## 5 Numerical Example
### Example 1  
Suppose that the coefficient matrices \(A\), \(B\), and \(C\) in (9) are given by  
\[A = \begin{bmatrix}1.03&0.6&0.2&0.2\\ 0&0.7&0&0.8\\ 0.5&0.4&1.01&0.2\\ 0.1&0.2&0.3&0.8\end{bmatrix} ,\;\;B= \begin{bmatrix}1.1&0.9\\ 0.6&-0.6\\ -0.5&1.2\\ 0.3&0.6\end{bmatrix} ,\] \[C = \begin{bmatrix}I_{2}&0_{2,2}\end{bmatrix} .\]  
The parameter \(\beta\) in (13) is given by \(\beta=0.1\). Suppose also that \(q\) (the number of communication channels) is given by \(q=2\), and \(\mathcal{I}_{1}^{a}=\{1\}\), \(\mathcal{I}_{1}^{s}=\{1\}\), \(\mathcal{I}_{2}^{a}=\{2\}\), and \(\mathcal{I}_{2}^{s}=\{2\}\) are given. In other words, the communication channel 1 is used in transmission of the first input and output. The communication channel 2 is used in transmission of the second input and output. First, consider solving Problem 1. Then, we can obtain  
\[X =1.0\times 10^{3} \begin{bmatrix}8.48&-1.13&-3.72&-5.48\\ -1.13&4.79&-1.06&-0.00\\ -3.72&-1.06&3.41&0.26\\ -5.48&-0.00&0.26&8.27\end{bmatrix} ,\] \[Y =1.0\times 10^{3} \begin{bmatrix}6.54&-3.81&-0.91&-0.33\\ -3.81&9.39&-3.06&-3.44\\ -0.91&-3.06&7.91&-1.81\\ -0.33&-3.44&-1.81&4.12\end{bmatrix} ,\] \[W_{1} = \begin{bmatrix}-1.44&-0.00\\ -1.44&-0.00\\ -1.44&-0.00\end{bmatrix} ,\] \[W_{2} = \begin{bmatrix}0.00&0.00&0.00&0.00\\ 1.44&1.44&1.44&1.44\end{bmatrix} ,\] \[W_{3} = \begin{bmatrix}490.06&-980.10&40.52&473.63\\ -477.73&930.80&-9.54&-460.47\\ -341.44&700.31&-7.04&-357.93\\ 255.72&-526.64&-45.31&301.49\end{bmatrix} .\]  
From these matrices, we can obtain coefficient matrices of the controller (10):  
\[A_{c} = \begin{bmatrix}-0.50&0.23&-0.08&-0.26\\ -0.34&1.08&0.29&1.27\\ -1.39&-0.11&0.62&-0.43\\ -0.87&-0.05&0.11&0.49\end{bmatrix} ,\] \[B_{c} = \begin{bmatrix}-1.25&-0.00\\ -0.63&-0.00\\ -1.51&-0.00\\ -0.78&-0.00\end{bmatrix} ,\] \[C_{c} = \begin{bmatrix}0.00&0.00&0.00\\ 0.40&0.54&0.41&0.66\end{bmatrix} .\]  
From this result, we see that \(B_{c}\) is column-sparse and \(C_{c}\) is row-sparse. It is theoretically guaranteed that the obtained controller can stabilize the closed-loop system. However, in implementing the controller, the second column of \(B_{c}\) and the first row of \(C_{c}\) must be replaced with a zero vector and a zero row vector, respectively. Then, by a numerical simulation from time 0 to 80 (the state almost converges to the origin by time 80), we canconfirm that owing to the truncation error, (13) with \(\beta=0.1\) does not hold, and (13) with \(\beta=0.99\) holds. Although a sparse controller can be obtained, two communication channels must be used. Next, consider solving Problem 2. We set \(w_{(1)}=1\) and \(w_{(2)}=10\) (i.e., the channel 1 is used preferentially). Then, we can obtain  
\[X =1.0\times 10^{4} \begin{bmatrix}1.11&-0.24&-0.60&-0.43\\ -0.24&0.73&-0.02&-0.17\\ -0.60&-0.02&0.68&-0.34\\ -0.43&-0.17&-0.34&1.48\end{bmatrix} ,\] \[Y =1.0\times 10^{4} \begin{bmatrix}1.06&-0.16&-0.55&-0.17\\ -0.16&0.72&-0.54&-0.15\\ -0.55&-0.54&1.49&-0.14\\ -0.17&-0.15&-0.14&0.31\end{bmatrix} ,\] \[W_{1} = \begin{bmatrix}-1.26&-0.00\\ -1.26&-0.00\\ -1.26&-0.00\\ -1.26&-0.00\end{bmatrix} ,\] \[W_{2} = \begin{bmatrix}1.26&1.26&1.26\\ 0.00&0.00&0.00\end{bmatrix} ,\] \[W_{3} =1.0\times 10^{3} \begin{bmatrix}-0.97&1.65&0.46&-1.03\\ 1.04&-1.32&-0.87&0.98\\ 1.30&-1.91&-0.98&1.36\\ -1.78&2.09&1.85&-1.75\end{bmatrix} \]  
From these matrices, we can obtain coefficient matrices of the controller (10),  
\[A_{c} = \begin{bmatrix}-0.68&-0.01&-0.26&-0.54\\ -0.88&0.36&-0.25&0.39\\ -0.66&0.86&1.36&0.77\\ -0.78&0.07&0.21&0.65\end{bmatrix} ,\] \[B_{c} = \begin{bmatrix}-1.25&-0.00\\ -0.63&-0.00\\ -1.51&-0.00\\ -0.78&-0.00\end{bmatrix} ,\] \[C_{c} = \begin{bmatrix}0.49&0.65&0.49&0.80\\ 0.00&0.00&0.00\end{bmatrix} .\]  
From this result, we see that \(B_{c}\) is column-sparse and \(C_{c}\) is row-sparse. In implementing the controller, the second column of \(B_{c}\) and the second row of \(C_{c}\) are replaced with a zero vector and a zero row vector, respectively. Then, by a numerical simulation from time 0 to 80, we can confirm that (13) with \(\beta=0.1\) holds.