Introduction
Consider the LTI system
{ x ˙ ( t ) = A x ( t ) + B u ( t ) , y ( t ) = C x ( t ) , (1) \begin{cases}
\dot{x}(t) = Ax(t) + Bu(t),\\
y(t) = Cx(t), \tag*{(1)}
\end{cases} { x ˙ ( t ) = A x ( t ) + B u ( t ) , y ( t ) = C x ( t ) , (1) where x ( t ) ∈ R n x(t) \in\mathbb{R}^{n} x ( t ) ∈ R n u ( t ) ∈ R m u(t) \in\mathbb{R}^{m} u ( t ) ∈ R m y ( t ) ∈ R p y(t) \in\mathbb{R}^{p} y ( t ) ∈ R p A ∈ R n × n A \in\mathbb{R}^{n \times n} A ∈ R n × n B ∈ R n × m B \in\mathbb{R}^{n \times m} B ∈ R n × m C ∈ R p × n C \in\mathbb{R}^{p \times n} C ∈ R p × n
Static output feedback is attractive because of its simple implementation and low computational cost, but its stabilizing capability is generally more limited than that of state feedback [1 , 2 ] . These limitations have motivated extensions of
For systems without delay, as one of the challenging open problems in control theory, Brockett posed the following time-varying output-feedback stabilization problem [3 ] .
Brockett problem. Given matrices A A A B B B C C C
u ( t ) = K ( t ) y ( t ) u(t) = K(t)y(t) u ( t ) = K ( t ) y ( t ) such that the closed-loop system is asymptotically stable?
This problem motivated the development of periodic output-feedback methods [4 , 5 , 6 ] . In this direction, [7 ] presented stabilization conditions and constructive methods using delay-free time-varying output feedback for a broad class of systems. For the constant-gain case, recent results [8 , 9 , 10 ] established that deciding whether a linear system admits a stabilizing constant output-feedback gain is NP-hard. For SISO systems, memoryless time-varying output feedback and delayed output feedback with a constant gain are two representative extensions of constant output feedback. The former employs a time-varying gain, u ( t ) = − k ( t ) y ( t ) u(t)=-k(t)y(t) u ( t ) = − k ( t ) y ( t ) u ( t ) = − k y ( t − τ ) u(t)=-ky(t-\tau) u ( t ) = − k y ( t − τ ) [11 ] , which provided a complete comparison for second-order SISO systems.
For systems with a given feedback delay, the Brockett problem was extended in [12 ] as follows.
Brockett problem with feedback delay. Given matrices A A A B B B C C C τ > 0 \tau> 0 τ > 0
u ( t ) = K ( t ) y ( t − τ ) u(t) = K(t)y(t-\tau) u ( t ) = K ( t ) y ( t − τ ) such that the closed-loop system is asymptotically stable?
To address this problem and the difficulty caused by the infinite-dimensional closed-loop dynamics, [12 ] investigated the act-and-wait control framework introduced in [13 ] , in which the feedback is periodically switched off and on. When the waiting interval is at least as long as the feedback delay, stability can be characterized by a finite-dimensional monodromy matrix. However, general conditions for the existence of a stabilizing gain within this framework were left as an open problem. Multiple artificial delays have also been investigated for static output feedback stabilization of special classes of linear systems using frequency-domain methods [14 , 15 , 16 ] and time-domain methods [17 , 18 , 19 , 20 , 21 , 22 , 23 ] . These results primarily establish asymptotic, exponential, or hyperexponential stability under restrictions on the system structure.
Recently, artificial time delays have been incorporated into smooth feedback via the act-and-wait method to achieve convergence within an initial-condition-independent time bound, yielding prescribed-time stabilization for systems without input delays and fixed-time stabilization for systems with input delays [24 ] . Using this method, [25 ] studied fixed-time stabilization of systems with nilpotent state matrices subject to large unknown delays. In addition, an alternative approach to fixed-time stabilization based on sampled-data control was proposed in [26 ] .
Prescribed-time static output-feedback design is particularly challenging because the full state information is unavailable. By employing Taylor expansions or finite-difference approximations, [27 ] developed such a prescribed-time static output-feedback controller for LTI systems. However, its applicability is restricted to systems without internal dynamics [28 ] , where observability assumption is not sufficient. A systematic construction of static output feedback controllers for stabilization of general controllable and observable linear systems remains largely unexplored.
In this paper, to the best of our knowledge, we for the first time solve the stabilization problem for every controllable and observable LTI system (1) by static output feedback. Beyond stabilization, the proposed design guarantees exact convergence to the origin by any prescribed time, independently of the initial condition. The following assumption is imposed.
Assumption 1 The pair ( A , B ) (A, B) ( A , B ) ( A , C ) (A, C) ( A , C )
Following [24 ] , we adopt the following definition of prescribed-time stability.
Definition 1 Consider system (1) under a given feedback controller, and let T ( x 0 ) T(x_0) T ( x 0 ) T > 0 T > 0 T > 0 x ( t ; x 0 ) = 0 x(t; x_0) = 0 x ( t ; x 0 ) = 0 ∀ t ≥ T \forall t \ge T ∀ t ≥ T x 0 x_0 x 0 T T T x 0 x_0 x 0
The prescribed-time stabilization problem is formulated as follows.
Problem 1 Consider system (1) under Assumption 1. Given a prescribed time T > 0 T > 0 T > 0
In this paper, we propose two static output-feedback methods for prescribed-time stabilization of linear systems. The first employs multiple artificial output delays and provides an explicit gain construction for every controllable and observable LTI system, based on a generalized controllability–observability Gramian introduced in this paper. The second combines constant-gain feedback of the current output with time-varying-gain feedback of a single artificially delayed output and also applies to every controllable and observable LTI system. It has a simpler structure but requires auxiliary parameters whose existence is guaranteed, while their explicit construction is not provided for general MIMO systems. For SISO systems, the constant current-output gain can be readily selected. The multiple-delay design is further extended to known bounded and fast-varying (without any constraints on the delay derivative) measurement delays, yielding fixed-time stabilization under a corresponding rank condition. For unknown constant measurement delays, fixed-time stabilization is established for systems with nilpotent state matrices. Fixed-time stabilization by state feedback for such systems was initiated in [25 ] .
The remainder of this paper is organized as follows. Section II presents the multiple-delay feedback design. Section III presents static feedback that depends on the current and delayed output. Section IV extends the multiple-delay design to systems with measurement delays. Section V presents a numerical implementation. Section VI provides numerical examples, and Section VII concludes the paper. The proofs of auxiliary results are collected in the Appendix.
Notation. Z 0 + \mathbb{Z}_0^+ Z 0 + N \mathbb{N} N ∥ ⋅ ∥ \lVert\cdot\rVert ∥ ⋅ ∥ M M M M T M^{\mathsf{T}} M T rank ( M ) \operatorname{rank}(M) rank ( M ) det ( M ) \det(M) det ( M ) adj ( M ) \operatorname{adj}(M) adj ( M ) λ min ( M ) \lambda_{\min}(M) λ m i n ( M ) λ max ( M ) \lambda_{\max}(M) λ m a x ( M ) σ min ( M ) \sigma_{\min}(M) σ m i n ( M ) Im ( M ) \operatorname{Im}(M) Im ( M ) span { ⋅ } \operatorname{span}\{\cdot\} span { ⋅ } M M M vec ( ⋅ ) \operatorname{vec}(\cdot) vec ( ⋅ ) vec − 1 ( ⋅ ) \operatorname{vec}^{-1}(\cdot) vec − 1 ( ⋅ ) ⌊ a ⌋ \lfloor a \rfloor ⌊ a ⌋ a a a I ⊂ R \mathcal{I} \subset\mathbb{R} I ⊂ R ν \nu ν C ν ( I ) C^\nu(\mathcal{I}) C ν ( I ) ν \nu ν I \mathcal{I} I ⊗ \otimes ⊗ ⊕ \oplus ⊕
Feedback with Multiple Output Delays
Given a prescribed time T > 0 T > 0 T > 0 h = T / 3 h = T/3 h = T /3
l = ⌊ t 3 h ⌋ , θ = t − 3 h l ∈ [ 0 , 3 h ) . l = \left\lfloor\frac{t}{3h} \right\rfloor,\qquad\theta= t - 3hl \in[0, 3h). l = ⌊ 3 h t ⌋ , θ = t − 3 h l ∈ [ 0 , 3 h ) . The act-and-wait function is defined as
g ( t ) = { 0 , θ ∈ [ 0 , 2 h ] , 1 , θ ∈ ( 2 h , 3 h ) . (2) g(t) =
\begin{cases}
0, & \theta\in[0, 2h],\\
1, & \theta\in(2h, 3h).
\end{cases} \tag*{(2)} g ( t ) = { 0 , 1 , θ ∈ [ 0 , 2 h ] , θ ∈ ( 2 h , 3 h ) . (2) Throughout the paper, each controller is defined to give u ( t ) = 0 u(t) = 0 u ( t ) = 0 g ( t ) = 0 g(t) = 0 g ( t ) = 0
Choose N N N h 1 , … , h N h_1,\ldots,h_N h 1 , … , h N h ≤ h 1 < h 2 < ⋯ < h N ≤ 2 h h \le h_1 < h_2 < \cdots< h_N \le2h h ≤ h 1 < h 2 < ⋯ < h N ≤ 2 h
O h = [ C e − A h 1 C e − A h 2 ⋮ C e − A h N ] ∈ R N p × n . (3) \mathcal{O}_h =
\begin{bmatrix}
Ce^{-Ah_1}\\
Ce^{-Ah_2}\\
\vdots\\
Ce^{-Ah_N}
\end{bmatrix}
\in\mathbb{R}^{Np \times n}. \tag*{(3)} O h = C e − A h 1 C e − A h 2 ⋮ C e − A h N ∈ R N p × n . (3) For SISO systems, one typically chooses N = n N=n N = n p > 1 p>1 p > 1 N < n N<n N < n
Lemma 1 Suppose that ( A , C ) (A,C) ( A , C ) N N N h i ∈ [ h , 2 h ] h_i \in[h,2h] h i ∈ [ h , 2 h ] i = 1 , … , N i=1,\ldots,N i = 1 , … , N
rank ( O h ) = n . (4) \operatorname{rank}(\mathcal{O}_h)=n. \tag*{(4)} rank ( O h ) = n . (4) Proof. Let I = [ h , 2 h ] \mathcal{I}=[h,2h] I = [ h , 2 h ]
span { Im ( e − A T s C T ) : s ∈ I } = R n . \operatorname{span}\left\{\operatorname{Im}\left(e^{-A^{\mathsf{T}}s}C^{\mathsf{T}}\right):s\in\mathcal{I}\right\}=\mathbb{R}^{n}. span { Im ( e − A T s C T ) : s ∈ I } = R n . Otherwise, there would exist a nonzero vector v ∈ R n v\in\mathbb{R}^{n} v ∈ R n C e − A s v = 0 Ce^{-As}v=0 C e − A s v = 0 s ∈ I s\in\mathcal{I} s ∈ I C e − A s v Ce^{-As}v C e − A s v s s s C A k e − A s 0 v = 0 CA^k e^{-As_0}v=0 C A k e − A s 0 v = 0 k = 0 , … , n − 1 k=0,\ldots,n-1 k = 0 , … , n − 1 s 0 ∈ ( h , 2 h ) s_0\in(h,2h) s 0 ∈ ( h , 2 h ) e − A s 0 e^{-As_0} e − A s 0 ( A , C ) (A,C) ( A , C ) R n \mathbb{R}^{n} R n
Since R n \mathbb{R}^{n} R n N N N h i ∈ I h_i\in\mathcal{I} h i ∈ I
Next, we provide a constructive method for selecting n n n h i h_i h i i = 1 , … , n i=1,\ldots,n i = 1 , … , n
Lemma 2 Suppose that ( A , C ) (A,C) ( A , C ) h 0 ∈ [ h , 2 h ) h_0\in[h,2h) h 0 ∈ [ h , 2 h ) h i = h 0 + ( i − 1 ) d h_i=h_0+(i-1)d h i = h 0 + ( i − 1 ) d i = 1 , … , n i=1,\ldots,n i = 1 , … , n d > 0 d>0 d > 0 h 0 + ( n − 1 ) d < 2 h h_0+(n-1)d<2h h 0 + ( n − 1 ) d < 2 h d ∗ > 0 d^*>0 d ∗ > 0 0 < d < d ∗ 0<d<d^* 0 < d < d ∗ N = n N=n N = n
For brevity, the proof is provided in Appendix A1.
For s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ]
Γ d , i ( s ) : = e A T ( s − h i − 2 h ) C T ⊗ e A ( 3 h − s ) B ∈ R n 2 × m p , (5) \Gamma_{d,i}(s):=e^{A^{\mathsf{T}}(s-h_i-2h)}C^{\mathsf{T}}\otimes e^{A(3h-s)}B\in\mathbb{R}^{n^2\times mp}, \tag*{(5)} Γ d , i ( s ) := e A T ( s − h i − 2 h ) C T ⊗ e A ( 3 h − s ) B ∈ R n 2 × m p , (5) where i = 1 , … , N i=1,\ldots,N i = 1 , … , N
Γ d ( s ) : = [ Γ d , 1 ( s ) ⋯ Γ d , N ( s ) ] ∈ R n 2 × N m p . (6) \Gamma_d(s):=\begin{bmatrix}\Gamma_{d,1}(s)&\cdots&\Gamma_{d,N}(s)\end{bmatrix}\in\mathbb{R}^{n^2\times Nmp}. \tag*{(6)} Γ d ( s ) := [ Γ d , 1 ( s ) ⋯ Γ d , N ( s ) ] ∈ R n 2 × N m p . (6) For a nonnegative integer ν \nu ν ρ \rho ρ
ρ ∈ C ν ( [ 2 h , 3 h ] ) \rho\in C^\nu([2h,3h]) ρ ∈ C ν ([ 2 h , 3 h ])
ρ ( s ) > 0 \rho(s)>0 ρ ( s ) > 0 s ∈ ( 2 h , 3 h ) s\in(2h,3h) s ∈ ( 2 h , 3 h )
ρ ( k ) ( 2 h ) = ρ ( k ) ( 3 h ) = 0 \rho^{(k)}(2h)=\rho^{(k)}(3h)=0 ρ ( k ) ( 2 h ) = ρ ( k ) ( 3 h ) = 0 k = 0 , 1 , … , ν k=0,1,\ldots,\nu k = 0 , 1 , … , ν
Define a generalized controllability-observability Gramian
G = ∫ 2 h 3 h ρ ( s ) Γ d ( s ) Q d Γ d T ( s ) d s ∈ R n 2 × n 2 , (7) G=\int_{2h}^{3h}\rho(s)\Gamma_d(s)Q_d\Gamma_d^{\mathsf{T}}(s)\,ds\in\mathbb{R}^{n^2\times n^2}, \tag*{(7)} G = ∫ 2 h 3 h ρ ( s ) Γ d ( s ) Q d Γ d T ( s ) d s ∈ R n 2 × n 2 , (7) where
Q d = Q d T ∈ R N m p × N m p , Q d > 0 , (8) Q_d=Q_d^{\mathsf{T}}\in\mathbb{R}^{Nmp\times Nmp},\qquad Q_d>0, \tag*{(8)} Q d = Q d T ∈ R N m p × N m p , Q d > 0 , (8) is a constant weighting matrix.
Lemma 3 Suppose that ( A , B ) (A,B) ( A , B ) G G G G > 0 G>0 G > 0
For brevity, the proof is provided in Appendix A2.
With the above preparations, we can establish the following result.
Theorem 1 Consider system (1) under Assumption 1. Let T = 3 h > 0 T=3h>0 T = 3 h > 0 h i ∈ [ h , 2 h ] h_i\in[h,2h] h i ∈ [ h , 2 h ] i = 1 , … , N i=1,\ldots,N i = 1 , … , N
u ( t ) = g ( t ) ∑ i = 1 N L i ( θ ) y ( t − h i ) , t ≥ 0 , (9) u(t)=g(t)\sum_{i=1}^{N}L_i(\theta)y(t-h_i),\qquad t\ge0, \tag*{(9)} u ( t ) = g ( t ) i = 1 ∑ N L i ( θ ) y ( t − h i ) , t ≥ 0 , (9) where g ( t ) g(t) g ( t ) L i ( s ) = vec − 1 ( ℓ i ( s ) ) ∈ R m × p L_i(s)=\operatorname{vec}^{-1}(\ell_i(s))\in\mathbb{R}^{m\times p} L i ( s ) = vec − 1 ( ℓ i ( s )) ∈ R m × p i = 1 , … , N i=1,\ldots,N i = 1 , … , N
ℓ d ( s ) : = col i = 1 N { ℓ i ( s ) } = [ ℓ 1 T ( s ) ⋯ ℓ N T ( s ) ] T ∈ R N m p , \ell_d(s):=\operatorname{col}_{i=1}^{N}\{\ell_i(s)\}=
\begin{bmatrix}
\ell_1^{\mathsf{T}}(s)&\cdots&\ell_N^{\mathsf{T}}(s)
\end{bmatrix}^{\mathsf{T}}\in\mathbb{R}^{Nmp}, ℓ d ( s ) := col i = 1 N { ℓ i ( s )} = [ ℓ 1 T ( s ) ⋯ ℓ N T ( s ) ] T ∈ R N m p , with
ℓ d ( s ) = − ρ ( s ) Q d Γ d T ( s ) G − 1 vec ( e A h ) , s ∈ [ 2 h , 3 h ] , (10) \ell_d(s)=-\rho(s)Q_d\Gamma_d^{\mathsf{T}}(s)G^{-1}\operatorname{vec}(e^{Ah}),\qquad s\in[2h,3h], \tag*{(10)} ℓ d ( s ) = − ρ ( s ) Q d Γ d T ( s ) G − 1 vec ( e A h ) , s ∈ [ 2 h , 3 h ] , (10) in which Γ d ( s ) \Gamma_d(s) Γ d ( s ) ρ ( s ) \rho(s) ρ ( s ) G G G Q d Q_d Q d T = 3 h T=3h T = 3 h u ( t ) = 0 u(t)=0 u ( t ) = 0 t ≥ 3 h t\ge3h t ≥ 3 h
Proof. For t ∈ [ 0 , 2 h ] t\in[0,2h] t ∈ [ 0 , 2 h ] g ( t ) = 0 g(t)=0 g ( t ) = 0 x ( t ) = e A t x ( 0 ) x(t)=e^{At}x(0) x ( t ) = e A t x ( 0 ) x ( 2 h ) = e 2 A h x ( 0 ) x(2h)=e^{2Ah}x(0) x ( 2 h ) = e 2 A h x ( 0 )
For t ∈ [ 2 h , 3 h ] t\in[2h,3h] t ∈ [ 2 h , 3 h ] h i ∈ [ h , 2 h ] h_i\in[h,2h] h i ∈ [ h , 2 h ] t − h i ∈ [ 0 , 2 h ] t-h_i\in[0,2h] t − h i ∈ [ 0 , 2 h ] [ 0 , 2 h ] [0,2h] [ 0 , 2 h ]
x ( t − h i ) = e A ( t − h i ) x ( 0 ) = e A ( t − h i − 2 h ) x ( 2 h ) , x(t-h_i)=e^{A(t-h_i)}x(0)=e^{A(t-h_i-2h)}x(2h), x ( t − h i ) = e A ( t − h i ) x ( 0 ) = e A ( t − h i − 2 h ) x ( 2 h ) , and thus
y ( t − h i ) = C e A ( t − h i − 2 h ) x ( 2 h ) . y(t-h_i)=Ce^{A(t-h_i-2h)}x(2h). y ( t − h i ) = C e A ( t − h i − 2 h ) x ( 2 h ) . By the variation-of-constants formula, we can obtain
x ( 3 h ) = [ e A h + ∑ i = 1 N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s ] x ( 2 h ) . x(3h)=
\left[
e^{Ah}+\sum_{i=1}^{N}\int_{2h}^{3h}e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\,ds
\right]x(2h). x ( 3 h ) = [ e A h + i = 1 ∑ N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s ] x ( 2 h ) . We next show that the matrix inside the square brackets is zero. Using the linearity of the vectorization operator and the identity vec ( X Y Z ) = ( Z T ⊗ X ) vec ( Y ) \operatorname{vec}(XYZ)=(Z^{\mathsf{T}}\otimes X)\operatorname{vec}(Y) vec ( X Y Z ) = ( Z T ⊗ X ) vec ( Y ) i ∈ { 1 , … , N } i\in\{1,\ldots,N\} i ∈ { 1 , … , N }
vec ( e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) ) = [ ( C e A ( s − h i − 2 h ) ) T ⊗ e A ( 3 h − s ) B ] vec ( L i ( s ) ) \operatorname{vec}\left(e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\right)
=
\left[
\left(Ce^{A(s-h_i-2h)}\right)^{\mathsf{T}}\otimes e^{A(3h-s)}B
\right]\operatorname{vec}(L_i(s)) vec ( e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) ) = [ ( C e A ( s − h i − 2 h ) ) T ⊗ e A ( 3 h − s ) B ] vec ( L i ( s )) = [ e A T ( s − h i − 2 h ) C T ⊗ e A ( 3 h − s ) B ] ℓ i ( s ) = Γ d , i ( s ) ℓ i ( s ) . =
\left[
e^{A^{\mathsf{T}}(s-h_i-2h)}C^{\mathsf{T}}\otimes e^{A(3h-s)}B
\right]\ell_i(s)
=\Gamma_{d,i}(s)\ell_i(s). = [ e A T ( s − h i − 2 h ) C T ⊗ e A ( 3 h − s ) B ] ℓ i ( s ) = Γ d , i ( s ) ℓ i ( s ) . Consequently, by the definitions of Γ d ( s ) \Gamma_d(s) Γ d ( s ) ℓ d ( s ) \ell_d(s) ℓ d ( s )
Γ d ( s ) ℓ d ( s ) = [ Γ d , 1 ( s ) ⋯ Γ d , N ( s ) ] [ ℓ 1 ( s ) ⋮ ℓ N ( s ) ] \Gamma_d(s)\ell_d(s)=
\begin{bmatrix}
\Gamma_{d,1}(s)&\cdots&\Gamma_{d,N}(s)
\end{bmatrix}
\begin{bmatrix}
\ell_1(s)\\
\vdots\\
\ell_N(s)
\end{bmatrix} Γ d ( s ) ℓ d ( s ) = [ Γ d , 1 ( s ) ⋯ Γ d , N ( s ) ] ℓ 1 ( s ) ⋮ ℓ N ( s ) = ∑ i = 1 N Γ d , i ( s ) ℓ i ( s ) = ∑ i = 1 N vec ( e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) ) . \begin{aligned}
= \sum_{i=1}^{N} \Gamma_{d,i}(s)\ell_i(s) \\
&= \sum_{i=1}^{N} \operatorname{vec}\left(e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\right).
