【Paper】2019_ Consensus Control of Multiple AUVs Recovery System Under Switching Topologies and Time D

Zhang W, Zeng J, Yan Z, et al. Consensus control of multiple AUVs recovery system under switching topologies and time delays[J]. IEEE Access, 2019, 7: 119965-119980.

link ：【Paper】2017_ Research on coordinated control method of underwater vehicle formation marine survey

I. INTRODUCTION

II. PROBLEM FORMULATION

A. GRAPH THEORY

B. THE AUV MODEL

combination Eq.(2.7) Sum formula (2.8), available AUV A feedback linearized dynamic model in standard second-order integral form :
$$\begin{array}{r}{\dot{x}}_i=v_i\\ {\dot{v}}_i=u_i\end{array}\text{\hspace{1em}(2.9)}$$

among ,$x_i\in {\mathbb{R}}^5,v_i\in {\mathbb{R}}^5,u_i\in {\mathbb{R}}^5$.

C. LEMMA

III. CONSENSUS CONTROL UNDER SWITCHING TOPOLOGIES AND TIME DELAYS

As mentioned earlier , Next, we will introduce the switching topology and the multi - node scheme with time delay AUV Restore the consistency control algorithm of the system . Mother ship and many AUV Run in the form of random switching topology . Suppose the interconnection topology is Markov switching , Markov chain describing switching process has a stationary probability distribution . Since the graph is allowed to be time-varying , We assume that there is $M$ Two possible different graphs , The network topology switches between them .

A. DESIGN PROCEDURE

In this section , Many assumptions AUV The consensus control algorithm of the recovery system can ensure that all members reach the common motion state and expectation point . The mother ship is considered to be all AUV The point that must be pursued , Multiple AUV The recovery problem of can also be regarded as the consensus problem of multi member systems .

Definition 1

In a AUV In a system composed of , The first $i$ individual AUV In time $t$ The motion state vector of is $x_i(t)$, The motion state of the mother ship is $x_m(t)$. If the system satisfies the following formula , The recycling system can achieve consistency , And guarantee that AUV Be able to follow the mother ship continuously and stably .

$$\underset{t\to \infty }{\lim }=\Vert x_i(t)-x_m(t)\Vert =0$$

According to dynamics （2.9）, The consistency algorithm considering mother ship dynamics is
$${\dot{x}}_m(t)=v_m(t)$$

Define a group of independent topologies auv The dynamics of is as follows :
$$\begin{array}{r}{\dot{x}}_i(t)=v_i(t)\\ {\dot{v}}_i(t)=u_i(t)\end{array}$$

Consider a reason for AUV Composed of many AUV Recycling system , Each one AUV Treat as a directed graph $G$ One of the nodes in , Utilization （2.9） in AUV Standard dual integrator dynamics , Based on the consensus algorithm, the following consensus control can be designed :

Define a group of independent topologies auv The dynamics of is as follows :
$$\begin{array}{rl}{u}_i(t)=& -K\sum _{j\in N_i}{a}_{ij}(t)((p_i(t)-p_j(t))+(v_i(t)-v_j(t)))\\ & -K{c}_i(t)((p_i(t)-p_m(t))+(v_i(t)-v_m(t)))\end{array}\text{\hspace{1em}(3.1)}$$

Due to limited communication , It is easy to cause delay . A consensus algorithm with delay is given
$$\begin{array}{rl}{u}_i(t)=& -K\sum _{j\in N_i}{a}_{ij}(t)((p_i(t-{\tau }_1)-p_j(t-{\tau }_1-{\tau }_2))+(v_i(t-{\tau }_1)-v_j(t-{\tau }_1-{\tau }_2)))\\ & -K{c}_i(t)((p_i(t-{\tau }_1)-p_m(t))+(v_i(t-{\tau }_1)-v_m(t)))\end{array}\text{\hspace{1em}(3.2)}$$

$u_i$： Is the control input
$K$： Yes protocol gain
$p_i\in {\mathbb{R}}^n$： Is the position state
$v_i\in {\mathbb{R}}^n$： Is the speed state
$p_m\in {\mathbb{R}}^n$： Is the mother ship position status
$v_m\in {\mathbb{R}}^n$： Is the speed status of the mother ship
$a_{ij}(t)$： Communication weight
$c_m(t)$： Communication weight with mother ship
$C=\text{diag}\{c_1,c_2,\cdots ,c_N\}$
${\tau }_1,{\tau }_2(t)$： Respectively represent the input （ Constant ） And communication weights （ Time varying in the interval ）

