# Stability and Closed Trajectory for Second-Order Control Systems with Single-Saturated Input

Chapter

## Abstract

In this chapter, based on a new method through defining new equilibrium points, the relationship criterion among equilibrium points is discussed for linear system with saturated inputs. The asymptotic stability of the origin of the linear system in the presence of a single saturation input is analyzed, and the existence equations of closed trajectory is also considered for the same control systems. Finally, characteristics of commutative matrices of MIMO linear systems are considered. And at the same time, we present the simple criterion equations of asymptotic stability of second-order control systems with single-saturated input when $$ABK=BKA$$.

## 3.1 Introduction

The main contribution of this chapter is that a new method for analyzing nonlinear systems is presented by defining the new equilibrium points. The main results of this chapter are as follows: First, the new equilibrium points are defined. Second, the criteria of the new equilibrium points are presented too. Third, the sufficient conditions of GAS of the origin are presented for second-order control systems with a single-saturated input by qualitative analysis. Fourth, the existence of the closed trajectory is also considered for the same control systems. Finally, the structure of the commutative matrices is considered. And at the same time, we present the simple criterion equations of asymptotic stability of second-order control systems with single-saturated input when $$ABK=BKA$$.

The continuous-time system in paper [1, 6, 11], which is as follows:
\begin{aligned} \left\{ \begin{array}{lll} \dot{x}(t) &{}=&{} Ax(t) + B{u_s}(t)\\ y(t) &{}=&{} {C^T}x(t)\\ {u_s}&{} =&{} sat( - Kx) \end{array} \right. \end{aligned}
(3.1)
is considered, where $$x(t)\in \mathbb {R}^2$$ is the state vector, $$y(t)\in \mathbb {R}^1$$ is the output vector, $$u_s(t)\in \mathbb {R}^1$$ is the control input. A, B, C, K are matrices of appropriate dimensions and the system is minimal.
Saturated input $$sat(-Kx)$$ can be expressed as
\begin{aligned} sat( - Kx) = \left\{ \begin{array}{l} \begin{array}{*{20}{c}} {{u_{\lim }},}&{}{if - Kx \ge {u_{\lim }},} \end{array}\\ \begin{array}{*{20}{c}} { - Kx,}&{}{if\left| { - Kx} \right| < {u_{\lim }},} \end{array}\\ \begin{array}{*{20}{c}} {-{u_{\lim }},}&{}{if - {K_i}x \le -{u_{\lim }},} \end{array} \end{array} \right. \end{aligned}
(3.2)
where $$u_{\lim }$$ represents the saturation limit of the feedback system and $$\Vert u_s\Vert \le u_{\lim }$$.
From the above saturating definition, it is reasonable that we transform the SISO system to the following control systems
\begin{aligned} \left\{ \begin{array}{l} \dot{x}(t) = (A - \lambda BK)x(t) + {\lambda ^ * }B{u_{\lim }},\\ y = Cx(t), \end{array} \right. \end{aligned}
(3.3)
where $$\lambda$$ demonstrates 0 or 1, and $${\lambda ^*}$$ is derived form according to
\begin{aligned} \left\{ \begin{array}{l} (\lambda ,{\lambda ^ * }) = \begin{array}{*{20}{c}} {(0,1)}&{}{ iff}&{}{ - Kx - {u_{\lim }} \ge 0,} \end{array}\\ (\lambda ,{\lambda ^ * }) = \begin{array}{*{20}{c}} {(0,-1)}&{}{ iff}&{}{ - Kx + {u_{\lim }} \le 0,} \end{array}\\ (\lambda ,{\lambda ^ * }) = \begin{array}{*{20}{c}} {(1,0)}&{}{ iff}&{}{\left| {-Kx} \right| < {u_{\lim }}.} \end{array} \end{array} \right. \end{aligned}
(3.4)
If $$A-\lambda BK$$ is nonsingular, it follows that
\begin{aligned} x_{eq,\lambda ,\lambda ^{*}}= -\lambda ^{*}(A -\lambda BK)^{-1}Bu_{\lim }. \end{aligned}
We explain its physical meaning as follows:
\begin{aligned} \begin{array}{l} {\mathbb {D}_{0,1}} = \left\{ {x\left| { - Kx \ge {u_{\lim }}} \right. } \right\} ,\\ {\mathbb {D}_{0,-1}} = \left\{ {x\left| { - Kx \le - {u_{\lim }}} \right. } \right\} ,\\ {\mathbb {D}_{1,0}} = \left\{ {x\left| {\left| { - Kx} \right| < {u_{\lim }}} \right. } \right\} . \end{array} \end{aligned}
For convenience, we give the following definitions.
For system (3.1), if matrix $$A - \lambda BK$$ is nonsingular, then
$${x_{eq,\lambda ,{\lambda ^ * }}} = - {\lambda ^ * }{(A - \lambda BK)^{ - 1}}B{u_{\lim }}.$$

## Definition 3.1

(I) For system (3.1), some pair $$(\lambda , {\lambda ^ * })$$ and its corresponding $${x_{eq,\lambda , {\lambda ^ * }}}$$ satisfying (3.3), if $${x_{eq,\lambda , {\lambda ^ * }}} \in {\mathbb {D}_{\lambda ,{\lambda ^ * }}}$$, we call $${x_{eq,\lambda ,{\lambda ^ * }}}$$ a genuine equilibrium point. For such an equilibrium, if $$\left( {A - \lambda BK} \right)$$ is a stable (unstable) matrix, then $${x_{eq,\lambda ,{\lambda ^ * }}}$$ is a genuine stable (unstable) equilibrium point.

(II) For system (3.1), some pair $$(\lambda , {\lambda ^ * })$$ and its corresponding $${x_{eq,\lambda , {\lambda ^ * }}}$$ satisfying (3.3), if $${x_{eq,\lambda ,{\lambda ^ * }}} \notin {\mathbb {D}_{\lambda ,{\lambda ^ * }}}$$, then we call $${x_{eq,\lambda ,{\lambda ^ * }}}$$ a spurious equilibrium point. For such spurious equilibria, if $$\left( {A - \lambda BK} \right)$$ is stable (unstable), then we say that $${x_{eq,\lambda ,{\lambda ^ * }}}$$ is a spurious stable (unstable) equilibrium point.

