## 7 Dynamic State-Feedback Synthesis
### Problem Solution  
The DSF synthesis problem can be solved in two steps. First, we find the matrix \(H\) such that  
\[\det(E+BH)\
eq 0. \tag{81}\]  
Premultiplication of (80) by \((E+BH)^{-1}\) gives  
\[\Delta^{\alpha}x_{i+1}=A_{C}x_{i}, \tag{82}\]  
where  
\[A_{C}=(E+BH)^{-1}(A-BK). \tag{83}\]  
Taking into consideration (2) and introducing the length \(L\) of practical implementation (the constraint on the upper limit of the summation), as in the case of (9), from (82) we obtain  
\[x_{i+1}=A_{C\alpha}x_{i}+\sum_{j=1}^{L}c_{j}x_{i-j}, \tag{84}\]  
where  
\[A_{C\alpha}=A_{C}+\mathbb{I}_{n}\alpha \tag{85}\]  
and \(x_{-k}=0\), \(k=1,2,\ldots\). In the second step, we find the matrix \(K\) such that the closed-loop system (84) has the desired properties. The stability of (84) can be tested using well-known methods. **Theorem 12** ([27]). : _The DDFL system (1) with the DSF (79) satisfying (81) is practically stable for given length \(L\) of practical implementation if and only if all roots of the characteristic equation_  
\[\det\left[\mathbb{I}_{n}z-A_{C\alpha}-\sum_{j=1}^{L}\mathbb{I}_{n}c_{j}z^{-j} \right]=0 \tag{86}\]  
_are located inside the unit circle._  
**Theorem 13** ([27]). : _The DDFL system (1) with the DSF (79) satisfying (81) is asymptotically stable (independent of \(L\)) if and only if all roots of the characteristic equation_  
\[\det\Bigl{[}\mathbb{I}_{n}(z-1)^{\alpha}z^{1-\alpha}-A_{C}\Bigr{]}=0 \tag{87}\]  
_are located inside the unit circle._  
The superstability of the closed-loop system (84) can be tested using the following approach. **Theorem 14**. : _The DDFL system (1) with the DSF (79) satisfying (81) is:_  
1. _practically superstable for given length_ \(L\) _of practical implementation if the conditions (_50_)-(_52_) are satisfied for the norm of the matrix (_85_);_
2. _superstable if the condition (_57_) is satisfied for the norm of the matrix (_85_)._  
Proof. : The proof follows immediately from Theorem 9. The analysis can be simplified by finding the matrix \(H\) such that  
\[E+BH=\mathbb{I}_{n}. \tag{88}\]  
Equation (88) has the solution if and only if  
\[\text{rank}B=\text{rank}\bigl{[} \begin{array}{cc}B&\mathbb{I}_{n}-E\end{array}  \bigr{]}. \tag{89}\]  
If \(\text{rank}B=m\), then there exists the left pseudoinverse of the matrix \(B\) given by [33]  
\[B_{L}=(B^{T}B)^{-1}B^{T}+H_{1}\Bigl{[}\mathbb{I}_{n}-B(B^{T}B)^{-1}B^{T} \Bigr{]}, \tag{90}\]  
where the matrix \(H_{1}\in\mathbb{R}^{m\times n}\) is arbitrary. Using (88) and (90) we obtain  
\[H=B_{L}(\mathbb{I}_{n}-E)=\Bigl{\{}(B^{T}B)^{-1}B^{T}+H_{1}\Bigl{[}\mathbb{I}_ {n}-B(B^{T}B)^{-1}B^{T}\Bigr{]}\Bigr{\}}(\mathbb{I}_{n}-E). \tag{91}\]  
In the particular case when \(H_{1}=0\) we have  
\[H=(B^{T}B)^{-1}B^{T}(\mathbb{I}_{n}-E). \tag{92}\]  
Thus, we obtain the closed-loop system (84) with \(A_{C}=A-BK\). However, in many cases it is impossible to fulfill the condition (89). **8. Numerical Example**  
Let us consider the DDFL system (1) with \(\alpha=0.4\) and [3]  
\[E=\left[ \begin{array}{ccc}0&-2&0\\ -3.3333&-5&0\\ 0&-1&0\end{array} \right],\quad A=\left[ \begin{array}{ccc}0&1&0\\ 1&0&0\\ 0&0&1\end{array} \right],\quad B=\left[ \begin{array}{ccc}1&0\\ 0&-2\\ 1&-1\end{array} \right], \tag{93}\]  
\[A_{\alpha}=A+0.4E=\left[ \begin{array}{ccc}0&0.2&0\\ -0.3333&-2&0\\ 0&-0.4&1\end{array} \right].\]  
The matrix pencil of the pair \((E,A)\) of (93) is regular since  
\[\det[E\lambda-A]=0.333(2\lambda+1)(10\lambda+3)\
eq 0. \tag{94}\]  
From (5) for \(c=0\) we have\[\bar{E}=[-A]^{-1}E=\left[ \begin{array}{ccc}3.3333&5&0\\ 0&2&0\\ 0&1&0\end{array} \right],\quad\bar{A}=[-A]^{-1}A=\left[ \begin{array}{ccc}-1&0&0 \\ 0&-1&0\\ 0&0&-1\end{array} \right], \tag{95}\] \[\bar{A}_{\alpha}=\bar{A}+0.4\bar{E}=\left[ \begin{array}{ccc}0.333 3&2&0\\ 0&-0.2&0\\ 0&0.4&-1\end{array} \right].\]  
Observe that \(\text{rank}\bar{E}=\text{rank}\bar{E}^{2}\) and \(q=1\). The Drazin inverse of the matrix \(\bar{E}\) can be computed using one of the methods from the literature; see, e.g., [3]. Thus, we obtain  
\[\bar{E}^{D}=\left[