Before defining the feedback convolutional equivalence between families of convolutional codes, we recall some results above the degree and the controllability indices of convolutional codes over finite fields (see [23]). These results will be used in the sequel:

The degree of the encoder *G*(*z*) of a convolutional code 𝓒 over a finite field, that is denoted by *δ*(*G*(*z*)), is defined by their column degrees, *ν*_{i} := max{deg (*g*_{i,j}(*z*) where *i* = 1, …, *n*}, as *δ*(*G*(*z*)): = $\sum _{l=1}^{k}$ *ν*_{l}.

The complexity of a convolutional code 𝓒, denoted by *δ*(𝓒), is defined as the highest degree of the full size minors of any encoder *G*(*z*).

An encoder matrix *G*(*z*) of 𝓒 is minimal if and only if *δ*(*G*(*z*))= *δ* (𝓒).

The set of column degrees of any minimal encoder of 𝓒 are known as the Forney or controllability indices of the code. We can reorder them if it is necessary such that *κ*_{1} ≥ … ≥ *κ*_{k}. The invariant *δ* = $\sum}_{i=1}^{k$ *κ*_{i} is the degree of the code 𝓒. The complexity of a convolutional code over a finite field equals its degree. In the following, we denote the set of controllability indices by 𝔎^{𝓒} = (*κ*_{1}, …, *κ*_{k}).

Note that the controllability indices of a convolutional code are unique and invariants of the code.

Since by first order representation we can compute an I/S/O representation for a convolutional code 𝓒 ([21, 22, 23]), the relation between controllability indices of the code and controllability (reachability) indices of the I/S/O representation associated as dynamical linear system is clear and it is given in the following theorem:

#### Theorem 3.2

(c.f. Theorem 2, [44].). *If Σ* = (*A*, *B*) *forms a controllable pair such that A* ∈ 𝔽^{δ×δ} *and B* ∈ 𝔽^{δ×k}, *then there exist positive integers κ*_{1} ≥ … ≥ *κ*_{k} (often referred to as the controllability or Kronecker indices of the pair Σ) only dependent on the GL_{δ} equivalence class of Σ having the following properties:

*κ*_{1} = *κ*, *the controllability index of Σ*.

$\sum}_{i=1}^{k$ *κ*_{i} = *δ*, *the size of matrix A*.

*There exists polynomial matrices X*(*z*), *Y*(*z*), *U*(*z*) *satisfying*
$$\begin{array}{}Ker(\begin{array}{ccc}zI-A& 0& -B\\ -C& I& -D\end{array})=Im\left(\begin{array}{c}X(z)\\ Y(z)\\ U(z)\end{array}\right)\end{array}$$

*and having the property that the i-th column degree of G*(*z*) = $(\begin{array}{c}Y(z)\\ U(z)\end{array})$ *is equal to κ*_{i}, and the ith column degree of X(*z*) *is equal to κ*_{i} − 1 *for i* = 1, …, *k*.

Since I/S/O representations over a finite field allow us to represent a convolutional code as 𝓒 ≃ *Ker*(*zK* + *L* | *M*) = $\begin{array}{}Ker(\begin{array}{ccc}zI-A& 0& -B\\ -C& I& -D\end{array}),\end{array}$ Theorem 3.2 could be applied and then controllability indices of 𝓒 equal the Kronecker indices of the pair *Σ*^{𝓒} = (*A*, *B*)^{𝓒}. In particular, one has that 𝔎^{𝓒} = 𝓚^{Σ𝓒}.

#### Example 3.3

*Consider the encoder of a* (3, 2, 2)-*convolutional code* 𝓒 *over* ℤ/2ℤ
$$\begin{array}{}G(z)=\left(\begin{array}{cc}z+1& 1\\ {z}^{2}& 0\\ 0& 1\end{array}\right)\end{array}$$

*whose controllability indices are* 𝔎^{𝓒} = (*κ*_{1}, *κ*_{2}) = (2, 0).