## Theorem 3.1

If $${x_{eq,\lambda ,{\lambda ^ * }}}$$ is a genuine stable (unstable) equilibrium point for system (3.1), then so is $${x_{eq,\lambda ,{-\lambda ^ * }}}$$. Similarly, if $${x_{eq,\lambda ,{\lambda ^ * }}}$$ is a spurious stable (unstable) equilibrium point, then so is $${x_{eq,\lambda ,{-\lambda ^ * }}}$$. In particular, the origin $$x=0$$ is always an equilibrium point for both the unsaturated and the saturated systems.

The proof, which is readily obtained via Definition 3.1 and noting that $${x_{eq,\lambda ,{-\lambda ^ * }}} =-{x_{eq,\lambda ,{\lambda ^ * }}}$$, is omitted here.

## 3.2 Criteria for Stability Analysis

We shall investigate the sufficient conditions on the globally asymptotical stability of the origin of the saturated control system (3.1) when $$n=2$$, $$m=1.$$

## Theorem 3.2

Suppose that $$n=2$$, $$m=1$$ of system (3.1), let

(a)  The other equilibrium points are spurious stable equilibrium points except the stable origin;

(b)  There does not exist any closed trajectory, then the origin of control system (3.1) with SISO-saturated input is globally asymptotical stable.

## Proof

That $$x = 0$$ is a genuine stable equilibrium point which implies that $$A - BK$$ is asymptotically stable. We deduce that there exists a positive definite symmetric matrix P in virtue of Lyapunov theorem such that
\begin{aligned} \begin{array}{l} {(A - BK)^T}P + P(A - BK) < 0. \end{array} \end{aligned}
(3.5)
Let
\begin{aligned} \begin{array}{l} \mathbb {S} = \{ x\left| {{x^T}Px \le d}\right\} . \end{array} \end{aligned}
The maximum d in $$\mathbb {S} \subset {\mathbb {D}_{I,0}}$$ is called the maximum invariant set and we define $${\mathbb {S}_{I,0}}$$ as $${\mathbb {S}_{I,0}}: = \{ x\left| {{x^T}Px \le {d_{\max }}} \right. \}.$$  Suppose that $$-Kx = u_{\lim }$$ and $$-Kx =-u_{\lim }$$ be tangent lines, whose tangent points are $${p^{'}_{0,1}}$$ and $${p^{''}_{0, - 1}}$$ (See Fig. 3.1).
Let $${x_0} \in {\mathbb {D}_{0,1}}$$, since $${x_{eq,0, - 1}}$$ and $${x_{eq,0,1}}$$ are the spurious stable equilibrium points, we get the following trajectory
\begin{aligned} \begin{array}{l} {x_1}(t) = {e^{A(t - {T_0})}}[{x_0} + \int _{{T_0}}^t {{e^{A(t - s)}}B{\Lambda ^*}{u_{\lim }}dt} ]\\ = {e^{A(t - {T_0})}}{x_0} + ({e^{A(t - {T_0})}} - I){A^{ - 1}}B{u_{\lim }}. \end{array} \end{aligned}
(3.6)
And that $${x_{eq,\lambda ,{\lambda ^*}}}$$ is a spurious stable equilibrium point implies that $${x_{eq,\lambda ,{\lambda ^*}}} \notin {\mathbb {D}_{0,1}}$$  $$({x_{eq,\lambda ,{\lambda ^*}}} \notin {\mathbb {D}_{0, - 1}})$$. Here take the former for example. We deduce that there exists some time point $$T_1$$ such as $$-K{x_1}({T_1}) = {u_{\lim }}.$$
Let $${x_1}({T_1}) = x({T_1},{T_0},{x_0})$$ be as an initial value, its trajectory is
\begin{aligned} \begin{array}{l} {x_2}(t) = {e^{(A - BK)(t - {T_1})}}{x_1}({T_1}). \end{array} \end{aligned}
If there exists a time point $$T_2^{'}$$ such as $${x_2}(T_2^{'}) \in \partial {\mathbb {S}_{(1,0)}},$$ then we obtain the limit relation
$$\mathop {\lim }\limits _{t \rightarrow \infty } x(t,T_2^{'},{x_2}(T_2^{'})) = 0.$$
If not, there exist a time point $$T_2$$ such that $$- Kx_1(T_2)=-u_{\lim }$$. Similarly, in the region $$\mathbb {D}_{0,-1}$$, there exists a time point $$T_3$$ such that
\begin{aligned} \left\{ \begin{array}{l} {x_3}({T_3}) = {e^{A({T_3} - {T_2})}}[{x_2}({T_2}) - \int _{{T_0}}^{{T_3}} {{e^{A({T_3} - s)}}{A^{ - 1}}B{u_{\lim }}dt} ]\\ K{x_3}({T_3}) = - {u_{\lim }}. \end{array} \right\} \end{aligned}
Otherwise, there is a genuine stable origin equilibrium point in the region $$\mathbb {D}_{1,0}$$, if there exists a $$T_3^{'}$$ such that
\begin{aligned} \begin{array}{l} {x_4}(T_3^{'}) = {e^{(A - BK)(t - {T_3})}}{x_{{T_3}}} \in \partial {\mathbb {S}_{I,0}}. \end{array} \end{aligned}
We come to conclusion that
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty } x(t,{t_0},{x_0}) = 0. \end{array} \end{aligned}
If not, a sequence of $$x_k$$ enters into the region $$\mathbb {D}_{0,1}$$, then there are three possibilities for $$x_k$$ with initial value $$x_4(T_4)$$, which are as follows:

(a$$x(t,t_{4},x_4)$$ is superposing with $$x(t,t_0,x_0)$$ in the region $$\mathbb {D}_{1,0}$$;

(b)  the joint point between $$x(t,t_4,x_4)$$ and $$-Kx=u_{\lim }$$ lies in the inner near $$x_1$$;

(c)  the joint between $$x(t,t_4,x_4)$$ and $$-Kx=u_{\lim }$$ lies in the outer near $$x_1$$.

By the Theorem 3.2 (II), we learn that the case (a) is impossible, so we only need to check (b) and (c). If the case (c) holds, there exist $$x_6^{''}$$ which lies outwards far from $$x_2$$, $$x_7^{''}$$ which lies outwards far from $$x_3$$, and etc, we will obtain a sequence of $$x_i^{''}(i = 6,7, \ldots )$$ ($$x_i^{''}(t)$$ denotes that the joint points between the trajectory $$x(t,t_{i-4},x_{i-4})$$  and lines $$-Kx =u_{\lim }$$ or $$-Kx=-u_{\lim }$$) which will be far away from point $$p^{'}_{0,1}$$ or from $$p_{0,-1}$$.

