# The Iterative Solution to Discrete-Time \(H_{\infty}\) Control Problems for Periodic Systems
## 3 Numerical Experiments
### Experiment 1  
Consider a discrete-time periodic linear system described by  
\[A(0)=\left(\begin{array}{ccc}-1.2&0&0.4\\ 4&2&0\\ 0.12&-8&1.6\end{array}\right),A(1)=\left(\begin{array}{ccc}-2&2&0\\ 4.2&6.4&0\\ 1.2&-10&2.8\end{array}\right),A(2)=\left(\begin{array}{ccc}-1.2&0.8&0\\ 8&0.3&0.5\\ 1.2&-8&4.5\end{array}\right)\]  
\[B_{1}(0)=\left(\begin{array}{ccc}1.8048&0.5877\\ -0.3916&-0.0175\\ 1.2778&0.5082\end{array}\right),B_{1}(1)=\left(\begin{array}{ccc}-0.4959&0. 1848\\ 2.4629&-1.4624\\ -0.9885&1.3702\end{array}\right),B_{1}(2)=\left(\begin{array}{ccc}-1.0077&- 0.1932\\ -0.4189&1.77\\ 0.4144&-1.1287\end{array}\right)\]  
\[B_{2}(0)=\left(\begin{array}{ccc}0.4089&-1.0077\\ -0.1932&-0.4189\\ 0.4144&-1.1287\end{array}\right),B_{2}(1)=\left(\begin{array}{ccc}-0.7202& 0.5488\\ 0.5385&0.1610\\ -0.4652&-1.0513\end{array}\right),B_{2}(2)=\left(\begin{array}{ccc}-2.107& -0.604\\ -0.4065&-0.075\\ -0.109&-0.807\end{array}\right)\]  
\[C(0)=\left(\begin{array}{ccc}0.233&1.5&0.3\\ 0.2&1.5&0.28\end{array}\right),C(1)=\left(\begin{array}{ccc}0.72&2.0&1.5\\ 0.3&1.5&0.3\end{array}\right),C(2)=\left(\begin{array}{ccc}1.5&0.06&2.5\\ 1.5&1.2&0.85\end{array}\right)\]  
\[D_{1}(0)=\left(\begin{array}{ccc}0.35&-0.45\\ 0.28&-0.13\end{array}\right),D_{1}(1)=\left(\begin{array}{ccc}0.1&0.4\\ 0.95&0.48\end{array}\right),D_{1}(2)=\left(\begin{array}{ccc}7.07&2.2\\ 1.22&-0.95\end{array}\right)\]  
\[D_{2}(0)=\left(\begin{array}{ccc}0.15&0.75\\ -0.18&-0.25\end{array}\right),D_{2}(1)=\left(\begin{array}{ccc}2&-1.85\\ 1.46&-0.98\end{array}\right),D_{2}(2)=\left(\begin{array}{ccc}3.4&4.82\\ 3.28&-0.35\end{array}\right)\]  
with \(\theta=2,\ t=0,1,2,\ \gamma=6.5,\ n=3\). For all of the discussed methods are executed consequently 100 and 200 runs with the same matrix coefficients. We use \(tol=10^{-5}\) (the number \(tol\) is the computational tolerance for the first sequence \(\{X^{(k)}(t)\}_{k=0^{\prime}}^{\infty}\) see [7]) and \(eps=10^{-6}\) for all executions. The results could be summarized as follows.  
Algorithm TMS with Block Successive Approximations Equation (8) requires 3 main iterations and it computes the stabilizing solution with \(\max_{t}\ \ \|\bar{X}^{(3)}(t)-\mathcal{R}(t,\bar{X}^{(3)})\|=8.1904\times 10^{-8}<\) tol. CPU time for execution of the algorithm is 0.4 s for 100 runs and 0.97 s for 200 runs.  
Algorithm TMS with Block Stein Iteration Equation (9) requires 3 main iterations and it computes the stabilizing solution with \(\max_{t}\ \|\hat{X}^{(3)}(t)-\mathcal{R}(t,\hat{X}^{(3)})\|=1.0794\times 10^{-9}\) < tol. CPU time for execution of the algorithm is 0.78 s for 100 runs and 1.57 s for 200 runs.  
Algorithm TMS with Successive Approximations Equation by Equation (10) requires 3 main iterations and it computes the stabilizing solution with \(\max_{t}\ \|\hat{X}^{(3)}(t)-\mathcal{R}(t,\hat{X}^{(3)})\|=4.0976\times\ 10^{-6}\) < tol. CPU time for execution of iteration Equation (10) is 0.3750 s for 100 runs and 0.75 s for 200 runs.  
Algorithm OMS with Block Stein Iteration Equation (13) requires 3 main iterations and it computes the stabilizing solution with \(\|\hat{X}^{(3)}-\mathbb{Q}(\hat{X}^{(3)})\|=5.8077\times 10^{-8}\). CPU time for execution of iteration Equation (13) is 1.15 s for 100 runs and 2.41 s for 200 runs.  
Algorithm OMS with Method of Successive Approximations Equation by Equation (14) requires 3 main iterations and it computes the stabilizing solution with \(\|\hat{X}^{(3)}-\mathbb{Q}(\hat{X}^{(3)})\|=2.5060\times 10^{-8}\). CPU time for execution of iteration Equation (14) is 0.5150 s for 100 runs and 1.14 s for 200 runs.