t)dt + Bσ(u(t))dt + Edw(t)
(3)
where the state x, the control u and the disturbance w are vector-valued signals of dimension n, m and , respectively. Here w is a Wiener process (a process of independent Brownian motions) with mean 0 and rate Q, that is, Var[W (t)] = Qt and the initial condition x0 of (3) is a Gaussian random vector which is independent of w. Its solution x is rigorously defined through Wiener integrals and is a Gauss-Markov process. Like in the discrete time case, σ denotes the standard saturation function and admissible feedbacks are possibly nonlinear static state feedbacks of the form (2) where f is a Lipschitz-continuous mapping with f (0) = 0. Problem II.3 Given the system (3), the semiglobal external stochastic stabilization problem is to find an admissible feedback (2) such that the following properties hold: 1) in the absence of the external input w, the equilibrium point x = 0 of the controlled system (3)-(2) is globally asymptotically stable. 2) the variance Var(x(t)) of the state of the controlled system (3)-(2) is bounded over t ≥ 0. Like in the discrete time, the fact that controllers exist that achieve global asymptotic stability in the absence of disturbances as described in condition (1) is well-known. The main objective of this paper is to look at the additional requirement on the variance. The following is the main result of this paper for continuous time systems: Theorem II.4 Consider the system (3) and suppose that (A, B) is stabilizable while A is neutrally stable, i.e. the eigenvalues of A are in the closed left half plane and the eigenvalues on the imaginary axis have equal geometric and algebraic multliplicity. Then for any given rate Q of the Wiener process w, there exists a linear feedback which solves the semiglobal external stochastic stabilization problem as defined in Problem II.3.
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We claim that the condition above is only sufficient and that we can weaken the condition of A neutrally stable to the requirement that all eigenvalues of A are in the closed left half plane. However, this does, in general, require a nonlinear feedback. Conjecture II.2 Consider the system (3). Then for any given variance Q of the white noise stochastic process w(t) there exists a feedback (2) which solves the semiglobal external stochastic stabilization problem as defined in Problem II.3 if and only if (A, B) is stabilizable while the eigenvalues of A are in the closed left half plane. III. P ROOFS
FOR THE DISCRETE - TIME CASE
We first present a little lemma that we need later: Lemma III.1 For any a ∈ Rm and b ∈ Rm and n > 1 we have: 1 σ(a)2 − 8nb2 σ(a + b)2 ≥ 1 − 2n where σ is the standard saturation function. Proof: It is easily verified that: σ(a + b) ≥ σ(a) − |b| where the inequality is componentwise and |b| ∈ Rm is defined as the absolute value of each component of b. This implies: σ(a + b)2 ≥ σ(a)2 − 2|b|T σ(a) + b2 Using |b|T σ(a) ≤ 4nb2 +
1 σ(a)2 4n
then yields the required result. Using a simple basis transformation we can assume without loss of generality that: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ A11 0 B E ⎠ , B = ⎝ 1⎠ , E = ⎝ 1⎠ A=⎝ B2 E2 0 A22 with A11 asymptotically stable while AT22 A22 = I. This is guaranteed by the fact that A is neutrally stable. Next we consider the subsystem: x2 (t + 1) = A22 x2 (t) + B2 u(t) + E2 w(t) December 18, 2006
(4) DRAFT
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If we find an admissible feedback of the form u(t) = f (x2 (t))
(5)
which solves the semiglobal stochastic stabilization problem for the system (4) then it is easily verified that this controller also solves the semiglobal stochastic stabilization problem for the original system (1). In other words we can restrict attention to the system (4). We will use a controller of the form: u(k) = −ρB2T A22 x2 (k) and we will establish that for any given variance Q of the white noise there exists ρ small enough such that the resulting system has a bounded variance for the state. The choice of ρ is independent of the variance of the initial condition x0 . Proof: [Proof of Theorem II.2] To keep notation simple, we assume the original system satisfies AT A = I and we consider a feedback of the form: u(k) = −ρB T Ax(k)