*We can compute a minimal first order representation of* 𝓒; *that is, the matrices K, L and M that characterize the encoder G*(*z*):
$$\begin{array}{}K=& \left(\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right),L=\left(\begin{array}{cc}0& 0\\ 1& 0\\ 1& 1\end{array}\right)\mathit{\text{\hspace{0.17em}and\hspace{0.17em}}}M=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 1& 0& 1\end{array}\right).\end{array}$$

*From the above triple of matrices, we compute its I/S/O representation Σ*^{𝓒} *that is equal to the system Σ obtained in the Example 2.3. Thus* 𝓚^{Σ𝓒} = (*κ*_{1}, *κ*_{2})=(2,0).

Let *R* be a noetherian von Neumann regular ring. Let 𝕮 ≃ ${\oplus}_{j=1}^{t}$ 𝓒_{j} be an (*n*, *k*, *δ*) systematic family of convolutional codes over *R* where 𝓒_{j} is an (*n*, *k*, *δ*) convolutional code over each *R*/𝔪_{j} = 𝔽_{j} for each maximal ideal 𝔪_{j} ∈ *Spec*(*R*).

#### Definition 3.4

*We define the set of controllability indices of* 𝕮, *and we denote them by* 𝔎^{𝕮}, *as the t-uple of vectors where each component is the set of controllability indices of each* 𝓒_{j}, *namely*
$$\begin{array}{}{\mathfrak{K}}^{\mathfrak{C}}=[({\kappa}_{1}^{1},\dots ,{\kappa}_{1}^{t}),\dots ,({\kappa}_{k}^{1},\dots ,{\kappa}_{k}^{t})],\end{array}$$

*where* $({\kappa}_{i}^{j})$ *is the i-th controllability index of the convolutional code* 𝓒_{j}.

#### Definition 3.6

*We define the set of Kronecker indices of Σ*^{𝕮}, *and we denote them by* 𝓚^{Σ𝕮}, *as the t-uple of vectors where each component is the set of Kronecker indices of each Σ*^{𝓒j}, *namely*
$$\begin{array}{}{\mathcal{K}}^{{\mathit{\Sigma}}^{\mathfrak{C}}}=[({\mathcal{K}}_{1}^{1},\dots ,{\mathcal{K}}_{1}^{t}),\dots ,({\mathcal{K}}_{k}^{1},\dots ,{\mathcal{K}}_{k}^{t})],\end{array}$$

where $({\mathcal{K}}_{i}^{j})$ *is the i-th Kronecker index of the associated I/S/O representation Σ*^{𝓒j} *of the convolutional code* 𝓒_{j}.

#### Example 3.7

*Let* 𝕮 *be the following encoder of a family of convolutional codes over* ℤ/6ℤ
$$\begin{array}{}\mathfrak{C}=\left(\begin{array}{cc}1+z& 2z+3\\ 3{z}^{2}+2z& 0\\ 0& 2z+3\end{array}\right)\mathit{\text{\hspace{0.17em}where\hspace{0.17em}}}{\mathfrak{K}}^{\mathfrak{C}}=[(2,1),(0,1)]\mathit{\text{\hspace{0.17em}because\hspace{0.17em}}}\\ \mathfrak{C}\otimes \mathbb{Z}/2\mathbb{Z}={\mathcal{C}}^{\mathbb{Z}/2\mathbb{Z}}=\left(\begin{array}{cc}1+z& 1\\ {z}^{2}& 0\\ 0& 1\end{array}\right)\mathit{\text{\hspace{0.17em}with\hspace{0.17em}}}{\mathfrak{K}}^{{\mathcal{C}}^{\mathbb{Z}/2\mathbb{Z}}}=[(2,0)]\mathit{\text{\hspace{0.17em}and\hspace{0.17em}}}\\ \mathfrak{C}\otimes \mathbb{Z}/3\mathbb{Z}={\mathcal{C}}^{\mathbb{Z}/3\mathbb{Z}}=\left(\begin{array}{cc}1+z& 2z\\ 2z& 0\\ 0& 2z\end{array}\right)\mathit{\text{\hspace{0.17em}with\hspace{0.17em}}}{\mathfrak{K}}^{{\mathcal{C}}^{\mathbb{Z}/3\mathbb{Z}}}=[(1,1)].\end{array}$$

