# Observer-Based Consensus Control for Heterogeneous Multi-Agent Systems with Output Saturations
## 3 Main Results
### Leaderless Case
\begin{split}\dot{e}_{i}=& T_{i}(\hat{x}_{i}-\dot{ \hat{x}}_{i})\\ =& T_{i}A_{i}(x_{i}-\hat{x}_{i})-T_{i}H_{i}\hat{z}_{i} \\ =&\bar{A}_{i}\dot{e}_{i}-T_{i}H_{i}\hat{z}_{i}, \end{split}  \tag{13}\]  
and, from Assumption 1, the regulation error dynamics is given by  
\[ \begin{split}\dot{e}_{i}=&\dot{\hat{x}}_{i}-\Pi_{i}w _{i}\\ =& A_{i}\hat{x}_{i}+B_{i}(K_{i}(\hat{x}_{i}-\Pi_{i}w _{i})+\Gamma_{i}w_{i})+H_{i}\hat{z}_{i}-\Pi_{i}Sw_{i}-\Pi_{i}F\left(\sum_{j=1} ^{N}\alpha_{ij}(w_{j}-w_{i})\right)\\ =& A_{i}\hat{x}_{i}+B_{i}(K_{i}\epsilon_{i}+\Gamma_{ i}w_{i})+H_{i}\hat{z}_{i}-A_{i}\Pi_{i}w_{i}-B_{i}\Gamma_{i}w_{i}-\Pi_{i}F \left(\sum_{j=1}^{N}\alpha_{ij}(w_{j}-w_{i})\right)\\ =&(A_{i}+B_{i}K_{i})\epsilon_{i}+H_{i}\hat{z}_{i}- \Pi_{i}F\left(\sum_{j=1}^{N}\alpha_{ij}(w_{j}-w_{i})\right).\end{split}  \tag{14}\]  
Let \(e=[e_{1}^{T},...,e_{N}^{T}]^{T}\), \(\epsilon=[e_{1}^{T},...,e_{N}^{T}]^{T}\), \(\hat{x}=[\hat{x}_{1}^{T},...,\hat{x}_{N}^{T}]^{T}\), \(\hat{z}=[\hat{z}_{1}^{T},...,\hat{z}_{N}^{T}]^{T}\), \(w=[w_{1}^{T},...,w_{N}^{T}]^{T}\), \(T=\text{blkdiag}(T_{i})_{i=1}^{N}\), \(A=\text{blkdiag}(A_{i})_{i=1}^{N}\), \(B=\text{blkdiag}(B_{i})_{i=1}^{N}\), \(K=\text{blkdiag}(K_{i})_{i=1}^{N}\), \(H=\text{blkdiag}(H_{i})_{i=1}^{N}\), \(\Pi=\text{blkdiag}(\Pi_{i})_{i=1}^{N}\), and \(\bar{A}=\text{blkdiag}(\bar{A}_{i})_{i=1}^{N}\). Then, from (12), (13), and (14), the overall error dynamics can be written as  
\[ \begin{split}\dot{\xi}=&(I_{N-1}\otimes S-\bar{ \Lambda}\otimes F)\xi,\\ \dot{e}=&\bar{A}e-TH\hat{z}\\ \dot{e}=&(A+BK)\epsilon+H\hat{z}+\Pi(L\otimes F)w. \end{split}  \tag{15}\]  
It is clear that we can achieve the consensus if the error dynamics (15) is asymptotically stable, i.e., if \((\xi,e,\epsilon)\rightarrow(0,0,0)\), then \(y_{i}-y_{j}=0\), \(\forall i,j\in\mathcal{V}\). Then, to prove the consensus, we first analyze the asymptotic convergence of \(\xi\) and \(e\) to the origin, applying Lemma 2. We define the continuously differentiable function\(V\) as follows:  
\[V=\xi^{T}(I_{N-1}\otimes W)\xi+e^{T}Pe, \tag{16}\]  
where \(W\) is the solution of (9) in Lemma 4 and \(P=\text{blkdiag}(P_{i})_{i=1}^{N}\) with \(P_{i}\) given in Lemma 1. Let \(\beta=\text{diag}(\beta_{1},...,\beta_{N})\) and \(\bar{C}=\text{blkdiag}(\bar{C}_{i})_{i=1}^{N}\). Then, the time-derivative of \(V\) is given by  
\[ \begin{split}\dot{V}=& 2\xi^{T}(I_{N-1}\otimes W)\xi ^{+}_{\xi}+2e^{T}Pe\\ =& 2\xi^{T}(I_{N-1}\otimes W)(I_{N-1}\otimes S-\bar{ \Lambda}\otimes F)\xi+2e^{T}P(\bar{A}e-TH\hat{z})\\ =&\xi^{T}(I_{N-1}\otimes WS+S^{T}W-2\bar{\Lambda} \otimes\tau QQ^{T}W)\xi+e^{T}(P\bar{A}+\bar{A}^{T}p)e\\ &-2e^{T}P\bar{P}\bar{P}^{-1}P^{-1}P^{-1}C^{T}\hat{z}\\ =&\xi^{T}(I_{N-1}\otimes WS+S^{T}W-2\tau\bar{\Lambda} \otimes WQQ^{T}W)\xi+e^{T}(P\bar{A}+\bar{A}^{T}p)e-2\beta e^{T}C^{T}\hat{z}\\ =&\sum_{i=2}^{N}\xi_{i}^{T}(WS+S^{T}W-2\tau\lambda_{ i}(L)WQQ^{T}W)\xi_{i}+\sum_{i=1}^{N}e_{i}^{T}(A_{s,i}^{T}+A_{s,i}+A_{h,i}^{T}p_{ h,i}+P_{h,i}A_{h,i})e\\ &-2\sum_{i=1}^{N}\beta_{i}(C_{i}x_{i}-C_{i}\delta_{i})^{T}( \sigma(C_{i}x_{i})-\sigma(C_{i}\delta_{i})).\end{split}  \tag{17}\]Then, from Lemma 1 and 4, and the fact that \(\tau\lambda_{2}(L)>1\), the time-derivative of \(V\) can be rewritten as  
\[ \begin{split}\dot{V}\leq&\sum_{i=2}^{N}\zeta_{i}^{T}( WS+S^{T}W-2 WQQ^{T}W)\zeta_{i}-2\sum_{i=1}^{N}\beta_{i}(C_{i}x_{i}-C_{i}\hat{x}_{i})^{T}( \sigma(C_{i}x_{i})-\sigma(C_{i}\hat{x}_{i}))\\ \leq& 0,\end{split}  \tag{18}\]  
where we have used the fact that, for any \(a,b\in\mathbf{R}\), \(\operatorname{sign}(a-b)=\operatorname{sign}(\sigma(a)-\sigma(b))\). Then, we have shown that \(V\geq 0\) is radially unbounded and \(\dot{V}\leq 0\), which implies that Conditions 1 and 2 in Lemma 2 hold. Then, to prove Condition 3, we define \(\mathcal{M}:=\{(e,\bar{\zeta}):\dot{V}=0\}\). \(\dot{V}=0\) implies \(\bar{\zeta}=0\), i.e., \((w_{i}-w_{j})=0\), \(\forall i,j\in\mathcal{V}\), and \((L\otimes I_{n_{0}})w=0\), and \(\hat{z}_{i}=(\sigma(C_{i}x_{i})-\sigma(C_{i}\hat{x}_{i}))=0\), \(\forall i\in\mathcal{V}\).