(6)
The general case of the theorem can then easily be established using the reduction to the system (4) as described earlier. If the system is not affected by noise, it is well-known that the feedback (6) achieves asymptotic stability which is also easily verified by noting that V (x) = xT x is a suitable Lyapunov function. To study the effect of the noise, we first note that due to the Markov property of the state, we have: E [x(k + r)T x(k + r) | x(k + s)] = E [E [x(k + r)T x(k + r) | x(k + t)] | x(k + s)] for r ≥ t ≥ s. Next we consider: E [x(k + 1)T x(k + 1) | x(k)]
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Using the feedback (6), the dynamics (1) and the fact that AT A = I we obtain: E [x(k + 1)T x(k + 1) | x(k)] = x(k)T x(k) − 2x(k)T AT Bσ(ρB T Ax(k)) + σ(ρB T Ax(k))T B T Bσ(ρB T Ax(k)) + trace EQE T ≤ x(k)T x(k) − ρ2 σ(ρB T Ax(k))T σ(ρB T Ax(k)) + σ(ρB T Ax(k))T B T Bσ(ρB T Ax(k)) + trace EQE T ≤ x(k)T x(k) − ρ1 σ(ρB T Ax(k))2 + trace EQE T
(7)
provided ρ is small enough such that ρB T B < I. Next, we consider:
2 E σ(ρB T At x(k + 1)) | x(k)
for some integer t ≥ 1. We get:
2 E σ(ρB T At x(k + 1)) | x(k)
2 = E σ(ρB T At+1 x(k) − ρB T At Bσ(ρB T Ax(k)) + ρB T At Ew(k)) | x(k) 1 )σ(ρB T At+1 x(k)2 − 8n E − ρB T At Bσ(ρB T Ax(k)) ≥ (1 − 2n +ρB T At Ew(k))2 | x(k) ≥ (1 −
1 )σ(ρB T At+1 x(k))2 2n
− 8nρ2 B T At B2 σ(ρB T Ax(k))2
− 8ρ2 B T At 2 trace EQE T where the first inequality follows from Lemma III.1. If we choose ρ∗ such that ρ∗ B T B < I [as used earlier in deriving (7)] and √ 8n ρ∗ B T At 2 ≤ 1,
√ 8n ρ∗ B T At B2 ≤ 1
for all t, then it is easy to see that for all ρ < ρ∗ and any positive integer t we have:
2 E σ(ρB T At x(k + 1)) | x(k) ≥ (1 −
1 )σ(ρB T At+1 x(k))2 2n
− ρ3/2 σ(ρB T Ax(k))2 − ρ3/2 trace EQE T
The following lemma presents a crucial inequality:
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Lemma III.2 We have for ρ small enough that E [x(k + s)T x(k + s) | x(k)]
s
1 t
σ(ρB T At x(k)) 2 + s 1 + s trace EQE T (8) ≤ x(k) x(k) − 1− ρ t=1 2n n T
Proof: We will establish this inequality recursively. Note that we already established this inequality for s = 1 given (7). Next, consider any s > 1 and assume the above inequality is true for s (and any k). Then we obtain: E [x(k + s + 1)T x(k + s + 1) | x(k)] = E [E [x(k + s + 1)T x(k + s + 1) | x(k + 1)] | x(k)] s
1 t T
σ(ρB T At x(k + 1)) 2 ≤ E x(k + 1) x(k + 1) − 1− ρ t=1 2n s +s 1 + trace EQE T | x(k) n Using the previously obtained inequality we have: s
t 2
σ(ρB T At x(k + 1)) | x(k) E 1− 2n t=1 s
1 t 2
σ(ρB T At+1 x(k)) − sρ3/2 σ(ρB T Ax(k)) ≥ 1− 1− 2n 2n t=1 − sρ3/2 trace EQE T s
t+1
σ(ρB T At+1 x(k)) 2 − sρ3/2 σ(ρB T Ax(k)) − sρ3/2 trace EQE T 1− ≥ 2n t=1
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We then obtain: E [x(k + s + 1)T x(k + s + 1) | x(k)] = E [E [x(k + s + 1)T x(k + s + 1) | x(k + 1)] | x(k)] ≤ x(k)T x(k) − 1ρ σ(ρB T Ax(k))2 + trace EQE T s
1 t+1
σ(ρB T At+1 x(k)) 2 − 1− ρ t=1 2n
√ √ s + s ρσ(ρB T Ax(k))2 + s ρ trace EQE T + s 1 + trace EQE T n s+1
2 1 s+1 t T T t
≤ x(k) x(k) − 1− σ(ρB A x(k)) + (s + 1) 1 + trace EQE T ρ t=1 2n n provided ρ < ρ∗ is such that
√
ρ≤
1 , n
sρ3/2 ≤
1 . 2n
Next we note that since (A, B) is controllable while A is invertible, the matrix ⎞ ⎛ B TA ⎟ ⎜ ⎜ B T A2 ⎟ ⎟ ⎜ ⎜ . ⎟ ⎜ .. ⎟ ⎠ ⎝ B T An is injective which implies that there exists α (the smallest singular value of this matrix divided by n) such that for any x there exists a positive integer t ≤ n such that B T At x ≥ αx and hence
1 σ(ρB T At x)2 ≥ min ρα2 x2 , ρ1 ρ
But then (8) yields: E [x(k + n)T x(k + n) | x(k)] ≤ x(k)T x(k) − 12 min ρα2 x(k)2 , 1ρ + 2n trace EQE T We define M=
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1 ρ2 α2
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and y(k) = max { x(k)T x(k), M } The following lemma presents another crucial result: Lemma III.3 We have: E [y(k + n) | y(k)] ≤ y(k)
(9)
provided ρ is small enough such that provided ρ is small enough such that the inequality of Lemma III.2 is valid for s = 1, . . . , n and: 1 > 2n trace EQE T 2ρ
and ρα2 < 1
Moreover, E[y(k + pn)] ≤ E[y(k)],
p = 1, 2, . . .