*Moreover, note that Σ*^{𝕮ℤ/6ℤ} *described by*
$$\begin{array}{}{\mathit{\Sigma}}^{{\mathfrak{C}}^{\mathbb{Z}/6\mathbb{Z}}}=[A=\left(\begin{array}{cc}0& 0\\ 3& 0\end{array}\right),B=\left(\begin{array}{cc}5& 0\\ 0& 2\end{array}\right),C=(\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}),D=(\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}})]\end{array}$$

*has Kronecker’s indices* 𝓚^{Σ𝕮} = [(2, 1), (0, 1)] *because* 𝓚^{Σ𝓒ℤ/2ℤ} = 𝓚^{Σ𝓒1} = [2, 0] *(see Example 3.3) and* 𝓚^{Σ𝓒ℤ/3ℤ} = 𝓚^{Σ𝓒2} = [1, 1] *because*
$$\begin{array}{}{\mathit{\Sigma}}^{{\mathcal{C}}_{2}}=[{A}_{2}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),{B}_{2}=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right),{C}_{2}=(\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\phantom{\rule{thinmathspace}{0ex}}),{D}_{2}=(\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}})]\end{array}$$

*and then ξ*_{1} = *rk*(*B*_{2}) = 2 *and ξ*_{2} = *rk* (*B*_{2} *A*_{2}*B*_{2}) − *rk*(*B*_{2}) = 2 − 2 = 0. *So, by *, 𝓚^{Σ𝓒2} = [1, 1].

Table 2 Kronecker’s indices over ℤ/3ℤ

Now we define the feedback convolutional relation between families of convolutional codes over a noetherian von Neumann regular ring *R*.

#### Definition 3.8

*Let* 𝕮 *and* 𝕮 *be* (*n*, *k*, *δ*) - *systematic families of convolutional codes over R. The feedback convolutional relation between families of convolutional codes*, $\stackrel{f.c.e.}{\sim}$, *is defined as*
$$\begin{array}{}\mathfrak{C}\stackrel{f.c.e}{\sim}\overline{\mathfrak{C}}\iff {\mathfrak{K}}^{\mathfrak{C}}={\mathfrak{K}}^{\overline{\mathfrak{C}}}.\end{array}$$

*Note that f.c.e. is an equivalence relation*.

#### Lemma 3.9

*The number of feedback equivalence classes of a systematic family of convolutional codes with degree δ under feedback equivalence is p*_{ℕ} (*δ*)^{t} *where p*_{ℕ}(*δ*) *denotes the number of partitions of the degree of the codes*.

#### Proof

Feedback equivalence classes of a family of convolutional codes 𝕮 such that the components of their controllability indices verify ${\sum}_{j=1}^{k}{\kappa}_{i}^{j}=\delta$ for each *i* = 1, …, *t*, are the possible ways to get *δ*, *t* times; that is, *p*_{ℕ}(*δ*)^{t}.□

We give our main result.

#### Theorem 3.11

*Let* 𝕮 *and* 𝕮 *be systematic families of convolutional codes over R. Let Σ*^{𝕮} *and* *Σ*^{𝕮} *be the corresponding I/S/O representations. Then*
$$\begin{array}{}\mathfrak{C}\stackrel{f.c.e}{\sim}\overline{\mathfrak{C}}\iff {\mathit{\Sigma}}^{\mathfrak{C}}\stackrel{f.i}{\simeq}{\overline{\mathit{\Sigma}}}^{\overline{\mathfrak{C}}},\end{array}$$

*where f.i denotes the feedback isomorphism between locally Brunovsky linear systems with state space of rank δ and Σ*^{𝕮} = (*A*, *B*)^{𝕮} (*respectively for* *Σ*^{𝕮}).