\end{aligned} = i = 1 ∑ N Γ d , i ( s ) ℓ i ( s ) = i = 1 ∑ N vec ( e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) ) . It follows from the linearity of the vectorization operator and integration that
vec ( ∑ i = 1 N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s ) = ∫ 2 h 3 h ∑ i = 1 N vec ( e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) ) d s = ∫ 2 h 3 h Γ d ( s ) ℓ d ( s ) d s . \begin{aligned}
\operatorname{vec}\left(\sum_{i=1}^{N}\int_{2h}^{3h}e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\,\mathrm{d}s\right) \\
&= \int_{2h}^{3h}\sum_{i=1}^{N}\operatorname{vec}\left(e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\right)\,\mathrm{d}s \\
&= \int_{2h}^{3h}\Gamma_d(s)\ell_d(s)\,\mathrm{d}s.
\end{aligned} vec ( i = 1 ∑ N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s ) = ∫ 2 h 3 h i = 1 ∑ N vec ( e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) ) d s = ∫ 2 h 3 h Γ d ( s ) ℓ d ( s ) d s . Substituting (11) into the preceding identity gives
∫ 2 h 3 h Γ d ( s ) ℓ d ( s ) d s = − ∫ 2 h 3 h ρ ( s ) Γ d ( s ) Q d Γ d T ( s ) G − 1 vec ( e A h ) d s = − [ ∫ 2 h 3 h ρ ( s ) Γ d ( s ) Q d Γ d T ( s ) d s ] G − 1 vec ( e A h ) = − G G − 1 vec ( e A h ) = − vec ( e A h ) , \begin{aligned}
\int_{2h}^{3h}\Gamma_d(s)\ell_d(s)\,\mathrm{d}s \\
&= -\int_{2h}^{3h}\rho(s)\Gamma_d(s)Q_d\Gamma_d^{\mathsf T}(s)G^{-1}\operatorname{vec}(e^{Ah})\,\mathrm{d}s \\
&= -\left[\int_{2h}^{3h}\rho(s)\Gamma_d(s)Q_d\Gamma_d^{\mathsf T}(s)\,\mathrm{d}s\right]G^{-1}\operatorname{vec}(e^{Ah}) \\
&= -GG^{-1}\operatorname{vec}(e^{Ah})=-\operatorname{vec}(e^{Ah}),
\end{aligned} ∫ 2 h 3 h Γ d ( s ) ℓ d ( s ) d s = − ∫ 2 h 3 h ρ ( s ) Γ d ( s ) Q d Γ d T ( s ) G − 1 vec ( e A h ) d s = − [ ∫ 2 h 3 h ρ ( s ) Γ d ( s ) Q d Γ d T ( s ) d s ] G − 1 vec ( e A h ) = − G G − 1 vec ( e A h ) = − vec ( e A h ) , where (8) has been used in the third equality. We have therefore proved that
vec ( ∑ i = 1 N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s ) = − vec ( e A h ) \operatorname{vec}\left(\sum_{i=1}^{N}\int_{2h}^{3h}e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\,\mathrm{d}s\right)
=-\operatorname{vec}(e^{Ah}) vec ( i = 1 ∑ N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s ) = − vec ( e A h ) Since the vectorization operator is injective, it follows that
∑ i = 1 N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s = − e A h . \sum_{i=1}^{N}\int_{2h}^{3h}e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\,\mathrm{d}s=-e^{Ah}. i = 1 ∑ N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s = − e A h . Consequently, x ( 3 h ) = 0 x(3h)=0 x ( 3 h ) = 0
We finally show that the state and control input remain zero after 3 h 3h 3 h [ 3 h , 5 h ] [3h,5h] [ 3 h , 5 h ] x ( 3 h ) = 0 x(3h)=0 x ( 3 h ) = 0 x ( t ) = 0 x(t)=0 x ( t ) = 0 t ∈ [ 3 h , 5 h ] t\in[3h,5h] t ∈ [ 3 h , 5 h ] [ 5 h , 6 h ] [5h,6h] [ 5 h , 6 h ] t ∈ [ 5 h , 6 h ] t\in[5h,6h] t ∈ [ 5 h , 6 h ] h i ∈ [ h , 2 h ] h_i\in[h,2h] h i ∈ [ h , 2 h ] t − h i ∈ [ 3 h , 5 h ] t-h_i\in[3h,5h] t − h i ∈ [ 3 h , 5 h ] y ( t − h i ) = 0 y(t-h_i)=0 y ( t − h i ) = 0 i = 1 , … , N i=1,\ldots,N i = 1 , … , N u ( t ) = 0 u(t)=0 u ( t ) = 0 x ( t ) = 0 x(t)=0 x ( t ) = 0 t ∈ [ 5 h , 6 h ] t\in[5h,6h] t ∈ [ 5 h , 6 h ] x ( t ) = 0 x(t)=0 x ( t ) = 0 u ( t ) = 0 u(t)=0 u ( t ) = 0 t ≥ 3 h t\ge3h t ≥ 3 h
Finally we prove that there exists a constant M > 0 M>0 M > 0 sup 0 ≤ t ≤ 3 h ∥ x ( t ) ∥ ≤ M ∥ x ( 0 ) ∥ \sup_{0\le t\le3h}\|x(t)\|\le M\|x(0)\| sup 0 ≤ t ≤ 3 h ∥ x ( t ) ∥ ≤ M ∥ x ( 0 ) ∥ ρ ( ⋅ ) \rho(\cdot) ρ ( ⋅ ) Γ d ( ⋅ ) \Gamma_d(\cdot) Γ d ( ⋅ ) [ 2 h , 3 h ] [2h,3h] [ 2 h , 3 h ] G > 0 G>0 G > 0 ℓ d ( ⋅ ) \ell_d(\cdot) ℓ d ( ⋅ ) x ( t ) = 0 x(t)=0 x ( t ) = 0 t ≥ 3 h t\ge3h t ≥ 3 h T = 3 h T=3h T = 3 h
Feedback with a current output and a single output delay
In this section, we design a stabilizing feedback controller that uses the current measurement and a single delayed measurement, yielding a simpler controller structure. However, the
controller gain depends on a matrix H H H v v v w w w
Let K ∈ R m × p K\in\mathbb{R}^{m\times p} K ∈ R m × p
F K = A + B K C ∈ R n × n , (11) F_K=A+BKC\in\mathbb{R}^{n\times n}, \tag*{(11)} F K = A + B K C ∈ R n × n , (11) and
A K = A T ⊗ I n − I n ⊗ F K ∈ R n 2 × n 2 , A_K=A^{\mathsf T}\otimes I_n-I_n\otimes F_K\in\mathbb{R}^{n^2\times n^2}, A K = A T ⊗ I n − I n ⊗ F K ∈ R n 2 × n 2 , B K = C T ⊗ B ∈ R n 2 × m p . B_K=C^{\mathsf T}\otimes B\in\mathbb{R}^{n^2\times mp}. B K = C T ⊗ B ∈ R n 2 × m p . For s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ]
Γ K ( s ) : = e A T ( s − 3 h ) C T ⊗ e F K ( 3 h − s ) B = e A K ( s − 3 h ) B K ∈ R n 2 × m p . (12) \Gamma_K(s):=e^{A^{\mathsf T}(s-3h)}C^{\mathsf T}\otimes e^{F_K(3h-s)}B
=e^{A_K(s-3h)}B_K\in\mathbb{R}^{n^2\times mp}. \tag*{(12)} Γ K ( s ) := e A T ( s − 3 h ) C T ⊗ e F K ( 3 h − s ) B = e A K ( s − 3 h ) B K ∈ R n 2 × m p . (12) Define a generalized controllability-observability Gramian
G K = ∫ 2 h 3 h ρ ( s ) Γ K ( s ) Q Γ K T ( s ) d s , (13) G_K=\int_{2h}^{3h}\rho(s)\Gamma_K(s)Q\Gamma_K^{\mathsf T}(s)\,\mathrm{d}s, \tag*{(13)} G K = ∫ 2 h 3 h ρ ( s ) Γ K ( s ) Q Γ K T ( s ) d s , (13) where Q = Q T ∈ R m p × m p > 0 Q=Q^{\mathsf T}\in\mathbb{R}^{mp\times mp}>0 Q = Q T ∈ R m p × m p > 0 ρ \rho ρ ( A K , B K ) (A_K,B_K) ( A K , B K )
K K = [ B K A K B K ⋯ A K n 2 − 1 B K ] ∈ R n 2 × n 2 m p . \mathcal{K}_K=\begin{bmatrix}B_K&A_KB_K&\cdots&A_K^{n^2-1}B_K\end{bmatrix}\in\mathbb{R}^{n^2\times n^2mp}. K K = [ B K A K B K ⋯ A K n 2 − 1 B K ] ∈ R n 2 × n 2 m p . The following lemma shows that a suitable constant current-output gain always exists under Assumption 1.
Lemma 4 Consider system (1) under Assumption 1. Then there exists a constant matrix K ∈ R m × p K\in\mathbb{R}^{m\times p} K ∈ R m × p rank K K = n 2 \operatorname{rank}\mathcal{K}_K=n^2 rank K K = n 2 H ∈ R m × p H\in\mathbb{R}^{m\times p} H ∈ R m × p v ∈ R m v\in\mathbb{R}^{m} v ∈ R m w ∈ R p w\in\mathbb{R}^{p} w ∈ R p
K α = H + α v w T , α ∈ R , (14) K_\alpha=H+\alpha vw^{\mathsf T},\qquad\alpha\in\mathbb{R}, \tag*{(14)} K α = H + α v w T , α ∈ R , (14) the set of rank-deficient parameter values
E H , v , w : = { α ∈ R : rank K K α < n 2 } \mathcal{E}_{H,v,w}:=\{\alpha\in\mathbb{R}:\operatorname{rank}\mathcal{K}_{K_\alpha}<n^2\} E H , v , w := { α ∈ R : rank K K α < n 2 } is finite. Moreover, for any constant matrix K K K rank K K = n 2 \operatorname{rank}\mathcal{K}_K=n^2 rank K K = n 2 G K > 0 G_K>0 G K > 0
For brevity, the proof is provided in Appendix A3.
Remark 1 The proof of Lemma 4 provides a way to select K K K H H H F H = A + B H C F_H=A+BHC F H = A + B H C [29 ] . Next, choose v v v w w w ( F H , B v ) (F_H,Bv) ( F H , B v ) ( F H , w T C ) (F_H,w^{\mathsf T}C) ( F H , w T C ) α \alpha α K = K α K=K_\alpha K = K α rank K K = n 2 \operatorname{rank}\mathcal{K}_K=n^2 rank K K = n 2 α \alpha α [30 ] is required, because the current-output gain remains constant throughout the act interval.
We next present the prescribed-time stabilization result for system (1).
Theorem 2 Consider system (1) under Assumption 1. Let T = 3 h > 0 T = 3h > 0 T = 3 h > 0 K ∈ R m × p K \in\mathbb{R}^{m \times p} K ∈ R m × p rank K K = n 2 \operatorname{rank} K_K = n^2 rank K K = n 2 Lemma 4 , and the Gramian G K G_K G K (13) is positive definite. Consider the time-varying static output feedback controller
u ( t ) = g ( t ) K y ( t ) + g ( t ) L K ( θ ) y ( t − h ) , t ≥ 0 , (15) u(t) = g(t)Ky(t) + g(t)L_K(\theta)y(t-h), \qquad t \ge0, \tag*{(15)} u ( t ) = g ( t ) K y ( t ) + g ( t ) L K ( θ ) y ( t − h ) , t ≥ 0 , (15) where g ( t ) g(t) g ( t ) K ∈ R m × p K \in\mathbb{R}^{m \times p} K ∈ R m × p L K ( s ) = vec − 1 ( ℓ K ( s ) ) ∈ R m × p L_K(s) = \operatorname{vec}^{-1}(\ell_K(s)) \in\mathbb{R}^{m \times p} L K ( s ) = vec − 1 ( ℓ K ( s )) ∈ R m × p ℓ K ( s ) \ell_K(s) ℓ K ( s ) s ∈ [ 2 h , 3 h ] s \in[2h,3h] s ∈ [ 2 h , 3 h ]
ℓ K ( s ) = − ρ ( s ) Q Γ K T ( s ) G K − 1 vec ( e F K h ) ∈ R m p , (16) \ell_K(s) = -\rho(s)Q\Gamma_K^{\mathsf T}(s)G_K^{-1}\operatorname{vec}\left(e^{F_Kh}\right) \in\mathbb{R}^{mp}, \tag*{(16)} ℓ K ( s ) = − ρ ( s ) Q Γ K T ( s ) G K − 1 vec ( e F K h ) ∈ R m p , (16) in which ρ ( s ) \rho(s) ρ ( s ) F K F_K F K Γ K ( s ) \Gamma_K(s) Γ K ( s ) T = 3 h T = 3h T = 3 h u ( t ) = 0 u(t) = 0 u ( t ) = 0 t ≥ 3 h t \ge3h t ≥ 3 h
Proof. For t ∈ [ 0 , 2 h ] t \in[0,2h] t ∈ [ 0 , 2 h ] g ( t ) = 0 g(t) = 0 g ( t ) = 0
x ( t ) = e A t x ( 0 ) , t ∈ [ 0 , 2 h ] . (17) x(t) = e^{At}x(0), \qquad t \in[0,2h]. \tag*{(17)} x ( t ) = e A t x ( 0 ) , t ∈ [ 0 , 2 h ] . (17) For t ∈ ( 2 h , 3 h ) t \in(2h,3h) t ∈ ( 2 h , 3 h )
x ˙ ( t ) = F K x ( t ) + B L K ( t ) y ( t − h ) . \dot{x}(t) = F_Kx(t) + BL_K(t)y(t-h). x ˙ ( t ) = F K x ( t ) + B L K ( t ) y ( t − h ) . Since t − h ∈ [ h , 2 h ] t-h \in[h,2h] t − h ∈ [ h , 2 h ] (17) that y ( t − h ) = C e A ( t − h ) x ( 0 ) = C e A ( t − 3 h ) x ( 2 h ) y(t-h) = Ce^{A(t-h)}x(0) = Ce^{A(t-3h)}x(2h) y ( t − h ) = C e A ( t − h ) x ( 0 ) = C e A ( t − 3 h ) x ( 2 h )
By the variation-of-constants formula,
x ( 3 h ) = e F K h x ( 2 h ) + ∫ 2 h 3 h e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) d s x ( 2 h ) . x(3h) = e^{F_Kh}x(2h) + \int_{2h}^{3h} e^{F_K(3h-s)}BL_K(s)Ce^{A(s-3h)}\,\mathrm{d}s\,x(2h). x ( 3 h ) = e F K h x ( 2 h ) + ∫ 2 h 3 h e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) d s x ( 2 h ) . Using the Kronecker–vectorization identity vec ( X Y Z ) = ( Z T ⊗ X ) vec ( Y ) \operatorname{vec}(XYZ) = (Z^{\mathsf T} \otimes X)\operatorname{vec}(Y) vec ( X Y Z ) = ( Z T ⊗ X ) vec ( Y ) X = e F K ( 3 h − s ) B X = e^{F_K(3h-s)}B X = e F K ( 3 h − s ) B Y = L K ( s ) Y = L_K(s) Y = L K ( s ) Z = C e A ( s − 3 h ) Z = Ce^{A(s-3h)} Z = C e A ( s − 3 h )
vec ( e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) ) = [ ( C e A ( s − 3 h ) ) T ⊗ e F K ( 3 h − s ) B ] vec ( L K ( s ) ) = [ e A T ( s − 3 h ) C T ⊗ e F K ( 3 h − s ) B ] ℓ K ( s ) = Γ K ( s ) ℓ K ( s ) . \begin{aligned}
\operatorname{vec}\left(e^{F_K(3h-s)}BL_K(s)Ce^{A(s-3h)}\right)
&= \left[\left(Ce^{A(s-3h)}\right)^{\mathsf T} \otimes e^{F_K(3h-s)}B\right]\operatorname{vec}(L_K(s)) \\
&= \left[e^{A^{\mathsf T}(s-3h)}C^{\mathsf T} \otimes e^{F_K(3h-s)}B\right]\ell_K(s) \\
&= \Gamma_K(s)\ell_K(s).
\end{aligned} vec ( e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) ) = [ ( C e A ( s − 3 h ) ) T ⊗ e F K ( 3 h − s ) B ] vec ( L K ( s )) = [ e A T ( s − 3 h ) C T ⊗ e F K ( 3 h − s ) B ] ℓ K ( s ) = Γ K ( s ) ℓ K ( s ) . Using this identity and (17), we obtain
vec ( ∫ 2 h 3 h e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) d s ) = ∫ 2 h 3 h Γ K ( s ) ℓ K ( s ) d s = − [ ∫ 2 h 3 h ρ ( s ) Γ K ( s ) Q Γ K T ( s ) d s ] G K − 1 vec ( e F K h ) = − vec ( e F K h ) . \begin{aligned}
\operatorname{vec}\left(\int_{2h}^{3h} e^{F_K(3h-s)}BL_K(s)Ce^{A(s-3h)}\,\mathrm{d}s\right)
&= \int_{2h}^{3h}\Gamma_K(s)\ell_K(s)\,\mathrm{d}s \\
&= -\left[\int_{2h}^{3h}\rho(s)\Gamma_K(s)Q\Gamma_K^{\mathsf T}(s)\,\mathrm{d}s\right]G_K^{-1}\operatorname{vec}\left(e^{F_Kh}\right) \\
&= -\operatorname{vec}\left(e^{F_Kh}\right).