Mother ship and the $i$ individual AUV The status of is
$$\begin{array}{r}{x}_m(t)={\left[\begin{array}{cc}{p}_m^{\text{T}}(t)& v_m^{\text{T}}(t)\end{array}\right]}^{\text{T}}\in {\mathbb{R}}^{10n}\\ x_i(t)={\left[\begin{array}{cc}{p}_i^{\text{T}}(t)& v_i^{\text{T}}(t)\end{array}\right]}^{\text{T}}\in {\mathbb{R}}^{10n}\end{array}$$

The above recovery problem is transformed into an error analysis problem , Definition No $i$ individual AUV The system state error vector relative to the motion state information of the Mothership is
${\epsilon }_i(t)=x_i(t)-x_m(t)$
Definition
$\epsilon (t)=({\epsilon }_1^{\text{T}}(t),{\epsilon }_2^{\text{T}}(t),\cdots ,{\epsilon }_N^{\text{T}}(t){)}^{\text{T}}$,
$\epsilon (t)={\left[\begin{array}{cc}{\epsilon }_p^{\text{T}}(t)& {\epsilon }_v^{\text{T}}(t)\end{array}\right]}^{\text{T}}$
here
${\epsilon }_p(t),{\epsilon }_v(t)$ Respectively represent the error of position state and speed state .

Then the system can be transformed into the following matrix form
$$\dot{\epsilon }(t)=(I_n\otimes A)\epsilon (t)-B\otimes K\epsilon (t-{\tau }_1)+C\otimes K\epsilon (t-\mu )\text{\hspace{1em}(3.3)}$$

The original formula is as follows , But there's a problem , The revised ones are as follows ：

$$\dot{\epsilon }(t)=(I_n\otimes A)\epsilon (t)-(I_n\otimes K_{i}B)\epsilon (t-{\tau }_1)+(I_n\otimes K_{i}C)\epsilon (t-\mu )\text{\hspace{1em}(3.3)}$$

here
$\mu ={\tau }_1+{\tau }_2$
$A=\left[\begin{array}{cc}0& I\\ 0& 0\end{array}\right]\in {\mathbb{R}}^{10\times 10}$
$B=\left[\begin{array}{cc}0& 0\\ \mathcal{D}+\mathcal{C}& \mathcal{D}+\mathcal{C}\end{array}\right]\in {\mathbb{R}}^{10\times 10}$
$C=\left[\begin{array}{cc}0& 0\\ \mathcal{A}& \mathcal{A}\end{array}\right]\in {\mathbb{R}}^{10\times 10}$
$\mathcal{A}=\left[\begin{array}{c}{a}_{ij}\end{array}\right]\in {\mathbb{R}}^{5\times 5}$
$\mathcal{C}=\text{diag}\{c_1,c_2,\cdots ,c_N\}\in {\mathbb{R}}^{5\times 5}$
$\mathcal{D}=\text{diag}\{{\sum }_{j\in N_1}{a}_{1j},{\sum }_{j\in N_2}{a}_{2j},\cdots ,{\sum }_{j\in N_N}{a}_{Nj}\}\in {\mathbb{R}}^{5\times 5}$