(10)
Proof: This follows from the fact that if y(k) = x(k)T x(k) and hence x(k)T x(k) ≥ M we have: E [x(k + n)T x(k + n) | x(k)] 2 2 1 1 ≤ y(k) − 2 min ρα x(k) , ρ + 2n trace EQE T ≤ y(k) −
1 2ρ
+ 2n trace EQE T
≤ y(k) while for the case y(k) = M and hence x(k)T x(k) ≤ M we have: E [x(k + n)T x(k + n) | x(k)]
≤ x(k)T x(k) − 12 min ρα2 x(k)2 , 1ρ + 2n trace EQE T
≤ (1 − 12 ρα2 )x(k)T x(k) + 2n trace EQE T ≤ (1 − 12 ρα2 )y(k) + 2n trace EQE T ≤ y(k) − 12 ρα2 M + 2n trace EQE T ≤ y(k) but then (using that y(k) ≥ M): E [y(k + n) | x(k)] ≤ y(k) December 18, 2006
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and finally, E [y(k + n) | y(k)] = E [E [y(k + n) | x(k)] | y(k)] ≤ E [y(k) | y(k)] = y(k) Similarly (where we have to be careful since y(k) is not Markov), E [y(k + 2n) | y(k)] = E [E [y(k + 2n) | x(k + n), x(k)] | y(k)] = E [E [E [y(k + 2n) | x(k + n), x(k)] | x(k)] | y(k)] = E [E [E [y(k + 2n) | x(k + n)] | x(k)] | y(k)] ≤ E [E [y(k + n) | x(k)] | y(k)] ≤ E [y(k) | y(k)] = y(k) Using a simple recursion, we obtain (10). We have, using (7): E[x(k + 1)T x(k + 1)|x(k)] ≤ x(k)T x(k) + trace EQE T which yields: E[x(k)T x(k)] ≤ Var[x(0)] + k trace EQE T for k = 1, . . . , n − 1 and hence: E[y(k)] ≤ max { M, Var[x(0)] + k trace EQE T } for k = 1, . . . , n − 1. This implies that the expectation of y(0), . . . , y(n − 1) is bounded. Using (10) we find that the expectation of y(k) is bounded in k. Using that y(k) ≥ x(k)T x(k) this yields that the variance of x(k) is bounded in k. IV. P ROOFS
FOR THE CONTINUOUS - TIME CASE
Using a simple basis transformation we can assume without loss of generality that: ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ B1 E A11 0 ⎠ , B = ⎝ ⎠ , E = ⎝ 1⎠ A=⎝ B2 E2 0 A22
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with A11 asymptotically stable while A22 + AT22 = 0. This is guaranteed by the fact that A is neutrally stable. Next we consider the subsystem: dx2 (t) = A22 x2 (t)dt + B2 σ(u(t))dt + E2 dw(t)
(11)
If we find an admissible feedback of the form u(t) = f (x2 (t))
(12)
which solves the semiglobal external stochastic stabilization problem for the system (11) then it is easily verified that this controller also solves the semiglobal stochastic stabilization problem for the original system (3). In other words we can restrict attention to the system (11). We will use a controller of the form: u(t) = −ρB2T x2 (t) and we will establish that for any given rate Q of the Wiener process there exists ρ small enough such that the resulting system has a bounded variance for the state. The choice of ρ is independent of the variance of the initial condition x0 . Proof: [Proof of Theorem II.4] To keep notation simple, we assume the original system satisfies A + AT = 0 and we consider a feedback of the form: u(t) = −ρB T x(t)
(13)
The general case of the theorem can then easily be established using the reduction to the system (11) as described earlier. If the system is not affected by noise, it is well-known that the feedback (13) achieves asymptotic stability which is also easily verified by noting that V (x) = xT x is a suitable Lyapunov function. To study the effect of the noise, we first note that: τ τ A(τ −t) A(τ −s) x(t) + e Bσ(u(s))ds + eA(τ −s) Edw(s) x(τ ) = e t
t
= eA(τ −t) x(t) + vt (τ ) We note that σ(u) is bounded by 1 and the second integral has bounded variance for all τ such that τ ∈ [t, t + 1]. This implies that there exists a constant N such that: Var[vt (τ )] < N December 18, 2006
(14) DRAFT
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for all τ such that τ ∈ [t, t + 1]. Next, we note that σ(ρB T x(τ ))2 = σ ρB T eA(τ −t) x(t) + ρB T vt (τ ) 2 ≥ 12 σ(ρB T eA(τ −t) x(t))2 − 8ρ2 B2 vt (τ )2
(15)
where we applied Lemma III.1 with n = 1. Next we note that: dxT x = (2xT Bσ(u) + trace E T E)dt + 2xT Edw using that A + AT = 0 and Itˆo’s lemma. This yields that x(t + 1)T x(t + 1) − x(t)T x(t) t+1 t+1 2x(τ )T Bσ(u(τ ))dτ + trace E T E + 2x(τ )T Edw(τ ) = t
t
and hence: E[x(t + 1)T x(t + 1) | x(t)]
⎡ t+1 ⎤ = x(t)T x(t) + E ⎣ 2x(τ )T Bσ(u(τ ))dτ | x(t)⎦ + trace E T E t
On the other hand, we have: 2x(τ )T Bσ(u(τ )) = 2x(τ )T Bσ(−ρB T x(τ )) ≤ − 2ρ σ(ρB T x(τ ))2 Combining the last two equations we obtain: E[x(t + 1)T x(t + 1) | x(t)]
⎡ t+1 ⎤ ≤ x(t)T x(t) + trace E T E − 2ρ E ⎣ σ(ρB T x(τ ))2 dτ | x(t)⎦ t
Using (14), (15) and the fact that x(t) and vt (τ ) are independent for τ ∈ [t, t + 1] we obtain a crucial inequality: E[x(t + 1)T x(t + 1) | x(t)] ≤ x(t)T x(t) + V −
1 ρ
t+1 σ(ρB T eA(τ −t) x(t))2 dτ
(16)
t December 18, 2006
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where V = 16ρB2 N + trace E T E. Lemma IV.1 Assume that (A, B) is controllable. Then there exist suitably defined constants α and R such that for all ρ < ρ¯ we have that: 1 ρ
t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ ≥ min{ραx(t)2 , R } ρ t
Proof: If the saturation does not get activated and hence we get: t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ = ρ2 x(t)T Qx(t) t
where:
1 Q=
eAτ BB T eAτ dτ
0
Controllability of (A, B) implies that Q is invertible and hence we obtain: x(t)Qx(t) ≥ αx(t)2 where α is the smallest singular value of Q. Hence we obtain: 1 ρ
t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ ≥ ραx(t)2
(17)
t
On the other hand, in case the saturation does get activated then it is easily seen that there must exist an N (independent of ρ) such that: x(t) >
N ρ
We obtain: 2 t+1 t+1
2
Nx(t) T A(τ −t) T A(τ −t)
σ(ρB e
dτ
σ ρB e x(t)) dτ ≥
ρx(t) t
We consider the integral:
t
2 t+1
σ ρB T eA(τ −t) Nx(t) dτ
ρx(t)
(18)
t
In case saturation does not arise in this integral we immediately find that the term is larger or equal to
Nx(t) 2
≥ αN 2
ρ α ρx(t) 2
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using the same techniques as for the case that the original integral did not contain a term which saturates. But then we find 1 ρ
t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ ≥ 1 αN 2 ρ
(19)
t
Finally we look at the case where the saturation in the integral (18) does get activated. In that case we note that there exists τ such that:
2
Nx(t) T A(τ −t)