\end{aligned} vec ( ∫ 2 h 3 h e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) d s ) = ∫ 2 h 3 h Γ K ( s ) ℓ K ( s ) d s = − [ ∫ 2 h 3 h ρ ( s ) Γ K ( s ) Q Γ K T ( s ) d s ] G K − 1 vec ( e F K h ) = − vec ( e F K h ) . Since the vectorization operator is injective, we obtain
∫ 2 h 3 h e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) d s = − e F K h . \int_{2h}^{3h} e^{F_K(3h-s)}BL_K(s)Ce^{A(s-3h)}\,\mathrm{d}s = -e^{F_Kh}. ∫ 2 h 3 h e F K ( 3 h − s ) B L K ( s ) C e A ( s − 3 h ) d s = − e F K h . Consequently, x ( 3 h ) = 0 x(3h) = 0 x ( 3 h ) = 0
Since x ( 3 h ) = 0 x(3h) = 0 x ( 3 h ) = 0 [ 3 h , 5 h ] [3h,5h] [ 3 h , 5 h ] [ 4 h , 5 h ] [4h,5h] [ 4 h , 5 h ] x ( t ) = 0 x(t) = 0 x ( t ) = 0 u ( t ) = 0 u(t) = 0 u ( t ) = 0 t ≥ 3 h t \ge3h t ≥ 3 h
Finally, K K K L K ( ⋅ ) L_K(\cdot) L K ( ⋅ ) [ 2 h , 3 h ] [2h,3h] [ 2 h , 3 h ] M > 0 M > 0 M > 0 x ( 0 ) x(0) x ( 0 ) sup 0 ≤ t ≤ 3 h ∥ x ( t ) ∥ ≤ M ∥ x ( 0 ) ∥ \sup_{0 \le t \le3h}\lVert x(t)\rVert\le M\lVert x(0)\rVert sup 0 ≤ t ≤ 3 h ∥ x ( t )∥ ≤ M ∥ x ( 0 )∥ x ( t ) = 0 x(t) = 0 x ( t ) = 0 t ≥ 3 h t \ge3h t ≥ 3 h T = 3 h T = 3h T = 3 h
Remark 2 For SISO systems, the constant current-output gain is a scalar K = q ∈ R K = q \in\mathbb{R} K = q ∈ R q 0 q_0 q 0 rank K q 0 = n 2 \operatorname{rank} K_{q_0} = n^2 rank K q 0 = n 2 K q K_q K q q q q
E : = { q ∈ R : rank K q < n 2 } \mathcal{E} := \{q \in\mathbb{R} : \operatorname{rank} K_q < n^2\} E := { q ∈ R : rank K q < n 2 } is finite. By Lemma 4 and Theorem 2, controller (16) with K = q K = q K = q q ∉ E q \notin\mathcal{E} q ∈ / E ∣ q ∣ \lvert q\rvert ∣ q ∣
Remark 3 The controller (16) can be modified by replacing g ( t ) g(t) g ( t )
g 0 ( t ) = { 0 , θ ∈ [ 0 , h ] , 1 , θ ∈ ( h , 2 h ) , g_0(t) =
\begin{cases}
0, & \theta\in[0,h],\\
1, & \theta\in(h,2h),
\end{cases} g 0 ( t ) = { 0 , 1 , θ ∈ [ 0 , h ] , θ ∈ ( h , 2 h ) , and adjusting the cycle length, local phase, and the corresponding gain formulas accordingly, yielding a settling time of 2 h 2h 2 h g ( t ) g(t) g ( t )
Remark 4 The two controller designs in (10) and (16) share the same principle: the output-feedback gains are chosen so that the accumulated contribution of the delayed feedback cancels the free terminal transition, namely, e A h x ( 2 h ) e^{Ah}x(2h) e A h x ( 2 h ) e F K h x ( 2 h ) e^{F_Kh}x(2h) e F K h x ( 2 h ) x ( 2 h ) x(2h) x ( 2 h )
To see this, define R i ( s ) = C e A ( s − h i − 2 h ) R_i(s) = Ce^{A(s-h_i-2h)} R i ( s ) = C e A ( s − h i − 2 h ) y ( s − h i ) = R i ( s ) x ( 2 h ) y(s-h_i) = R_i(s)x(2h) y ( s − h i ) = R i ( s ) x ( 2 h )
vec ( Ψ ( 3 h , s ) B L i ( s ) R i ( s ) ) = ( R i T ( s ) ⊗ Ψ ( 3 h , s ) B ) vec ( L i ( s ) ) , \operatorname{vec}\left(\Psi(3h,s)BL_i(s)R_i(s)\right) = \left(R_i^{\mathsf T}(s) \otimes\Psi(3h,s)B\right)\operatorname{vec}(L_i(s)), vec ( Ψ ( 3 h , s ) B L i ( s ) R i ( s ) ) = ( R i T ( s ) ⊗ Ψ ( 3 h , s ) B ) vec ( L i ( s )) , where Ψ ( 3 h , s ) \Psi(3h,s) Ψ ( 3 h , s ) s s s 3 h 3h 3 h
For the multiple-delay design, Ψ ( 3 h , s ) = e A ( 3 h − s ) \Psi(3h,s) = e^{A(3h-s)} Ψ ( 3 h , s ) = e A ( 3 h − s ) R i ( s ) R_i(s) R i ( s )
For the current-plus-single-delay design, only the delay h h h R ( s ) = C e A ( s − 3 h ) R(s) = Ce^{A(s-3h)} R ( s ) = C e A ( s − 3 h ) Γ K ( s ) \Gamma_K(s) Γ K ( s ) K K K e F K ( 3 h − s ) e^{F_K(3h-s)} e F K ( 3 h − s )
Under the respective conditions established above, the columns of Γ d ( s ) \Gamma_d(s) Γ d ( s ) Γ K ( s ) \Gamma_K(s) Γ K ( s ) s ∈ ( 2 h , 3 h ) s \in(2h,3h) s ∈ ( 2 h , 3 h ) R n 2 \mathbb{R}^{n^2} R n 2 x ( 3 h ) = 0 x(3h)=0 x ( 3 h ) = 0
Measurement Delays
In this section, we extend the multiple-delay design to known time-varying measurement delays and unknown constant measurement delays.
Known Time-Varying Measurement Delays
Consider the LTI system
x ˙ ( t ) = A x ( t ) + B u ( t ) , y ( t ) = { C x ( t − τ ( t ) ) , t − τ ( t ) ≥ 0 , 0 , t − τ ( t ) < 0 , (18) \begin{aligned}
\dot{x}(t) &= Ax(t)+Bu(t),\\
y(t) &=
\begin{cases}
Cx(t-\tau(t)), & t-\tau(t)\ge0,\\
0, & t-\tau(t)<0,
\end{cases} \tag*{(18)}
\end{aligned} x ˙ ( t ) y ( t ) = A x ( t ) + B u ( t ) , = { C x ( t − τ ( t )) , 0 , t − τ ( t ) ≥ 0 , t − τ ( t ) < 0 , (18) where A A A B B B C C C x ( t ) x(t) x ( t ) u ( t ) u(t) u ( t ) y ( t ) y(t) y ( t ) τ ( t ) \tau(t) τ ( t ) 0 ≤ τ ( t ) ≤ τ M 0\le\tau(t)\le\tau_M 0 ≤ τ ( t ) ≤ τ M
For the design developed below, the settling-time bound can be assigned a priori subject to T > 3 τ M T>3\tau_M T > 3 τ M [24 ] , [25 ] , [27 ] . We thus adopt the following definition of fixed-time stability [31 ] .
Definition 2 Consider system (19) under a given feedback controller. The resulting closed-loop system is said to be fixed-time stable if it is Lyapunov stable and there exists a constant T f > 0 T_f>0 T f > 0 x 0 x_0 x 0 x ( t ; x 0 ) = 0 x(t;x_0)=0 x ( t ; x 0 ) = 0 t ≥ T f t\ge T_f t ≥ T f x 0 ∈ R n x_0\in\mathbb{R}^n x 0 ∈ R n T f T_f T f
The stabilization problem is formulated as follows.
Problem 2 Consider system (19) under Assumption 1. Given a time bound T > 3 τ M T>3\tau_M T > 3 τ M T T T
Let h = T / 3 > τ M h=T/3>\tau_M h = T /3 > τ M N N N
h ≤ h i ≤ 2 h − τ M , i = 1 , … , N . (19) h\le h_i\le2h-\tau_M,\qquad i=1,\ldots,N. \tag*{(19)} h ≤ h i ≤ 2 h − τ M , i = 1 , … , N . (19) For s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ]
Γ v , i ( s ) : = e A T ( s − h i − 2 h − τ ( s − h i ) ) C T ⊗ e A ( 3 h − s ) B ∈ R n 2 × m p , \Gamma_{v,i}(s):=e^{A^{\mathsf{T}}(s-h_i-2h-\tau(s-h_i))}C^{\mathsf{T}}\otimes e^{A(3h-s)}B\in\mathbb{R}^{n^2\times mp}, Γ v , i ( s ) := e A T ( s − h i − 2 h − τ ( s − h i )) C T ⊗ e A ( 3 h − s ) B ∈ R n 2 × m p , where i = 1 , … , N i=1,\ldots,N i = 1 , … , N
Γ v ( s ) : = [ Γ v , 1 ( s ) ⋯ Γ v , N ( s ) ] ∈ R n 2 × N m p . (20) \Gamma_v(s):=\begin{bmatrix}\Gamma_{v,1}(s)&\cdots&\Gamma_{v,N}(s)\end{bmatrix}\in\mathbb{R}^{n^2\times Nmp}. \tag*{(20)} Γ v ( s ) := [ Γ v , 1 ( s ) ⋯ Γ v , N ( s ) ] ∈ R n 2 × N m p . (20) Define
G v = ∫ 2 h 3 h ρ ( s ) Γ v ( s ) Q v Γ v T ( s ) d s , (21) G_v=\int_{2h}^{3h}\rho(s)\Gamma_v(s)Q_v\Gamma_v^{\mathsf{T}}(s)\,\mathrm{d}s, \tag*{(21)} G v = ∫ 2 h 3 h ρ ( s ) Γ v ( s ) Q v Γ v T ( s ) d s , (21) where
Q v = Q v T ∈ R N m p × N m p , Q v > 0. (22) Q_v=Q_v^{\mathsf{T}}\in\mathbb{R}^{Nmp\times Nmp},\qquad Q_v>0. \tag*{(22)} Q v = Q v T ∈ R N m p × N m p , Q v > 0. (22) Define the effective sampled observability matrix
O h , τ ( s ) = [ C e − A ( h 1 + τ ( s − h 1 ) ) C e − A ( h 2 + τ ( s − h 2 ) ) ⋮ C e − A ( h N + τ ( s − h N ) ) ] . \mathcal{O}_{h,\tau}(s)=
\begin{bmatrix}
Ce^{-A(h_1+\tau(s-h_1))}\\
Ce^{-A(h_2+\tau(s-h_2))}\\
\vdots\\
Ce^{-A(h_N+\tau(s-h_N))}
\end{bmatrix}. O h , τ ( s ) = C e − A ( h 1 + τ ( s − h 1 )) C e − A ( h 2 + τ ( s − h 2 )) ⋮ C e − A ( h N + τ ( s − h N )) . Lemma 5 Suppose that ( A , B ) (A,B) ( A , B ) I ⊂ ( 2 h , 3 h ) \mathcal{I}\subset(2h,3h) I ⊂ ( 2 h , 3 h ) s ↦ τ ( s − h i ) s\mapsto\tau(s-h_i) s ↦ τ ( s − h i ) i = 1 , … , N i=1,\ldots,N i = 1 , … , N
rank ( O h , τ ( s ) ) = n , s ∈ I . (23) \operatorname{rank}(\mathcal{O}_{h,\tau}(s))=n,\qquad s\in\mathcal{I}. \tag*{(23)} rank ( O h , τ ( s )) = n , s ∈ I . (23) Then the Gramian defined in (22) is positive definite, namely, G v > 0 G_v>0 G v > 0
For brevity, the proof is provided in Appendix A4.
The rank condition in Lemma 5 depends on the specific delay function τ ( t ) \tau(t) τ ( t )
Lemma 6 Suppose that ( A , B ) (A,B) ( A , B ) A ≠ 0 A\ne0 A = 0
0 ≤ τ M < 1 ∥ A ∥ ln ( 1 + σ min ( O h ) ( ∑ i = 1 N ∥ C e − A h i ∥ 2 ) 1 / 2 ) . (24) 0\le\tau_M<\frac{1}{\lVert A\rVert}\ln\left(1+\frac{\sigma_{\min}(\mathcal{O}_h)}{\left(\sum_{i=1}^{N}\lVert Ce^{-Ah_i}\rVert^2\right)^{1/2}}\right). \tag*{(24)} 0 ≤ τ M < ∥ A ∥ 1 ln 1 + ( ∑ i = 1 N ∥ C e − A h i ∥ 2 ) 1/2 σ m i n ( O h ) . (24) where O h \mathcal{O}_h O h A = 0 A=0 A = 0 τ M \tau_M τ M 0 ≤ τ ( t ) ≤ τ M 0\le\tau(t)\le\tau_M 0 ≤ τ ( t ) ≤ τ M
rank ( O h , τ ( s ) ) = n , s ∈ [ 2 h , 3 h ] . \operatorname{rank}(\mathcal{O}_{h,\tau}(s))=n,\qquad s\in[2h,3h]. rank ( O h , τ ( s )) = n , s ∈ [ 2 h , 3 h ] . Consequently, G v > 0 G_v>0 G v > 0
Proof. First consider A ≠ 0 A\ne0 A = 0 i i i s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ] i i i O h , τ ( s ) − O h \mathcal{O}_{h,\tau}(s)-\mathcal{O}_h O h , τ ( s ) − O h
C e − A ( h i + τ ( s − h i ) ) − C e − A h i = C e − A h i ( e − A τ ( s − h i ) − I n ) . Ce^{-A(h_i+\tau(s-h_i))}-Ce^{-Ah_i}=Ce^{-Ah_i}\left(e^{-A\tau(s-h_i)}-I_n\right). C e − A ( h i + τ ( s − h i )) − C e − A h i = C e − A h i ( e − A τ ( s − h i ) − I n ) . Since 0 ≤ τ ( s − h i ) ≤ τ M 0\le\tau(s-h_i)\le\tau_M 0 ≤ τ ( s − h i ) ≤ τ M
∥ e − A τ ( s − h i ) − I n ∥ ≤ ∑ k = 1 ∞ ∥ A k ∥ τ ( s − h i ) k k ! ≤ ∑ k = 1 ∞ ∥ A ∥ k τ ( s − h i ) k k ! = e ∥ A ∥ τ ( s − h i ) − 1 ≤ e ∥ A ∥ τ M − 1. \begin{aligned}
\left\|e^{-A\tau(s-h_i)}-I_n\right\|
\le\sum_{k=1}^{\infty}\frac{\lVert A^k\rVert\tau(s-h_i)^k}{k!} \\
&\le\sum_{k=1}^{\infty}\frac{\lVert A\rVert^k\tau(s-h_i)^k}{k!}
=e^{\lVert A\rVert\tau(s-h_i)}-1\le e^{\lVert A\rVert\tau_M}-1.
