## 6 Static State-Feedback Synthesis
### Problem Solution  
**Lemma 4** ([31]). : _The matrix \(\bar{K}\) can be chosen so that_  
\[\bar{K}x_{1,i}\
eq 0\quad\text{and}\quad\bar{K}x_{2,i}=0\]  
_or_  
\[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]  
If we choose the matrix \(\bar{K}\) such that \(\bar{K}x_{2,i}=\bar{K}(\mathbb{I}_{n}-\bar{E}\bar{E}^{D})x_{i}=0\), i.e., \(\bar{K}(\mathbb{I}_{n}-\bar{E}\bar{E}^{D})=0\), then from (61) and (62) we obtain  
\[x_{1,i+1}=\bar{A}_{\text{C1}}x_{1,i}+\sum_{j=1}^{L}c_{j}\bar{E}\bar{E}^{D}x_{1,i-j}, \tag{63}\]  
\[\bar{N}x_{2,i+1}=(\mathbb{I}_{n}+\bar{N}\alpha)x_{2,i}+\sum_{j=1}^{L}c_{j} \bar{N}x_{2,i-j}-\bar{A}_{\text{C2}}x_{1,i}, \tag{64}\]  
where  
\[\bar{A}_{\text{C1}}=\bar{A}_{1\alpha}-\bar{B}_{1}\bar{K},\quad\bar{A}_{\text{ C2}}=\bar{B}_{2}\bar{K}. \tag{65}\]  
Taking into account the considerations presented in Sections 3 and 4, the solution to Equation (63) is given by \[x_{1,i}=\Phi_{i}^{(A_{\rm C1})}\bar{E}\bar{E}^{D}x_{10}=\Phi_{i}^{(A_{\rm C1})} \bar{E}\bar{E}^{D}x_{0}, \tag{66}\]  
where \(x_{10}=\bar{E}\bar{E}^{D}x_{0}\) and \(\Phi_{i}^{(A_{\rm C1})}\) is defined analogously to (49) for the matrix \(\bar{A}_{\rm C1}\) given by (65). The solution to Equation (64) has the form [3]  
\[x_{2,i}=\sum_{k=0}^{q-1}\bar{N}^{k}\bar{A}_{\rm C2}\theta_{i,k}, \tag{67}\]  
where  
\[\theta_{i,0} =x_{1,i}, \tag{68}\] \[\theta_{i,1} =\theta_{i+1,0}-\alpha\theta_{i,0}-\sum_{j=1}^{L}c_{j}\theta_{i-j,0}=x_{1,i+1}-\alpha x_{1,i}-\sum_{j=1}^{L}c_{j}x_{1,i-j},\] \[\vdots\] \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]0  
and \(\theta_{i,-k}=0\), \(x_{1,-k}=0\), \(k=1,2,\ldots\). Given that  
\[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]1 \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]2 \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]3 \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]4 \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]5  
from (68) and (69) we obtain  
\[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]6 \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]7 \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]8 \[\bar{K}x_{1,i}=0\quad\text{and}\quad\bar{K}x_{2,i}\
eq 0.\]9 \[x_{1,i+1}=\bar{A}_{\text{C1}}x_{1,i}+\sum_{j=1}^{L}c_{j}\bar{E}\bar{E}^{D}x_{1,i-j}, \tag{63}\]0  
and thus the solution (67) takes the form  
\[x_{1,i+1}=\bar{A}_{\text{C1}}x_{1,i}+\sum_{j=1}^{L}c_{j}\bar{E}\bar{E}^{D}x_{1,i-j}, \tag{63}\]1  
**Lemma 5**. : _For the DDFL system (1) with the SSF (60) such that \(\bar{K}x_{2,i}=0\) we have \(\lim_{i\to\infty}x_{2,i}=0\) if and only if \(\lim_{i\to\infty}x_{1,i}=0\)._  
Proof. : The proof follows immediately from (71). In consequence, the stability of the CL-DDFL system (63)-(65) depends only on Equation (63) and for its analysis we can use Theorems 2-5, substituting:  
1. \((\bar{A}_{n}-\bar{B}\bar{K})\) for \(\bar{A}_{n}\) in Theorems 2 and 4;
2. \(\bar{A}_{\rm C1}\) for \(\bar{A}_{1n}\) in Theorems 3 and 5. The superstability of the CL-DDFL system (63)-(65) can be tested using the following approach. **Theorem 10**. : _The DDFL system (1) with the SSF (60) such that \(\bar{K}x_{2,i}=0\) is practically superstable for given length \(L\) of practical implementation if there exists an arbitrary matrix \(G\) such that:_  
1. _the conditions (_50_)-(_52_) are satisfied for the norm of the matrix_ \[x_{1,i+1}=\bar{A}_{\text{C1}}x_{1,i}+\sum_{j=1}^{L}c_{j}\bar{E}\bar{E}^{D}x_{1,i-j}, \tag{63}\]2 (72)
2. _the following inequality is true_ \[x_{1,i+1}=\bar{A}_{\text{C1}}x_{1,i}+\sum_{j=1}^{L}c_{j}\bar{E}\bar{E}^{D}x_{1,i-j}, \tag{63}\]3 (73)  
Proof. : The proof for the first condition follows immediately from Theorem 9. If the norm of (72) satisfies the conditions (50)-(52), then from (66) we have  
\[x_{1,i+1}=\bar{A}_{\text{C1}}x_{1,i}+\sum_{j=1}^{L}c_{j}\bar{E}\bar{E}^{D}x_{1,i-j}, \tag{63}\]4  
where \(\bar{E}\bar{E}^{D}x_{0}=x_{0}\) since \(x_{0}\in{\rm Im}\bar{E}^{D}\). Using (22), (66) and (71) we obtain  
\[