## 7 Dynamic State-Feedback Synthesis
### Problem Solution
\begin{array}{ccc}0.3&-0.75&0\\ 0&0.5&0\\ 0&0.25&0\end{array} \right],\quad\bar{A}^{D}=\bar{A}^{-1}=\bar{A} \tag{96}\]  
and  
\[\bar{A}_{1\alpha}=\bar{E}^{D}\bar{A}_{\alpha}=\left[ \begin{array}{ccc}0.1&0.75&0\\ 0&-0.1&0\\ 0&-0.05&0\end{array} \right],\quad\bar{B}_{1}=\bar{E}^{D}\bar{A}=\left[ \begin{array}{ccc}0.75&0.6\\ -0.5&0\\ -0.25&0\end{array} \right],\]  
\[\bar{B}_{2}=(\mathbb{I}_{3}-\bar{E}\bar{E}^{D})\bar{A}^{D}\bar{B}=\left[ \begin{array}{ccc}0&0\\ 0&0\\ 0.5&-1\end{array} \right] \tag{97}\]  
\[\bar{N}=(\mathbb{I}_{3}-\bar{E}\bar{E}^{D})\bar{A}^{D}\bar{E}=\left[ \begin{array}{ccc}0&0&0\\ 0&0&0\\ 0&0&0\end{array} \right].\]  
Assuming \(\bar{G}=0\) the norm of the matrix  
\[\|\bar{F}\|=\left\|\bar{A}_{1\alpha}+\bar{G}(\mathbb{I}_{3}-\bar{E}\bar{E}^{D })\right\|=\left\|\left[ \begin{array}{ccc}0.1&0.75&0\\ 0&-0.1&0\\ 0&-0.05&0\end{array} \right]\right\|=0.85 \tag{98}\]  
takes its minimal value. The desired values of the norm (98) for superstable systems can be determined using (50)-(52). Therefore, we have  
\[\|\bar{F}\|\in(0;1) \text{for}\quad L=0, \tag{99}\] \[\|\bar{F}\|\in(0.1464;0.8536) \text{for}\quad L=1,\] \[\|\bar{F}\|\in(0.0785;0.7965) \text{for}\quad L=2,\] \[\vdots\] \[\|\bar{F}\|\in(0;0.5) \text{for}\quad L\rightarrow\infty.\]  
From (99) it follows that the considered system is practically superstable for \(L=1\). The state vector norms for \(x_{0}=\left[ \begin{array}{ccc}1&4&2\end{array} \right]\) and different values of \(L\) are plotted in Figure 1. We can see the monotonic decrease only for \(L=1\). Now let us consider the SSF (60) with  
\[\bar{K}=\left[ \begin{array}{ccc}k_{11}&k_{12}&k_{13}\\ k_{21}&k_{22}&k_{23}\end{array} \right] \tag{100}\]  
and  
\[\bar{K}\bar{E}^{D}=\left[ \begin{array}{ccc}k_{11}&k_{12}+0.5k_{13}&0\\ k_{21}&k_{22}+0.5k_{23}&0\end{array} \right], \tag{101}\] \[\bar{K}(\mathbb{I}_{3}-\bar{E}\bar{E}^{D})=\left[ \begin{array}{ ccc}0&-0.5k_{13}&k_{13}\\ 0&-0.5k_{23}&k_{23}\end{array} \right].\]  
Choosing \(k_{13}=0\), \(k_{23}=0\) yields \(\bar{K}(\mathbb{I}_{3}-\bar{E}\bar{E}^{D})=0\). From (97) and  
\[\bar{K}=\left[ \begin{array}{ccc}0&0&0\\ 0&1&0\end{array} \right] \tag{102}\]  
we have  
\[\bar{A}_{C1}=\bar{A}_{1\alpha}-\bar{B}_{1}\bar{K}=\left[ \begin{array}{ccc}0.1&0.15&0\\ 0&-0.1&0\\ 0&-0.05&0\end{array} \right],\quad\bar{A}_{C2}=\bar{B}_{2}\bar{K}=\left[ \begin{array}{ccc}0&0&0\\ 0&0&0\\ 0&-1&0\end{array} \right] \tag{103}\]  
and from (72) \(\bar{F}_{C}=\bar{A}_{C1}\) for \(\bar{G}=0\). The desired values of the norm \(\|\bar{F}_{C}\|\) are also given by (99) since the superstability conditions of Theorems 7 and 8 depend only on the fractional order \(\alpha\) of the system. Thus, we have  
\[\|\bar{F}_{C}\|=\|\bar{A}_{C1}\|=\left\|\left[ \begin{array}{ccc}0.1&0.15&0\\ 0&-0.1&0\\ 0&-0.05&0\end{array} \right]\right\|=0.25, \tag{104}\] \[\|\bar{A}_{C2}\|=\left\|\left[ \begin{array}{ccc}0&0&0\\ 0&0&0\\ 0&-1&0\end{array} \right]\right\|=1.\]  
Therefore, by Theorem 11 the DDFL system (1), (93) with \(\alpha=0.4\) and the SSF (60), (102) is superstable (for \(L\rightarrow\infty\)) since \(\|\bar{F}_{C}\|<0.4\) and \(\|\bar{A}_{C2}\|\leq 1\). The state vector norms for \(x_{0}=\left[ \begin{array}{cc}1&4&-2\end{array} \right]\) and different values of \(L\) are plotted in Figure 2. We can see the monotonic decrease in every considered case. Similar results can also be obtained for any  
Figure 1: The state vector norms of the DDFL system (1) with \(\alpha=0.4\) and (93) for \(x_{0}=\left[ \begin{array}{cc}1&4&2\end{array} \right]\) and different values of \(L\). \(L>10\). The set of consistent initial conditions of the CL-DDFL system with SSF is different since from (75) for \(i=0\) we have \(x_{0}\in\text{Im}(\mathbb{I}_{3}+\bar{A}_{\mathbb{C}2})\bar{E}\bar{E}^{D}\). Finally, let us consider the DSF (79). In this case we cannot find the matrix \(H\) such that (88) holds since \(\text{rank}B=2\), \(\text{rank}[ \begin{array}{cc}B&\mathbb{I}_{3}-E\end{array} ]=3\) and the condition (89) is not satisfied. Using (83), (85), (93) and  
\[H=\left[