Let
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty } \left\| {x_i^{''} - {p_{0,I}}} \right\| =\alpha < \infty ,\alpha >0. \end{array} \end{aligned}
It is a contradiction with the fact that there does not exist any closed trajectory. It follows from the uniqueness of the solution of (3.1) that
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty } \left\| {x_i^{''}-p_{0,I}} \right\| = \infty \,\,{ or}\,\, \mathop {\lim }\limits _{t \rightarrow \infty } \left\| {x_i^{''}-p_{0, - 1}} \right\| = \infty . \end{array} \end{aligned}
So
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty } \left\| x_i^{''} \right\| \ge \mathop {\lim }\limits _{t \rightarrow \infty } \left\| x_i^{''} - p_{0,1} \right\| - \left\| p_{0,1} \right\| = \infty . \end{array} \end{aligned}
Further
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty } \left\| {x_i^{''}} \right\| = \infty . \end{array} \end{aligned}
With regard to the Theorem 3.2 (I), it follows that
\begin{aligned} \begin{array}{l} \max {{\mathrm Re}} (\lambda (A - BK))< 0,\max {{\mathrm Re}} (\lambda (A)) < 0. \end{array} \end{aligned}
where $$\lambda (.)$$ denotes the eigenvalue of the matrix.
It follows from Theorem 3.2 (II) that there exists $$T(\varepsilon )>0$$ such that
\begin{aligned} \begin{array}{l} \left| {\left\| {{e^{(A - \lambda BK)(t - {t_0})}}{x_0} + {\lambda ^*}[{e^{(A - \lambda BK)(t - {t_0})}} - I]{{(A - \lambda BK)}^{ - 1}}B{u_{\lim }}} \right\| } \right| \\ -\left\| {\lambda ^{*}{{(A - \lambda BK)}^{ - 1}}B{u_{\lim }}} \right\| < \varepsilon ,t \ge T(\varepsilon ). \end{array} \end{aligned}
So
\begin{aligned} \begin{array}{l} \left\| {x(t,{t_0},{x_0})} \right\| \\ \le \mathop {\sup }\limits _{\forall {x_0} \in {\mathbb {R}^n},\forall t \in \left[ {0,\infty } \right) } \{ \left\| {{e^{(A - \lambda BK)(t - {t_0})}}{x_0} + {\lambda ^*}\left[ {{e^{(A - \lambda BK)(t - {t_0})}} - I} \right] {{(A - \lambda BK)}^{ - 1}}B{u_{\lim }}} \right\| \} \\ = \mathop {\max }\limits _{\forall {x_0} \in {\mathbb {R}^n},\forall \varepsilon ({x_0})> 0} \{ \mathop {\sup }\limits _{\forall {x_0} \in {\mathbb {R}^n},{t_0} \le t \le T(\varepsilon )} \{\Vert e^{(A - \lambda BK)(t -t_0)}x_0 + {\lambda ^*}[e^{(A-\lambda BK)(t-t_0)}-I]\\ *(A - \lambda BK)^{-1}Bu_{\lim }\Vert , \mathop {\max }\limits _{(\lambda ,{\lambda ^*}),\forall \varepsilon ({x_0})> 0} \{ \left\| {{\lambda ^*}{{(A - \lambda BK)}^{ - 1}}B{u_{\lim }}} \right\| + \varepsilon \} \} \\ = \mathop {\max }\limits _{\forall {x_0} \in {\mathbb {R}^n},\forall \varepsilon ({x_0}) > 0,{t_0} \le t \le T(\varepsilon )} \{ \left\| {{e^{(A - BK)(t - {t_0})}}{x_0}} \right\| ,\\ \left\| {{e^{A(t - {t_0})}}{x_0} \pm \left[ {{e^{A(t - {t_0})}} - I} \right] {A^{ - 1}}B{u_{\lim }}} \right\| ,\left\| {{A^{ - 1}}B{u_{\lim }}} \right\| + \varepsilon \} \\ < \infty . \end{array} \end{aligned}
It follows from the above discussions that the case (c) is impossible.

For the case  (b), there are two possibilities below.

(I)  In the region $$\mathbb {D}_{1,0}$$, the trajectory $$x(t,t_4,x_4)$$ will be intersect with $$\partial {\mathbb {S}_{1,0}}$$;

(II)  Trajectory $$x(t,t_4,x_4)$$ will at last reach to the tangent points $$p^{'}_{0,1}$$ or $$p^{''}_{0,-1}$$.

If the case (I) holds, it is clear that $$x(t,{t_0},{x_0}) = 0$$ is globally asymptotically stable. If not, the sequence of the joint points $$x_{4k + 1}$$ on the line $$-Kx=u_{\lim }$$ satisfies
\begin{aligned} \begin{array}{l} 0< \overline{r({x_{4k + 1}}p_{1,0}^{'})} < \overline{r({x_1}p_{1,0}^{'})}, 0> \overline{r({x_{4k}}p_{1,0}^{'})}> \cdots > \overline{r({x_4}p_{1,0}^{'})}. \end{array} \end{aligned}
where $$\overline{r}$$ denotes the distance between two points.
Similarly, it follows the same result for a sequence of the joint points in the line $$-Kx=-u_{\lim }$$. The sequence $$\overline{r({x_{4k + 1}}p_{1,0}^{'})}$$ monotonically decreases as k increases, while the sequence $$\overline{r({x_{4k}}p_{1,0}^{'})}$$ monotonously increases. Since the limit relation $$\overline{r({x_{4k + 1}}p_{1,0}^{'})}$$ is the existence and the uniqueness, it follows that
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty } \overline{r({x_{4k + 1}}p_{1,0}^{'})} = 0,\mathop {\lim }\limits _{t \rightarrow \infty } \overline{r({x_{4k}}p_{1,0}^{'})}= 0. \end{array} \end{aligned}
That is to say that the trajectory $$x(t,{t_0},{x_0}) = 0$$ will pass through the point $$p^{'}_{0,1}$$ or $$p^{''}_{0,-1}$$.
Considering the positively invariant and contractive of the maximum ellipsoid $$\mathbb {S}_{0,1}$$, it follows that
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty ,\,x_0 \in \mathbb {S}_{0,1} } x(t,t_0,x_0)=0. \end{array} \end{aligned}
Similarly, we obtain the same results with respect to $$\forall x_0 \in \mathbb {D}_{0,-1}$$ or $$\forall x_0 \in \mathbb {D}_{0,1}.$$
In short, it follows that
\begin{aligned} \begin{array}{l} \mathop {\lim }\limits _{t \rightarrow \infty } x(t,{t_0},{x_0}) = 0, \end{array} \end{aligned}
with respect to $$\forall x_0 \in {\mathbb {R}^2}.$$    $$\square$$

## 3.3 Criteria to the Closed Trajectory

From the above discussions, we know that for Theorem 3.2 (II), that there does not exist any closed trajectory, is a precondition. More importantly, how to judge the closed trajectory is a meaningful problem. Now, we shall discuss some criteria about the existence of closed trajectories.