≥1
σ ρB e
ρx(t) But the derivative with respect to τ of this term is bounded by some constant γ independent of our choice for τ ∈ [t, t + 1] and independent of our choice for x(t). This implies that 2 t+1
σ ρB T eA(τ −t) Nx(t) dτ ≥ 1
ρx(t) γ t
This yields 1 ρ
2 t+1 t+1
2
1 Nx(t) T A(τ −t) T A(τ −t)
dτ
σ ρB e
σ(ρB e x(t)) dτ ≥
ρ ρx(t) t
t
1 ≥ ργ
(20)
Combining (17) with (19) and (20), the result of the lemma follows immediately for 1 R = min{N 2 , }. γ The above lemma yields, with the help of (16), that: E[x(t + 1)T x(t + 1) | x(t)] ≤ x(t)T x(t) − min{ραx(t)2 , Rρ } + V We define: M=
R ρ2 α
and y(t) = max { x(t)T x(t), M } The following lemma presents another crucial result:
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Lemma IV.2 We have: E [y(t + 1) | y(t)] ≤ y(t)
(21)
provided ρ is small enough such that: ρ
On external semi-global stochastic stabilization of linear systems with input saturation Anton A. Stoorvogel1 , Ali Saberi2 and Siep Weiland3
Abstract This paper introduces the notion of external semi-global stochastic stabilization for linear plants with saturating actuators, driven by a stochastic external disturbance, and having random Gaussiandistributed initial conditions. The aim of this stabilization is to control such plants by a possibly nonlinear static state feedback law that achieves global asymptotic stability in the absence of disturbances, while guaranteeing a bounded variance of the state vector for all time in the presence of disturbances and Gaussian distributed initial conditions. We report complete results for both continuous- and discrete-time open-loop critically stable plants.
I. I NTRODUCTION Internal and external stabilization of linear plants with actuators subject to saturation has been the subject of intense renewed interest among the control research community for the past two decades. The number of recent books and special issues of control journals devoted to this subject matter evidence this intense research focus, see for instance [1], [11], [4], [15] and the references therein. It is now considered a classical fact that both continuous- and discrete-time linear plants with saturating actuators can be globally internally stabilized if and only if all of the open-loop poles are located in the closed left half plane (in the continuous-time case), and are in inside 1
Department of Mathematics, and Computing Science, Eindhoven Univ. of Technology, P.O. Box 513, 5600 MB Eindhoven,
The Netherlands. Department of Electrical Engineering, Mathematics and Computer Science, Delft Univ. of Technology, P.O. Box 5031, 2600 GA Delft, The Netherlands. E-mail: [email protected]. 2
School of Electrical Engineering and Computer Science, Washington State University, Pullman, WA 99164-2752, U.S.A.
E-mail: [email protected]. 3
Department of Electrical Engineering, Eindhoven University of Technology, P.O. Box 513, 5600 MB Eindhoven, The
Netherlands, E-mail: [email protected].
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and/or on the unit circle (in the discrete-time case). These conditions on open-loop plants can be equivalently stated as a requirement that the plant be asymptotically null controllable with bounded control (ANCBC). It is also a classical fact that, in general, global internal stabilization of ANCBC plants requires nonlinear feedback laws. See for instance. [10], [2], [16]. By weakening the notion of global internal stabilization to a semiglobal framework, Saberi et al. showed that ANCBC plants with saturating actuators can be internally stabilized in the semiglobal sense using only linear feedback laws. Such a relaxation from a global to semiglobal framework is, from an engineering standpoint, both sensible and attractive. A body of work exists proposing design methodologies (low-gain, low-high gain, and variations) for both continuousand discrete-time ANCBC plants with saturating actiators. See for instance [3], [17], [6], [5] With regard to external stabilization, the picture is complicated. Unlike linear plants, internal stability does not necessarily guarantee external stability when saturation is present. Hence, stabilization must be done simultaneously in both the internal and external sense. There is a body of work that examines the standard notion of Lp stability for ANCBC plants with saturating actuators when the external input (disturbance) is input-additive and at the same time requiring either global or semi global internal stability in the absence of disturbnce, see [7]. For the more general case in which the disturbance is not necessarily input-additive, there are surprising results in the literature which point to the complexity of the notion of external stability for ANCBC plants with saturating actuators. Notable among these is the result from [13] of Stoorvogel et.al., that for a double integrator with saturating actuators, “any linear internally stabilizing control law can achieve Lp stability without finite gain for p ∈ [1, 2], but no linear internally stabilizing control law can achieve Lp stability for any p ∈ (2, ∞]”. Moreover, considering the recently developed notion of input-to-state stability (ISS) as a framework for simutaneous external and internal stability, such a double-integrator system cannot achieve ISS stability with a linear feedback law, see [8]. All these results point toward the delicacy of the external stability concept for linear plants with saturating actuators. Moreover, this literature also points out that, for sustained disturbances, the induced L∞ norm does not yield a suitable problem formulation. It contains mathematical functions that are not reasonable models for disturbances, and it is impossible to get a good “stable” response if we use these functions as models for disturbance. In fact one would need to consider a class of sensible sustained disturbances from an engineering point of view. All these considerations lead us to believe that a suitable notion of external December 18, 2006