#### Proof

An I/S/O representation of a family of convolutional codes over *R* is a reachable linear system, thus *R* is a Locally Brunovsky ring by [36]. Therefore, we can apply the results of feedback classification of locally Brunovsky linear systems described in Subsection 2.1. Since **P**(*R*) = ℕ^{t} is cancellative, the invariants *I*_{i} classify too, and thus,
$$\begin{array}{}{\mathit{\Sigma}}^{\mathfrak{C}}\stackrel{f.i}{\simeq}{\overline{\mathit{\Sigma}}}^{\overline{\mathfrak{C}}}\iff [{Z}_{i}^{{\mathit{\Sigma}}^{\mathfrak{C}}}]=[{Z}_{i}^{{\overline{\mathit{\Sigma}}}^{\overline{\mathfrak{C}}}}]\iff [{I}_{i}^{{\mathit{\Sigma}}^{\mathfrak{C}}}]=[{I}_{i}^{{\overline{\mathit{\Sigma}}}^{\overline{\mathfrak{C}}}}].\end{array}$$(8)

From **P**(*R*) ≃ **P**(𝔽_{1}) × … × **P**(𝔽_{t}), Equation (8) implies that
$$\begin{array}{}[{I}_{i}^{{\mathit{\Sigma}}^{\mathfrak{C}}}{\otimes}_{R}{\mathbb{F}}_{j}]=[{I}_{i}^{{\overline{\mathit{\Sigma}}}^{\overline{\mathfrak{C}}}}{\otimes}_{R}{\mathbb{F}}_{j}]\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}j=1,\dots ,t\Rightarrow [{I}_{i}^{{\mathit{\Sigma}}^{{\mathcal{C}}_{j}}}]=[{I}_{i}^{{\overline{\mathit{\Sigma}}}^{{\overline{\mathcal{C}}}_{j}}}].\end{array}$$

Now, *Σ*^{𝓒j} and *Σ*^{𝓒j} are I/S/O representations (reachable dynamical linear systems) over 𝔽_{j}. The set of finitely generated projective modules of a finite field is the cancellative monoid **P**(𝔽_{j}) = ℕ. So,
$$\begin{array}{}[{I}_{i}^{{\mathit{\Sigma}}^{{\mathcal{C}}_{j}}}]=[{I}_{i}^{{\overline{\mathit{\Sigma}}}^{{\overline{\mathcal{C}}}_{j}}}]\iff {\mathit{\Sigma}}^{{\mathcal{C}}_{j}}\stackrel{f.e.}{\simeq}{\overline{\mathit{\Sigma}}}^{{\overline{\mathcal{C}}}_{j}}\iff {\mathcal{K}}^{{\mathit{\Sigma}}^{{\mathcal{C}}_{j}}}={\mathcal{K}}^{{\mathit{\Sigma}}^{{\overline{\mathcal{C}}}_{j}}}.\end{array}$$

Then, by [44], 𝔎^{𝓒j} = 𝔎^{𝓒j} and
$$\begin{array}{}[({\kappa}_{1}^{1},\dots ,{\kappa}_{1}^{t}),\dots ,({\kappa}_{k}^{1},\dots ,{\kappa}_{k}^{t})]=[({\overline{\kappa}}_{1}^{1},\dots ,{\overline{\kappa}}_{1}^{t}),\dots ,({\overline{\kappa}}_{k}^{1},\dots ,{\overline{\kappa}}_{k}^{t})]\Rightarrow {\mathfrak{K}}^{\mathfrak{C}}={\mathfrak{K}}^{\overline{\mathfrak{C}}}\iff \mathfrak{C}\stackrel{f.c.e.}{\simeq}\overline{\mathfrak{C}}.\end{array}$$□

#### Theorem 3.12

*Let R be a noetherian von Neumann regular ring that decomposes in finite fields. Then*
$$\begin{array}{}\{{\mathit{\Sigma}}^{\mathfrak{C}}/\sim \}=\{Br(R)/\sim \},\end{array}$$

*where* {*Br*(*R*)} *denotes the set of locally Brunovsky linear systems over R with state space of rank δ*, ∼ *denotes the feedback isomorphism, and Σ*^{𝕮} = (*A*, *B*)^{𝕮} (*respectively for* *Σ*^{𝕮}).

