### cs代写|复杂网络代写complex network代考|Cooperative Control of Complex Network Systems with Dynamic Topologies

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## cs代写|复杂网络代写complex network代考|CNSS WITH STATIC COUPLING AND SWITCHING TOPOLOGIES

This section studies the consensus disturbance rejection problem for CNSs under directed switching topologies. Before moving forward, the definition of consensus disturbance rejection is given.

Definition 4.1 The consensus disturbance rejection of CNSs (4.1) and (4.3) with disturbances generated by (4.2) is said to be achieved if
$$\ lim {t \rightarrow \infty}\left|x{i}(t)-x_{0}(t)\right|=0, \lim {t \rightarrow \infty}\left|\hat{d}{i}(t)-d_{i}(t)\right|=0,$$
hold for arbitrary initial values $x_{i}\left(t_{0}\right), x_{0}\left(t_{0}\right), \hat{d}{i}\left(t{0}\right), d_{i}\left(t_{0}\right), i=1, \ldots, N$.
Theorem 4.2 Suppose Assumptions 4.1-4.3 hold. If the ADT $\tau_{a}>\ln \nu$, then the consensus disturbance rejection of CNSs (4.1) and (4.3) with the disturbances generated by (4.2) can be achieved by adopting the consensus error estimator (4.5), the state estimator (4.9), and the disturbance observer (4.10) based controller (4.11) with $K=-B^{T} P^{-1}, Q=\mu R^{-1} D^{T}, \rho \geq 4 \alpha / \lambda_{0}, \mu \geq 4 / \lambda_{0}$, where $\alpha$ is a positive constant, $\lambda_{0}$ is given by (4.4), $P>0$ and $R>0$ are, respectively, obtained by solving the LMIs (4.18) and (4.19),
$$\begin{gathered} A P+P A^{T}-\alpha B B^{T}+P<0, \ W^{T} R+R W-D^{T} D+2 R<0 . \end{gathered}$$
Proof $4.2$ For any $t \in\left[t_{j}, t_{j+1}\right), j=0,1,2, \ldots$, we construct the following $M L F s$
$$V_{1}(t)=V_{11}(t)+V_{12}(t)+V_{13}(t)+V_{14}(t),$$
where
\begin{aligned} &V_{11}(t)=\zeta^{T}(t)\left(I_{N} \otimes P^{-1}\right) \zeta(t), \ &V_{12}(t)=\frac{\gamma_{1}}{2} \sum_{i=1}^{N} \phi_{i}^{\sigma(t)}\left(2+\varrho_{i}(t)\right) \varrho_{i}(t), \ &V_{13}(t)=\gamma_{1} \gamma_{2} \tilde{d}^{T}(t)\left(\Phi^{\sigma(t)} \otimes R\right) \tilde{d}(t), \ &V_{14}(t)=\gamma_{1} \gamma_{3} \tilde{\delta}^{T}(t)\left(\Phi^{\sigma(t)} \otimes S\right) \tilde{\delta}(t), \end{aligned}