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stability for linear plants with saturating actuators, and indeed for general nonlinear systems in the presence of disturbances and nonzero initial conditions, is yet to be developed. In this paper we look at the simultaneous external and internal stabilization of ANCBC plants with saturating actuators when the external input is a stochastic disturbance. Specifically, we consider a linear time-invariant system subject to input saturation, stochastic external disturbances and random Gaussian distributed initial conditions, independent of the external disturbances. The aim will be to control this system by a possibly nonlinear static state feedback law that achieves global asymptotic stability in the absence of disturbances, while guaranteeing a bounded variance of the state vector for all time. This problem seems a natural extension of the results in [9], [12]. In [14], a result was claimed. In this paper we give a complete proof for the case that the system is neutrally stable, i.e., the eigenvalues of the system matrix of the continuous-time (discrete-time) system are in the closed left half plane (open unit disc) and the eigenvalues on the imaginary axis (unit circle) have equal algebraic and geometric multiplicity. The paper is organized as follows. For both discrete- and continuous-time systems, a formal problem formulation and the main results are stated in Section II. A formal proof of the discretetime case for neutrally stable systems is given in Section III. A discussion on the implications of the main results, some extensions and conclusions are collected in Section V. II. P ROBLEM
FORMULATION AND MAIN RESULTS
A. The discrete-time case In discrete time, we consider systems of the form x(k + 1) = Ax(k) + Bσ(u(k)) + Ew(k)
(1)
where the state x, the control u and the disturbance w are vector-valued signals of dimension n, m and , respectively. Here, k ∈ Z+ , w is a white noise stochastic process with variance Q and mean 0, the initial condition x0 of (1) is a Gaussian random vector independent of w(k) for all k ≥ 0. Moreover σ is the standard saturation ⎧ ⎪ ⎪ −1 ⎪ ⎪ ⎨ σ(u) = u ⎪ ⎪ ⎪ ⎪ ⎩1 December 18, 2006
function given by: u < −1 −1 ≤ u ≤ 1 u>1 DRAFT
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An admissible feedback is a nonlinear feedback of the form u(k) = f (x(k))
(2)
where f : Rn → Rm is a continuous map with f (0) = 0. We therefore consider nonlinear static state feedbacks. We will be interested in the following problem. Problem II.1 Given the system (1), the semiglobal external stochastic stabilization problem is to find an admissible feedback (2) such that the following properties hold: 1) in the absence of the external input w, the equilibrium point x = 0 of the controlled system (1)-(2) is globally asymptotically stable. 2) the variance Var(x(k)) of the state of the controlled system (1)-(2) is bounded over k ≥ 0. The fact that controllers exist that achieve global asymptotic stability in the absence of disturbances as described in condition (1) is well-known. The main objective of this paper is to look at the additional requirement on the variance. The following is the main result of this paper for discrete time systems: Theorem II.2 Consider the system (1) and suppose that (A, B) is stabilizable while A is neutrally stable, i.e. the eigenvalues of A are in the closed unit disc and the eigenvalues on the unit circle have equal geometric and algebraic multliplicity. In the above case, there exists a linear feedback which solves the semiglobal external stochastic stabilization problem as defined in Problem II.1. We claim that the condition above is only sufficient and that we can weaken the condition of A neutrally stable to the requirement that all eigenvalues of A are in the closed unit disc. However, this does, in general, require a nonlinear feedback. Conjecture II.1 Consider the system (1). Then for any given variance Q of the white noise stochastic process w there exists a feedback (2) which solves the semiglobal external stochastic stabilization problem as defined in Problem II.1 if and only if (A, B) is stabilizable while the eigenvalues of A are in the closed unit disc.
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B. The continuous-time case In continuous time we consider the stochastic differential equation dx(t) = Ax(t)dt + Bσ(u(t))dt + Edw(t)
(3)
where the state x, the control u and the disturbance w are vector-valued signals of dimension n, m and , respectively. Here w is a Wiener process (a process of independent Brownian motions) with mean 0 and rate Q, that is, Var[W (t)] = Qt and the initial condition x0 of (3) is a Gaussian random vector which is independent of w. Its solution x is rigorously defined through Wiener integrals and is a Gauss-Markov process. Like in the discrete time case, σ denotes the standard saturation function and admissible feedbacks are possibly nonlinear static state feedbacks of the form (2) where f is a Lipschitz-continuous mapping with f (0) = 0. Problem II.3 Given the system (3), the semiglobal external stochastic stabilization problem is to find an admissible feedback (2) such that the following properties hold: 1) in the absence of the external input w, the equilibrium point x = 0 of the controlled system (3)-(2) is globally asymptotically stable. 2) the variance Var(x(t)) of the state of the controlled system (3)-(2) is bounded over t ≥ 0. Like in the discrete time, the fact that controllers exist that achieve global asymptotic stability in the absence of disturbances as described in condition (1) is well-known. The main objective of this paper is to look at the additional requirement on the variance. The following is the main result of this paper for continuous time systems: Theorem II.4 Consider the system (3) and suppose that (A, B) is stabilizable while A is neutrally stable, i.e. the eigenvalues of A are in the closed left half plane and the eigenvalues on the imaginary axis have equal geometric and algebraic multliplicity. Then for any given rate Q of the Wiener process w, there exists a linear feedback which solves the semiglobal external stochastic stabilization problem as defined in Problem II.3.