#### Proof

Since *R* is a Locally Brunovsky ring, and by Theorem 3.11,
$$\begin{array}{}\mathrm{\#}\{\text{feedback classes of\hspace{0.17em}}\mathfrak{C}\}=\mathrm{\#}\{[{\mathit{\Sigma}}^{\mathfrak{C}}]\}\subseteq \mathrm{\#}\{[Br(R)]\}=f{e}_{R}(\delta ).\end{array}$$

By Lemma 3.9, # { feedback classes of 𝕮 } = *p*(*δ*)^{t} and since *R* is a noetherian von Neuman regular ring, this number is equal to *fe*_{R}(*δ*), so we conclude the proof.□

Note that Kronecker indices of a linear system *Σ* are obtained from the pair (*A*, *B*) and hence, they are independent of *C* and *D*. Therefore, since controllability invariant indices of an (*n*, *k*, *δ*) - family of convolutional codes, 𝕮, are obtained from its I/S/O representation *Σ*^{𝕮}, it follows that controllability indices of 𝕮 are obtained from pair (*A*, *B*)^{𝕮}, and they are independent of (*C*, *D*)^{𝕮}. We highlight this property in the following example.

#### Example 3.13

*Let R be the ring of modular integers* ℤ/6ℤ ≃ ℤ/2ℤ × ℤ/3ℤ. *Let Σ*_{i} be the following classes of feedback isomorphisms of locally Brunovsky linear systems over ℤ/6ℤ
$$\begin{array}{}{\mathit{\Sigma}}_{1}=[(\begin{array}{cc}0& 0\\ 1& 0\end{array}),(\begin{array}{cc}1& 0\\ 0& 0\end{array})],{\mathit{\Sigma}}_{2}=[(\begin{array}{cc}0& 0\\ 0& 0\end{array}),(\begin{array}{cc}1& 0\\ 0& 1\end{array})],\\ {\mathit{\Sigma}}_{3}=[(\begin{array}{cc}0& 0\\ 3& 0\end{array}),(\begin{array}{cc}5& 0\\ 0& 2\end{array})]\mathit{\text{\hspace{0.17em}and\hspace{0.17em}}}{\mathit{\Sigma}}_{4}=[(\begin{array}{cc}0& 0\\ 2& 0\end{array}),(\begin{array}{cc}5& 0\\ 0& 3\end{array})].\end{array}$$

*By Theorem 3.12 we can consider the above systems as I/S/O representations over* ℤ/6ℤ *obtaining the corresponding classes of feedback convolutional equivalence of families of convolutional codes. In order to compute them, we have added random pairs of matrices C and D*,
$$\begin{array}{}{\mathit{\Sigma}}_{1}=[A=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),B=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),C=(\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\phantom{\rule{thinmathspace}{0ex}}),D=(\phantom{\rule{thinmathspace}{0ex}}4\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}})].\end{array}$$

*Then, we obtain the minimal first order representation*
$$\begin{array}{}{K}_{1}=\left(\begin{array}{cc}5& 0\\ 0& 5\\ 0& 0\end{array}\right),{L}_{1}=\left(\begin{array}{cc}0& 0\\ 1& 0\\ 1& 0\end{array}\right),{M}_{1}=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 5& 4& 1\end{array}\right)\end{array}$$

*and the family of convolutional codes*
$$\begin{array}{}{\mathfrak{C}}_{1}({\mathit{\Sigma}}_{1})=\left(\begin{array}{cc}1z+4{z}^{2}& 1\\ 5{z}^{2}& 0\\ 0& 1\end{array}\right)\mathit{\text{\hspace{0.17em}where\hspace{0.17em}}}{\kappa}^{{\mathfrak{C}}_{1}({\mathit{\Sigma}}_{1})}=[(2,2),(0,0)].\end{array}$$