Suppose that there exists the feedback control law $$u =- Kx$$ such that matrices A, $$A-BK$$ are Hurwitz stable for (1). If there exists the closed trajectory in $$\mathbb {R}^2$$, then it must intersect with line $$-Kx(t)=u_{\lim }$$ and line $$-Kx(t)=-u_{\lim }$$. Since $$x_{eq,0,1}$$ and $$x_{eq,0,-1}$$ are the spurious equilibrium points, there exists some time point $$T_1$$ such that $$-Kx({T_1})=u_{\lim }$$ with respect to any initial point $$x_0 \in \mathbb {D}_{1,0}$$ and any initial time $$t_0$$. Let $$\Delta T_1^{'} = {T_1} - {t_0}$$, if the trajectory $$x(t,{T_1},{x_1})$$ does not intersect with the maximum ellipsoid $$\mathbb {S}_{1,0} \in \mathbb {D}_{1,0}$$, then there exists time point $$T_2$$ where $$x(t,{T_1},{x_1})$$ intersects with $$-Kx(t)=-u_{\lim }$$, whose joint point is $$x_2=x(T_2,{T_1},{x_1})$$. Let $$\Delta T_2= T_2- T_1$$. The trajectory $$x(t,{T_1},{x_1})$$ intersects with $$- Kx(t)=-u_{\lim }$$ in $$\mathbb {D}_{1,0}$$ whose passing time is $$\Delta T_4$$ and whose joint point is $$x_4$$.

In the region $$\mathbb {D}_{1,0}$$, the trajectory $$x(t,{T_1},{x_1})$$ intersects with $$-Kx(t) = {u_{\lim }}$$ after time $$\Delta T_1^{''}$$ whose joint point is $$x_1$$. From the above discussions, it follows that
\begin{aligned} \begin{array}{l} \left\{ {\begin{array}{*{20}{l}} {\Delta {T_1}}&{} = &{}\Delta T_1^{'} + \Delta T_1^{''},\\ {\Delta {T_2}}&{} = &{}{T_2} - {T_1},\\ {\Delta {T_3}}&{} = &{}{T_3} - {T_2},\\ {\Delta {T_4}}&{} = &{}{T_4} - {T_3}. \end{array}} \right. \end{array} \end{aligned}
Suppose that the closed trajectory is $$\overline{{x_1}{x_2}{x_3}{x_4}}$$ where $$x_1$$, $$x_2$$, $$x_3$$, $$x_4$$ are joint points with lines $$-Kx(t)=u_{\lim }$$, and $$-Kx(t)=-u_{\lim }$$ (See Fig. 3.2).
The closed trajectory with the initial value $$x_0$$ is as follows:
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {x_1}({T_1}) = {e^{A\Delta T_1^{'}}}{x_0} + ({e^{A\Delta {T_1}}} - I){A^{ - 1}}B{u_{\lim }},\\ - K{x_1}({T_1}) = {u_{\lim }}. \end{array} \right. \\ \left\{ \begin{array}{l} {x_2}({T_2}) = {e^{(A - BK)\Delta {T_2}}}{x_1},\\ - K{x_2}({T_2}) = - {u_{\lim }}. \end{array} \right. \\ \left\{ \begin{array}{l} {x_3}({T_3}) = {e^{A\Delta T_3^{}}}{x_2} - ({e^{A\Delta {T_3}}} - I){A^{ - 1}}B{u_{\lim }},\\ - K{x_3}({T_3}) = - {u_{\lim }}. \end{array} \right. \\ \left\{ \begin{array}{l} {x_4}({T_4}) = {e^{(A - BK)\Delta {T_4}}}{x_3},\\ - K{x_4}({T_4}) = -{u_{\lim }}. \end{array} \right. \\ \left\{ \begin{array}{l} {x_1}({T_1}) = e^{A\Delta T_1}{x_4} + ({e^{A\Delta {T_1}}} - I){A^{ - 1}}B{u_{\lim }},\\ - K{x_1}({T_1}) = {u_{\lim }}. \end{array} \right. \end{array} \end{aligned}
Since the closed trajectory $$\overline{{x_1}{x_2}{x_3}{x_4}}$$ is periodic, the closed trajectories with initial values $$x_1$$, $$x_2$$, $$x_3$$, $$x_4$$ are defined as follows.
(1) The closed trajectories with an initial value $$x_1$$ is
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {x_2}({T_2}) = {e^{(A - BK)\Delta {T_2}}}{x_1},\\ - K{x_2}({T_2}) = {u_{\lim }}. \end{array} \right. \\ \left\{ \begin{array}{l} {x_3}({T_3}) = {e^{A\Delta T_3^{}}}{x_2} - ({e^{A\Delta {T_3}}} - I){A^{ - 1}}B{u_{\lim }},\\ - K{x_3}({T_3}) = - {u_{\lim }}. \end{array} \right. \\ \left\{ \begin{array}{l} {x_4}({T_4}) = {e^{(A - BK)\Delta {T_4}}}{x_3},\\ - K{x_4}({T_4}) = - {u_{\lim }}. \end{array} \right. \\ \left\{ \begin{array}{l} {x_1}({T_1}) = {e^{A\Delta T_1^{}}}{x_4} + ({e^{A\Delta {T_1}}} - I){A^{ - 1}}B{u_{\lim }},\\ - K{x_1}({T_1}) = {u_{\lim }}. \end{array} \right. \end{array} \end{aligned}
That is to say
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} ({e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}} - I){x_1} = {e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}}{A^{ - 1}}B{u_{\lim }}\\ - {e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}{A^{ - 1}}B{u_{\lim }} - {e^{A\Delta {T_1}}}^{}{A^{ - 1}}B{u_{\lim }} + {A^{ - 1}}B{u_{\lim }},\\ - K{x_1} = {u_{\lim }}. \end{array} \right. \end{array} \end{aligned}
Let
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_1}{x_1} = {B_1}{A^{ - 1}}B{u_{\lim }},\\ - K{x_1} = {u_{\lim }}, \end{array} \right. \end{array} \end{aligned}
(3.7)
where
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_1} = {e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}} - I,\\ {B_1} = {e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}} - {e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}} - {e^{A\Delta {T_1}}} + I. \end{array} \right. \end{array} \end{aligned}
Similarly, we give the following closed trajectories equations with different initial values.
(2) The closed trajectories with an initial value $$x_2$$ is
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_2}{x_2} = {B_2}{A^{ - 1}}B{u_{\lim }},\\ - K{x_2}({T_2}) = - {u_{\lim }}, \end{array} \right. \end{array} \end{aligned}
(3.8)
where
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_2} = {e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}} - I,\\ {B_2} = {e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}} - {e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}}\\ - {e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}} + {e^{(A - BK)\Delta {T_2}}}. \end{array} \right. \end{array} \end{aligned}
(3) The closed trajectories with an initial value $$x_3$$ is
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_3}{x_3} = {B_3}{A^{ - 1}}B{u_{\lim }},\\ - K{x_3} = - {u_{\lim }}, \end{array} \right. \end{array} \end{aligned}
(3.9)
where
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_3}={e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}}{e^{(A - BK)\Delta {T_4}}} - I,\\ {B_3}=-{e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}} + {e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}} + {e^{A\Delta {T_3}}} - I. \end{array} \right. \end{array} \end{aligned}
(4) The closed trajectories with an initial value $$x_4$$ is
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_4}{x_4} = {B_4}{A^{ - 1}}B{u_{\lim }},\\ - K{x_4} = {u_{\lim }}, \end{array} \right. \end{array} \end{aligned}
(3.10)
where
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} {A_4} = {e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}} - I,\\ {B_4} = - {e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}}{e^{A\Delta {T_1}}} + {e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}}{e^{(A - BK)\Delta {T_2}}}\\ - {e^{(A - BK)\Delta {T_4}}}{e^{A\Delta {T_3}}} + {e^{(A - BK)\Delta {T_4}}}. \end{array} \right. \end{array} \end{aligned}
If the trajectory $$\overline{{x_1}{x_2}{x_3}{x_4}}$$ is closed, then (3.7)–(3.10) simultaneously hold. On the other hand, if points $$x_1$$, $$x_2$$, $$x_3$$, $$x_4$$ which satisfy (3.7) as initial values must simultaneously satisfy (3.7)–(3.10) with regard to uniqueness of the solution.