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We claim that the condition above is only sufficient and that we can weaken the condition of A neutrally stable to the requirement that all eigenvalues of A are in the closed left half plane. However, this does, in general, require a nonlinear feedback. Conjecture II.2 Consider the system (3). Then for any given variance Q of the white noise stochastic process w(t) there exists a feedback (2) which solves the semiglobal external stochastic stabilization problem as defined in Problem II.3 if and only if (A, B) is stabilizable while the eigenvalues of A are in the closed left half plane. III. P ROOFS
FOR THE DISCRETE - TIME CASE
We first present a little lemma that we need later: Lemma III.1 For any a ∈ Rm and b ∈ Rm and n > 1 we have: 1 σ(a)2 − 8nb2 σ(a + b)2 ≥ 1 − 2n where σ is the standard saturation function. Proof: It is easily verified that: σ(a + b) ≥ σ(a) − |b| where the inequality is componentwise and |b| ∈ Rm is defined as the absolute value of each component of b. This implies: σ(a + b)2 ≥ σ(a)2 − 2|b|T σ(a) + b2 Using |b|T σ(a) ≤ 4nb2 +
1 σ(a)2 4n
then yields the required result. Using a simple basis transformation we can assume without loss of generality that: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ A11 0 B E ⎠ , B = ⎝ 1⎠ , E = ⎝ 1⎠ A=⎝ B2 E2 0 A22 with A11 asymptotically stable while AT22 A22 = I. This is guaranteed by the fact that A is neutrally stable. Next we consider the subsystem: x2 (t + 1) = A22 x2 (t) + B2 u(t) + E2 w(t) December 18, 2006
(4) DRAFT
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If we find an admissible feedback of the form u(t) = f (x2 (t))
(5)
which solves the semiglobal stochastic stabilization problem for the system (4) then it is easily verified that this controller also solves the semiglobal stochastic stabilization problem for the original system (1). In other words we can restrict attention to the system (4). We will use a controller of the form: u(k) = −ρB2T A22 x2 (k) and we will establish that for any given variance Q of the white noise there exists ρ small enough such that the resulting system has a bounded variance for the state. The choice of ρ is independent of the variance of the initial condition x0 . Proof: [Proof of Theorem II.2] To keep notation simple, we assume the original system satisfies AT A = I and we consider a feedback of the form: u(k) = −ρB T Ax(k)
(6)
The general case of the theorem can then easily be established using the reduction to the system (4) as described earlier. If the system is not affected by noise, it is well-known that the feedback (6) achieves asymptotic stability which is also easily verified by noting that V (x) = xT x is a suitable Lyapunov function. To study the effect of the noise, we first note that due to the Markov property of the state, we have: E [x(k + r)T x(k + r) | x(k + s)] = E [E [x(k + r)T x(k + r) | x(k + t)] | x(k + s)] for r ≥ t ≥ s. Next we consider: E [x(k + 1)T x(k + 1) | x(k)]
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Using the feedback (6), the dynamics (1) and the fact that AT A = I we obtain: E [x(k + 1)T x(k + 1) | x(k)] = x(k)T x(k) − 2x(k)T AT Bσ(ρB T Ax(k)) + σ(ρB T Ax(k))T B T Bσ(ρB T Ax(k)) + trace EQE T ≤ x(k)T x(k) − ρ2 σ(ρB T Ax(k))T σ(ρB T Ax(k)) + σ(ρB T Ax(k))T B T Bσ(ρB T Ax(k)) + trace EQE T ≤ x(k)T x(k) − ρ1 σ(ρB T Ax(k))2 + trace EQE T
(7)
provided ρ is small enough such that ρB T B < I. Next, we consider:
2 E σ(ρB T At x(k + 1)) | x(k)
for some integer t ≥ 1. We get:
2 E σ(ρB T At x(k + 1)) | x(k)
2 = E σ(ρB T At+1 x(k) − ρB T At Bσ(ρB T Ax(k)) + ρB T At Ew(k)) | x(k) 1 )σ(ρB T At+1 x(k)2 − 8n E − ρB T At Bσ(ρB T Ax(k)) ≥ (1 − 2n +ρB T At Ew(k))2 | x(k) ≥ (1 −
1 )σ(ρB T At+1 x(k))2 2n
− 8nρ2 B T At B2 σ(ρB T Ax(k))2
− 8ρ2 B T At 2 trace EQE T where the first inequality follows from Lemma III.1. If we choose ρ∗ such that ρ∗ B T B < I [as used earlier in deriving (7)] and √ 8n ρ∗ B T At 2 ≤ 1,
√ 8n ρ∗ B T At B2 ≤ 1
for all t, then it is easy to see that for all ρ < ρ∗ and any positive integer t we have:
2 E σ(ρB T At x(k + 1)) | x(k) ≥ (1 −
1 )σ(ρB T At+1 x(k))2 2n
− ρ3/2 σ(ρB T Ax(k))2 − ρ3/2 trace EQE T
The following lemma presents a crucial inequality:
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Lemma III.2 We have for ρ small enough that E [x(k + s)T x(k + s) | x(k)]
s
1 t
σ(ρB T At x(k)) 2 + s 1 + s trace EQE T (8) ≤ x(k) x(k) − 1− ρ t=1 2n n T
