## 4 Solution Method  
### Solution Method for Problem 1  
First, we derive a solution method for Problem 1.  
Figure 1: Example of SANs, where \(s_{i}\), \(i\in\{1,2,\ldots,6\}\) and \(a_{i}\), \(i\in\{1,2,3,4\}\) are sensors and actuators, respectively. There are three routers.  
From the system (9) and the controller (10), the closed-loop system is given by  
\[\bar{x}(k+1)=\bar{A}\bar{x}(k), \tag{11}\]  
where  
\[\bar{x}=\begin{bmatrix}x\\ \hat{x}\end{bmatrix},\ \ \bar{A}=\begin{bmatrix}A&B&C_{c}\\ B_{c}C&A_{c}\end{bmatrix}.\]  
To guarantee the stability of the closed-loop system, we introduce the following quadratic Lyapunov function:  
\[V(k)=\bar{x}^{\top}(k)P\bar{x}(k), \tag{12}\]  
where \(P=P^{\top}\in\mathcal{R}^{2n\times 2n}\) is a positive-definite matrix. To guarantee the stability of the closed-loop system, consider the problem of finding a controller such that  
\[V(k+1)-V(k)<-\beta V(k), \tag{13}\]  
where \(\beta\in[0,1)\) is a given parameter.  
We have the following theorem.  
Theorem 1: _Problem 1 is reduced to the following LMI optimization problem._  
Problem 3: \[\begin{split}\text{find}\ \ X\succ 0,Y\succ 0,W_{1},W_{2},W_{3} \\ \min\ \left\|\begin{bmatrix}W_{1}&W_{2}^{\top}\end{bmatrix} \right\|_{c_{1}}\\ \text{subject to}\ \begin{bmatrix}\bar{\beta}X&*&*&*\\ \bar{B}I&\bar{\beta}Y&*&*\\ XA+W_{1}C&W_{3}&X&*\\ A&AY-BW_{2}&I&Y\end{bmatrix}\succ 0,\end{split}\] (14) _where \(\bar{\beta}=1-\beta\)._  
Using the solution to Problem 3, the coefficient matrices \(A_{c}\), \(B_{c}\), and \(C_{c}\) in the controller (10) are derived as  
\[A_{c} =Z^{-1}(XAY+ZB_{c}CY-XBC_{c}Y-W_{3})Y^{-1}, \tag{16}\] \[B_{c} =Z^{-1}W_{1},\] (17) \[C_{c} =W_{2}Y^{-1}, \tag{18}\]  
where \(Z=X-Y^{-1}\).  
Proof: First, without of loss of generality, the positive definite matrix \(P\) can be replaced with  
\[P=\begin{bmatrix}X&Z\\ Z&Z\end{bmatrix}\]  
(see [44]). By applying the Schur complement [33] to \(P\succ 0\), we can obtain \(X-Z\succ 0\). Then, the matrix \(Y:=(X-Z)^{-1}\succ 0\) is defined.  
Next, from (13), we can obtain  
\[\bar{\beta}P-\bar{A}^{\top}P\bar{A}\succ 0. \tag{19}\]  
By applying the Schur complement to (19), we can obtain  
\[\begin{bmatrix}\bar{\beta}P&*\\ P\bar{A}&P\end{bmatrix}\succ 0. \tag{20}\]We define the matrix \(T\) as follows:  
\[T:=\begin{bmatrix}I_{n}&0\\ Y&-Y\end{bmatrix}.\]  
Pre-multiplying by block-diag\((T,T)\), and post-multiplying by block-diag\((T^{\top},T^{\top})\), we can obtain  
\[\begin{bmatrix}\tilde{B}TP^{\top}&*\\ TP\tilde{A}T^{\top}&TP^{\top}\end{bmatrix}\succ 0, \tag{21}\]  
where  
\[TPT^{\top} =\begin{bmatrix}X&I\\ I&Y\end{bmatrix},\] \[TP\tilde{A}T^{\top} =\begin{bmatrix}XA+W_{1}C&W_{3}\\ A&AY-BW_{2}\end{bmatrix}\] \[W_{1} =ZB_{c}, \tag{22}\] \[W_{2} =C_{c}Y,\] (23) \[W_{3} =XAY+ZB_{c}CY-XBC_{c}Y-ZA_{c}Y. \tag{24}\]  
From (21), we can obtain (15). From (15), \(P\succ 0\) is guaranteed. In addition, from (22)-(24), we can obtain (16)-(18).  
Finally, from (17), if the matrix \(W_{1}\) is column-sparse, then the matrix \(B_{c}\) is also column-sparse. In a similar way, from (18), if the matrix \(W_{2}\) is row-sparse, then the matrix \(C_{c}\) is also row-sparse. Hence, we introduce the objective function (14), which includes decision variables.  
This completes the proof.