We again consider the fact of uniqueness of the solution and nonsingular matrices $$e^{(A - BK)\Delta {T}}$$, $$e^{A\Delta {T}}$$, it follows that the above four trajectories are superposing.

Hence,
\begin{aligned} \begin{array}{l} \left\{ {\begin{array}{*{20}{c}} {{x_1}(t)}\\ { - K{x_1}(t)}\\ t\\ {{x_1}({T_1})}\\ { - K{x_1}({T_1})} \end{array}} \right. \begin{array}{*{20}{c}} = \\ \ge \\ \in \\ = \\ = \end{array}\begin{array}{*{20}{c}} {{e^{A(t - {T_0})}}{x_0} + ({e^{A(t - {T_0})}} - I){A^{ - 1}}B{u_{\lim }},}\\ {{u_l}_{im},}\\ {\left[ {{t_0},{T_1}} \right] ,}\\ {{e^{A\Delta T_1^{'}}}{x_0} + ({e^{A\Delta {T_1}}} - I){A^{ - 1}}B{u_{\lim }},}\\ {{u_l}_{im}.} \end{array} \\ \left\{ {\begin{array}{*{20}{c}} {{x_2}(t)}\\ {\left\| { - K{x_2}({T_2})} \right\| }\\ t\\ {{x_2}({T_2})}\\ { - K{x_2}({T_2})} \end{array}} \right. \begin{array}{*{20}{c}} = \\< \\ \in \\ = \\ = \end{array}\begin{array}{*{20}{c}} {{e^{(A - BK)(t - {T_1})}}{x_1},}\\ {{u_l}_{im},}\\ {[{T_1},{T_2}),}\\ {{e^{(A - BK)\Delta {T_2}}}{x_1},}\\ { - {u_l}_{im}.} \end{array} \\ \left\{ {\begin{array}{*{20}{c}} {{x_3}(t)}\\ { - K{x_3}(t)}\\ t\\ {{x_3}({T_3})}\\ { - K{x_3}({T_3})} \end{array}} \right. \begin{array}{*{20}{c}} = \\ \le \\ \in \\ = \\ = \end{array}\begin{array}{*{20}{c}} {{e^{A(t - {T_2})}}{x_2} - ({e^{A(t - {T_2})}} - I){A^{ - 1}}B{u_{\lim }},}\\ { - {u_l}_{im},}\\ {\left[ {{T_2},{T_3}} \right] ,}\\ {{e^{A\Delta T_3^{}}}{x_2} - ({e^{A\Delta {T_3}}} - I){A^{ - 1}}B{u_{\lim }},}\\ { - {u_l}_{im}.} \end{array}\\ \left\{ {\begin{array}{*{20}{c}} {{x_4}(t)}\\ {\left\| { - K{x_4}(t)} \right\| }\\ t\\ {{x_4}({T_4})}\\ { - K{x_4}({T_4})} \end{array}} \right. \begin{array}{*{20}{c}} = \\ < \\ \in \\ = \\ = \end{array}\begin{array}{*{20}{c}} {{e^{(A - BK)(t - {T_3})}}{x_3}}\\ {{u_l}_{im}}\\ {[{T_3},{T_4}]}\\ {{e^{(A - BK)\Delta {T_4}}}{x_3}}\\ {{u_l}_{im}} \end{array}\\ \left\{ {\begin{array}{*{20}{c}} {{x_1}(t)}\\ { - K{x_1}({T_1})}\\ t\\ {{x_1}(\Delta {T_1} + {T_4})}\\ { - K{x_1}(\Delta {T_1} + {T_4})} \end{array}} \right. \begin{array}{*{20}{c}} = \\ \ge \\ \in \\ = \\ = \end{array}\begin{array}{*{20}{c}} {{e^{A(t - {T_4})}}{x_4} + ({e^{A\Delta {T_1}}} - I){A^{ - 1}}B{u_{\lim }},}\\ {{u_l}_{im},}\\ {\left[ {{T_4},\Delta {T_1} + {T_4}} \right] ,}\\ {{e^{A\Delta T_1^{}}}{x_4} + ({e^{A\Delta {T_1}}} - I){A^{ - 1}}B{u_{\lim }},}\\ {{u_l}_{im}.} \end{array} \end{array} \end{aligned}
(3.11)
We consider the simplicities of (3.7)–(3.11) below. If $$A_1$$, $$A_2$$, $$A_3$$, $$A_4$$ are invertible, (3.7)–(3.11) transforms to
\begin{aligned} \left\{ \begin{array}{l} - KA_1^{ - 1}{B_1}{A^{ - 1}}B{u_{\lim }} = {u_{\lim }},\\ - KA_2^{ - 1}{B_2}{A^{ - 1}}B{u_{\lim }} = - {u_{\lim }},\\ - KA_3^{ - 1}{B_3}{A^{ - 1}}B{u_{\lim }} = - {u_{\lim }},\\ - KA_4^{ - 1}{B_4}{A^{ - 1}}B{u_{\lim }} = {u_{\lim }}. \end{array} \right. \end{aligned}
(3.12)
Because of the constant $$u_{\lim }>0$$, (3.12) transforms to
\begin{aligned} \left\{ \begin{array}{l} - KA_1^{-1}{B_1}{A^{-1}}B = 1,\\ - KA_2^{-1}{B_2}{A^{-1}}B = -1,\\ - KA_3^{-1}{B_3}{A^{-1}}B = -1,\\ - KA_4^{-1}{B_4}{A^{-1}}B = 1. \end{array} \right. \end{aligned}
(3.13)
For (3.13), if there exists a group of positive constants $$\Delta {T_1}$$, $$\Delta {T_2}$$, $$\Delta {T_3}$$, $$\Delta {T_4}$$ such that (3.13) holds, then there exists closed trajectory. If for every group of positive constants $$\Delta {T_1}$$, $$\Delta {T_2}$$, $$\Delta {T_3}$$, $$\Delta {T_4}$$, at least one equality of (3.13) does not hold, then there does not exist any closed trajectory.
If matrices $${A_1}$$, $${A_2}$$, $${A_3}$$, $${A_4}$$ are singular, we consider inequalities
\begin{aligned} \left\{ \begin{array}{l} Rank({A_1},{B_1}{A^{ - 1}}B)> Rank{A_1},\\ Rank({A_2},{B_2}{A^{ - 1}}B)> Rank{A_2},\\ Rank({A_3},{B_3}{A^{ - 1}}B)> Rank{A_3},\\ Rank({A_4},{B_4}{A^{ - 1}}B) > Rank{A_4}. \end{array} \right. \end{aligned}
(3.14)
For any group of positive constants $$\Delta {T_1}$$, $$\Delta {T_2}$$, $$\Delta {T_3}$$, $$\Delta {T_4}$$, if at least one inequalities of (3.14) does not hold, then there does not exist any solution for algebraic inequalities (3.14) and any closed trajectory of (3.1).   $$\square$$

