# Observer-Based Consensus Control for Heterogeneous Multi-Agent Systems with Output Saturations
## 3 Main Results
### Leader-Following Case
Let \(\zeta=[\zeta_{1}^{T},...,\zeta_{N}^{T}]^{T}\) and \(\bar{L}=L+\text{diag}(\kappa_{10},...,\kappa_{N0})\), and define \(e,e,T,H,A,B,K\), and \(\Pi\) as in the proof of Theorem 1. Then, the overall error dynamics can be written as  
\[ \begin{split}\zeta=&(I_{N}\otimes S-L\otimes F)\zeta \\ \dot{e}=&\bar{A}e-TH\dot{z}\\ \dot{\epsilon}=&(A+BK)\epsilon+H\dot{z}+\Pi(\bar{L }\otimes F)\zeta.\end{split}  \tag{28}\]  
It is clear that we can achieve the leader-following output consensus if the error dynamics (28) is asymptotically stable; i.e., if \((\zeta,e,\epsilon)\rightarrow(0,0,0)\), then \(y_{i}-y_{0}=0\), \(\forall i\in\mathcal{V}\). Then, to prove the consensus, we define the continuously differentiable function\(V\) as follows:  
\[V=\zeta^{T}(I_{N}\otimes W)\zeta+e^{T}Pe, \tag{29}\]  
where \(W\) is the solution of (25) 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  
\[ \begin{split}\dot{V}=& 2\zeta^{T}(I_{N}\otimes W) \zeta+2e^{T}P\dot{e}\\ =& 2\zeta^{T}(I_{N}\otimes W)(I_{N}\otimes S-L \otimes F)\zeta+2e^{T}P(\bar{A}e-TH\dot{z})\\ =&\xi^{T}(I_{N}\otimes WS+S^{T}W-2\bar{L}\otimes \tau R^{T}R)\zeta+e^{T}(P\bar{A}+\bar{A}^{T}P)e-2\beta e^{T}\bar{C}^{T}\dot{z} \\ \leq&\sum_{i=1}^{N}\zeta_{i}^{T}(WS+S^{T}W-2\tau \lambda_{1}(\bar{L})R^{T}R)\zeta_{i}+e^{T}(P\bar{A}+\bar{A}^{T}P)e\\ &-2\sum_{i=1}^{N}\beta_{i}(\bar{C}_{i}e_{i})^{T}(\sigma(\bar{C}_ {i}x_{i})-\sigma(C_{i}\dot{x}_{i}))\\ \leq&\sum_{i=1}^{N}\zeta_{i}^{T}(WS+S^{T}W-2R^{T}R) \zeta_{i}-2\sum_{i=1}^{N}\beta_{i}(\bar{C}_{i}x_{i}-\bar{C}_{i}\dot{x}_{i})^{ T}(\sigma(\bar{C}_{i}x_{i})-\sigma(C_{i}\dot{x}_{i}))\\ \leq& 0,\end{split}  \tag{30}\]  
where we have used the fact that \(\tau\lambda_{1}(\bar{L})>1\). 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 \(\mathcal{M}:=\{(e,\zeta):\dot{V}=0\}\). \(\dot{V}=0\) implies \(\zeta=0\), i.e., \((w_{i}-x_{0})=0\), \(\forall i\in\mathcal{V}\), and \(\dot{z}_{i}=(\sigma(C_{i}x_{i})-\sigma(C_{i}\dot{x}_{i}))=0\), \(\forall i\in\mathcal{V}\). Then, in \(\mathcal{M}\), the error dynamics (28) can be rewritten as  
\[ \begin{split}\dot{e}=&\bar{A}e\\ \dot{\epsilon}=&(A+BK)\epsilon.\end{split}  \tag{31}\]  
Then, since \((A+BK)\) is Hurwitz and \(\|Rx_{0}\|_{\infty}<1\), we have  
\[\lim_{t\rightarrow\infty}\|e_{i}\|=\lim_{t\rightarrow\infty}\|C_{i}\dot{x}_{i }-Rx_{0}\|=0,\ \ \forall i\in\mathcal{V}, \tag{32}\]  
and, thus, there exists a finite time \(t_{1}\) such that \(\sigma(C_{i}\dot{x}_{i})=C_{i}\dot{x}_{i},\forall t\geq t_{1}\). Therefore, following the proof of Theorem 1, we can prove that \(e\) converges to \(0\). In summary, we have shown that \(\dot{V}\leq 0\), and \(e=\zeta=0\) is a unique, asymptotically stable equilibrium point in \(\mathcal{M}\). Therefore, according to Lemma 2, we have \(\lim_{t\rightarrow\infty}\|e_{i}\|=0\) and \(\lim_{t\rightarrow\infty}\|\dot{x}_{i}\|=0\), \(\forall i\in\mathcal{V}\). Moreover, since \((A+BK)\) is Hurwitz, the regulation error dynamics is input-to-state stable. Then, from Lemma 3, we have \(\lim_{t\rightarrow\infty}\|e_{i}\|=0\). Finally, we can conclude that \((\zeta,e,\epsilon)\rightarrow(0,0,0)\) and \(\lim_{t\rightarrow\infty}\|C_{i}x_{i}-Rx_{0}\|=0\), \(\forall i\in\mathcal{V}\), which completes the proof. **Remark 1**. : _Note that Theorems 1 and 2 give the sufficient conditions to achieve the output consensus. The existence of control gains follows from Assumptions 1-4, which are the standard assumptions to achieve the consensus for the heterogeneous agents. Moreover, the control gains can be constructed from the knowledge of the system matrices except \(\tau\), which requires the global information, i.e., \(\lambda_{2}(L)\) in Theorem 1 and \(\lambda_{1}(\bar{L})\) in Theorem 2.