*The following system is Σ*_{2}:
$$\begin{array}{}{\mathit{\Sigma}}_{2}=[A=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),B=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),C=(\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}),D=(\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}})].\end{array}$$

*Then, we obtain the following minimal first order representation*
$$\begin{array}{}{K}_{2}=\left(\begin{array}{cc}5& 0\\ 0& 5\\ 0& 0\end{array}\right),{L}_{2}=\left(\begin{array}{cc}0& 0\\ 0& 0\\ 3& 2\end{array}\right),{M}_{2}=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 5& 1& 2\end{array}\right)\end{array}$$

*and the family of convolutional codes*
$$\begin{array}{}{\mathfrak{C}}_{2}({\mathit{\Sigma}}_{2})=\left(\begin{array}{cc}3+z& 2+2z\\ z& 0\\ 0& z\end{array}\right)\mathit{\text{\hspace{0.17em}where\hspace{0.17em}}}\phantom{\rule{thinmathspace}{0ex}}{\kappa}^{{\mathfrak{C}}_{2}({\mathit{\Sigma}}_{2})}=[(1,1),(1,1)].\end{array}$$

*We are going to compute the results for Σ*_{3}
$$\begin{array}{}{\mathit{\Sigma}}_{3}=[A=\left(\begin{array}{cc}0& 0\\ 3& 0\end{array}\right),B=\left(\begin{array}{cc}5& 0\\ 0& 2\end{array}\right),C=(\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}),D=(\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}})].\end{array}$$

*Then, we obtain the minimal first order representation*
$${K}_{3}=\left(\begin{array}{cc}5& 0\\ 0& 5\\ 0& 0\end{array}\right),{L}_{3}=\left(\begin{array}{cc}0& 0\\ 3& 0\\ 1& 3\end{array}\right),{M}_{3}=\left(\begin{array}{ccc}0& 5& 0\\ 0& 0& 2\\ 5& 2& 1\end{array}\right)$$

*and the family of convolutional codes*
$$\begin{array}{}{\mathfrak{C}}_{3}({\mathit{\Sigma}}_{3})=\left(\begin{array}{cc}1+z& 2z+3\\ 3{z}^{2}+2z& 0\\ 0& 2z+3\end{array}\right)\mathit{\text{\hspace{0.17em}where\hspace{0.17em}}}{\mathfrak{K}}^{{\mathfrak{C}}_{3}({\mathit{\Sigma}}_{3})}=[(2,1),(0,1)].\end{array}$$

*Finally*,
$$\begin{array}{}{\mathit{\Sigma}}_{4}=[A=\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right),B=\left(\begin{array}{cc}5& 0\\ 0& 3\end{array}\right),C=(\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}),D=(\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}})].\end{array}$$

*Then, we obtain its minimal first order representation*
$$\begin{array}{}{K}_{4}=\left(\begin{array}{cc}5& 0\\ 0& 5\\ 0& 0\end{array}\right),{L}_{4}=\left(\begin{array}{cc}0& 0\\ 2& 0\\ 1& 1\end{array}\right),{M}_{4}=\left(\begin{array}{ccc}0& 5& 0\\ 0& 0& 3\\ 5& 3& 2\end{array}\right)\end{array}$$

*and the following family of convolutional codes*
$$\begin{array}{}{\mathfrak{C}}_{4}({\mathit{\Sigma}}_{4})=\left(\begin{array}{cc}1+5z& 5\\ 3z+4{z}^{2}& 0\\ 0& 4+3z\end{array}\right)\mathit{\text{\hspace{0.17em}where\hspace{0.17em}}}{\mathfrak{K}}^{{\mathfrak{C}}_{4}({\mathit{\Sigma}}_{4})}=[(1,2),(1,0)].\end{array}$$

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