From the above discussions, we can easily obtain the following results.

## Theorem 3.3

For system (3.1), if

(I) There exist a group of positive constants $$\Delta {T_1}$$, $$\Delta {T_2}$$, $$\Delta {T_3}$$, $$\Delta {T_4}$$ such that (3.14) holds, then there exists a closed trajectory.

(II) For any group of positive constants $$\Delta {T_1}$$, $$\Delta {T_2}$$, $$\Delta {T_3}$$, $$\Delta {T_4}$$, at least one equalities of (3.13) does not hold or satisfies at least one of (3.14), then there does exist any closed trajectory.

## 3.4 The Commutative Case

In this section, we analyze that when state matrices are commuting, the more simplified sufficient conditions on the GAS of the zero solution and the more simplified existence of the closed trajectory for the second-order control systems with single input compared to the above section. In particular, we transform the determination condition of the closed trajectory into the existence of the common solution for a set of higher order equations satisfying certain conditions. The above proposed conclusions provide us a more comprehensive understanding of the GAS of the zero solution for the second-order control systems with single input.

First, we give the following results.

## Theorem 3.4

$$AB = BA$$, $$rank (A)=n,$$ $$rank (B)=1$$ is only and if only that there exist nonsingular matrices $${Q_1}$$,C such that
\begin{aligned} A = Q_1\left( \begin{array}{*{20}{cc}} l_{11}&{}L_{12}\\ 0&{}L_{22} \end{array} \right) Q_1^{ - 1}, B = {Q_1}\left( \begin{array}{*{20}{c}} c_{11}&{}C_{12}\\ 0&{}0 \end{array} \right) Q_1^{-1}, \end{aligned}
where $$l_{11},\,c_{11}$$ are constants, $$L_{12}$$$$C_{12} \in \mathbb {R}^{1 \times (n - 1)}.$$

The above results are easily obtained, and we omit the above proof process.

