## 5 Conclusions
\begin{array}{c}x_{i}(t+1)=x_{i}(t)+u_{i}(t)\\ =(\hat{x}_{i}(t)-e_{i}(t))+u_{i}(t)\\ =(\hat{x}_{i}(t)-e_{i}(t))+(\sigma\;a_{ij}(\hat{x}_{j}(t)-\hat{x}_{i}(t))\\ =-e_{i}(t)+\sigma\;a_{ij}\hat{x}_{j}(t)+\left(1-\;\sigma\;a_{ij}\right)\hat{x}_ {i}(t))\end{array} \] (A1)  
Based on the normalized Laplacian matrix, Equation (1) is represented by a Perron matrix in discrete time as Equation (A2), where  
\[ \begin{array}{c}x_{i}(t+1)=x_{i}(t)+u_{i}(t)\\ =x_{i}(t)+\frac{1}{1+d_{i}}a_{ij}\sum\limits_{j\in N_{i}}(x_{j}(t)-x_{i}(t)) \end{array} \]since \(a_{ij}\sum_{j\in N_{i}}(x_{j}(t)-x_{i}(t))\) is equivalent to \(-Lx(t)\)  
\[ \begin{split}&=x_{i}(t)+\frac{1}{1+d_{i}}(-L\ x(t))\\ &=x_{i}(t)+(-(I+D)^{-1})L\ x(t))\\ &=x_{i}(t)-(I+D)^{-1})L\ x(t))\\ &=I-(I+D)^{-1}(I+A)x(t)\\ &\qquad=(I+D)^{-1}(I+A)x(t)\\ &\qquad\qquad=P_{e}x(t)\end{split}  \tag{10}\]  
Substituting Equation (15) into Equation (10), where \(x_{i}(t_{k}^{i})=\hat{x_{i}}(t)\) is the latest broadcast state  
\[x(t+1)=P_{e}(e_{i}(t)+x_{i}(t)) \tag{11}\]  
If the conditions of Equation (14) are satisfied, there is no event occurring at this time with the error of Equation (15), and \(e_{i}(t)\) is equal to 0, which means that there is no transmission and control update happening with the agent and its neighbours. This means that the sampled data are not updated, and the agent's state value will remain the same. The opposite occurs for the situation where the conditions of Equation (14) are violated. In this scenario, an event occurs, in which the error is equal to Equation (15) and the new state position of Equation (11) is sent to the connected neighbors for a control update. Even though the state measurement error gives a significant effect to the next agent's state, as stated in Equation (10), the results of the closed loop system are guaranteed asymptotically to reach average consensus, Equation (6), among agents in the group when the eigenvalues of square matrix \(P\) are strictly contained in the Gershgorin circle criterion. The eigenvalues of the Perron matrix for a topology of 10 agents are strongly connected with an undirected graph--these are \(\lambda_{1}\ =\ 0\,\ \lambda_{2}\ =\ 0.5\pm 0.5i\) and \(\lambda_{10}\ =\ 1\). The eigenvalues inside the Gershgorin circle unit are expected to be \(\lambda_{10}\ =\ 1\). \(\square\)  
Proof of Theorem 1.: Consider a Lyapunov function (Equation (11)), which is represented by Equation (12) when \(Lx\triangleq y=[y_{1},\ y_{2},\ldots y_{N}]^{T}\). \[ \begin{split}& Vx(t)=\frac{1}{2}x(t)^{T}Lx(t)\\ &\dot{V}=\frac{\partial V}{\partial x}\frac{dx}{dt}=x^{T}L\dot{x }\\ &=-x^{T}L(Lx+Le)\\ &=-LLxx^{T}-LLex^{T}\\ &=-y^{T}y-y^{T}Le\end{split}  \tag{12}\]  
With the Laplacian matrix, Equation (12) will be  
\[ \begin{split}&\dot{V}=\sum\limits_{i}y_{i}^{2}-\sum\limits_{i} \sum\limits_{j\in N_{i}}y_{i}\big{(}e_{i}-e_{j}\big{)}\\ &=\sum\limits_{i}y_{i}^{2}-\sum\limits_{i}|N_{i}|y_{i}e_{i}+\sum \limits_{i}\sum\limits_{j\in N_{i}}y_{i}e_{j}\\ \end{split}  \tag{13}\]  
For \(a>0,\ \dot{V}\) can bound using inequality \(|xy|\leq\frac{a}{2}x^{2}+\frac{1}{2a}y^{2}\), thus  
\[\dot{V}\leq-\sum\limits_{i}y_{i}^{2}+\sum\limits_{i}a|N_{i}|y_{i}^{2}+\sum \limits_{i}\frac{1}{2a}|N_{i}|y_{i}^{2}+\sum\limits_{i}\sum\limits_{j\in N_{i} }\frac{1}{2a}y_{j}^{2} \tag{14}\]  
Since the communication graph is undirected,  
\[\sum\limits_{i}\sum\limits_{j\in N_{i}}\frac{1}{2a}e_{i}^{2}=\sum\limits_{i} \sum\limits_{j\in N_{i}}\frac{1}{2a}e_{i}^{2}=\sum\limits_{i}\frac{1}{2a}|N_{i }|\ e_{i}^{2}\]Therefore,  
\[\dot{V}\leq-\sum_{i}(1-a|N_{i}|)y_{i}^{2}+\sum_{i}\frac{1}{a}\ |N_{i}|e_{i}^{2} \tag{10}\]  
\[\dot{V}\leq\sum_{i}(\sigma_{i}-1)(1-a\ |N_{i}|)y_{i}^{2}\]  
which shows that the system is stable when the derivative of the Lypunov function is negative definite for \(0<\ \sigma_{i}\ <\ 1\). \(\square\)  
**Lemma 1. **_Each state position of the agent, \(x_{i}\) is sampled at every specific odd time, \(t\) bounded by \(t+1\).