## 2 System Description and Preliminaries
### Physical Model of Flexible Robotic Arm
\right] \tag{4}\]  
where matrix \(\mathbf{T}_{m\times m}=(\mathbf{T}_{ij})\) satisfies \(\mathbf{T}_{ij}<0,\mathbf{T}_{ij}\geq 0,i\
eq j\), the non-negative column vector \(\mathbf{T}^{0}\) meets \(\mathbf{T}e+\mathbf{T}^{0}=0\). While \((\mathbf{a},m+1)\) is represented as the initial distribution vector, where \(\mathbf{a}=(a_{1},a_{2},\ldots,a_{m}),\mathbf{a}e+a_{m+1}=1\), and \(e\) stands for an m-dimensional column vector made up only of \(1\). Before moving on, let us review the following fundamental claims and definitions:  
**Proposition 1**. : _The rate at which \(\hat{r}(t)\) is absorbed in \(\hat{r}(t)\) is distributed as_  
\[F(t)\;=\;1-\mathbf{a}\exp(\mathbf{T}t)e,\quad t\geq 0. \tag{5}\]  
**Definition 1**. : _The phase of the distribution \(F(\cdot)\) at time \(t\) is the state that \(\hat{r}(t)\) reaches at that instant. Its representation of order \(m\) is \((\mathbf{a},e)\), and the distribution \(F(\cdot)\) described in version (5) on \([0,\infty)\) is known as a phase-type \((PH)\) distribution._  
If \(\mathcal{G}\) is a finite set, then, Markov chain is the denumerable phase semi-Markov process, where \(F_{i}(t)(i\in\mathcal{G})\) has a negative exponential distribution. However, the characteristic of Markov chain that the sojourn time obey negative exponential distribution can be overcome by a denumerable phase semi-Markov process. The main issue is whether or not the denumerable phase semi-Markov process can be changed into a Markov chain. Moreover, a finite Markov chain can be created from a finite phase semi-Markov process. The remainder of this paragraph will provide evidence supporting the validity of the aforementioned statement. Let \(\hat{r}(t)\) be a denumerable phase semi-Markov process on the state space \(\mathcal{G}\). Denote the \(n\)th jump point of process \(\hat{r}(t)\) by \(l_{n}(n=0,1,2,\cdots)\), where \(\iota_{0}\equiv 0<\iota_{1}<\iota_{2},\cdots,<\iota_{n},\cdots\). Let \((\mathbf{a}^{i},\mathbf{T}^{i})(i\in\mathcal{G})\) stand for the representation of \(F_{i}(t)\) in the \(m  
**Definition 2**. : _The pair of the semi-Markov process \(\hat{r}(t)\) is referred to as \(\{P,(\mathbf{a},\mathbf{T})\}\). Define \(\chi(t)=\) the phase of \(F_{P(t)}(\cdot)\) at time \(t-t_{n}\) for every \(n(n=0,1,\cdots)\), \(t_{n}\leq t<t_{n+1}\). For any \(i\in\mathcal{G}\), we define_  
\[\mathbf{T}_{j}^{i,0} = 1-\sum_{k=1}^{m(i)}\mathbf{T}_{jk}^{i}\quad(j=1,2,\cdots,m^{i}),\] \[\mathcal{U} = \{(i,k^{i})\mid i\in\mathcal{G},k^{i}=1,2,\cdots,m^{i}\}.\]  
We can readily obtain the following conclusions from the aforementioned analysis. **Lemma 1**. : _Markov chain \(Z(t)=(\hat{r}(t),\chi(t))\) has state space \(\mathcal{U}\). The pair of \(t(t)\) given by \(P,(\mathbf{a},\mathbf{T})\) determines the infinitesimal generator of \(Z(t)\) exclusively, as shown in the following_  
\[\left\{ \begin{array}{ll}\beta_{(i\dot{k})(i\dot{k})}&=\ \mathbf{T}_{\dot{k} \dot{k}^{i}}^{i},\quad(i,k^{i}\in\mathcal{U}),\\ \beta_{(i\dot{k})(i\dot{k})}&=\ \mathbf{T}_{\dot{k}\dot{k}^{i}}^{i},\quad k^{i} \
eq\dot{k},(i,k^{i})\in\mathcal{U}\text{ and }(i,\dot{k}^{i})\in\mathcal{U},\\ \beta_{(i\dot{k}^{i})(j\dot{k}^{i})}&=\ \lambda_{ij}\mathbf{T}_{\dot{k}^{i}}^{(i 0)}a_{\dot{k}^{j}}^{i},\quad i\
eq j,(i,k^{i})\in\mathcal{U}\text{ and }(j,k^{i})\in\mathcal{U}.\end{array} \right.\]  
Suppose that \(\mathcal{U}\) has \(s=\sum_{i\in\mathcal{G}}m^{i}\) elements, which results in \(s\) elements in the state space of \(Z(t)\). The \(s\) elements are numbered using the following procedure: the number of \((i,k)\) by \(\sum_{r=1}^{i-1}m^{r}+k(1\leq k\leq m^{i})\). Adding the letter \(\psi\) to this transformation as well, and one has  
\[\psi((i,k)) = \sum_{r=1}^{i-1}m^{r}+k\quad(i\in\mathcal{G},1\leq k\leq m^{i}). \tag{8}\]  
Furthermore, we define  
\[v_{\psi((i,k))\psi((i^{\prime},k^{\prime}))} = \beta_{(i\dot{k})(i^{\prime},k^{\prime})}, \tag{9}\]  
\[r(t)=\psi(Z(t)).\]  
The infinitesimal generator \(\Xi=(v_{im},1\leq i,m\leq s)\) with a state space \(S=\{1,2,\cdots,s\}\) make \(r(t)\) be a Markov chain.