Consider the system (3.1) or (3.3), we can simplify the identification process using Theorem 3.3 when $$ABK = BKA$$.
\begin{aligned} \begin{array}{rcl} A_1&{} =&{} A_2 \\ &{} = &{} A_3\\ &{} =&{} A_4 \\ &{}= &{} e^{A(\Delta T_1+\Delta T_3)}e^{(A-BK)(\Delta T_2+\Delta T_4)}-I. \end{array} \end{aligned}
(3.15)
Similar to the above conclusions, (3.12) can be replaced into
\begin{aligned} \left\{ \begin{array}{c} -KA_1B_1A^{-1}B=1,\\ -KA_1B_2A^{-1}B=-1,\\ -KA_1B_3A^{-1}B=-1,\\ -KA_1B_4A^{-1}B=1. \end{array} \right. \end{aligned}
(3.16)
Combining with Theorem 3.3, we can simplify the form of $$A_1$$, $$B_1$$, K, $$A^{-1}B$$ when $$ABK = BKA$$,
\begin{aligned} \begin{array}{l} (e^{A(\Delta T_1+\Delta T_3)} e^{(A-BK)(\Delta T_2+\Delta T_4)}\\ =Q_1e^{\left( \begin{array}{cc} l_{11} &{} L_{12} \\ 0 &{} L_{22} \end{array} \right) (\Delta T_1+\Delta T_3)} e^{\left( \begin{array}{cc} l_{11}-c_{11} &{} L_{12}-C_{12} \\ 0 &{} L_{22} \end{array} \right) (\Delta T_2+\Delta T_4)} Q_1^{-1}. \end{array} \end{aligned}
Now considering the particularity of the (1, 1) element of an upper triangular matrix which is similar to A, $$A-BK$$, we easily obtain
\begin{aligned} \begin{array}{rcl} e^{A(\Delta T_1+\Delta T_3)} e^{(A-BK)(\Delta T_2+\Delta T_4)}= Q_1\left( \begin{array}{cc} e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)} &{} * \\ 0 &{} * \end{array} \right) Q_1^{-1}. \end{array} \end{aligned}
So
\begin{aligned} \begin{array}{rcl} A_1 &{}=&{} Q_1(\left( \begin{array}{cc} e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)} &{} * \\ 0 &{} * \end{array} \right) -I)Q_1^{-1} \\ &{}=&{} Q_1\left( \begin{array}{cc} e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)}-1 &{} * \\ 0 &{} * \end{array} \right) Q_1^{-1}. \end{array} \end{aligned}
(3.17)
Similarly, we can get
\begin{aligned} \begin{array}{rcl} B_1 &{}=&{} Q_1\left( \begin{array}{cc} \begin{array}{r} e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)} \\ -e^{l_{11}\Delta T_1+(l_{11}-c_{11})\Delta T_4} \\ -e^{l_{11}\Delta T_1}+1 \end{array} &{} * \\ 0 &{} * \end{array} \right) Q_1^{-1}, \end{array} \end{aligned}
\begin{aligned} \begin{array}{rcl} B_2 &{}=&{} Q_1\left( \begin{array}{cc} \begin{array}{r} e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)}\\ -e^{l_{11}\Delta T_1+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)} \\ -e^{l_{11}\Delta T_1+(l_{11}-c_{11})\Delta T_2} \\ +e^{(l_{11}-c_{11})\Delta T_2} \end{array} &{} * \\ 0 &{} * \end{array} \right) Q_1^{-1}, \end{array} \end{aligned}
\begin{aligned} \begin{array}{l} B_3 =Q_1\left( \begin{array}{cc} \begin{array}{r} -e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})\Delta T_2}\\ +e^{l_{11}\Delta T_3+(l_{11}-c_{11})\Delta T_2}\\ +e^{l_{11}\Delta T_3} -1 \end{array} &{} * \\ 0 &{} * \end{array} \right) Q_1^{-1}, \end{array} \end{aligned}
\begin{aligned} \begin{array}{rcl} B_4 &{}=&{} Q_1\left( \begin{array}{cc} \begin{array}{r} -e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4} \\ +e^{l_{11}\Delta T_3+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)} \\ +e^{l_{11}\Delta T_3+(l_{11}-c_{11})\Delta T_4}\\ -e^{(l_{11}-c_{11})\Delta T_4} \end{array} &{} * \\ 0 &{} * \end{array} \right) Q_1^{-1}. \end{array} \end{aligned}
Therefore, (3.16) may be equivalent to
\begin{aligned} \begin{array}{l} -c_{11} [e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11}) (\Delta T_2+\Delta T_4)}-1]^{-1} (e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})\Delta T_4}\\ -e^{l_{11} \Delta T_1+(l_{11}-c_{11})\Delta T_4 } -e^{l_{11} \Delta T_1} +1)]l_{11}^{-1} =1, \\ -c_{11} [e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)}-1]^{-1} (e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_4+\Delta T_2)}\\ -e^{l_{11} \Delta T_1+(l_{11}-c_{11})(\Delta T_4+\Delta T_2)} -e^{l_{11} \Delta T_1+(l_{11}-c_{11})\Delta T_2} +e^{(l_{11}-c_{11})\Delta T_2})]l_{11}^{-1} =-1. \\ \end{array} \end{aligned}
\begin{aligned} \begin{array}{l} -c_{11} [e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)}-1]^{-1} (-e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})\Delta T_2} \\ +e^{l_{11} \Delta T_3+(l_{11}-c_{11})\Delta T_2} +e^{l_{11} \Delta T_3} -1)]l_{11}^{-1} =-1, \\ -c_{11} [e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)}-1] ^{-1} (-e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})\Delta T_4}\\ +e^{l_{11} \Delta T_3+(l_{11}-c_{11})(\Delta T_4+\Delta T_2)} +e^{l_{11}\Delta T_3+(l_{11}-c_{11})\Delta T_4} -e^{(l_{11}-c_{11})\Delta T_4})] l_{11}^{-1} =1. \end{array} \end{aligned}
(3.18)
We know that (3.16) may be equivalent with the following question if
\begin{aligned} \begin{array}{l} e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)}-1 \end{array} \end{aligned}
(3.19)
is equal to 0.

Since A, $$A-BK$$ are Hurwitz matrix, $$l_{11}$$, $$l_{11}-c_{11}$$ are obviously negative.