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Expand it for easy understanding ：
$$\begin{array}{rl}{\dot{\epsilon }}_1(t)& =A{\epsilon }_1-K_{i}B{\epsilon }_1+K_{i}C{\epsilon }_1\\ & =\left[\begin{array}{cccccccccc}0& 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ & & & & & & & & & \\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{p}_1^x-p_m^x\\ p_1^y-p_m^y\\ p_1^z-p_m^z\\ p_1^{\theta }-p_m^{\theta }\\ p_1^{\psi }-p_m^{\psi }\\ \\ v_1^u-v_m^u\\ v_1^v-v_m^v\\ v_1^w-v_m^w\\ v_1^q-v_m^q\\ v_1^r-v_m^r\end{array}\right]\\ & -K_i\left[\begin{array}{cccccccccc}0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ & & & & & & & & & \\ c_1+d_1& 0& 0& 0& 0& c_1+d_1& 0& 0& 0& 0\\ 0& c_2+d_2& 0& 0& 0& 0& c_2+d_2& 0& 0& 0\\ 0& 0& c_3+d_3& 0& 0& 0& 0& c_3+d_3& 0& 0\\ 0& 0& 0& c_4+d_4& 0& 0& 0& 0& c_4+d_4& 0\\ 0& 0& 0& 0& c_5+d_5& 0& 0& 0& 0& c_5+d_5\end{array}\right]\left[\begin{array}{c}{p}_1^x-p_m^x\\ p_1^y-p_m^y\\ p_1^z-p_m^z\\ p_1^{\theta }-p_m^{\theta }\\ p_1^{\psi }-p_m^{\psi }\\ \\ v_1^u-v_m^u\\ v_1^v-v_m^v\\ v_1^w-v_m^w\\ v_1^q-v_m^q\\ v_1^r-v_m^r\end{array}\right]\\ & +K_i\left[\begin{array}{cccccccccc}0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ & & & & & & & & & \\ 0& a_{12}& a_{13}& a_{14}& a_{15}& 0& a_{12}& a_{13}& a_{14}& a_{15}\\ a_{21}& 0& a_{23}& a_{24}& a_{25}& a_{21}& 0& a_{23}& a_{24}& a_{25}\\ a_{31}& a_{32}& 0& a_{34}& a_{35}& a_{31}& a_{32}& 0& a_{34}& a_{35}\\ a_{41}& a_{42}& a_{43}& 0& a_{45}& a_{41}& a_{42}& a_{43}& 0& a_{45}\\ a_{51}& a_{52}& a_{53}& a_{54}& 0& a_{51}& a_{52}& a_{53}& a_{54}& 0\end{array}\right]\left[\begin{array}{c}{p}_1^x-p_m^x\\ p_1^y-p_m^y\\ p_1^z-p_m^z\\ p_1^{\theta }-p_m^{\theta }\\ p_1^{\psi }-p_m^{\psi }\\ \\ v_1^u-v_m^u\\ v_1^v-v_m^v\\ v_1^w-v_m^w\\ v_1^q-v_m^q\\ v_1^r-v_m^r\end{array}\right]\end{array}$$

$${\dot{x}}_1(t)-{\dot{x}}_m(t)=A\epsilon -B\epsilon +C\epsilon$$

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Mother ship and many auv Run as a random switching topology . This paper gives the common probability space of all random variables $(\Omega ,\mathcal{F},\mathcal{P})$,$\Omega$ Is the space for basic events ,$\mathcal{F}$ Is in $\Omega$ On the potential $\sigma$- site ,$\mathcal{P}$ yes $\mathcal{F}$ Probability measure on .

In this paper, we consider using consistency algorithm （3.3） And Markov switching interconnection topology
$$\dot{\epsilon }(t)=(I_n\otimes A)\epsilon (t)-B({\rho }_t)\otimes K({\rho }_t)\epsilon (t-{\tau }_1)+C({\rho }_t)\otimes K({\rho }_t)\epsilon (t-\mu )\text{\hspace{1em}(3.4)}$$

System （3.4） Is a multiple with Markov switching topology and time delay AUVs Restore system consistency control , It reflects the mother ship and AUVs Between or AUVs Random communication process between . Communication is described by Markov random process ,${\rho }_t$ Is defined as a Markov process , In state space $\mathbb{M}=\{1,2,\cdot ,\mathbf{M}\}$ The inner value is . matrix $B({\rho }_t)$ and $C({\rho }_t)$ Random change from one mode to another by Markov jump process , The control gain changes from one mode to another in a Markov switching topology . Protocol gain $K({\rho }_t)$ To ensure the consensus of the recovery system .

B. CONVERGENCE ANALYSIS

IV. NUMERICAL EXAMPLES

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This is using the topology diagram （a） The simulation results

Through the speed state , It can be concluded that there is something wrong with the result in the original text

This is the result of modifying the topology

I feel that there is not only a problem with the result in the original text , And there are problems with the communication topology