Proof: We will establish this inequality recursively. Note that we already established this inequality for s = 1 given (7). Next, consider any s > 1 and assume the above inequality is true for s (and any k). Then we obtain: E [x(k + s + 1)T x(k + s + 1) | x(k)] = E [E [x(k + s + 1)T x(k + s + 1) | x(k + 1)] | x(k)] s
1 t T
σ(ρB T At x(k + 1)) 2 ≤ E x(k + 1) x(k + 1) − 1− ρ t=1 2n s +s 1 + trace EQE T | x(k) n Using the previously obtained inequality we have: s
t 2
σ(ρB T At x(k + 1)) | x(k) E 1− 2n t=1 s
1 t 2
σ(ρB T At+1 x(k)) − sρ3/2 σ(ρB T Ax(k)) ≥ 1− 1− 2n 2n t=1 − sρ3/2 trace EQE T s
t+1
σ(ρB T At+1 x(k)) 2 − sρ3/2 σ(ρB T Ax(k)) − sρ3/2 trace EQE T 1− ≥ 2n t=1
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We then obtain: E [x(k + s + 1)T x(k + s + 1) | x(k)] = E [E [x(k + s + 1)T x(k + s + 1) | x(k + 1)] | x(k)] ≤ x(k)T x(k) − 1ρ σ(ρB T Ax(k))2 + trace EQE T s
1 t+1
σ(ρB T At+1 x(k)) 2 − 1− ρ t=1 2n
√ √ s + s ρσ(ρB T Ax(k))2 + s ρ trace EQE T + s 1 + trace EQE T n s+1
2 1 s+1 t T T t
≤ x(k) x(k) − 1− σ(ρB A x(k)) + (s + 1) 1 + trace EQE T ρ t=1 2n n provided ρ < ρ∗ is such that
√
ρ≤
1 , n
sρ3/2 ≤
1 . 2n
Next we note that since (A, B) is controllable while A is invertible, the matrix ⎞ ⎛ B TA ⎟ ⎜ ⎜ B T A2 ⎟ ⎟ ⎜ ⎜ . ⎟ ⎜ .. ⎟ ⎠ ⎝ B T An is injective which implies that there exists α (the smallest singular value of this matrix divided by n) such that for any x there exists a positive integer t ≤ n such that B T At x ≥ αx and hence
1 σ(ρB T At x)2 ≥ min ρα2 x2 , ρ1 ρ
But then (8) yields: E [x(k + n)T x(k + n) | x(k)] ≤ x(k)T x(k) − 12 min ρα2 x(k)2 , 1ρ + 2n trace EQE T We define M=
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1 ρ2 α2
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and y(k) = max { x(k)T x(k), M } The following lemma presents another crucial result: Lemma III.3 We have: E [y(k + n) | y(k)] ≤ y(k)
(9)
provided ρ is small enough such that provided ρ is small enough such that the inequality of Lemma III.2 is valid for s = 1, . . . , n and: 1 > 2n trace EQE T 2ρ
and ρα2 < 1
Moreover, E[y(k + pn)] ≤ E[y(k)],
p = 1, 2, . . .
(10)
Proof: This follows from the fact that if y(k) = x(k)T x(k) and hence x(k)T x(k) ≥ M we have: E [x(k + n)T x(k + n) | x(k)] 2 2 1 1 ≤ y(k) − 2 min ρα x(k) , ρ + 2n trace EQE T ≤ y(k) −
1 2ρ
+ 2n trace EQE T
≤ y(k) while for the case y(k) = M and hence x(k)T x(k) ≤ M we have: E [x(k + n)T x(k + n) | x(k)]
≤ x(k)T x(k) − 12 min ρα2 x(k)2 , 1ρ + 2n trace EQE T
≤ (1 − 12 ρα2 )x(k)T x(k) + 2n trace EQE T ≤ (1 − 12 ρα2 )y(k) + 2n trace EQE T ≤ y(k) − 12 ρα2 M + 2n trace EQE T ≤ y(k) but then (using that y(k) ≥ M): E [y(k + n) | x(k)] ≤ y(k) December 18, 2006
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and finally, E [y(k + n) | y(k)] = E [E [y(k + n) | x(k)] | y(k)] ≤ E [y(k) | y(k)] = y(k) Similarly (where we have to be careful since y(k) is not Markov), E [y(k + 2n) | y(k)] = E [E [y(k + 2n) | x(k + n), x(k)] | y(k)] = E [E [E [y(k + 2n) | x(k + n), x(k)] | x(k)] | y(k)] = E [E [E [y(k + 2n) | x(k + n)] | x(k)] | y(k)] ≤ E [E [y(k + n) | x(k)] | y(k)] ≤ E [y(k) | y(k)] = y(k) Using a simple recursion, we obtain (10). We have, using (7): E[x(k + 1)T x(k + 1)|x(k)] ≤ x(k)T x(k) + trace EQE T which yields: E[x(k)T x(k)] ≤ Var[x(0)] + k trace EQE T for k = 1, . . . , n − 1 and hence: E[y(k)] ≤ max { M, Var[x(0)] + k trace EQE T } for k = 1, . . . , n − 1. This implies that the expectation of y(0), . . . , y(n − 1) is bounded. Using (10) we find that the expectation of y(k) is bounded in k. Using that y(k) ≥ x(k)T x(k) this yields that the variance of x(k) is bounded in k. IV. P ROOFS
FOR THE CONTINUOUS - TIME CASE
Using a simple basis transformation we can assume without loss of generality that: ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ B1 E A11 0 ⎠ , B = ⎝ ⎠ , E = ⎝ 1⎠ A=⎝ B2 E2 0 A22
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with A11 asymptotically stable while A22 + AT22 = 0. This is guaranteed by the fact that A is neutrally stable. Next we consider the subsystem: dx2 (t) = A22 x2 (t)dt + B2 σ(u(t))dt + E2 dw(t)
(11)
If we find an admissible feedback of the form u(t) = f (x2 (t))
(12)
which solves the semiglobal external stochastic stabilization problem for the system (11) then it is easily verified that this controller also solves the semiglobal stochastic stabilization problem for the original system (3). In other words we can restrict attention to the system (11). We will use a controller of the form: u(t) = −ρB2T x2 (t) and we will establish that for any given rate Q of the Wiener process there exists ρ small enough such that the resulting system has a bounded variance for the state. The choice of ρ is independent of the variance of the initial condition x0 . Proof: [Proof of Theorem II.4] To keep notation simple, we assume the original system satisfies A + AT = 0 and we consider a feedback of the form: u(t) = −ρB T x(t)