Therefore, for any positive number $$\Delta T_1$$, $$\Delta T_2$$, $$\Delta T_3$$, $$\Delta T_1$$, there must be
\begin{aligned} \begin{array}{l} e^{l_{11}(\Delta T_1+\Delta T_3)+(l_{11}-c_{11})(\Delta T_2+\Delta T_4)}-1<0. \end{array} \end{aligned}
(3.20)
Therefore, we only need to study the case of (3.18).
Also, we know that $$c_{11}, l_{11} \in R^1$$ from the above results, and suppose
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{rcl} e^{l_{11}\Delta T_1} &{} :=&{} x_1, \\ e^{l_{11}\Delta T_3}&{} :=&{} x_3, \\ e^{(l_{11}-c_{11})\Delta T_2}&{} :=&{} x_2, \\ e^{(l_{11}-c_{11})\Delta T_4}&{} :=&{} x_4. \end{array} \right. \end{array} \end{aligned}
(3.21)
Then, (3.18) may be changed into the following equations
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{c} \frac{l_{11}}{c_{11}}x_1x_2x_3x_4+x_1x_3x_4 -x_1x_4+(1-\frac{l_{11}}{c_{11}})=0,\\ (1-\frac{l_{11}}{c_{11}})x_1x_2x_3x_4-x_1x_2x_4-x_1x_2+x_2=0,\\ -\frac{l_{11}}{c_{11}}x_1x_2x_3x_4+x_1x_3x_4 -x_1x_2x_3+x_2x_3+x_3-(1-\frac{l_{11}}{c_{11}})=0, \\ \frac{l_{11}}{c_{11}}x_1x_2x_3x_4+x_1x_3x_4 -x_1x_3x_4+x_2x_3x_4+x_3x_4-x_4=0. \end{array} \right. \end{array} \end{aligned}
(3.22)
whether there exist a group of solutions $$(x_1, x_2, x_3, x_4)$$, $$0<x_i <1, i = 1,2,3,4$$.
Further, (3.22) can also be transformed into the following higher order equations
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{c} \frac{l_{11}}{c_{11}}x^4+x^3 -x^2+(1-\frac{l_{11}}{c_{11}})=0,\\ (1-\frac{l_{11}}{c_{11}})x^4-x^3-x^2+x=0,\\ -\frac{l_{11}}{c_{11}}x^4+x^2+x-(1-\frac{l_{11}}{c_{11}})=0, \\ \frac{l_{11}}{c_{11}}x^4+x^3+x^2-x=0, \end{array} \right. \end{array} \end{aligned}
(3.23)
whether there exists a common positive number solution x whose less than 1.
If we suppose that
\begin{aligned} \begin{array}{l} x=e^{l_{11}\Delta T}=0, \end{array} \end{aligned}
or
\begin{aligned} \begin{array}{l} x=e^{(l_{11}-c_{11})\Delta T}= 0. \end{array} \end{aligned}
And taking into account the following facts
\begin{aligned} \begin{array}{l} \max Re(\lambda (A))<0, \max Re(\lambda (A-BK))<0, \end{array} \end{aligned}
then there must be
\begin{aligned} \begin{array}{l} \Delta T=\infty . \end{array} \end{aligned}
While the closed trajectory requires
\begin{aligned} \begin{array}{l} \Delta T<\infty , \end{array} \end{aligned}
so we conclude that
\begin{aligned} \begin{array}{l} x=e^{l_{11}\Delta T}\ne 0, x=e^{(l_{11}-c_{11})\Delta T} \ne 0. \end{array} \end{aligned}
Therefore, (3.23) is equivalent to the following problem
\begin{aligned} \begin{array}{l} \left\{ \begin{array}{c} \frac{l_{11}}{c_{11}}x^4+x^3-x^2+(1-\frac{l_{11}}{c_{11}})=0,\\ (1-\frac{l_{11}}{c_{11}})x^3-x^2-x+1=0,\\ -\frac{l_{11}}{c_{11}}x^4+x^2+x-(1-\frac{l_{11}}{c_{11}})=0, \\ \frac{l_{11}}{c_{11}}x^3+x^2+x-1=0. \end{array} \right. \end{array} \end{aligned}
(3.24)
whether there exists a common root which is a positive number less than 1.
(By contradiction) If there exists a positive root which is less than 1, then, for (3.24), the first equation plus the third equation, we obtain
\begin{aligned} \begin{array}{l} x^3+x=0. \end{array} \end{aligned}
Similarly, the second equation plus the fourth equation, we obtain
\begin{aligned} \begin{array}{l} x^3=0. \end{array} \end{aligned}
Obviously, it is impossible. So, (3.24) does not exist any positive root which is less than 1.    $$\square$$

To sum up the above discussions, we outline the following results.

## Theorem 3.5

For (3.1), suppose A, $$A-BK$$ are Hurwitz, and $$ABK=BKA$$, then

(1) There do not exist any closed trajectory for (3.1);

(2) The zero solution of the system (3.1) is globally asymptotically stable.

Based on Theorems 3.1, 3.2 and 3.3, we can be obtain the above conclusions.

## 3.5 Example and Simulation

Consider the following second-order single-saturated input control system
\begin{aligned} \left\{ \begin{array}{rcl} \dot{y}_1(t)&{}=&{}-y_1(t)+2y_2(t)+u_s,\\ \dot{y}_2(t)&{}=&{}-3y_2(t),\\ u_s&{}=&{} sat(-0.5y_1(t)-y_2(t)), \end{array} \right. \end{aligned}
(3.25)
where saturated input $$u_s$$ is defined as
\begin{aligned} sat(-0.5y_1(t)-y_2(t))=\left\{ \begin{array}{lr} 1, &{} {if}\,\, -0.5y_1(t)-y_2(t) \ge 1,\\ -0.5y_1(t)-y_2(t),&{} {if}\,\,|-0.5y_1(t)-y_2(t)|<1,\\ -1,&{} {if}\,\, -0.5y_1(t)-y_2(t) \le -1. \end{array} \right. \end{aligned}
Clearly, A,B,K satisfy the commutative condition and A, $$A-BK$$ are Hurwitz stable. According to the conditions of Theorem 3.4, the zero solution of (3.25) is GAS. Through the following simulation, we illustrate the effectiveness of the above conclusions.
(1) Figures 3.3 and 3.4 show phase portrait $$x_2 (x_1)$$ and trajectories $$x_1 (t)$$, $$x_2 (t)$$ passing initial the points (5, 5) ($$-0.5*5-5)=-7.5\le -1$$), which denote that the zero solution of the system (3.25) is asymptotically stable.
(2) Figures 3.5 and 3.6 show phase portrait $$x_2 (x_1)$$ and trajectories $$x_1 (t)$$, $$x_2 (t)$$ passing initial the points $$(-5,-5)$$($$-0.5*5-(-5))=2.5\ge 1$$), which denote that the zero solution of the system (3.25) is asymptotically stable.
(3) Figures 3.7 and 3.8 show phase portrait $$x_2 (x_1)$$ and trajectories $$x_1 (t)$$, $$x_2 (t)$$ passing initial the points $$(5,-2)$$($$-0.5*5-(-2))=-0.5\le 1$$), which denote the zero solution of the system (3.25) is asymptotically stable.

## 3.6 Conclusion

We have introduced the spatial structure of control systems with saturated inputs and described its equations by $$0-1$$ structure. Commutative matrices of MIMO linear systems are considered, and the existence of the feedback matrices of commutative state matrix set in the MIMO closed loops is reduced to invariance subspace of matrix A, and the existence of feedback matrices in the open-loop systems is equivalent to the existence of the solution of Lyapunov matrix equation. The relationship among equilibrium points is discussed for linear system with single-saturated input under commutative matrices. The new method for control system with saturation inputs is presented. The sufficient conditions on the asymptotic stability of the origin of the linear system in the presence of a single saturation input is presented, and the existence of closed trajectory is for the same control systems also converted into the existence of solution for the algebraic equations.

## Notes

### Acknowledgements

This work is Supported by National Key Research and Development Program of China (2017YFF0207400).

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