\end{aligned} e − A τ ( s − h i ) − I n ≤ k = 1 ∑ ∞ k ! ∥ A k ∥ τ ( s − h i ) k ≤ k = 1 ∑ ∞ k ! ∥ A ∥ k τ ( s − h i ) k = e ∥ A ∥ τ ( s − h i ) − 1 ≤ e ∥ A ∥ τ M − 1. Hence,
∥ C e − A ( h i + τ ( s − h i ) ) − C e − A h i ∥ ≤ ∥ C e − A h i ∥ ( e ∥ A ∥ τ M − 1 ) , \left\|Ce^{-A(h_i+\tau(s-h_i))}-Ce^{-Ah_i}\right\|
\le\left\|Ce^{-Ah_i}\right\|\left(e^{\lVert A\rVert\tau_M}-1\right), C e − A ( h i + τ ( s − h i )) − C e − A h i ≤ C e − A h i ( e ∥ A ∥ τ M − 1 ) , which yields
∥ O h , τ ( s ) − O h ∥ ≤ ( ∑ i = 1 N ∥ C e − A h i ∥ 2 ) 1 / 2 ( e ∥ A ∥ τ M − 1 ) . \left\|\mathcal{O}_{h,\tau}(s)-\mathcal{O}_h\right\|
\le\left(\sum_{i=1}^{N}\left\|Ce^{-Ah_i}\right\|^2\right)^{1/2}
\left(e^{\lVert A\rVert\tau_M}-1\right). ∥ O h , τ ( s ) − O h ∥ ≤ ( i = 1 ∑ N C e − A h i 2 ) 1/2 ( e ∥ A ∥ τ M − 1 ) . Condition (25) is equivalent to
e ∥ A ∥ τ M − 1 < σ min ( O h ) ( ∑ i = 1 N ∥ C e − A h i ∥ 2 ) 1 / 2 . e^{\lVert A\rVert\tau_M}-1 < \frac{\sigma_{\min}(\mathcal{O}_h)}{\left(\sum_{i=1}^{N}\lVert C e^{-A h_i}\rVert^2\right)^{1/2}}. e ∥ A ∥ τ M − 1 < ( ∑ i = 1 N ∥ C e − A h i ∥ 2 ) 1/2 σ m i n ( O h ) . Consequently, ∥ O h , τ ( s ) − O h ∥ < σ min ( O h ) \lVert\mathcal{O}_{h,\tau}(s)-\mathcal{O}_h\rVert<\sigma_{\min}(\mathcal{O}_h) ∥ O h , τ ( s ) − O h ∥ < σ m i n ( O h ) s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ] [32 ] ,
σ min ( O h , τ ( s ) ) ≥ σ min ( O h ) − ∥ O h , τ ( s ) − O h ∥ > 0. \sigma_{\min}(\mathcal{O}_{h,\tau}(s))\geq\sigma_{\min}(\mathcal{O}_h)-\lVert\mathcal{O}_{h,\tau}(s)-\mathcal{O}_h\rVert>0. σ m i n ( O h , τ ( s )) ≥ σ m i n ( O h ) − ∥ O h , τ ( s ) − O h ∥ > 0. Thus, rank ( O h , τ ( s ) ) = n \operatorname{rank}(\mathcal{O}_{h,\tau}(s))=n rank ( O h , τ ( s )) = n [ 2 h , 3 h ] [2h,3h] [ 2 h , 3 h ]
If A = 0 A=0 A = 0 e − A ( h i + τ ( s − h i ) ) = e − A h i = I n e^{-A(h_i+\tau(s-h_i))}=e^{-Ah_i}=I_n e − A ( h i + τ ( s − h i )) = e − A h i = I n O h , τ ( s ) = O h \mathcal{O}_{h,\tau}(s)=\mathcal{O}_h O h , τ ( s ) = O h s s s
Since τ \tau τ I ⊂ ( 2 h , 3 h ) \mathcal{I}\subset(2h,3h) I ⊂ ( 2 h , 3 h ) s ↦ τ ( s − h i ) s\mapsto\tau(s-h_i) s ↦ τ ( s − h i ) i = 1 , … , N i=1,\ldots,N i = 1 , … , N G v > 0 G_v>0 G v > 0
Theorem 3 Consider system (19) under Assumption 1. Let T = 3 h > 3 τ M T=3h>3\tau_M T = 3 h > 3 τ M
u ( t ) = g ( t ) ∑ i = 1 N L i ( θ ) y ( t − h i ) , θ = t − 3 h ⌊ t 3 h ⌋ , (25) u(t)=g(t)\sum_{i=1}^{N}L_i(\theta)y(t-h_i),\qquad\theta=t-3h\left\lfloor\frac{t}{3h}\right\rfloor, \tag*{(25)} u ( t ) = g ( t ) i = 1 ∑ N L i ( θ ) y ( t − h i ) , θ = t − 3 h ⌊ 3 h t ⌋ , (25) where g ( t ) g(t) g ( t ) L i ( s ) = vec − 1 ( ℓ i , v ( s ) ) ∈ R m × p L_i(s)=\operatorname{vec}^{-1}(\ell_{i,v}(s))\in\mathbb{R}^{m\times p} L i ( s ) = vec − 1 ( ℓ i , v ( s )) ∈ R m × p i = 1 , … , N i=1,\ldots,N i = 1 , … , N
ℓ v ( s ) : = [ ℓ 1 , v T ( s ) ⋯ ℓ N , v T ( s ) ] T ∈ R N m p \ell_v(s):=\begin{bmatrix}\ell_{1,v}^{\mathsf{T}}(s)&\cdots&\ell_{N,v}^{\mathsf{T}}(s)\end{bmatrix}^{\mathsf{T}}\in\mathbb{R}^{Nmp} ℓ v ( s ) := [ ℓ 1 , v T ( s ) ⋯ ℓ N , v T ( s ) ] T ∈ R N m p is defined by
ℓ v ( s ) = − ρ ( s ) Q v Γ v T ( s ) G v − 1 vec ( e A h ) , s ∈ [ 2 h , 3 h ] . (26) \ell_v(s)=-\rho(s)Q_v\Gamma_v^{\mathsf{T}}(s)G_v^{-1}\operatorname{vec}(e^{Ah}),\qquad s\in[2h,3h]. \tag*{(26)} ℓ v ( s ) = − ρ ( s ) Q v Γ v T ( s ) G v − 1 vec ( e A h ) , s ∈ [ 2 h , 3 h ] . (26) Here, Γ v ( s ) \Gamma_v(s) Γ v ( s ) G v G_v G v Q v Q_v Q v T = 3 h T=3h T = 3 h x ( t ) = u ( t ) = 0 x(t)=u(t)=0 x ( t ) = u ( t ) = 0 t ≥ 3 h t\geq3h t ≥ 3 h
Proof. Since g ( t ) = 0 g(t)=0 g ( t ) = 0 t ∈ [ 0 , 2 h ] t\in[0,2h] t ∈ [ 0 , 2 h ] x ( t ) = e A t x ( 0 ) x(t)=e^{At}x(0) x ( t ) = e A t x ( 0 ) t ∈ [ 0 , 2 h ] t\in[0,2h] t ∈ [ 0 , 2 h ] s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ] 0 ≤ τ ( s − h i ) ≤ τ M 0\leq\tau(s-h_i)\leq\tau_M 0 ≤ τ ( s − h i ) ≤ τ M 0 ≤ s − h i − τ ( s − h i ) ≤ 2 h 0\leq s-h_i-\tau(s-h_i)\leq2h 0 ≤ s − h i − τ ( s − h i ) ≤ 2 h y ( s − h i ) = C e A ( s − h i − 2 h − τ ( s − h i ) ) x ( 2 h ) y(s-h_i)=Ce^{A(s-h_i-2h-\tau(s-h_i))}x(2h) y ( s − h i ) = C e A ( s − h i − 2 h − τ ( s − h i )) x ( 2 h )
By the variation-of-constants formula,
x ( 3 h ) = e A h x ( 2 h ) + [ ∑ i = 1 N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h − τ ( s − h i ) ) d s ] x ( 2 h ) . x(3h)=e^{Ah}x(2h)+\left[\sum_{i=1}^{N}\int_{2h}^{3h}e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h-\tau(s-h_i))}\,\mathrm{d}s\right]x(2h). x ( 3 h ) = e A h x ( 2 h ) + [ i = 1 ∑ N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h − τ ( s − h i )) d s ] x ( 2 h ) . The Kronecker–vectorization identity gives
vec ( ∑ i = 1 N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h − τ ( s − h i ) ) d s ) = ∫ 2 h 3 h Γ v ( s ) ℓ v ( s ) d s . \operatorname{vec}\left(\sum_{i=1}^{N}\int_{2h}^{3h}e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h-\tau(s-h_i))}\,\mathrm{d}s\right)
=\int_{2h}^{3h}\Gamma_v(s)\ell_v(s)\,\mathrm{d}s. vec ( i = 1 ∑ N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h − τ ( s − h i )) d s ) = ∫ 2 h 3 h Γ v ( s ) ℓ v ( s ) d s . Substituting (27) and using (22) yields
∫ 2 h 3 h Γ v ( s ) ℓ v ( s ) d s = − vec ( e A h ) . \int_{2h}^{3h}\Gamma_v(s)\ell_v(s)\,\mathrm{d}s=-\operatorname{vec}(e^{Ah}). ∫ 2 h 3 h Γ v ( s ) ℓ v ( s ) d s = − vec ( e A h ) . Hence, x ( 3 h ) = 0 x(3h)=0 x ( 3 h ) = 0
During the next waiting interval [ 3 h , 5 h ] [3h,5h] [ 3 h , 5 h ] t ∈ ( 5 h , 6 h ) t\in(5h,6h) t ∈ ( 5 h , 6 h ) 3 h ≤ t − h i − τ ( t − h i ) ≤ 5 h 3h\leq t-h_i-\tau(t-h_i)\leq5h 3 h ≤ t − h i − τ ( t − h i ) ≤ 5 h i = 1 , … , N i=1,\ldots,N i = 1 , … , N x ( t ) = u ( t ) = 0 x(t)=u(t)=0 x ( t ) = u ( t ) = 0 t ≥ 3 h t\geq3h t ≥ 3 h
Since the gains are bounded on [ 2 h , 3 h ] [2h,3h] [ 2 h , 3 h ] M > 0 M>0 M > 0 sup 0 ≤ t ≤ 3 h ∥ x ( t ) ∥ ≤ M ∥ x ( 0 ) ∥ \sup_{0\leq t\leq3h}\lVert x(t)\rVert\leq M\lVert x(0)\rVert sup 0 ≤ t ≤ 3 h ∥ x ( t )∥ ≤ M ∥ x ( 0 )∥ T = 3 h T=3h T = 3 h
Similarly to the sampled-data approach in [26 ] , the following corollary extends the measurement-delay design to sampled-data output feedback with time-varying gains.
Corollary 1 Consider system (1) under Assumption 1, with output samples
y ( t k ) = C x ( t k ) , t k = k d s , k ∈ Z + 0 , y(t_k)=Cx(t_k),\qquad t_k=kd_s,\qquad k\in\mathbb{Z}_+^0, y ( t k ) = C x ( t k ) , t k = k d s , k ∈ Z + 0 , where d s > 0 d_s>0 d s > 0 T = 3 h T=3h T = 3 h h > d s h>d_s h > d s h i = ν i d s h_i=\nu_i d_s h i = ν i d s ν i ∈ N \nu_i\in\mathbb{N} ν i ∈ N i = 1 , … , N i=1,\ldots,N i = 1 , … , N h ≤ h 1 < ⋯ < h N ≤ 2 h − d s h\leq h_1<\cdots<h_N\leq2h-d_s h ≤ h 1 < ⋯ < h N ≤ 2 h − d s
u ( t ) = g ( t ) ∑ i = 1 N L i ( θ ) y ( t k − ν i ) , t ∈ [ t k , t k + 1 ) , (27) u(t)=g(t)\sum_{i=1}^{N}L_i(\theta)y(t_{k-\nu_i}),\qquad t\in[t_k,t_{k+1}), \tag*{(27)} u ( t ) = g ( t ) i = 1 ∑ N L i ( θ ) y ( t k − ν i ) , t ∈ [ t k , t k + 1 ) , (27) where g ( t ) g(t) g ( t ) θ = t − 3 h ⌊ t / ( 3 h ) ⌋ \theta=t-3h\lfloor t/(3h)\rfloor θ = t − 3 h ⌊ t / ( 3 h )⌋ L i ( s ) = vec − 1 ( ℓ i , v ( s ) ) ∈ R m × p L_i(s)=\operatorname{vec}^{-1}(\ell_{i,v}(s))\in\mathbb{R}^{m\times p} L i ( s ) = vec − 1 ( ℓ i , v ( s )) ∈ R m × p i = 1 , … , N i=1,\ldots,N i = 1 , … , N τ ( t ) = t − t k \tau(t)=t-t_k τ ( t ) = t − t k t ∈ [ t k , t k + 1 ) t\in[t_k,t_{k+1}) t ∈ [ t k , t k + 1 ) τ M = d s \tau_M=d_s τ M = d s T = 3 h T=3h T = 3 h u ( t ) = 0 u(t)=0 u ( t ) = 0 t ≥ 3 h t\geq3h t ≥ 3 h
Proof. Since τ ( t ) = t − t k ∈ [ 0 , d s ) \tau(t)=t-t_k\in[0,d_s) τ ( t ) = t − t k ∈ [ 0 , d s ) h i = ν i d s h_i=\nu_i d_s h i = ν i d s y ˉ ( t − h i ) = y ( t k − ν i ) \bar y(t-h_i)=y(t_{k-\nu_i}) y ˉ ( t − h i ) = y ( t k − ν i ) t ∈ [ t k , t k + 1 ) t\in[t_k,t_{k+1}) t ∈ [ t k , t k + 1 ) y ˉ ( t ) = y ( t k ) \bar y(t)=y(t_k) y ˉ ( t ) = y ( t k ) τ \tau τ d s d_s d s τ ( s − h i ) = τ ( s ) \tau(s-h_i)=\tau(s) τ ( s − h i ) = τ ( s ) O h , τ ( s ) = O h e − A τ ( s ) \mathcal{O}_{h,\tau}(s)=\mathcal{O}_h e^{-A\tau(s)} O h , τ ( s ) = O h e − A τ ( s ) rank ( O h , τ ( s ) ) = rank ( O h ) = n \operatorname{rank}(\mathcal{O}_{h,\tau}(s))=\operatorname{rank}(\mathcal{O}_h)=n rank ( O h , τ ( s )) = rank ( O h ) = n G v > 0 G_v>0 G v > 0 h > d s = τ M h>d_s=\tau_M h > d s = τ M h i ∈ [ h , 2 h − τ M ] h_i\in[h,2h-\tau_M] h i ∈ [ h , 2 h − τ M ] u ( t ) = 0 u(t)=0 u ( t ) = 0 t ≥ 3 h t\geq3h t ≥ 3 h
Unknown Constant Measurement Delays
Consider the LTI system
{ x ˙ ( t ) = A x ( t ) + B u ( t ) , y ( t ) = { C x ( t − τ ) , t − τ ≥ 0 , 0 , t − τ < 0 , (28) \begin{cases}
\dot{x}(t)=Ax(t)+Bu(t),\\
y(t)=
\begin{cases}
Cx(t-\tau),&t-\tau\geq0,\\
0,&t-\tau<0,
\end{cases} \tag*{(28)}
\end{cases} ⎩ ⎨ ⎧ x ˙ ( t ) = A x ( t ) + B u ( t ) , y ( t ) = { C x ( t − τ ) , 0 , t − τ ≥ 0 , t − τ < 0 , (28) where τ ∈ [ 0 , τ M ] \tau\in[0,\tau_M] τ ∈ [ 0 , τ M ] τ M ≥ 0 \tau_M\geq0 τ M ≥ 0 ( A , B ) (A,B) ( A , B ) ( A , C ) (A,C) ( A , C ) A r = 0 A^r=0 A r = 0 r ≤ n r\leq n r ≤ n
Related fixed-time stabilization problems were studied using state feedback in [25 ] and static output feedback in [27 ] , with the latter restricted to systems without internal dynamics.
Similarly to the sampl…
[28 ] . The following proposition extends the static output-feedback result to systems with internal dynamics, including nonminimum-phase systems.
Proposition 1 Consider system (28) . Choose h > τ M h > \tau_M h > τ M N = r N = r N = r h i ∈ [ h , 2 h − τ M ] h_i \in[h, 2h-\tau_M] h i ∈ [ h , 2 h − τ M ] i = 1 , … , r i = 1,\ldots,r i = 1 , … , r (9) , with the gains defined in (10) , renders the closed-loop system fixed-time stable with settling-time bound 3 r h 3rh 3 r h x ( t ) = u ( t ) = 0 x(t) = u(t) = 0 x ( t ) = u ( t ) = 0 t ≥ 3 r h t \ge3rh t ≥ 3 r h
Proof. Since A r = 0 A^r = 0 A r = 0
O h = ( V h ⊗ I p ) col ( C , C A , … , C A r − 1 ) , O_h = (V_h \otimes I_p)\operatorname{col}(C,CA,\ldots,CA^{r-1}), O h = ( V h ⊗ I p ) col ( C , C A , … , C A r − 1 ) , where
V h = [ 1 − h 1 ⋯ ( − h 1 ) r − 1 ( r − 1 ) ! ⋮ ⋮ ⋮ 1 − h r ⋯ ( − h r ) r − 1 ( r − 1 ) ! ] . V_h =
\begin{bmatrix}
1 & -h_1 & \cdots& \dfrac{(-h_1)^{r-1}}{(r-1)!}\\
\vdots& \vdots& & \vdots\\
1 & -h_r & \cdots& \dfrac{(-h_r)^{r-1}}{(r-1)!}
\end{bmatrix}. V h = 1 ⋮ 1 − h 1 ⋮ − h r ⋯ ⋯ ( r − 1 )! ( − h 1 ) r − 1 ⋮ ( r − 1 )! ( − h r ) r − 1 . The matrix V h V_h V h ( A , C ) (A,C) ( A , C ) A r = 0 A^r = 0 A r = 0 rank ( col ( C , C A , … , C A r − 1 ) ) = n \operatorname{rank}(\operatorname{col}(C,CA,\ldots,CA^{r-1})) = n rank ( col ( C , C A , … , C A r − 1 )) = n rank O h = n \operatorname{rank} O_h = n rank O h = n (10) are well defined.
During each acting interval, every measurement used by the controller refers to the preceding waiting interval, since 0 ≤ s − h i − τ ≤ 2 h 0 \le s-h_i-\tau\le2h 0 ≤ s − h i − τ ≤ 2 h s ∈ [ 2 h , 3 h ] s \in[2h,3h] s ∈ [ 2 h , 3 h ]
x ( 3 h ( l + 1 ) ) = Δ τ x ( 3 h l ) , Δ τ = e 3 A h ( I − e − A τ ) , l ∈ Z + . x(3h(l+1)) = \Delta_\tau x(3hl),\qquad\Delta_\tau= e^{3Ah}(I-e^{-A\tau}),\qquad l \in\mathbb{Z}_{+}. x ( 3 h ( l + 1 )) = Δ τ x ( 3 h l ) , Δ τ = e 3 A h ( I − e − A τ ) , l ∈ Z + . Since A r = 0 A^r = 0 A r = 0 P τ \mathcal{P}_\tau P τ I − e − A τ = A P τ ( A ) I-e^{-A\tau} = A\mathcal{P}_\tau(A) I − e − A τ = A P τ ( A )
Δ τ r = e 3 r h A A r P τ ( A ) r = 0. \Delta_\tau^r = e^{3rhA}A^r\mathcal{P}_\tau(A)^r = 0. Δ τ r = e 3 r h A A r P τ ( A ) r = 0. Consequently, x ( 3 r h ) = 0 x(3rh) = 0 x ( 3 r h ) = 0 u ( t ) = 0 u(t) = 0 u ( t ) = 0 t ≥ 3 r h t \ge3rh t ≥ 3 r h
The bounded gains and the variation-of-constants formula give a constant M > 0 M > 0 M > 0 x ( 0 ) x(0) x ( 0 ) sup 0 ≤ t ≤ 3 r h ∥ x ( t ) ∥ ≤ M ∥ x ( 0 ) ∥ \sup_{0 \le t \le3rh}\|x(t)\| \le M\|x(0)\| sup 0 ≤ t ≤ 3 r h ∥ x ( t ) ∥ ≤ M ∥ x ( 0 ) ∥ x ( t ) = 0 x(t) = 0 x ( t ) = 0 t ≥ 3 r h t \ge3rh t ≥ 3 r h
Remark 5 The lifted directions described in Remark 4 also determine the numerical conditioning of the associated Gramians. Although the conditions established above guarantee G > 0 G > 0 G > 0 G K > 0 G_K > 0 G K > 0 d 1 = [ 1 , 0 ] T d_1 = [1,0]^{\mathsf{T}} d 1 = [ 1 , 0 ] T d 2 = [ cos ϑ , sin ϑ ] T d_2 = [\cos\vartheta,\sin\vartheta]^{\mathsf{T}} d 2 = [ cos ϑ , sin ϑ ] T D = [ d 1 , d 2 ] D = [d_1,d_2] D = [ d 1 , d 2 ] 0 < ϑ ≤ π / 2 0 < \vartheta\le\pi/2 0 < ϑ ≤ π /2
