## 5 Stability and Superstability Analysis
### Stability Analysis  
**Definition 3**. : _The DDFL system (1) with \(u_{i}=0\), \(i\in\mathbb{Z}_{+}\) is called asymptotically stable if_  
\[\lim_{i\to\infty}x_{i}=0 \tag{28}\]  
_for all consistent initial conditions \(x_{0}\in\mathbb{X}_{0}\)._  
From the solution to Equation (26), which is a third component of (16), it follows that for \(u_{i}=0\) the vector \(x_{2,i}\) is equal to zero for any \(i\in\mathbb{Z}_{+}\). Taking into account (22), the stability of the DDFL system (1) depends only on the vector \(x_{1,i}\), which is a solution to Equation (25). The stability of the DDFL system (1) can be tested using well-known methods in the literature; see, e.g., [27]. For the analysis, either Equation (6) or (25) can be used. **Definition 4**. : _The DDFL system (1) is called practically stable for given length \(L\) of practical implementation if the DDFL system (9) (or equivalently (25)) is asymptotically stable. If the DDFL system (9) (or equivalently (25)) is asymptotically stable for \(L\to\infty\), then the DDFL system (1) is called asymptotically stable (independent of \(L\))._  
**Theorem 2** ([3]). : _The DDFL system (1) with given length \(L\) of practical implementation is practically stable if and only if all roots of the characteristic equation_  
\[\det\left[\bar{E}z-\bar{A}_{\alpha}-\sum_{j=1}^{L}\bar{E}c_{j}z^{-j}\right]=0 \tag{29}\]  
_are located inside the unit circle._  
Taking into account that [32]  
\[\sum_{j=1}^{\infty}c_{j}z^{-j}=z-\alpha-(z-1)^{\alpha}z^{1-\alpha} \tag{30}\]from (29) we obtain the following. **Theorem 3** ([3]). : _The DDFL system (1) is asymptotically stable (independent of \(L\)) if and only if all roots of the characteristic equation_  
\[\det\Bigl{[}\bar{E}(z-1)^{\alpha}z^{1-\alpha}-\bar{A}\Bigr{]}=0 \tag{31}\]  
_are located inside the unit circle._  
The stability of the DDFL system (1) can also be tested using the approach based on the Equation (25). **Lemma 3**. : _The characteristic equation of (25) has the form_  
\[\det\left[\mathbb{I}_{n}z-\bar{A}_{1\alpha}-\sum_{j=1}^{L}\bar{E}\bar{E}^{D}cj ^{-j}\right]=z^{n-r}\det\left[\bar{E}z-\bar{A}_{\alpha}-\sum_{j=1}^{L}\bar{E}cj ^{-j}\right]=0 \tag{32}\]  
_and it has \(r=\text{rank}\bar{A}_{1}\) roots of the characteristic Equation (29) along with additional \(n-r\) zero-valued eigenvalues._  
Proof. : Using (15) we have  
\[ \begin{split}\det\left[\bar{E}z-\bar{A}_{\alpha}-\sum_{j=1}^{L} \bar{E}cj^{-j}\right]&=\det\biggl{\{}T\left[\begin{array}{cc}I \lambda-A_{1}&0\\ 0&N\lambda-A_{2}\end{array}\right]T^{-1}\biggr{\}}\\ &=\det T\text{det}[J\lambda-A_{1}]\text{det}[N\lambda-A_{2}]\text{det}T^{-1} \\ &=\det[J\lambda-A_{1}]\text{det}[N\lambda-A_{2}]\end{split}  \tag{33}\]  
since \(\text{det}T\text{det}T^{-1}=\mathbb{I}_{n}\) and  
\[\lambda=\left(z-\alpha-\sum_{j=1}^{L}cj^{-j}\right). \tag{34}\]  
Again using (15), we can write  
\[ \begin{split}\det\left[\mathbb{I}_{n}z-\bar{A}_{1\alpha}-\sum_{j= 1}^{L}\bar{E}\bar{E}^{D}cj^{-j}\right]&=\det\biggl{\{}T\left[ \begin{array}{cc}\mathbb{I}_{n_{1}}\lambda-J^{-1}A_{1}&0\\ 0&\mathbb{I}_{n_{2}}z\end{array}\right]T^{-1}\biggr{\}}\\ &=\det T\text{det}[\mathbb{I}_{n_{1}}\lambda-J^{-1}A_{1}]\text{det}[ \mathbb{I}_{n_{2}}z]\text{det}T^{-1}\\ &=z^{n_{2}}\text{det}[\mathbb{I}_{n_{1}}\lambda-J^{-1}A_{1}]\\ \end{split}  \tag{35}\]  
since \(\text{det}T\text{det}T^{-1}=\mathbb{I}_{n}\) and \(\lambda\) is given by (34). By (12) we have \(\bar{E}\bar{A}=\bar{A}\bar{E}\) and from (15) it follows that \(NA_{2}=A_{2}N\) if and only if \(A_{2}=\gamma\mathbb{I}_{n_{2}}\), where \(\gamma\in\mathbb{R}\), i.e., \(A_{2}\) is a scalar matrix. Therefore, Equation (33) can be written as  
\[ \begin{split}\det[J\lambda-A_{1}]\text{det}[N\lambda-\gamma \mathbb{I}_{n_{2}}]&=(-\gamma)^{n_{2}}\text{det}[J\lambda-A_{1}] \\ &=(-\gamma)^{n_{2}}\text{det}[\mathbb{I}_{n_{1}}\lambda-J^{-1}A_{1}] \end{split}  \tag{36}\]  
since \(\text{det}[N\lambda-\gamma\mathbb{I}_{n_{2}}]=(-\gamma)^{n_{2}}\). Equating (33) and (36) to zero and denoting \(n_{1}=r\) and \(n_{2}=n-r\) we obtain (32).