# The Iterative Solution to Discrete-Time \(H_{\infty}\) Control Problems for Periodic Systems
## 3 Numerical Experiments
### Experiment II  
We take \(\theta=2\), \(t=0,1,2\), and \(n=5,m_{1}=m_{2}=2\), \(\gamma=7.5\). The matrix coefficients are described below:  
\[A(0)=1.8\times\left( \begin{array}{cccc}-0.799&-0.01&-0.0326&0.1441&-0.3071 \\ -0.1931&-0.419&-0.4564&-0.2086&-0.1385\\ -0.1015&0.3199&-0.37&-0.1572&-0.1701\\ -0.1383&0.0281&-0.1879&0.0640&0.18\\ -0.3382&0.1764&-0.1084&-0.14&0.0555\\ \end{array} \right)\]  
\[A(1)=1.9\times\left( \begin{array}{cccc}0.182&-0.4412&-0.5427&0.4532&0.1874 \\ -0.1532&0.0329&0.475&0.097&-0.0548\\ 0.7056&-0.1355&0.6479&-0.7397&0.3912\\ 0.0364&-0.4506&0.2467&0.0482&-0.4428\\ -0.3066&-0.1566&-0.1077&0.0262&0.82\\ \end{array} \right)\]  
\[A(2)=1.7\times\left( \begin{array}{cccc}0.8507&0.0254&-0.0743&-0.2159&0.009 \\ -0.0823&0.5286&-0.9114&-0.4291&-0.3859\\ 0.0960&0.3086&0.20&-0.1370&0.3882\\ -0.2909&0.7105&0.0487&0.7335&0.2621\\ -0.3103&-0.1086&0.0126&-0.3097&0.034\\ \end{array} \right)\]  
\[B_{1}(0)=full(sprand(5,2,0.6))/7;\quad B_{2}(0)=\frac{1}{3}\left( \begin{array} []{cccc}0.4089&-1.0077\\ -0.1932&-0.4189\\ 0.4144&-1.1287\\ 0.19&-0.41\\ 1.24&0.86\\ \end{array} \right)\]  
\[B_{1}(1)=full(sprand(5,2,0.6))/7;\quad B_{2}(1)=\frac{1}{3}\left( \begin{array} []{cccc}-0.7202&0.5488\\ 0.5385&0.1610\\ -0.4652&-1.0513\\ 2.4629&-1.4624\\ -0.9885&1.37\\ \end{array} \right)\]\[B_{1}(2)=full(sprand(5,2,0.6))/7;\quad B_{2}(2)=\frac{1}{3}\left( \begin{array}{ c c c}-2.107&-0.604\\ -0.4065&-0.075\\ -0.109&-0.807\\ 0.5877&-0.3916\\ -0.0175&1.2778\end{array} \right)\]  
\[C(0)=0.9\left( \begin{array}{c c c c}0.233&1.5&0.3&1.5&0.3\\ 1.5&0.2&1.5&0.28&0.15\end{array} \right),\quad C(1)=0.9\left( \begin{array}{ c c c c}0.72&2.0&1.5&0.3&1.25\\ 0.3&0.06&2.5&1.04&1.15\end{array} \right)\]  
\[C(2)=0.9\left( \begin{array}{c c c c}1.5&0.06&2.5&1.04&1.15\\ 1.2&0.85&0.06&2.5&1.04\end{array} \right)\]  
\[D_{1}(0)=randn(2,2);\quad D_{1}(1)=randn(2,2);\quad D_{1}(2)=randn(2,2);\]  
\[D_{2}(0)=\left( \begin{array}{c c c}0.7615&0.4508\\ -0.6894&-1.565\end{array} \right),\;D_{2}(1)=\left( \begin{array}{c c}-0.0630& -0.5223\\ -0.7534&0.226\end{array} \right),D_{2}(2)=\left( \begin{array}{c c}-1.0126& -1.3643\\ -0.8901&-2.0057\end{array} \right)\]  
Two experimental trials are carried out. These are Test 1 and Test 2, where Test 1 consists of 100 examples and Test 2 of 200 examples. The matrix coefficients \(B_{1}(0),B_{1}(1),B_{1}(2)\) and \(D_{1}(0),D_{1}(1),D_{1}(2)\) differ for all of the examples as the random Matlab functions \(sprand\) and \(randn\) are used to define them. We execute Tests 1 and 2 successively. Our results might be summarized as follows. Block Successive Approximations Equation (8) with tol = \(10^{-7}\) and eps = \(10^{-7}\) requires 2 main iterations and it computes the stabilizing solution with \(\|\hat{X}^{(2)}-\mathbb{W}(\hat{X}^{(2)})\|=5.2147\times 10^{-8}\). CPU time for execution of iteration Equation (8) is 2.5 s for Test 1 and 5.1 s for Test 2. Block Stein Iteration Equation (9) with tol = \(10^{-6}\) and eps = \(10^{-6}\) requires 2 main iterations and it computes the stabilizing solution with \(\|\hat{X}^{(2)}-\mathbb{W}(\hat{X}^{(2)})\|=1.8306\times 10^{-10}\). CPU time for execution of iteration Equation (9) is 1.8 s for Test 1 and 3.7 s for Test 2. Method of Successive Approximations Equation by Equation (10) with tol = \(10^{-7}\) and eps = \(10^{-7}\) requires 2 main iterations and it computes the stabilizing solution with \(\|\hat{X}^{(2)}-\mathbb{W}(\hat{X}^{(2)})\|=5.7561\times 10^{-9}\). CPU time for execution of iteration Equation (10) is 1.5 s for Test 1 and 3.1 s for Test 2. Block Stein Iteration Equation (13) with tol = \(10^{-6}\) and eps = \(10^{-7}\) requires 2 main iterations and it computes the stabilizing solution with \(\|\hat{X}^{(2)}-\mathbb{Q}(\hat{X}^{(2)})\|=1.2059\times 10^{-8}\). CPU time for execution of iteration Equation (13) is 2.9 s for Test 1 and 5.1 s for Test 2.