κ 2 ( D D T ) = λ max ( D D T ) λ min ( D D T ) = 1 + cos ϑ 1 − cos ϑ ⟶ ∞ as ϑ → 0. \kappa_2(DD^{\mathsf{T}}) = \frac{\lambda_{\max}(DD^{\mathsf{T}})}{\lambda_{\min}(DD^{\mathsf{T}})} = \frac{1+\cos\vartheta}{1-\cos\vartheta} \longrightarrow\infty\quad\text{as } \vartheta\to0. κ 2 ( D D T ) = λ m i n ( D D T ) λ m a x ( D D T ) = 1 − cos ϑ 1 + cos ϑ ⟶ ∞ as ϑ → 0. Alternatively, taking d 1 = [ 1 , 0 ] T d_1 = [1,0]^{\mathsf{T}} d 1 = [ 1 , 0 ] T d 2 = [ 0 , δ ] T d_2 = [0,\delta]^{\mathsf{T}} d 2 = [ 0 , δ ] T 0 < δ ≤ 1 0 < \delta\le1 0 < δ ≤ 1 D D T = diag ( 1 , δ 2 ) DD^{\mathsf{T}} = \operatorname{diag}(1,\delta^2) D D T = diag ( 1 , δ 2 ) κ 2 ( D D T ) = δ − 2 \kappa_2(DD^{\mathsf{T}}) = \delta^{-2} κ 2 ( D D T ) = δ − 2
Numerical Implementation
The gain functions in Theorem 1 are explicitly constructed using the Gramian G G G G − 1 G^{-1} G − 1
Let { ϕ 0 , ϕ 1 , … , ϕ N b − 1 } \{\phi_0,\phi_1,\ldots,\phi_{N_b-1}\} { ϕ 0 , ϕ 1 , … , ϕ N b − 1 } C ν ( [ 2 h , 3 h ] ) C^{\nu}([2h,3h]) C ν ([ 2 h , 3 h ]) h i h_i h i i = 1 , … , N i = 1,\ldots,N i = 1 , … , N
vec ( L i ( s ) ) = ρ ( s ) ∑ k = 0 N b − 1 ϕ k ( s ) a i , k , s ∈ [ 2 h , 3 h ] , (29) \operatorname{vec}(L_i(s)) = \rho(s)\sum_{k=0}^{N_b-1}\phi_k(s)a_{i,k},\qquad s \in[2h,3h], \tag*{(29)} vec ( L i ( s )) = ρ ( s ) k = 0 ∑ N b − 1 ϕ k ( s ) a i , k , s ∈ [ 2 h , 3 h ] , (29) where a i , k ∈ R m p a_{i,k} \in\mathbb{R}^{mp} a i , k ∈ R m p ρ \rho ρ ρ \rho ρ
Following the derivation in the proof of Theorem 1, define the terminal map F F F
x ( 3 h ) = F x ( 2 h ) . x(3h) = Fx(2h). x ( 3 h ) = F x ( 2 h ) . By the variation-of-constants formula, it is given by
F = e A h + ∑ i = 1 N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s . F = e^{Ah} + \sum_{i=1}^{N}\int_{2h}^{3h}e^{A(3h-s)}BL_i(s)Ce^{A(s-h_i-2h)}\,\mathrm{d}s. F = e A h + i = 1 ∑ N ∫ 2 h 3 h e A ( 3 h − s ) B L i ( s ) C e A ( s − h i − 2 h ) d s . Using the Kronecker–vectorization identity and the definition of Γ d , i ( s ) \Gamma_{d,i}(s) Γ d , i ( s )
vec ( F ) = vec ( e A h ) + ∑ i = 1 N ∫ 2 h 3 h Γ d , i ( s ) vec ( L i ( s ) ) d s . \operatorname{vec}(F) = \operatorname{vec}(e^{Ah}) + \sum_{i=1}^{N}\int_{2h}^{3h}\Gamma_{d,i}(s)\operatorname{vec}(L_i(s))\,\mathrm{d}s. vec ( F ) = vec ( e A h ) + i = 1 ∑ N ∫ 2 h 3 h Γ d , i ( s ) vec ( L i ( s )) d s . Substituting (29) gives
vec ( F ) = vec ( e A h ) + ∑ i = 1 N ∑ k = 0 N b − 1 [ ∫ 2 h 3 h ρ ( s ) ϕ k ( s ) Γ d , i ( s ) d s ] a i , k . \operatorname{vec}(F) = \operatorname{vec}(e^{Ah})
+ \sum_{i=1}^{N}\sum_{k=0}^{N_b-1}
\left[\int_{2h}^{3h}\rho(s)\phi_k(s)\Gamma_{d,i}(s)\,\mathrm{d}s\right]a_{i,k}. vec ( F ) = vec ( e A h ) + i = 1 ∑ N k = 0 ∑ N b − 1 [ ∫ 2 h 3 h ρ ( s ) ϕ k ( s ) Γ d , i ( s ) d s ] a i , k . For each i = 1 , … , N i = 1,\ldots,N i = 1 , … , N k = 0 , … , N b − 1 k = 0,\ldots,N_b-1 k = 0 , … , N b − 1
K i , k : = ∫ 2 h 3 h ρ ( s ) ϕ k ( s ) Γ d , i ( s ) d s ∈ R n 2 × m p . K_{i,k} := \int_{2h}^{3h}\rho(s)\phi_k(s)\Gamma_{d,i}(s)\,\mathrm{d}s \in\mathbb{R}^{n^2 \times mp}. K i , k := ∫ 2 h 3 h ρ ( s ) ϕ k ( s ) Γ d , i ( s ) d s ∈ R n 2 × m p . Collect all coefficient vectors in the order a : = col { a 1 , 0 , … , a 1 , N b − 1 , … , a N , 0 , … , a N , N b − 1 } ∈ R N m p N b a := \operatorname{col}\{a_{1,0},\ldots,a_{1,N_b-1},\ldots,a_{N,0},\ldots,a_{N,N_b-1}\} \in\mathbb{R}^{NmpN_b} a := col { a 1 , 0 , … , a 1 , N b − 1 , … , a N , 0 , … , a N , N b − 1 } ∈ R N m p N b
K b : = [ K 1 , 0 ⋯ K 1 , N b − 1 ⋯ K N , 0 ⋯ K N , N b − 1 ] , (30) K_b := [K_{1,0}\ \cdots\ K_{1,N_b-1}\ \cdots\ K_{N,0}\ \cdots\ K_{N,N_b-1}], \tag*{(30)} K b := [ K 1 , 0 ⋯ K 1 , N b − 1 ⋯ K N , 0 ⋯ K N , N b − 1 ] , (30) with K b ∈ R n 2 × N m p N b K_b \in\mathbb{R}^{n^2 \times N_m p N_b} K b ∈ R n 2 × N m p N b K b a K_b a K b a
vec ( F ) = vec ( e A h ) + K b a . (31) \operatorname{vec}(F)=\operatorname{vec}\left(e^{Ah}\right)+K_ba. \tag*{(31)} vec ( F ) = vec ( e A h ) + K b a . (31) Thus, exact cancellation of the one-cycle terminal map is equivalent to the finite-dimensional linear equation
K b a = − vec ( e A h ) . (32) K_ba=-\operatorname{vec}\left(e^{Ah}\right). \tag*{(32)} K b a = − vec ( e A h ) . (32) Suppose that
rank ( K b ) = n 2 . (33) \operatorname{rank}(K_b)=n^2. \tag*{(33)} rank ( K b ) = n 2 . (33) Since K b K_b K b n 2 n^2 n 2 K b K_b K b K b K b T K_bK_b^{\mathsf{T}} K b K b T (32) is solvable for every right-hand side. Among all its solutions, the minimum-Euclidean-norm solution is
a = − K b T ( K b K b T ) − 1 vec ( e A h ) = − K b † vec ( e A h ) , (34) a=-K_b^{\mathsf{T}}\left(K_bK_b^{\mathsf{T}}\right)^{-1}\operatorname{vec}\left(e^{Ah}\right)=-K_b^{\dagger}\operatorname{vec}\left(e^{Ah}\right), \tag*{(34)} a = − K b T ( K b K b T ) − 1 vec ( e A h ) = − K b † vec ( e A h ) , (34) where K b † = K b T ( K b K b T ) − 1 K_b^{\dagger}=K_b^{\mathsf{T}}\left(K_bK_b^{\mathsf{T}}\right)^{-1} K b † = K b T ( K b K b T ) − 1 K b K_b K b
Substitution of (34) into (31) gives vec ( F ) = 0 \operatorname{vec}(F)=0 vec ( F ) = 0 K b K b † = I n 2 K_bK_b^{\dagger}=I_{n^2} K b K b † = I n 2 x ( 3 h ) = 0 x(3h)=0 x ( 3 h ) = 0 x ( t ) = 0 x(t)=0 x ( t ) = 0 u ( t ) = 0 u(t)=0 u ( t ) = 0 t ≥ 3 h t\geq3h t ≥ 3 h
The minimum-norm solution above ensures terminal cancellation but does not directly optimize transient performance. To retain a precomputed coefficient vector a 0 a_0 a 0 a 0 a_0 a 0 r 0 : = vec ( e A h ) + K b a 0 r_0:=\operatorname{vec}\left(e^{Ah}\right)+K_ba_0 r 0 := vec ( e A h ) + K b a 0
a = a 0 − K b † r 0 , a=a_0-K_b^{\dagger}r_0, a = a 0 − K b † r 0 , or, equivalently,
a = a 0 + K b † [ − vec ( e A h ) − K b a 0 ] . (35) a=a_0+K_b^{\dagger}\left[-\operatorname{vec}\left(e^{Ah}\right)-K_ba_0\right]. \tag*{(35)} a = a 0 + K b † [ − vec ( e A h ) − K b a 0 ] . (35) Indeed, using K b K b † = I n 2 K_bK_b^{\dagger}=I_{n^2} K b K b † = I n 2
vec ( e A h ) + K b a = r 0 − K b K b † r 0 = 0. \operatorname{vec}\left(e^{Ah}\right)+K_ba=r_0-K_bK_b^{\dagger}r_0=0. vec ( e A h ) + K b a = r 0 − K b K b † r 0 = 0. Therefore, the correction in (35) exactly removes the residual of the discretized terminal-map equation. Moreover, − K b † r 0 -K_b^{\dagger}r_0 − K b † r 0
For the numerical examples, the shifted Legendre basis is selected as
ϕ k ( s ) = P k ( 2 ( s − 2 h ) h − 1 ) , k = 0 , 1 , … , N b − 1 , \phi_k(s)=P_k\left(\frac{2(s-2h)}{h}-1\right),\qquad k=0,1,\ldots,N_b-1, ϕ k ( s ) = P k ( h 2 ( s − 2 h ) − 1 ) , k = 0 , 1 , … , N b − 1 , where P k P_k P k k k k s ↦ 2 ( s − 2 h ) / h − 1 s\mapsto2(s-2h)/h-1 s ↦ 2 ( s − 2 h ) / h − 1 [ 2 h , 3 h ] [2h,3h] [ 2 h , 3 h ] [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] C ν ( [ 2 h , 3 h ] ) C^\nu([2h,3h]) C ν ([ 2 h , 3 h ]) K b K_b K b (33) .
Numerical Examples
SISO: Nonminimum-Phase Systems
Consider the SISO system [11 , 30 ]
A = [ 0 1 1 − 0.5 ] , B = [ 0 1 ] , C = [ 1 − 1 ] . (36) A=\begin{bmatrix}0&1\\1&-0.5\end{bmatrix},\qquad B=\begin{bmatrix}0\\1\end{bmatrix},\qquad C=\begin{bmatrix}1&-1\end{bmatrix}. \tag*{(36)} A = [ 0 1 1 − 0.5 ] , B = [ 0 1 ] , C = [ 1 − 1 ] . (36) Its transfer function is
G ( s ) = C ( s I − A ) − 1 B = 1 − s s 2 + 1 2 s − 1 . G(s)=C(sI-A)^{-1}B=\frac{1-s}{s^2+\frac{1}{2}s-1}. G ( s ) = C ( s I − A ) − 1 B = s 2 + 2 1 s − 1 1 − s . Hence, the system possesses a right-half-plane transmission zero at s z = 1 s_z=1 s z = 1 [33 ] .
We take x ( 0 ) = [ 1 , − 1 ] T x(0)=[1,-1]^{\mathsf{T}} x ( 0 ) = [ 1 , − 1 ] T T = 3 T=3 T = 3 h = T / 3 = 1 h=T/3=1 h = T /3 = 1 (9) , the optimized artificial delays are h 1 = 1.0310 h_1=1.0310 h 1 = 1.0310 h 2 = 1.9534 h_2=1.9534 h 2 = 1.9534
Q d = [ 4.7275 − 1.8539 − 1.8539 0.7329 ] . Q_d=\begin{bmatrix}4.7275&-1.8539\\-1.8539&0.7329\end{bmatrix}. Q d = [ 4.7275 − 1.8539 − 1.8539 0.7329 ] . For the current-plus-single-delay controller in (15) , following Remark 2, we take K = q K=q K = q q = − 3.3118 q=-3.3118 q = − 3.3118 [ 2 h , 3 h ] [2h,3h] [ 2 h , 3 h ] [25 , 27 ] , we use ρ ( t ) = sin 2 ( π ( t − 2 h ) / h ) \rho(t)=\sin^2\left(\pi(t-2h)/h\right) ρ ( t ) = sin 2 ( π ( t − 2 h ) / h ) t ∈ [ 2 h , 3 h ] t\in[2h,3h] t ∈ [ 2 h , 3 h ] ν = 1 \nu=1 ν = 1
For comparison, we also consider the square-wave dither controller proposed in [30 ] , u ( t ) = ( k / ϵ ) sq ( t / ϵ ) y ( t ) u(t)=(k/\epsilon)\operatorname{sq}(t/\epsilon)y(t) u ( t ) = ( k / ϵ ) sq ( t / ϵ ) y ( t ) k = 9 k=9 k = 9 ϵ = 3.8 \epsilon=3.8 ϵ = 3.8 sq ( t / ϵ ) = sgn ( cos ( 2 π t / ϵ ) ) \operatorname{sq}(t/\epsilon)=\operatorname{sgn}(\cos(2\pi t/\epsilon)) sq ( t / ϵ ) = sgn ( cos ( 2 π t / ϵ )) [30 ] .
The simulation results are shown in Figure 1 . Both proposed controllers drive the state to the origin at the prescribed time T = 3 T=3 T = 3
Figure 1. State and controller trajectories of system (37)
MIMO: A Nonminimum-Phase System with a Nilpotent State Matrix
Consider system (1) , where the matrices A A A B B B [28 ] , and the output matrix C C C
A = [ 0 1 − 1 2 − 1 − 1 0 − 1 0 − 1 0 1 0 1 0 2 0 − 1 0 0 0 0 0 0 0 0 0 0 0 0 0 − 1 0 − 1 0 1 ] , B = [ 1 0 0 0 − 1 0 1 1 0 0 1 1 − 1 0 1 0 1 1 ] A=\begin{bmatrix}
0&1&-1&2&-1&-1\\
0&-1&0&-1&0&1\\
0&1&0&2&0&-1\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&-1&0&-1&0&1
\end{bmatrix},\quad
B=\begin{bmatrix}
1&0&0\\
0&-1&0\\
1&1&0\\
0&1&1\\
-1&0&1\\
0&1&1
\end{bmatrix} A = 0 0 0 0 0 0 1 − 1 1 0 0 − 1 − 1 0 0 0 0 0 2 − 1 2 0 0 − 1 − 1 0 0 0 0 0 − 1 1 − 1 0 0 1 , B = 1 0 1 0 − 1 0 0 − 1 1 1 0 1 0 0 0 1 1 1 and
C = [ − 1 3 − 1 7 − 4 − 3 0 2 0 3 − 2 − 1 0 0 0 1 0 − 1 ] . C=\begin{bmatrix}
-1&3&-1&7&-4&-3\\
0&2&0&3&-2&-1\\
0&0&0&1&0&-1
\end{bmatrix}. C = − 1 0 0 3 2 0 − 1 0 0 7 3 1 − 4 − 2 0 − 3 − 1 − 1 . Clearly, the system is controllable and observable, and has a right-half-plane transmission zero at s = 1 s=1 s = 1 A 2 ≠ 0 A^2 \ne0 A 2 = 0 A 3 = 0 A^3=0 A 3 = 0
For both controller designs below, we consider T = 9 T=9 T = 9 T = 18 T=18 T = 18 x ( 0 ) = [ 1 , 1 , 1 , 1 , 1 , 1 ] T / 6 x(0)=[1,1,1,1,1,1]^{\mathsf{T}}/\sqrt{6} x ( 0 ) = [ 1 , 1 , 1 , 1 , 1 , 1 ] T / 6 ∥ x ( 0 ) ∥ = 1 \lVert x(0)\rVert=1 ∥ x ( 0 )∥ = 1 ρ ( s ) = sin 2 ( π ( s − 2 h ) / h ) \rho(s)=\sin^2(\pi(s-2h)/h) ρ ( s ) = sin 2 ( π ( s − 2 h ) / h ) s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ] h h h
1) Current-Plus-Single-Delay Output Feedback: We first consider the system without measurement delays and apply controller (15) with h = T / 3 h=T/3 h = T /3 F H F_H F H H = diag ( 1 , 2 , 3 ) H=\operatorname{diag}(1,2,3) H = diag ( 1 , 2 , 3 ) v = [ 1 , 0 , 0 ] T v=[1,0,0]^{\mathsf{T}} v = [ 1 , 0 , 0 ] T w = [ 0 , 0 , 1 ] T w=[0,0,1]^{\mathsf{T}} w = [ 0 , 0 , 1 ] T b = B v = [ 1 , 0 , 1 , 0 , − 1 , 0 ] T b=Bv=[1,0,1,0,-1,0]^{\mathsf{T}} b = B v = [ 1 , 0 , 1 , 0 , − 1 , 0 ] T c = w T C = [ 0 , 0 , 0 , 1 , 0 , − 1 ] c=w^{\mathsf{T}}C=[0,0,0,1,0,-1] c = w T C = [ 0 , 0 , 0 , 1 , 0 , − 1 ] det [ b , F H b , … , F H 5 b ] = − 27648 ≠ 0 \det[b,F_Hb,\ldots,F_H^5b]=-27648\ne0 det [ b , F H b , … , F H 5 b ] = − 27648 = 0 det [ c T , F T T c T , … , ( F T T ) 5 c T ] = 256 ≠ 0 \det[c^{\mathsf{T}},F_T^{\mathsf{T}}c^{\mathsf{T}},\ldots,(F_T^{\mathsf{T}})^5c^{\mathsf{T}}]=256\ne0 det [ c T , F T T c T , … , ( F T T ) 5 c T ] = 256 = 0 ( F H , b , c ) (F_H,b,c) ( F H , b , c )
We then select α = 1 \alpha=1 α = 1 F K = A + B K C F_K=A+BKC F K = A + B K C
K = H + α v w T = [ 1 0 1 0 2 0 0 0 3 ] . K=H+\alpha vw^{\mathsf{T}}=
\begin{bmatrix}
1&0&1\\
0&2&0\\
0&0&3
\end{bmatrix}. K = H + α v w T = 1 0 0 0 2 0 1 0 3 . Since the only distinct eigenvalue of A A A
difference condition in the proof of Lemma 4 is satisfied. Consequently, the lifted pair ( A K , B K ) (A_K,B_K) ( A K , B K ) G K > 0 G_K>0 G K > 0
We take Q = I 9 Q=I_9 Q = I 9 (16) . For T = 9 T=9 T = 9 T = 18 T=18 T = 18 h = 3 h=3 h = 3 h = 6 h=6 h = 6
2) Unknown Constant Measurement Delays: We next consider system (28) with the same matrices A A A B B B C C C τ ∈ [ 0 , τ M ] \tau\in[0,\tau_M] τ ∈ [ 0 , τ M ] A 3 = 0 A^3=0 A 3 = 0 T = 9 h T=9h T = 9 h
For T = 9 T=9 T = 9 h = 1 h=1 h = 1 τ = τ M = 0.5 \tau=\tau_M=0.5 τ = τ M = 0.5
h = [ 1 1.25 1.5 ] T . h=\begin{bmatrix}1&1.25&1.5\end{bmatrix}^{\mathsf{T}}. h = [ 1 1.25 1.5 ] T . For T = 18 T=18 T = 18 h = 2 h=2 h = 2
τ = τ M = 1 , h = [ 2 2.5 3 ] T . \tau=\tau_M=1,\qquad h=\begin{bmatrix}2&2.5&3\end{bmatrix}^{\mathsf{T}}. τ = τ M = 1 , h = [ 2 2.5 3 ] T . Both choices satisfy h > τ M h>\tau_M h > τ M h ≤ h 1 < h 2 < h 3 ≤ 2 h − τ M h\le h_1<h_2<h_3\le2h-\tau_M h ≤ h 1 < h 2 < h 3 ≤ 2 h − τ M