(13)
The general case of the theorem can then easily be established using the reduction to the system (11) as described earlier. If the system is not affected by noise, it is well-known that the feedback (13) achieves asymptotic stability which is also easily verified by noting that V (x) = xT x is a suitable Lyapunov function. To study the effect of the noise, we first note that: τ τ A(τ −t) A(τ −s) x(t) + e Bσ(u(s))ds + eA(τ −s) Edw(s) x(τ ) = e t
t
= eA(τ −t) x(t) + vt (τ ) We note that σ(u) is bounded by 1 and the second integral has bounded variance for all τ such that τ ∈ [t, t + 1]. This implies that there exists a constant N such that: Var[vt (τ )] < N December 18, 2006
(14) DRAFT
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for all τ such that τ ∈ [t, t + 1]. Next, we note that σ(ρB T x(τ ))2 = σ ρB T eA(τ −t) x(t) + ρB T vt (τ ) 2 ≥ 12 σ(ρB T eA(τ −t) x(t))2 − 8ρ2 B2 vt (τ )2
(15)
where we applied Lemma III.1 with n = 1. Next we note that: dxT x = (2xT Bσ(u) + trace E T E)dt + 2xT Edw using that A + AT = 0 and Itˆo’s lemma. This yields that x(t + 1)T x(t + 1) − x(t)T x(t) t+1 t+1 2x(τ )T Bσ(u(τ ))dτ + trace E T E + 2x(τ )T Edw(τ ) = t
t
and hence: E[x(t + 1)T x(t + 1) | x(t)]
⎡ t+1 ⎤ = x(t)T x(t) + E ⎣ 2x(τ )T Bσ(u(τ ))dτ | x(t)⎦ + trace E T E t
On the other hand, we have: 2x(τ )T Bσ(u(τ )) = 2x(τ )T Bσ(−ρB T x(τ )) ≤ − 2ρ σ(ρB T x(τ ))2 Combining the last two equations we obtain: E[x(t + 1)T x(t + 1) | x(t)]
⎡ t+1 ⎤ ≤ x(t)T x(t) + trace E T E − 2ρ E ⎣ σ(ρB T x(τ ))2 dτ | x(t)⎦ t
Using (14), (15) and the fact that x(t) and vt (τ ) are independent for τ ∈ [t, t + 1] we obtain a crucial inequality: E[x(t + 1)T x(t + 1) | x(t)] ≤ x(t)T x(t) + V −
1 ρ
t+1 σ(ρB T eA(τ −t) x(t))2 dτ
(16)
t December 18, 2006
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where V = 16ρB2 N + trace E T E. Lemma IV.1 Assume that (A, B) is controllable. Then there exist suitably defined constants α and R such that for all ρ < ρ¯ we have that: 1 ρ
t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ ≥ min{ραx(t)2 , R } ρ t
Proof: If the saturation does not get activated and hence we get: t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ = ρ2 x(t)T Qx(t) t
where:
1 Q=
eAτ BB T eAτ dτ
0
Controllability of (A, B) implies that Q is invertible and hence we obtain: x(t)Qx(t) ≥ αx(t)2 where α is the smallest singular value of Q. Hence we obtain: 1 ρ
t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ ≥ ραx(t)2
(17)
t
On the other hand, in case the saturation does get activated then it is easily seen that there must exist an N (independent of ρ) such that: x(t) >
N ρ
We obtain: 2 t+1 t+1
2
Nx(t) T A(τ −t) T A(τ −t)
σ(ρB e
dτ
σ ρB e x(t)) dτ ≥
ρx(t) t
We consider the integral:
t
2 t+1
σ ρB T eA(τ −t) Nx(t) dτ
ρx(t)
(18)
t
In case saturation does not arise in this integral we immediately find that the term is larger or equal to
Nx(t) 2
≥ αN 2
ρ α ρx(t) 2
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using the same techniques as for the case that the original integral did not contain a term which saturates. But then we find 1 ρ
t+1
σ(ρB T eA(τ −t) x(t)) 2 dτ ≥ 1 αN 2 ρ
(19)
t
Finally we look at the case where the saturation in the integral (18) does get activated. In that case we note that there exists τ such that:
2
Nx(t) T A(τ −t)
≥1
σ ρB e
ρx(t) But the derivative with respect to τ of this term is bounded by some constant γ independent of our choice for τ ∈ [t, t + 1] and independent of our choice for x(t). This implies that 2 t+1
σ ρB T eA(τ −t) Nx(t) dτ ≥ 1
ρx(t) γ t
This yields 1 ρ
2 t+1 t+1
2
1 Nx(t) T A(τ −t) T A(τ −t)
dτ
σ ρB e
σ(ρB e x(t)) dτ ≥
ρ ρx(t) t
t
1 ≥ ργ
(20)
Combining (17) with (19) and (20), the result of the lemma follows immediately for 1 R = min{N 2 , }. γ The above lemma yields, with the help of (16), that: E[x(t + 1)T x(t + 1) | x(t)] ≤ x(t)T x(t) − min{ραx(t)2 , Rρ } + V We define: M=
R ρ2 α
and y(t) = max { x(t)T x(t), M } The following lemma presents another crucial result:
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Lemma IV.2 We have: E [y(t + 1) | y(t)] ≤ y(t)
(21)
provided ρ is small enough such that: ρ