## cs代写|复杂网络代写complex network代考|CNSS WITH DYNAMIC COUPLING AND FIXED TOPOLOGY

We could learn from Theorem $4.2$ that the coupling strength $\rho$ depends on the smallest eigenvalue $\lambda_{0}$ which is a global information associated with all the possible communication graphs. Consequently, the controller (4.11) can not be implemented in a distributed way. Motivated by this observation, we give a new state estimator with dynamic coupling strengths upon which a fully distributed controller can be reconstructed. While, unlike the last subsection, the directed topology of the CNSs considered in this subsection is assumed to be fixed. The state estimator is given as follows.
\begin{aligned} \dot{\hat{\xi}}{i}(t) &=A \hat{\xi}{i}(t)+\alpha B K \hat{\xi}{i}(t)+\left(\rho{i}+\varrho_{i}\right) B K\left(\hat{\zeta}{i}(t)-\hat{\delta}{i}(t)\right) \ \dot{\rho}{i} &=\left(\hat{\zeta}{i}(t)-\hat{\delta}{i}(t)\right)^{T} \Theta\left(\hat{\zeta}{i}(t)-\hat{\delta}{i}(t)\right) \ \varrho{i} &=\left(\hat{\zeta}{i}(t)-\hat{\delta}{i}(t)\right)^{T} P^{-1}\left(\hat{\zeta}{i}(t)-\hat{\delta}{i}(t)\right) \end{aligned}
where $\hat{\zeta}{i}(t)=\sum{j=1}^{N} a_{i j}\left(\hat{\xi}{i}(t)-\hat{\xi}{j}(t)\right)+a_{i 0} \hat{\xi}{i}(t), \Theta=P^{-1} B B^{T} P^{-1}, P>0$ will be given later, and the initial value $\rho{i}\left(t_{0}\right)>0$. Based on the estimator (4.34), the disturbance observer and the controller are then given by (4.35) and (4.36), respectively.
$$\begin{gathered} \hat{d}{i}(t)=z{i}(t)+Q \hat{\delta}{i}(t) \ \dot{z}{i}(t)=W z_{i}(t)+(W Q-Q A) \hat{\delta}{i}(t)-\alpha Q B K \hat{\zeta}{i}(t) \ u_{i}(t)=\alpha K \hat{\xi}{i}(t)-E \hat{d}{i}(t) \end{gathered}$$
By using the same analyses to those presented in Section $4.2$, we get
\begin{aligned} \dot{\hat{\delta}}(t)=&\left(I_{N} \otimes A\right) \hat{\delta}(t)+\alpha\left(I_{N} \otimes B K\right) \hat{\zeta}(t) \ &-(\overline{\mathcal{L}} \otimes D) \tilde{d}(t)+\left[I_{N} \otimes(G A-F C-A)\right] \tilde{\delta}(t) \end{aligned}$\dot{\tilde{\delta}}(t)=\left[I_{N} \otimes(G A-F C)\right] \tilde{\delta}(t)$, $\dot{\tilde{d}}(t)=\left(I_{N} \otimes W\right) \tilde{d}(t)-(\overline{\mathcal{L}} \otimes Q D) \tilde{d}(t)+\left[I_{N} \otimes Q(G A-F C-A)\right] \tilde{\delta}(t)$ $\dot{\hat{\zeta}}(t)=\left[I_{N} \otimes(A+\alpha B K)\right] \hat{\zeta}(t)+\overline{\mathcal{L}}(\rho+\varrho) \otimes B K)$ where $\hat{\zeta}(t)=\left[\hat{\zeta}{1}^{T}(t), \ldots, \hat{\zeta}{2}^{T}(t)\right]^{T}, \rho=\operatorname{diag}\left{\rho_{1}, \ldots, \rho_{N}\right}$, and the other symbols are the same as those defined in Section 4.2. the same as those defined in Section 4.2.

## cs代写|复杂网络代写complex network代考|NUMERICAL SIMULATIONS

We perform two examples to validate Theorems $4.2$ and $4.3$, respectively. The CNSs under consideration consist of five YF-22 research UAVs [34] whose longitudinal

dynamics satisfy (4.1) with
$$A=\left[\begin{array}{cccc} -0.284 & -23.096 & 2.420 & 9.913 \ 0 & -4.117 & 0.843 & 0.272 \ 0 & -33.884 & -8.263 & -19.543 \ 0 & 0 & 1 & 0 \end{array}\right]$$
$$B=\left[\begin{array}{c} 20.168 \ 0.544 \ -39.085 \ 0 \end{array}\right], D=B\left[\begin{array}{ll} 1 & 0 \end{array}\right], C=\left[\begin{array}{llll} 1 & 1 & 0 & 0 \ 0 & 0 & 1 & 1 \end{array}\right] \text {, }$$
where $x_{i}(t)=\left[x_{i 1}(t), x_{i 2}(t), x_{i 3}(t), x_{i 4}(t)\right]^{T}$ and $x_{i 1}(t), x_{i 2}(t), x_{i 3}(t), x_{i 4}(t)$ represent, respectively, the speed, the attack angle, the pitch rate, and the pitch angle, $i=0,1, \ldots, 4$. The harmonic disturbances are generated by (4.2) with $d_{i}(t)=$ $\left[d_{i 1}(t), d_{i 2}(t)\right]^{T}$ and
$$W=\left[\begin{array}{cc} 0 & 1.5 \ -1.5 & 0 \end{array}\right]$$
It is not difficult to verify that Assumptions 4.1, 4.2 and the conditions (1) and (2) in Remark $4.2$ hold. Then, we get
\begin{aligned} H &=\left[\begin{array}{cccc} -0.2135 & -0.0058 & 0.4137 & 0 \ 0.4029 & 0.0109 & -0.7808 & 0 \end{array}\right]^{T}, \ G &=\left[\begin{array}{cccc} 0.7865 & -0.2135 & 0.4029 & 0.4029 \ -0.0058 & 0.9942 & 0.0109 & 0.0109 \ 0.4137 & 0.4137 & 0.2192 & -0.7808 \ 0 & 0 & 0 & 1 \end{array}\right] . \end{aligned}
Solving the LMI (4.8) gives that
$$F=\left[\begin{array}{rrrr} 11.3220 & 1.5912 & 5.9677 & 0.0839 \ 5.2910 & 0.7606 & 3.1369 & 0.3498 \end{array}\right]^{T} .$$