Since A 3 = 0 A^3=0 A 3 = 0 Q = I 27 Q=I_{27} Q = I 27 (9) , with the gains constructed from (10) . The controller periods are 3 h = 3 3h=3 3 h = 3 3 h = 6 3h=6 3 h = 6 Δ τ 3 = 0 \Delta_\tau^3=0 Δ τ 3 = 0 T = 9 T=9 T = 9 T = 18 T=18 T = 18 [27 ] , because the system has internal dynamics with an unstable zero at s = 1 s=1 s = 1
The simulation results are shown in Figure 2 . Both controller designs drive the state to the origin by the corresponding time bounds T = 9 T=9 T = 9 T = 18 T=18 T = 18
Figure 2. State and control-input norms of the nonminimum-phase MIMO system under the two controller designs with T = 9 T = 9 T = 9 T = 18 T = 18 T = 18
Flexible-Structure System
To illustrate the applicability of the proposed method to a high-order nonminimum-phase system, consider the experimental flexible-structure system studied in [34 ] . We adopt the state-space model reported therein, which is described by (1) with
A = [ − 0.996 − 463 − 97.8 − 12131 − 8.11 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 ] , A = \begin{bmatrix} -0.996 & -463 & -97.8 & -12131 & -8.11 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}, A = − 0.996 1 0 0 0 0 − 463 0 1 0 0 0 − 97.8 0 0 1 0 0 − 12131 0 0 0 1 0 − 8.11 0 0 0 0 1 0 0 0 0 0 0 , and
B = [ 1 , 0 , 0 , 0 , 0 , 0 ] T , C = [ 0 , 1.65 , − 0.331 , − 576 , 90.6 , 19080 ] . B = [1,0,0,0,0,0]^{\mathsf{T}}, \quad C = [0,1.65,-0.331,-576,90.6,19080]. B = [ 1 , 0 , 0 , 0 , 0 , 0 ] T , C = [ 0 , 1.65 , − 0.331 , − 576 , 90.6 , 19080 ] . Thus, the system order is n = 6 n = 6 n = 6
The pair ( A , B ) (A,B) ( A , B ) ( A , C ) (A,C) ( A , C ) x ( 0 ) = [ 1 , 1 , 1 , 1 , 1 , 1 ] T / 6 x(0) = [1,1,1,1,1,1]^{\mathsf{T}}/\sqrt{6} x ( 0 ) = [ 1 , 1 , 1 , 1 , 1 , 1 ] T / 6 h = T / 3 h = T/3 h = T /3 ρ ( t ) = sin 2 ( π ( t − 2 h ) / h ) \rho(t) = \sin^{2}(\pi(t-2h)/h) ρ ( t ) = sin 2 ( π ( t − 2 h ) / h ) t ∈ [ 2 h , 3 h ] t \in[2h,3h] t ∈ [ 2 h , 3 h ]
1) Without Measurement Delays: To investigate the effect of the prescribed time on transient performance, we consider T = 60 T = 60 T = 60 T = 30 T = 30 T = 30 Q d = I 6 Q_d = I_6 Q d = I 6 [ h , 2 h ] [h,2h] [ h , 2 h ] h i u = h + i h / 7 h_i^{\mathrm{u}} = h + ih/7 h i u = h + ih /7 i = 1 , … , 6 i = 1,\ldots,6 i = 1 , … , 6 T = 60 T = 60 T = 60
h 60 u = [ 22.8571 , 25.7143 , 28.5714 , 31.4286 , 34.2857 , 37.1429 ] T , h_{60}^{\mathrm{u}} = [22.8571,25.7143,28.5714,31.4286,34.2857,37.1429]^{\mathsf{T}}, h 60 u = [ 22.8571 , 25.7143 , 28.5714 , 31.4286 , 34.2857 , 37.1429 ] T , whereas, for T = 30 T = 30 T = 30
h 30 u = [ 11.4286 , 12.8571 , 14.2857 , 15.7143 , 17.1429 , 18.5714 ] T . h_{30}^{\mathrm{u}} = [11.4286,12.8571,14.2857,15.7143,17.1429,18.5714]^{\mathsf{T}}. h 30 u = [ 11.4286 , 12.8571 , 14.2857 , 15.7143 , 17.1429 , 18.5714 ] T . The artificial-delay locations are then optimized directly with respect to the closed-loop transient response using the Nelder–Mead simplex method with multiple initial points. For T = 60 T = 60 T = 60
h 60 o = [ 20.0000 , 20.2849 , 20.7428 , 39.5348 , 39.7399 , 39.9999 ] T , h_{60}^{\mathrm{o}} = [20.0000,20.2849,20.7428,39.5348,39.7399,39.9999]^{\mathsf{T}}, h 60 o = [ 20.0000 , 20.2849 , 20.7428 , 39.5348 , 39.7399 , 39.9999 ] T , For T = 30 T = 30 T = 30
h 30 o = [ 10.0005 , 10.1012 , 10.7990 , 19.7476 , 19.8979 , 20.0000 ] T . h_{30}^{\mathrm{o}} = [10.0005,10.1012,10.7990,19.7476,19.8979,20.0000]^{\mathsf{T}}. h 30 o = [ 10.0005 , 10.1012 , 10.7990 , 19.7476 , 19.8979 , 20.0000 ] T . The resulting trajectories are shown in Figure 3 . For both prescribed times, the state converges to the origin by t = T t = T t = T T = 60 T = 60 T = 60
Figure 3. State and controller trajectories of flexible-structure system
2) With Measurement Delays: Sampled Outputs: We next apply the known time-varying measurement-delay design to the same flexible-structure system. Consider (19) with
t k = k d s , τ ( t ) = t − t k , t ∈ [ t k , t k + 1 ) , t_k = kd_s,\quad\tau(t) = t-t_k,\quad t \in[t_k,t_{k+1}), t k = k d s , τ ( t ) = t − t k , t ∈ [ t k , t k + 1 ) , where k ∈ Z 0 + k \in\mathbb{Z}_{0}^{+} k ∈ Z 0 + d s > 0 d_s > 0 d s > 0 0 ≤ τ ( t ) < d s 0 \leq\tau(t) < d_s 0 ≤ τ ( t ) < d s y ( t ) = C x ( t k ) y(t) = Cx(t_k) y ( t ) = C x ( t k ) t ∈ [ t k , t k + 1 ) t \in[t_k,t_{k+1}) t ∈ [ t k , t k + 1 )
We select T = 60 T = 60 T = 60 h = 20 h = 20 h = 20 d s = 2 / 7 d_s = 2/7 d s = 2/7 τ M = d s \tau_M = d_s τ M = d s h i 60 h_{i}^{60} h i 60 h i = 20 + 20 i / 7 = ( 70 + 10 i ) d s h_i = 20 + 20i/7 = (70 + 10i)d_s h i = 20 + 20 i /7 = ( 70 + 10 i ) d s i = 1 , … , 6 i = 1,\ldots,6 i = 1 , … , 6 h ≤ h i ≤ 2 h − τ M h \leq h_i \leq2h-\tau_M h ≤ h i ≤ 2 h − τ M i = 1 , … , 6 i = 1,\ldots,6 i = 1 , … , 6 G v > 0 G_v > 0 G v > 0 Q v = I 6 Q_v = I_6 Q v = I 6
The red curves in Figure 3 show the trajectories under sampled-output measurements. The state converges to the origin by T = 60 T = 60 T = 60
Cart–Pole System With Different Output Configurations
To further demonstrate the applicability of the proposed method to both minimum-phase and nonminimum-phase systems, consider the linearized cart–pole system adopted from [35 ] , which is described by (1) with x ( t ) = [ x c ( t ) x ˙ c ( t ) θ ( t ) θ ˙ ( t ) ] T x(t) = [x_c(t)\ \dot{x}_c(t)\ \theta(t)\ \dot{\theta}(t)]^{\mathsf{T}} x ( t ) = [ x c ( t ) x ˙ c ( t ) θ ( t ) θ ˙ ( t ) ] T
A = [ 0 1 0 0 0 − ( I + m l 2 ) b c p m 2 g l 2 p 0 0 0 0 1 0 − m l b c p m g l ( M + m ) p 0 ] , A = \begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & -\frac{(I+ml^{2})b_c}{p} & \frac{m^{2}gl^{2}}{p} & 0 \\
0 & 0 & 0 & 1 \\
0 & -\frac{mlb_c}{p} & \frac{mgl(M+m)}{p} & 0
\end{bmatrix}, A = 0 0 0 0 1 − p ( I + m l 2 ) b c 0 − p m l b c 0 p m 2 g l 2 0 p m g l ( M + m ) 0 0 1 0 , B = [ 0 I + m l 2 p 0 m l p ] T , B = \begin{bmatrix} 0 & \frac{I+ml^{2}}{p} & 0 & \frac{ml}{p} \end{bmatrix}^{\mathsf{T}}, B = [ 0 p I + m l 2 0 p m l ] T , where p = I ( M + m ) + M m l 2 p = I(M+m)+Mml^{2} p = I ( M + m ) + M m l 2 x c ( t ) x_c(t) x c ( t ) θ ( t ) \theta(t) θ ( t ) [35 ] , the physical parameters are selected as M = 0.5 M = 0.5 M = 0.5 m = 0.2 m = 0.2 m = 0.2 b c = 0.1 b_c = 0.1 b c = 0.1 I = 0.006 I = 0.006 I = 0.006 2 ^{2} 2 l = 0.3 l = 0.3 l = 0.3 g = 9.8 g = 9.8 g = 9.8 2 ^{2} 2
We consider the following two output configurations:
y p ( t ) = C p x ( t ) , C p = [ 1 0 0 0 ] , (37) y_p(t) = C_p x(t), \qquad C_p = [1\ \ 0\ \ 0\ \ 0], \tag*{(37)} y p ( t ) = C p x ( t ) , C p = [ 1 0 0 0 ] , (37) y p a ( t ) = C p a x ( t ) , C p a = [ 1 0 0 0 0 0 1 0 ] . (38) y_{\mathrm{pa}}(t)=C_{\mathrm{pa}}x(t),\qquad C_{\mathrm{pa}}=\begin{bmatrix}1&0&0&0\\0&0&1&0\end{bmatrix}. \tag*{(38)} y pa ( t ) = C pa x ( t ) , C pa = [ 1 0 0 0 0 1 0 0 ] . (38) For the position-only output in (38), the transfer function from u u u x c x_{\mathrm{c}} x c
G p ( s ) = 1.8182 s 2 − 44.5455 s 4 + 0.1818 s 3 − 31.1818 s 2 − 4.4545 s . G_{\mathrm{p}}(s)=\frac{1.8182s^{2}-44.5455}{s^{4}+0.1818s^{3}-31.1818s^{2}-4.4545s}. G p ( s ) = s 4 + 0.1818 s 3 − 31.1818 s 2 − 4.4545 s 1.8182 s 2 − 44.5455 . Its zeros are
s = ± m g l I + m l 2 = ± 4.9497. (39) s=\pm\sqrt{\frac{mgl}{I+ml^{2}}}=\pm4.9497. \tag*{(39)} s = ± I + m l 2 m g l = ± 4.9497. (39) Therefore, G p ( s ) G_{\mathrm{p}}(s) G p ( s ) s = 4.9497 s=4.9497 s = 4.9497
For the position-and-angle output in (39), the transfer matrix is
G p a ( s ) = 1 D ( s ) [ 1.8182 s 2 − 44.5455 4.5455 s 2 ] , (40) G_{\mathrm{pa}}(s)=\frac{1}{D(s)}
\begin{bmatrix}
1.8182s^{2}-44.5455\\
4.5455s^{2}
\end{bmatrix}, \tag*{(40)} G pa ( s ) = D ( s ) 1 [ 1.8182 s 2 − 44.5455 4.5455 s 2 ] , (40) where D ( s ) = s 4 + 0.1818 s 3 − 31.1818 s 2 − 4.4545 s D(s)=s^{4}+0.1818s^{3}-31.1818s^{2}-4.4545s D ( s ) = s 4 + 0.1818 s 3 − 31.1818 s 2 − 4.4545 s
Both ( A , C p ) (A,C_{\mathrm{p}}) ( A , C p ) ( A , C p a ) (A,C_{\mathrm{pa}}) ( A , C pa ) x ( 0 ) = [ 0.01 0 0.1 π / 180 0 ] T x(0)=[0.01\quad0\quad0.1\pi/180\quad0]^{\mathsf{T}} x ( 0 ) = [ 0.01 0 0.1 π /180 0 ] T T = 0.9 T=0.9 T = 0.9 h = T / 3 = 0.3 h=T/3=0.3 h = T /3 = 0.3 h 1 = 0.3 h_{1}=0.3 h 1 = 0.3 h 2 = 0.4 h_{2}=0.4 h 2 = 0.4 h 3 = 0.5 h_{3}=0.5 h 3 = 0.5 h 4 = 0.6 h_{4}=0.6 h 4 = 0.6 N b = 8 N_{\mathrm{b}}=8 N b = 8 ρ ( s ) = sin 2 ( π ( s − 2 h ) / h ) \rho(s)=\sin^{2}(\pi(s-2h)/h) ρ ( s ) = sin 2 ( π ( s − 2 h ) / h ) s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ]
For both output configurations, the matrices K b K_{\mathrm{b}} K b rank ( K b ) = n 2 = 16 \operatorname{rank}(K_{\mathrm{b}})=n^{2}=16 rank ( K b ) = n 2 = 16
The simulation results are shown in Figure 4 . Both output configurations achieve convergence to the origin by the prescribed time T = 0.9 T=0.9 T = 0.9
Figure 4. State and control trajectories of the cart–pole system under the position-only output and the position-and-angle output configurations.
Conclusion
This paper has provided a constructive solution to the static output-feedback stabilization problem and further achieved prescribed-time stabilization for every controllable and observable LTI system using two methods. The first employs multiple artificial output delays with smooth time-varying gains and stabilizes every controllable and observable LTI system using a generalized controllability–observability Gramian. The second, simpler method combines constant-gain feedback of the current output with time-varying-gain feedback of a single artificially delayed output and also applies to every controllable and observable LTI system. The multiple-delay design has been extended to known bounded and fast-varying (without any constraints on the delay derivative) measurement delays, yielding fixed-time stabilization under a corresponding rank condition. For unknown constant measurement delays, fixed-time stabilization has been established for systems with nilpotent state matrices. A numerical implementation has also been presented. Numerical examples involving nonminimum-phase, flexible-structure, and cart–pole systems have illustrated the effectiveness of the proposed methods. Future work will address robustness to model uncertainty, measurement noise, and imperfect delay information.
Appendix
The Proof of Lemma 2
Define Φ d : = e − A d \Phi_{d}:=e^{-Ad} Φ d := e − A d h i = h 0 + ( i − 1 ) d h_{i}=h_{0}+(i-1)d h i = h 0 + ( i − 1 ) d C e − A h i = C e − A [ h 0 + ( i − 1 ) d ] = C Φ d i − 1 e − A h 0 Ce^{-Ah_{i}}=Ce^{-A[h_{0}+(i-1)d]}=C\Phi_{d}^{i-1}e^{-Ah_{0}} C e − A h i = C e − A [ h 0 + ( i − 1 ) d ] = C Φ d i − 1 e − A h 0 N = n N=n N = n
O h = [ C e − A h 1 C e − A h 2 ⋮ C e − A h n ] = [ C C Φ d ⋮ C Φ d n − 1 ] e − A h 0 . (41) \mathcal{O}_{h}=
\begin{bmatrix}
Ce^{-Ah_{1}}\\
Ce^{-Ah_{2}}\\
\vdots\\
Ce^{-Ah_{n}}
\end{bmatrix}
=
\begin{bmatrix}
C\\
C\Phi_{d}\\
\vdots\\
C\Phi_{d}^{n-1}
\end{bmatrix}e^{-Ah_{0}}. \tag*{(41)} O h = C e − A h 1 C e − A h 2 ⋮ C e − A h n = C C Φ d ⋮ C Φ d n − 1 e − A h 0 . (41) Since e − A h 0 e^{-Ah_{0}} e − A h 0
rank O h = rank ( col ( C , C Φ d , … , C Φ d n − 1 ) ) . \operatorname{rank}\mathcal{O}_{h}=\operatorname{rank}\left(\operatorname{col}(C,C\Phi_{d},\ldots,C\Phi_{d}^{n-1})\right). rank O h = rank ( col ( C , C Φ d , … , C Φ d n − 1 ) ) . Therefore, it is sufficient to prove that the discrete-time pair ( Φ d , C ) (\Phi_{d},C) ( Φ d , C ) d > 0 d>0 d > 0
Let λ 1 , … , λ s \lambda_{1},\ldots,\lambda_{s} λ 1 , … , λ s A A A Φ d \Phi_{d} Φ d e − λ 1 d , … , e − λ s d e^{-\lambda_{1}d},\ldots,e^{-\lambda_{s}d} e − λ 1 d , … , e − λ s d λ j \lambda_{j} λ j λ k \lambda_{k} λ k A A A Φ d \Phi_{d} Φ d e − λ j d = e − λ k d e^{-\lambda_{j}d}=e^{-\lambda_{k}d} e − λ j d = e − λ k d ( λ j − λ k ) d = 2 π i ℓ (\lambda_{j}-\lambda_{k})d=2\pi\mathrm{i}\ell ( λ j − λ k ) d = 2 π i ℓ ℓ \ell ℓ
Because A A A d A > 0 d_{A}>0 d A > 0
e − λ j d ≠ e − λ k d , j ≠ k , 0 < d < d A . (42) e^{-\lambda_{j}d}\ne e^{-\lambda_{k}d},\qquad j\ne k,\qquad0<d<d_{A}. \tag*{(42)} e − λ j d = e − λ k d , j = k , 0 < d < d A . (42) Under condition (43), distinct eigenvalues of A A A − d e − λ d ≠ 0 -d e^{-\lambda d} \ne0 − d e − λ d = 0 Φ d \Phi_d Φ d A A A λ \lambda λ A A A A v = λ v Av = \lambda v A v = λ v ( A , C ) (A,C) ( A , C ) C v ≠ 0 Cv \ne0 C v = 0 v v v A A A Φ d \Phi_d Φ d C C C ( Φ d , C ) (\Phi_d,C) ( Φ d , C ) 0 < d < d A 0 < d < d_A 0 < d < d A
Finally, because h 0 < 2 h h_0 < 2h h 0 < 2 h
d ∗ < min { d A , 2 h − h 0 n − 1 } d^* < \min\left\{d_A,\frac{2h-h_0}{n-1}\right\} d ∗ < min { d A , n − 1 2 h − h 0 } when n ≥ 2 n \ge2 n ≥ 2 0 < d < d ∗ 0 < d < d^* 0 < d < d ∗ h ≤ h 1 < h 2 < ⋯ < h n < 2 h h \le h_1 < h_2 < \cdots< h_n < 2h h ≤ h 1 < h 2 < ⋯ < h n < 2 h ( Φ d , C ) (\Phi_d,C) ( Φ d , C ) n n n rank ( col ( C , C Φ d , … , C Φ d n − 1 ) ) = n \operatorname{rank}(\operatorname{col}(C,C\Phi_d,\ldots,C\Phi_d^{n-1}))=n rank ( col ( C , C Φ d , … , C Φ d n − 1 )) = n rank O h = n \operatorname{rank}O_h=n rank O h = n n = 1 n=1 n = 1 ( A , C ) (A,C) ( A , C )
The Proof of Lemma 3
For s ∈ [ 2 h , 3 h ] s \in[2h,3h] s ∈ [ 2 h , 3 h ] L ( s ) : = e A ( 3 h − s ) B ∈ R n × m \mathcal{L}(s):=e^{A(3h-s)}B \in\mathbb{R}^{n\times m} L ( s ) := e A ( 3 h − s ) B ∈ R n × m R i ( s ) : = C e A ( s − h i − 2 h ) ∈ R p × n \mathcal{R}_i(s):=Ce^{A(s-h_i-2h)} \in\mathbb{R}^{p\times n} R i ( s ) := C e A ( s − h i − 2 h ) ∈ R p × n i = 1 , … , N i=1,\ldots,N i = 1 , … , N i i i
Γ d , i ( s ) = e A T ( s − h i − 2 h ) C T ⊗ e A ( 3 h − s ) B = R i T ( s ) ⊗ L ( s ) ∈ R n 2 × m p . \Gamma_{d,i}(s)=e^{A^\mathsf{T}(s-h_i-2h)}C^\mathsf{T}\otimes e^{A(3h-s)}B
=\mathcal{R}_i^\mathsf{T}(s)\otimes\mathcal{L}(s)\in\mathbb{R}^{n^2\times mp}. Γ d , i ( s ) = e A T ( s − h i − 2 h ) C T ⊗ e A ( 3 h − s ) B = R i T ( s ) ⊗ L ( s ) ∈ R n 2 × m p . Consequently,
Γ d ( s ) = [ R 1 T ( s ) ⊗ L ( s ) ⋯ R N T ( s ) ⊗ L ( s ) ] . \Gamma_d(s)=\left[\mathcal{R}_1^\mathsf{T}(s)\otimes\mathcal{L}(s)\ \cdots\ \mathcal{R}_N^\mathsf{T}(s)\otimes\mathcal{L}(s)\right]. Γ d ( s ) = [ R 1 T ( s ) ⊗ L ( s ) ⋯ R N T ( s ) ⊗ L ( s ) ] . Collect all the delayed-output matrices in
R ( s ) : = [ R 1 ( s ) ⋮ R N ( s ) ] = [ C e A ( s − h 1 − 2 h ) ⋮ C e A ( s − h N − 2 h ) ] = O h e A ( s − 2 h ) ∈ R N p × n . \mathcal{R}(s):=
\begin{bmatrix}
\mathcal{R}_1(s)\\
\vdots\\
\mathcal{R}_N(s)
\end{bmatrix}
=
\begin{bmatrix}
Ce^{A(s-h_1-2h)}\\
\vdots\\
Ce^{A(s-h_N-2h)}
\end{bmatrix}
=O_he^{A(s-2h)}\in\mathbb{R}^{Np\times n}. R ( s ) := R 1 ( s ) ⋮ R N ( s ) = C e A ( s − h 1 − 2 h ) ⋮ C e A ( s − h N − 2 h ) = O h e A ( s − 2 h ) ∈ R N p × n . By condition (4), rank ( O h ) = n \operatorname{rank}(O_h)=n rank ( O h ) = n e A ( s − 2 h ) e^{A(s-2h)} e A ( s − 2 h ) s s s
rank ( R ( s ) ) = rank ( O h e A ( s − 2 h ) ) = n , s ∈ [ 2 h , 3 h ] . \operatorname{rank}(\mathcal{R}(s))=\operatorname{rank}\left(O_he^{A(s-2h)}\right)=n,\qquad s\in[2h,3h]. rank ( R ( s )) = rank ( O h e A ( s − 2 h ) ) = n , s ∈ [ 2 h , 3 h ] . Thus, R ( s ) \mathcal{R}(s) R ( s ) R T ( s ) \mathcal{R}^\mathsf{T}(s) R T ( s ) s ∈ [ 2 h , 3 h ] s\in[2h,3h] s ∈ [ 2 h , 3 h ] J ( s ) ∈ R N p × n \mathcal{J}(s)\in\mathbb{R}^{Np\times n} J ( s ) ∈ R N p × n R T ( s ) J ( s ) = I n \mathcal{R}^\mathsf{T}(s)\mathcal{J}(s)=I_n R T ( s ) J ( s ) = I n
We next characterize the null space of Γ d T ( s ) \Gamma_d^\mathsf{T}(s) Γ d T ( s ) z ∈ R n 2 z\in\mathbb{R}^{n^2} z ∈ R n 2 Z : = vec − 1 ( z ) ∈ R n × n Z:=\operatorname{vec}^{-1}(z)\in\mathbb{R}^{n\times n} Z := vec − 1 ( z ) ∈ R n × n Γ d T ( s ) z = 0 \Gamma_d^\mathsf{T}(s)z=0 Γ d T ( s ) z = 0 s ∈ ( 2 h , 3 h ) s\in(2h,3h) s ∈ ( 2 h , 3 h ) Γ d ( s ) \Gamma_d(s) Γ d ( s ) Γ d , i ( s ) \Gamma_{d,i}(s) Γ d , i ( s )
Γ d , i T ( s ) z = 0 , i = 1 , … , N . \Gamma_{d,i}^\mathsf{T}(s)z=0,\qquad i=1,\ldots,N. Γ d , i T ( s ) z = 0 , i = 1 , … , N . Using Γ d , i ( s ) = R i T ( s ) ⊗ L ( s ) \Gamma_{d,i}(s)=\mathcal{R}_i^\mathsf{T}(s)\otimes\mathcal{L}(s) Γ d , i ( s ) = R i T ( s ) ⊗ L ( s )
Γ d , i T ( s ) z = ( R i ( s ) ⊗ L T ( s ) ) vec ( Z ) . \Gamma_{d,i}^\mathsf{T}(s)z=\left(\mathcal{R}_i(s)\otimes\mathcal{L}^\mathsf{T}(s)\right)\operatorname{vec}(Z). Γ d , i T ( s ) z = ( R i ( s ) ⊗ L T ( s ) ) vec ( Z ) . Applying the Kronecker–vectorization identity ( B T ⊗ A ) vec ( X ) = vec ( A X B ) (B^\mathsf{T}\otimes A)\operatorname{vec}(X)=\operatorname{vec}(AXB) ( B T ⊗ A ) vec ( X ) = vec ( A X B ) A = L T ( s ) A=\mathcal{L}^\mathsf{T}(s) A = L T ( s ) X = Z X=Z X = Z B = R i T ( s ) B=\mathcal{R}_i^\mathsf{T}(s) B = R i T ( s )
Γ d , i T ( s ) z = vec ( L T ( s ) Z R i T ( s ) ) . \Gamma_{d,i}^\mathsf{T}(s)z=\operatorname{vec}\left(\mathcal{L}^\mathsf{T}(s)Z\mathcal{R}_i^\mathsf{T}(s)\right). Γ d , i T ( s ) z = vec ( L T ( s ) Z R i T ( s ) ) . Therefore, Γ d , i T ( s ) z = 0 \Gamma_{d,i}^\mathsf{T}(s)z=0 Γ d , i T ( s ) z = 0
L T ( s ) Z R i T ( s ) = 0 , i = 1 , … , N . \mathcal{L}^\mathsf{T}(s)Z\mathcal{R}_i^\mathsf{T}(s)=0,\qquad i=1,\ldots,N. L T ( s ) Z R i T ( s ) = 0 , i = 1 , … , N . Collecting these equalities for all i i i L T ( s ) Z R T ( s ) = 0 \mathcal{L}^\mathsf{T}(s)Z\mathcal{R}^\mathsf{T}(s)=0 L T ( s ) Z R T ( s ) = 0 J ( s ) \mathcal{J}(s) J ( s )
L T ( s ) Z = L T ( s ) Z R T ( s ) J ( s ) = 0 , s ∈ ( 2 h , 3 h ) . \mathcal{L}^\mathsf{T}(s)Z=\mathcal{L}^\mathsf{T}(s)Z\mathcal{R}^\mathsf{T}(s)\mathcal{J}(s)=0,\qquad s\in(2h,3h). L T ( s ) Z = L T ( s ) Z R T ( s ) J ( s ) = 0 , s ∈ ( 2 h , 3 h ) . It remains to use the controllability of ( A , B ) (A,B) ( A , B )
span { Im L ( s ) : s ∈ ( 2 h , 3 h ) } = R n . (43) \operatorname{span}\{\operatorname{Im}\mathcal{L}(s):s\in(2h,3h)\}=\mathbb{R}^n. \tag*{(43)} span { Im L ( s ) : s ∈ ( 2 h , 3 h )} = R n . (43) Suppose, to the contrary, that a vector v ∈ R n v\in\mathbb{R}^n v ∈ R n
v T L ( s ) = v T e A ( 3 h − s ) B = 0 , s ∈ ( 2 h , 3 h ) . v^\mathsf{T}\mathcal{L}(s)=v^\mathsf{T}e^{A(3h-s)}B=0,\qquad s\in(2h,3h). v T L ( s ) = v T e A ( 3 h − s ) B = 0 , s ∈ ( 2 h , 3 h ) . The left-hand side is an analytic matrix-valued function of s s s s 0 ∈ ( 2 h , 3 h ) s_0\in(2h,3h) s 0 ∈ ( 2 h , 3 h ) s = s 0 s=s_0 s = s 0
d k d s k e A ( 3 h − s ) = ( − 1 ) k e A ( 3 h − s ) A k , \frac{\mathrm{d}^k}{\mathrm{d}s^k}e^{A(3h-s)}=(-1)^ke^{A(3h-s)}A^k, d s k d k e A ( 3 h − s ) = ( − 1 ) k e A ( 3 h − s ) A k , and hence
( − 1 ) k v T e A ( 3 h − s 0 ) A k B = 0 , k = 0 , 1 , … , n − 1. (-1)^kv^\mathsf{T}e^{A(3h-s_0)}A^kB=0,\qquad k=0,1,\ldots,n-1. ( − 1 ) k v T e A ( 3 h − s 0 ) A k B = 0 , k = 0 , 1 , … , n − 1. Define
v ~ T : = v T e A ( 3 h − s 0 ) . \widetilde{v}^\mathsf{T}:=v^\mathsf{T}e^{A(3h-s_0)}. v T := v T e A ( 3 h − s 0 ) . Then
v ~ T [ B A B ⋯ A n − 1 B ] = 0. \widetilde{v}^\mathsf{T}\begin{bmatrix}B&AB&\cdots&A^{n-1}B\end{bmatrix}=0. v T [ B A B ⋯ A n − 1 B ] = 0. Since ( A , B ) (A,B) ( A , B )
rank [ B A B ⋯ A n − 1 B ] = n . \operatorname{rank}\begin{bmatrix}B&AB&\cdots&A^{n-1}B\end{bmatrix}=n. rank [ B A B ⋯ A n − 1 B ] = n . Therefore, v ~ = 0 \widetilde{v}=0 v = 0 e A ( 3 h − s 0 ) e^{A(3h-s_0)} e A ( 3 h − s 0 ) v = 0 v=0 v = 0
Recall that
L T ( s ) Z = 0 , s ∈ ( 2 h , 3 h ) . \mathcal{L}^\mathsf{T}(s)Z=0,\qquad s\in(2h,3h). L T ( s ) Z = 0 , s ∈ ( 2 h , 3 h ) . This means that x T Z = 0 x^\mathsf{T}Z=0 x T Z = 0
x ∈ span { Im L ( s ) : s ∈ ( 2 h , 3 h ) } . x\in\operatorname{span}\{\operatorname{Im}\mathcal{L}(s):s\in(2h,3h)\}. x ∈ span { Im L ( s ) : s ∈ ( 2 h , 3 h )} . By (44), this span is equal to R n \mathbb{R}^n R n
x T Z = 0 , ∀ x ∈ R n , x^\mathsf{T}Z=0,\qquad\forall x\in\mathbb{R}^n, x T Z = 0 , ∀ x ∈ R n , which implies Z = 0 Z=0 Z = 0 z = vec ( Z ) z=\operatorname{vec}(Z) z = vec ( Z ) z = 0 z=0 z = 0
[ Γ d T ( s ) z = 0 for every s ∈ ( 2 h , 3 h ) ] ⟹ z = 0. (44) \left[\Gamma_d^\mathsf{T}(s)z=0\text{ for every }s\in(2h,3h)\right]\Longrightarrow z=0. \tag*{(44)} [ Γ d T ( s ) z = 0 for every s ∈ ( 2 h , 3 h ) ] ⟹ z = 0. (44) We now prove the positive definiteness of G G G R n 2 \mathbb{R}^{n^2} R n 2 (7) that
z T G z = ∫ 2 h 3 h ρ ( s ) z T Γ d ( s ) Q d Γ d T ( s ) z d s ≥ λ min ( Q d ) ∫ 2 h 3 h ρ ( s ) ∥ Γ d T ( s ) z ∥ 2 d s ≥ 0. \begin{aligned}
z^{\mathsf{T}}Gz=\int_{2h}^{3h}\rho(s)z^{\mathsf{T}}\Gamma_d(s)Q_d\Gamma_d^{\mathsf{T}}(s)z\,\mathrm{d}s \\
&\geq\lambda_{\min}(Q_d)\int_{2h}^{3h}\rho(s)\left\|\Gamma_d^{\mathsf{T}}(s)z\right\|^2\mathrm{d}s\geq0.
\end{aligned} z T G z = ∫ 2 h 3 h ρ ( s ) z T Γ d ( s ) Q d Γ d T ( s ) z d s ≥ λ m i n ( Q d ) ∫ 2 h 3 h ρ ( s ) Γ d T ( s ) z 2 d s ≥ 0. Thus, G G G
Suppose that z T G z = 0 z^{\mathsf{T}}Gz=0 z T G z = 0 Q d > 0 Q_d>0 Q d > 0 ρ ( s ) > 0 \rho(s)>0 ρ ( s ) > 0 ( 2 h , 3 h ) (2h,3h) ( 2 h , 3 h ) Γ d T ( s ) z \Gamma_d^{\mathsf{T}}(s)z Γ d T ( s ) z s s s
Γ d T ( s ) z = 0 , s ∈ ( 2 h , 3 h ) . \Gamma_d^{\mathsf{T}}(s)z=0,\qquad s\in(2h,3h). Γ d T ( s ) z = 0 , s ∈ ( 2 h , 3 h ) . It then follows from (44) that z = 0 z=0 z = 0
z T G z > 0 , ∀ z ∈ R n 2 ∖ { 0 } , z^{\mathsf{T}}Gz>0,\qquad\forall z\in\mathbb{R}^{n^2}\setminus\{0\}, z T G z > 0 , ∀ z ∈ R n 2 ∖ { 0 } , and hence G > 0 G>0 G > 0
The Proof of Lemma 4
The eigenvalue arguments are carried out over C \mathbb{C} C R \mathbb{R} R C \mathbb{C} C λ 1 , … , λ ς \lambda_1,\ldots,\lambda_{\varsigma} λ 1 , … , λ ς A A A
Step 1: Selection of H H H v v v w w w By the output-feedback result in [29 ] , Assumption 1 guarantees the existence of a real matrix H ∈ R m × p H\in\mathbb{R}^{m\times p} H ∈ R m × p F H = A + B H C F_H=A+BHC F H = A + B H C n n n ζ 1 , … , ζ n \zeta_1,\ldots,\zeta_n ζ 1 , … , ζ n ( F H , B ) (F_H,B) ( F H , B ) ( F H , C ) (F_H,C) ( F H , C )
Let r i r_i r i l i l_i l i F H F_H F H ζ i \zeta_i ζ i
l i T B ≠ 0 , C r i ≠ 0 , i = 1 , … , n . l_i^{\mathsf{T}}B\ne0,\qquad Cr_i\ne0,\qquad i=1,\ldots,n. l i T B = 0 , C r i = 0 , i = 1 , … , n . Consequently, the real vectors v ∈ R m v\in\mathbb{R}^{m} v ∈ R m l i T B v = 0 l_i^{\mathsf{T}}Bv=0 l i T B v = 0 i i i R m \mathbb{R}^{m} R m v v v l i T B v ≠ 0 l_i^{\mathsf{T}}Bv\ne0 l i T B v = 0 i i i w ∈ R p w\in\mathbb{R}^{p} w ∈ R p w T C r i ≠ 0 w^{\mathsf{T}}Cr_i\ne0 w T C r i = 0 i i i
b = B v ∈ R n , c = w T C ∈ R 1 × n , b=Bv\in\mathbb{R}^{n},\qquad c=w^{\mathsf{T}}C\in\mathbb{R}^{1\times n}, b = B v ∈ R n , c = w T C ∈ R 1 × n , the PBH criterion shows that ( F H , b ) (F_H,b) ( F H , b ) ( F H , c ) (F_H,c) ( F H , c )
For K α = H + α v w T K_\alpha=H+\alpha vw^{\mathsf{T}} K α = H + α v w T
F K α = F H + α b c . (45) F_{K_\alpha}=F_H+\alpha bc. \tag*{(45)} F K α = F H + α b c . (45) Thus, a rank-one output-feedback family can be used after the preliminary selection of H H H
Step 2: Eigenvalue conditions outside a finite set. Define
D ( z ) = det ( z I n − F H ) , U ( z ) = c adj ( z I n − F H ) b . D(z)=\det(zI_n-F_H),\qquad U(z)=c\operatorname{adj}(zI_n-F_H)b. D ( z ) = det ( z I n − F H ) , U ( z ) = c adj ( z I n − F H ) b . Since ( F H , b , c ) (F_H,b,c) ( F H , b , c ) D D D U U U U ≢ 0 U\not\equiv0 U ≡ 0 deg D = n \deg D=n deg D = n deg U ≤ n − 1 \deg U\leq n-1 deg U ≤ n − 1 (45) and the rank-one determinant identity det ( M + u v T ) = det ( M ) + v T adj ( M ) u \det(M+uv^{\mathsf{T}})=\det(M)+v^{\mathsf{T}}\operatorname{adj}(M)u det ( M + u v T ) = det ( M ) + v T adj ( M ) u M ∈ C n × n M\in\mathbb{C}^{n\times n} M ∈ C n × n u , v ∈ C n u,v\in\mathbb{C}^{n} u , v ∈ C n M = z I n − F H M=zI_n-F_H M = z I n − F H u = − α b u=-\alpha b u = − α b v = c T v=c^{\mathsf{T}} v = c T
P α ( z ) : = det ( z I n − F K α ) = D ( z ) − α U ( z ) . P_\alpha(z):=\det(zI_n-F_{K_\alpha})=D(z)-\alpha U(z). P α ( z ) := det ( z I n − F K α ) = D ( z ) − α U ( z ) . (i) Distinct eigenvalues of F K α F_{K_\alpha} F K α If z z z P α P_\alpha P α
D ( z ) − α U ( z ) = 0 , D ′ ( z ) − α U ′ ( z ) = 0. D(z)-\alpha U(z)=0,\qquad D'(z)-\alpha U'(z)=0. D ( z ) − α U ( z ) = 0 , D ′ ( z ) − α U ′ ( z ) = 0. Coprimeness implies U ( z ) ≠ 0 U(z)\ne0 U ( z ) = 0
α = D ( z ) U ( z ) , V ( z ) : = D ′ ( z ) U ( z ) − D ( z ) U ′ ( z ) = 0. \alpha=\frac{D(z)}{U(z)},\qquad V(z):=D'(z)U(z)-D(z)U'(z)=0. α = U ( z ) D ( z ) , V ( z ) := D ′ ( z ) U ( z ) − D ( z ) U ′ ( z ) = 0. The polynomial V V V ( D / U ) ′ ≡ 0 (D/U)'\equiv0 ( D / U ) ′ ≡ 0 D / U D/U D / U deg D > deg U \deg D>\deg U deg D > deg U
Z 0 : = { D ( z ) U ( z ) : z ∈ C , V ( z ) = 0 , U ( z ) ≠ 0 } ∩ R Z_0:=\left\{\frac{D(z)}{U(z)}:z\in\mathbb{C},\ V(z)=0,\ U(z)\ne0\right\}\cap\mathbb{R} Z 0 := { U ( z ) D ( z ) : z ∈ C , V ( z ) = 0 , U ( z ) = 0 } ∩ R is finite, and F K α F_{K_\alpha} F K α n n n α ∉ Z 0 \alpha\notin Z_0 α ∈ / Z 0
(ii) Exclusion of eigenvalue-difference coincidences. For α ∉ Z 0 \alpha\notin Z_0 α ∈ / Z 0 μ 1 , … , μ n \mu_1,\ldots,\mu_n μ 1 , … , μ n F K α F_{K_\alpha} F K α A K α = A T ⊗ I n − I n ⊗ F K α A_{K_\alpha}=A^{\mathsf{T}}\otimes I_n-I_n\otimes F_{K_\alpha} A K α = A T ⊗ I n − I n ⊗ F K α λ j − μ i \lambda_j-\mu_i λ j − μ i ( j , i ) (j,i) ( j , i ) j ≠ k j\ne k j = k d j k = λ k − λ j ≠ 0 d_{jk}=\lambda_k-\lambda_j\ne0 d j k = λ k − λ j = 0 λ j − μ i = λ k − μ ℓ \lambda_j-\mu_i=\lambda_k-\mu_\ell λ j − μ i = λ k − μ ℓ μ ℓ = μ i + d j k \mu_\ell=\mu_i+d_{jk} μ ℓ = μ i + d j k P α = D − α U P_\alpha=D-\alpha U P α = D − α U