## cs代写|复杂网络代写complex network代考|CNSS WITH STATIC COUPLING AND SWITCHING TOPOLOGIES

$$\ lim {t \rightarrow \infty}\left|x{i}(t) – x_{0}(t)\right|=0, \lim {t \rightarrow \infty}\left|\hat{d}{i}(t)-d_{i}(t)\right|=0, H○ldF○r一个rb一世吨r一个r是一世n一世吨一世一个l在一个l在和sX一世(吨0),X0(吨0),d^一世(吨0),d一世(吨0),一世=1,…,ñ.吨H和○r和米4.2小号在pp○s和一个ss在米p吨一世○ns4.1−4.3H○ld.我F吨H和一个D吨τ一个>ln⁡ν,吨H和n吨H和C○ns和ns在sd一世s吨在rb一个nC和r和j和C吨一世○n○FCñ小号s(4.1)一个nd(4.3)在一世吨H吨H和d一世s吨在rb一个nC和sG和n和r一个吨和db是(4.2)C一个nb和一个CH一世和在和db是一个d○p吨一世nG吨H和C○ns和ns在s和rr○r和s吨一世米一个吨○r(4.5),吨H和s吨一个吨和和s吨一世米一个吨○r(4.9),一个nd吨H和d一世s吨在rb一个nC和○bs和r在和r(4.10)b一个s和dC○n吨r○ll和r(4.11)在一世吨Hķ=−乙吨磷−1,问=μR−1D吨,ρ≥4一个/λ0,μ≥4/λ0,在H和r和一个一世s一个p○s一世吨一世在和C○ns吨一个n吨,λ0一世sG一世在和nb是(4.4),磷>0一个ndR>0一个r和,r和sp和C吨一世在和l是,○b吨一个一世n和db是s○l在一世nG吨H和大号米我s(4.18)一个nd(4.19), 一个磷+磷一个吨−一个乙乙吨+磷<0, 在吨R+R在−D吨D+2R<0. 磷r○○F4.2F○r一个n是吨∈[吨j,吨j+1),j=0,1,2,…,在和C○ns吨r在C吨吨H和F○ll○在一世nG米大号Fs V_{1}(t)=V_{11}(t)+V_{12}(t)+V_{13}(t)+V_{14}(t)， 在H和r和 在11(吨)=G吨(吨)(我ñ⊗磷−1)G(吨), 在12(吨)=C12∑一世=1ñφ一世σ(吨)(2+ϱ一世(吨))ϱ一世(吨), 在13(吨)=C1C2d~吨(吨)(披σ(吨)⊗R)d~(吨), 在14(吨)=C1C3d~吨(吨)(披σ(吨)⊗小号)d~(吨),$$

## cs代写|复杂网络代写complex network代考|CNSS WITH DYNAMIC COUPLING AND FIXED TOPOLOGY

X^˙一世(吨)=一个X^一世(吨)+一个乙ķX^一世(吨)+(ρ一世+ϱ一世)乙ķ(G^一世(吨)−d^一世(吨)) ρ˙一世=(G^一世(吨)−d^一世(吨))吨θ(G^一世(吨)−d^一世(吨)) ϱ一世=(G^一世(吨)−d^一世(吨))吨磷−1(G^一世(吨)−d^一世(吨))

d^一世(吨)=和一世(吨)+问d^一世(吨) 和˙一世(吨)=在和一世(吨)+(在问−问一个)d^一世(吨)−一个问乙ķG^一世(吨) 在一世(吨)=一个ķX^一世(吨)−和d^一世(吨)

d^˙(吨)=(我ñ⊗一个)d^(吨)+一个(我ñ⊗乙ķ)G^(吨) −(大号¯⊗D)d~(吨)+[我ñ⊗(G一个−FC−一个)]d~(吨)d~˙(吨)=[我ñ⊗(G一个−FC)]d~(吨), d~˙(吨)=(我ñ⊗在)d~(吨)−(大号¯⊗问D)d~(吨)+[我ñ⊗问(G一个−FC−一个)]d~(吨)$\dot{\hat{\zeta}}(t)=\left[I_{N} \otimes(A+\alpha BK)\right] \hat{\zeta}(t)+ \overline{\mathcal{L }}(\rho+\varrho) \otimes BK )在H和r和\hat{\zeta}(t)=\left[\hat{\zeta}{1}^{T}(t), \ldots, \hat{\zeta}{2}^{T}(t)\ right]^{T}、\rho=\operatorname{diag}\left{\rho_{1}、\ldots、\rho_{N}\right}$，其他符号同4.2节定义. 与第 4.2 节中定义的相同。

## cs代写|复杂网络代写complex network代考|NUMERICAL SIMULATIONS

H=[−0.2135−0.00580.41370 0.40290.0109−0.78080]吨, G=[0.7865−0.21350.40290.4029 −0.00580.99420.01090.0109 0.41370.41370.2192−0.7808 0001].

F=[11.32201.59125.96770.0839 5.29100.76063.13690.3498]吨.

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