D ( μ i ) = α U ( μ i ) , D ( μ i + d j k ) = α U ( μ i + d j k ) . D(\mu_i)=\alpha U(\mu_i),\qquad D(\mu_i+d_{jk})=\alpha U(\mu_i+d_{jk}). D ( μ i ) = α U ( μ i ) , D ( μ i + d j k ) = α U ( μ i + d j k ) . Eliminating α \alpha α
D ( μ i ) U ( μ i + d j k ) − D ( μ i + d j k ) U ( μ i ) = 0. D(\mu_i)U(\mu_i+d_{jk})-D(\mu_i+d_{jk})U(\mu_i)=0. D ( μ i ) U ( μ i + d j k ) − D ( μ i + d j k ) U ( μ i ) = 0. This motivates defining
R j k ( z ) : = D ( z ) U ( z + d j k ) − D ( z + d j k ) U ( z ) . R_{jk}(z):=D(z)U(z+d_{jk})-D(z+d_{jk})U(z). R j k ( z ) := D ( z ) U ( z + d j k ) − D ( z + d j k ) U ( z ) . We next show that R j k R_{jk} R j k R j k ≡ 0 R_{jk}\equiv0 R j k ≡ 0
D ( z ) U ( z + d j k ) = D ( z + d j k ) U ( z ) . D(z)U(z+d_{jk})=D(z+d_{jk})U(z). D ( z ) U ( z + d j k ) = D ( z + d j k ) U ( z ) . If U ( z ) U ( z + d j k ) = 0 U(z)U(z+d_{jk})=0 U ( z ) U ( z + d j k ) = 0 z z z z + d j k z+d_{jk} z + d j k P α P_\alpha P α D D D U U U U ( z ) U ( z + d j k ) ≠ 0 U(z)U(z+d_{jk})\ne0 U ( z ) U ( z + d j k ) = 0
D ( z + d j k ) U ( z + d j k ) = D ( z ) U ( z ) . \frac{D(z+d_{jk})}{U(z+d_{jk})}=\frac{D(z)}{U(z)}. U ( z + d j k ) D ( z + d j k ) = U ( z ) D ( z ) . Since U U U f = D / U f=D/U f = D / U d j k d_{jk} d j k
If f f f z 0 z_0 z 0 z 0 + ℓ d j k z_0+\ell d_{jk} z 0 + ℓ d j k ℓ ∈ Z + 0 \ell\in\mathbb{Z}_+^0 ℓ ∈ Z + 0 d j k ≠ 0 d_{jk}\ne0 d j k = 0 D / U D/U D / U R j k ≠ 0 R_{jk}\ne0 R j k = 0
As shown above, any row coincidence λ j − μ i = λ k − μ ℓ \lambda_j-\mu_i=\lambda_k-\mu_\ell λ j − μ i = λ k − μ ℓ j ≠ k j \ne k j = k R j k ( μ i ) = 0 R_{jk}(\mu_i)=0 R j k ( μ i ) = 0 U ( μ i ) ≠ 0 U(\mu_i)\ne0 U ( μ i ) = 0 P α ( μ i ) = 0 P_\alpha(\mu_i)=0 P α ( μ i ) = 0 D ( μ i ) = 0 D(\mu_i)=0 D ( μ i ) = 0 D D D U U U U ( μ i + d j k ) = U ( μ ℓ ) ≠ 0 U(\mu_i+d_{jk})=U(\mu_\ell)\ne0 U ( μ i + d j k ) = U ( μ ℓ ) = 0 α = D ( μ i ) / U ( μ i ) \alpha=D(\mu_i)/U(\mu_i) α = D ( μ i ) / U ( μ i )
Since R j k R_{jk} R j k α \alpha α
Z j k : = { D ( z ) U ( z ) : z ∈ C , R j k ( z ) = 0 , U ( z ) U ( z + d j k ) ≠ 0 } ∩ R . \mathcal{Z}_{jk}:=\left\{\frac{D(z)}{U(z)}:z\in\mathbb{C},\ R_{jk}(z)=0,\ U(z)U(z+d_{jk})\ne0\right\}\cap\mathbb{R}. Z j k := { U ( z ) D ( z ) : z ∈ C , R j k ( z ) = 0 , U ( z ) U ( z + d j k ) = 0 } ∩ R . Define
Z = Z 0 ∪ ⋃ 1 ≤ j , k ≤ ς j ≠ k Z j k . \mathcal{Z}=\mathcal{Z}_0\cup\bigcup_{\substack{1\le j,k\le\varsigma\\j\ne k}}\mathcal{Z}_{jk}. Z = Z 0 ∪ 1 ≤ j , k ≤ ς j = k ⋃ Z j k . This set is finite because it is a finite union of finite sets. For every α ∈ R ∖ Z \alpha\in\mathbb{R}\setminus\mathcal{Z} α ∈ R ∖ Z F K α F_{K_\alpha} F K α n n n μ 1 , … , μ n \mu_1,\ldots,\mu_n μ 1 , … , μ n j ≠ k j\ne k j = k j = k j=k j = k μ i = μ ℓ \mu_i=\mu_\ell μ i = μ ℓ i = ℓ i=\ell i = ℓ
λ j − μ i = λ k − μ ℓ ⟹ ( j , i ) = ( k , ℓ ) . (46) \lambda_j-\mu_i=\lambda_k-\mu_\ell\quad\Longrightarrow\quad(j,i)=(k,\ell). \tag*{(46)} λ j − μ i = λ k − μ ℓ ⟹ ( j , i ) = ( k , ℓ ) . (46) Controllability through a Sylvester equation
Fix α ∉ Z \alpha\notin\mathcal{Z} α ∈ / Z K = K α K=K_\alpha K = K α η ∈ C \eta\in\mathbb{C} η ∈ C
A K T z = η z , B K T z = 0 A_K^{\mathsf{T}}z=\eta z,\qquad B_K^{\mathsf{T}}z=0 A K T z = η z , B K T z = 0 imply z = 0 z=0 z = 0 z = vec ( Z ) z=\operatorname{vec}(Z) z = vec ( Z ) Z ∈ C n × n Z\in\mathbb{C}^{n\times n} Z ∈ C n × n
A K T = A ⊗ I n − I n ⊗ F K T , B K T = C ⊗ B T , A_K^{\mathsf{T}}=A\otimes I_n-I_n\otimes F_K^{\mathsf{T}},\qquad B_K^{\mathsf{T}}=C\otimes B^{\mathsf{T}}, A K T = A ⊗ I n − I n ⊗ F K T , B K T = C ⊗ B T , writing z = vec ( Z ) z=\operatorname{vec}(Z) z = vec ( Z ) vec ( M X N ) = ( N T ⊗ M ) vec ( X ) \operatorname{vec}(MXN)=(N^{\mathsf{T}}\otimes M)\operatorname{vec}(X) vec ( M X N ) = ( N T ⊗ M ) vec ( X )
A K T z = ( A ⊗ I n ) vec ( Z ) − ( I n ⊗ F K T ) vec ( Z ) = vec ( Z A T ) − vec ( F K T Z ) = vec ( Z A T − F K T Z ) , \begin{aligned}
A_K^{\mathsf{T}}z&=(A\otimes I_n)\operatorname{vec}(Z)-(I_n\otimes F_K^{\mathsf{T}})\operatorname{vec}(Z)\\
&=\operatorname{vec}(ZA^{\mathsf{T}})-\operatorname{vec}(F_K^{\mathsf{T}}Z)=\operatorname{vec}(ZA^{\mathsf{T}}-F_K^{\mathsf{T}}Z),
\end{aligned} A K T z = ( A ⊗ I n ) vec ( Z ) − ( I n ⊗ F K T ) vec ( Z ) = vec ( Z A T ) − vec ( F K T Z ) = vec ( Z A T − F K T Z ) , and
B K T z = ( C ⊗ B T ) vec ( Z ) = vec ( B T Z C T ) . B_K^{\mathsf{T}}z=(C\otimes B^{\mathsf{T}})\operatorname{vec}(Z)=\operatorname{vec}(B^{\mathsf{T}}ZC^{\mathsf{T}}). B K T z = ( C ⊗ B T ) vec ( Z ) = vec ( B T Z C T ) . Since the vectorization operator is injective, the conditions A K T z = η z A_K^{\mathsf{T}}z=\eta z A K T z = η z B K T z = 0 B_K^{\mathsf{T}}z=0 B K T z = 0
Z A T − F K T Z = η Z , B T Z C T = 0. (47) ZA^{\mathsf{T}}-F_K^{\mathsf{T}}Z=\eta Z,\qquad B^{\mathsf{T}}ZC^{\mathsf{T}}=0. \tag*{(47)} Z A T − F K T Z = η Z , B T Z C T = 0. (47) Since F K F_K F K n n n R ∈ C n × n R\in\mathbb{C}^{n\times n} R ∈ C n × n
F K T R = R Λ , (48) F_K^{\mathsf{T}}R=R\Lambda, \tag*{(48)} F K T R = R Λ , (48) where Λ = diag ( μ 1 , … , μ n ) \Lambda=\operatorname{diag}(\mu_1,\ldots,\mu_n) Λ = diag ( μ 1 , … , μ n ) Z = R Y Z=RY Z = R Y Y ∈ C n × n Y\in\mathbb{C}^{n\times n} Y ∈ C n × n
Y A T − Λ Y = η Y . YA^{\mathsf{T}}-\Lambda Y=\eta Y. Y A T − Λ Y = η Y . Let y i ∈ C 1 × n y_i\in\mathbb{C}^{1\times n} y i ∈ C 1 × n i i i Y Y Y i i i Λ Y \Lambda Y Λ Y μ i y i \mu_i y_i μ i y i
y i A T = ( η + μ i ) y i , i = 1 , … , n . (49) y_iA^{\mathsf{T}}=(\eta+\mu_i)y_i,\qquad i=1,\ldots,n. \tag*{(49)} y i A T = ( η + μ i ) y i , i = 1 , … , n . (49) If y i ≠ 0 y_i\ne0 y i = 0 y i T y_i^{\mathsf{T}} y i T A A A η = λ j − μ i \eta=\lambda_j-\mu_i η = λ j − μ i j j j Y Y Y
Suppose that Z ≠ 0 Z\ne0 Z = 0 Y Y Y y i y_i y i R = [ r 1 , … , r n ] R=[r_1,\ldots,r_n] R = [ r 1 , … , r n ] Z = R Y = r i y i Z=RY=r_i y_i Z = R Y = r i y i r i r_i r i i i i R R R r i r_i r i F K F_K F K ( F K , B ) (F_K,B) ( F K , B ) B T r i ≠ 0 B^{\mathsf{T}}r_i\ne0 B T r i = 0 y i T y_i^{\mathsf{T}} y i T A A A ( A , C ) (A,C) ( A , C ) C y i T ≠ 0 Cy_i^{\mathsf{T}}\ne0 C y i T = 0 y i C T ≠ 0 y_iC^{\mathsf{T}}\ne0 y i C T = 0
B T Z C T = ( B T r i ) ( y i C T ) ≠ 0 , B^{\mathsf{T}}ZC^{\mathsf{T}}=(B^{\mathsf{T}}r_i)(y_iC^{\mathsf{T}})\ne0, B T Z C T = ( B T r i ) ( y i C T ) = 0 , contradicting (48). Thus Z = 0 Z=0 Z = 0 ( A K , B K ) (A_K,B_K) ( A K , B K ) rank K K α = n 2 \operatorname{rank}K_{K_\alpha}=n^2 rank K K α = n 2 α ∉ Z \alpha\notin\mathcal{Z} α ∈ / Z E H , v , ω ⊆ Z \mathcal{E}_{H,v,\omega}\subseteq\mathcal{Z} E H , v , ω ⊆ Z
Positive definiteness of the Gramian
For any constant matrix K K K rank K K = n 2 \operatorname{rank}K_K=n^2 rank K K = n 2 z ∈ R n 2 z\in\mathbb{R}^{n^2} z ∈ R n 2
z T G K z ≥ λ min ( Q ) ∫ 2 h 3 h ρ ( s ) ∥ Γ K T ( s ) z ∥ 2 d s ≥ 0. z^{\mathsf{T}}G_Kz\ge\lambda_{\min}(Q)\int_{2h}^{3h}\rho(s)\left\|\Gamma_K^{\mathsf{T}}(s)z\right\|^2\,\mathrm{d}s\ge0. z T G K z ≥ λ m i n ( Q ) ∫ 2 h 3 h ρ ( s ) Γ K T ( s ) z 2 d s ≥ 0. If z T G K z = 0 z^{\mathsf{T}}G_Kz=0 z T G K z = 0 ρ \rho ρ ( 2 h , 3 h ) (2h,3h) ( 2 h , 3 h )
z T e A K ( s − 3 h ) B K = 0 , s ∈ ( 2 h , 3 h ) . z^{\mathsf{T}}e^{A_K(s-3h)}B_K=0,\qquad s\in(2h,3h). z T e A K ( s − 3 h ) B K = 0 , s ∈ ( 2 h , 3 h ) . Fix s 0 ∈ ( 2 h , 3 h ) s_0\in(2h,3h) s 0 ∈ ( 2 h , 3 h )
z T e A K ( s 0 − 3 h ) A K j B K = 0 , j = 0 , … , n 2 − 1. z^{\mathsf{T}}e^{A_K(s_0-3h)}A_K^jB_K=0,\qquad j=0,\ldots,n^2-1. z T e A K ( s 0 − 3 h ) A K j B K = 0 , j = 0 , … , n 2 − 1. Thus z T e A K ( s 0 − 3 h ) K K = 0 z^{\mathsf{T}}e^{A_K(s_0-3h)}\mathcal{K}_K=0 z T e A K ( s 0 − 3 h ) K K = 0 K K \mathcal{K}_K K K z = 0 z=0 z = 0 G K > 0 G_K>0 G K > 0
Proof of Lemma 5
Define
L v ( s ) = e A ( 3 h − s ) B \mathcal{L}_v(s)=e^{A(3h-s)}B L v ( s ) = e A ( 3 h − s ) B and
R v , i ( s ) = C e A ( s − h i − 2 h − τ ( s − h i ) ) , i = 1 , … , N . R_{v,i}(s)=Ce^{A(s-h_i-2h-\tau(s-h_i))},\qquad i=1,\ldots,N. R v , i ( s ) = C e A ( s − h i − 2 h − τ ( s − h i )) , i = 1 , … , N . Then
Γ v , i ( s ) = R v , i T ( s ) ⊗ L v ( s ) . \Gamma_{v,i}(s)=R_{v,i}^{\mathsf{T}}(s)\otimes\mathcal{L}_v(s). Γ v , i ( s ) = R v , i T ( s ) ⊗ L v ( s ) . Collecting all effective observation directions gives
R v ( s ) = [ C e A ( s − h 1 − 2 h − τ ( s − h 1 ) ) ⋮ C e A ( s − h N − 2 h − τ ( s − h N ) ) ] = O h , τ ( s ) e A ( s − 2 h ) . \mathcal{R}_v(s)=
\begin{bmatrix}
Ce^{A(s-h_1-2h-\tau(s-h_1))}\\
\vdots\\
Ce^{A(s-h_N-2h-\tau(s-h_N))}
\end{bmatrix}
=\mathcal{O}_{h,\tau}(s)e^{A(s-2h)}. R v ( s ) = C e A ( s − h 1 − 2 h − τ ( s − h 1 )) ⋮ C e A ( s − h N − 2 h − τ ( s − h N )) = O h , τ ( s ) e A ( s − 2 h ) . Since e A ( s − 2 h ) e^{A(s-2h)} e A ( s − 2 h )
rank ( R v ( s ) ) = n , s ∈ I . \operatorname{rank}(\mathcal{R}_v(s))=n,\qquad s\in\mathcal{I}. rank ( R v ( s )) = n , s ∈ I . For any z ∈ R n 2 z\in\mathbb{R}^{n^2} z ∈ R n 2
z T G v z = ∫ 2 h 3 h ρ ( s ) z T Γ v ( s ) Q v Γ v T ( s ) z d s ≥ λ min ( Q v ) ∫ 2 h 3 h ρ ( s ) ∥ Γ v T ( s ) z ∥ 2 d s ≥ 0. \begin{aligned}
z^{\mathsf{T}}G_vz=\int_{2h}^{3h}\rho(s)z^{\mathsf{T}}\Gamma_v(s)Q_v\Gamma_v^{\mathsf{T}}(s)z\,\mathrm{d}s \\
&\ge\lambda_{\min}(Q_v)\int_{2h}^{3h}\rho(s)\left\|\Gamma_v^{\mathsf{T}}(s)z\right\|^2\,\mathrm{d}s\ge0.
\end{aligned} z T G v z = ∫ 2 h 3 h ρ ( s ) z T Γ v ( s ) Q v Γ v T ( s ) z d s ≥ λ m i n ( Q v ) ∫ 2 h 3 h ρ ( s ) Γ v T ( s ) z 2 d s ≥ 0. Thus, G v G_v G v z T G v z = 0 z^{\mathsf{T}}G_vz=0 z T G v z = 0 z ∈ R n 2 z\in\mathbb{R}^{n^2} z ∈ R n 2 Z = vec − 1 ( z ) Z=\operatorname{vec}^{-1}(z) Z = vec − 1 ( z ) ρ ( s ) > 0 \rho(s)>0 ρ ( s ) > 0 ( 2 h , 3 h ) (2h,3h) ( 2 h , 3 h ) Γ v \Gamma_v Γ v I \mathcal{I} I Γ v T ( s ) z = 0 \Gamma_v^{\mathsf{T}}(s)z=0 Γ v T ( s ) z = 0 s ∈ I s\in\mathcal{I} s ∈ I i = 1 , … , N i=1,\ldots,N i = 1 , … , N
Γ v , i T ( s ) z = ( R v , i ( s ) ⊗ L v T ( s ) ) vec ( Z ) = vec ( L v T ( s ) Z R v , i T ( s ) ) . \Gamma_{v,i}^{\mathsf{T}}(s)z=\left(R_{v,i}(s)\otimes L_v^{\mathsf{T}}(s)\right)\operatorname{vec}(Z)=\operatorname{vec}\left(L_v^{\mathsf{T}}(s)ZR_{v,i}^{\mathsf{T}}(s)\right). Γ v , i T ( s ) z = ( R v , i ( s ) ⊗ L v T ( s ) ) vec ( Z ) = vec ( L v T ( s ) Z R v , i T ( s ) ) . Thus, collecting the equalities Γ v , i T ( s ) z = 0 \Gamma_{v,i}^{\mathsf{T}}(s)z=0 Γ v , i T ( s ) z = 0 L v T ( s ) Z R v T ( s ) = 0 L_v^{\mathsf{T}}(s)ZR_v^{\mathsf{T}}(s)=0 L v T ( s ) Z R v T ( s ) = 0 s ∈ I s\in\mathcal{I} s ∈ I
Because R v ( s ) R_v(s) R v ( s ) I \mathcal{I} I R v T ( s ) R_v^{\mathsf{T}}(s) R v T ( s ) L v T ( s ) Z = 0 L_v^{\mathsf{T}}(s)Z=0 L v T ( s ) Z = 0 s ∈ I s\in\mathcal{I} s ∈ I
It remains to show that the column spaces of L v ( s ) L_v(s) L v ( s ) s s s I \mathcal{I} I R n \mathbb{R}^n R n v ∈ R n v\in\mathbb{R}^n v ∈ R n
v T L v ( s ) = v T e A ( 3 h − s ) B = 0 , s ∈ I . v^{\mathsf{T}}L_v(s)=v^{\mathsf{T}}e^{A(3h-s)}B=0,\qquad s\in\mathcal{I}. v T L v ( s ) = v T e A ( 3 h − s ) B = 0 , s ∈ I . The left-hand side is analytic in s s s s 0 ∈ I s_0\in\mathcal{I} s 0 ∈ I s = s 0 s=s_0 s = s 0
v T e A ( 3 h − s 0 ) A k B = 0 , k = 0 , … , n − 1. v^{\mathsf{T}}e^{A(3h-s_0)}A^kB=0,\qquad k=0,\ldots,n-1. v T e A ( 3 h − s 0 ) A k B = 0 , k = 0 , … , n − 1. Let v ~ T = v T e A ( 3 h − s 0 ) \widetilde{v}^{\mathsf{T}}=v^{\mathsf{T}}e^{A(3h-s_0)} v T = v T e A ( 3 h − s 0 )
v ~ T [ B A B ⋯ A n − 1 B ] = 0. \widetilde{v}^{\mathsf{T}}\begin{bmatrix}B&AB&\cdots&A^{n-1}B\end{bmatrix}=0. v T [ B A B ⋯ A n − 1 B ] = 0. Controllability of ( A , B ) (A,B) ( A , B ) v ~ = 0 \widetilde{v}=0 v = 0 e A ( 3 h − s 0 ) e^{A(3h-s_0)} e A ( 3 h − s 0 ) v = 0 v=0 v = 0
span { Im L v ( s ) : s ∈ I } = R n . \operatorname{span}\{\operatorname{Im}L_v(s):s\in\mathcal{I}\}=\mathbb{R}^n. span { Im L v ( s ) : s ∈ I } = R n . Now, L v T ( s ) Z = 0 L_v^{\mathsf{T}}(s)Z=0 L v T ( s ) Z = 0 s ∈ I s\in\mathcal{I} s ∈ I x T Z = 0 x^{\mathsf{T}}Z=0 x T Z = 0 x ∈ Im L v ( s ) x\in\operatorname{Im}L_v(s) x ∈ Im L v ( s ) s ∈ I s\in\mathcal{I} s ∈ I R n \mathbb{R}^n R n x T Z = 0 x^{\mathsf{T}}Z=0 x T Z = 0 x ∈ R n x\in\mathbb{R}^n x ∈ R n Z = 0 Z=0 Z = 0 z = 0 z=0 z = 0
Thus, the null space of G v G_v G v G v G_v G v G v > 0 G_v>0 G v > 0
