Algebraic and Polynomial Description

1 Division of the generator matrix into partial matrices
2 Generator matrix of a convolutional encoder with memory $m$
3 Generator matrix for convolutional encoder of rate $1/n$
4 GF(2) description forms of a digital filter
5 Application of the D–transform to rate $1/n$ convolution encoders
6 Transfer Function Matrix
7 Systematic convolutional codes
8 Equivalent systematic convolutional code
9 Filter structure with fractional–rational transfer function
10 Exercises for the chapter

Division of the generator matrix into partial matrices

Following the discussion in the earlier section "Linear Codes and Cyclic Codes" the code word $\underline{x}$ of a linear block code can be determined from the information word $\underline{u}$ and the generator matrix $\mathbf{G}$ in a simple way: $\underline{x} = \underline{u} \cdot { \boldsymbol{\rm G}}$. The following holds:

The vectors $\underline{u}$ and $\underline{x}$ have length $k$ $($bit count of an info word$)$ resp. $n$ $($bit count of a code word$)$ and $\mathbf{G}$ has dimension $k × n$ $(k$ rows and $n$ columns$)$.
In convolutional coding, on the other hand $\underline{u}$ and $\underline{x}$ denote sequences with $k\hspace{0.05cm}' → ∞$ and $n\hspace{0.05cm}' → ∞$.
Therefore, the generator matrix $\mathbf{G}$ will also be infinitely extended in both directions.

In preparation for the introduction of the generator matrix $\mathbf{G}$ on the next page,

we define $m + 1$ "partial matrices", each with $k$ rows and $n$ columns, which we denote by $\mathbf{G}_l$

where $0 ≤ l ≤ m$ holds.

$\text{Definition:}$ The »partial matrix« $\mathbf{G}_l$ describes the following fact:

If the matrix element $\mathbf{G}_l(\kappa, j) = 1$, this says that the code bit $x_i^{(j)}$ is influenced by the information bit $u_{i-l}^{(\kappa)}$.

Otherwise, this matrix element is $\mathbf{G}_l(\kappa, j) =0$.

This definition will now be illustrated by an example.

$\text{Example 1:}$ We again consider the convolutional encoder according to the diagram with the following code bits:

Convolutional encoder with $k = 2, \ n = 3, \ m = 1$

\[x_i^{(1)} = u_{i}^{(1)} + u_{i-1}^{(1)}+ u_{i-1}^{(2)} \hspace{0.05cm},\]

\[x_i^{(2)} = u_{i}^{(2)} + u_{i-1}^{(1)} \hspace{0.05cm},\]

\[x_i^{(3)} = u_{i}^{(1)} + u_{i}^{(2)}+ u_{i-1}^{(1)} \hspace{0.05cm}.\]

Because of the memory $m = 1$ this encoder is fully characterized by the partial matrices $\mathbf{G}_0$ and $\mathbf{G}_1$ :

\[{ \boldsymbol{\rm G} }_0 = \begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.5cm} { \boldsymbol{\rm G} }_1 = \begin{pmatrix} 1 & 1 & 1\\ 1 & 0 & 0 \end{pmatrix}\hspace{0.05cm}.\]

These matrices are to be interpreted as follows:

First row of $\mathbf{G}_0$, red arrows: $\hspace{1.3cm}u_i^{(1)}$ affects both $x_i^{(1)}$ and $x_i^{(3)}$, but not $x_i^{(2)}$.

Second row of $\mathbf{G}_0$, blue arrows: $\hspace{0.6cm}u_i^{(2)}$ affects $x_i^{(2)}$ and $x_i^{(3)}$, but not $x_i^{(1)}$.

First row of $\mathbf{G}_1$, green arrows: $\hspace{0.9cm}u_{i-1}^{(1)}$ affects all three encoder outputs.

Second row of $\mathbf{G}_1$, brown arrow: $\hspace{0.45cm}u_{i-1}^{(2)}$ affects only $x_i^{(1)}$.

Generator matrix of a convolutional encoder with memory $m$

The $n$ code bits at time $i$ can be expressed with the partial matrices $\mathbf{G}_0, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \mathbf{G}_m$ as follows:

\[\underline{x}_i = \sum_{l = 0}^{m} \hspace{0.15cm}\underline{u}_{i-l} \cdot { \boldsymbol{\rm G}}_l = \underline{u}_{i} \cdot { \boldsymbol{\rm G}}_0 + \underline{u}_{i-1} \cdot { \boldsymbol{\rm G}}_1 +\hspace{0.05cm} \text{...} \hspace{0.05cm} + \underline{u}_{i-m} \cdot { \boldsymbol{\rm G}}_m \hspace{0.05cm}.\]

The following vectorial quantities must be taken into account:

\[\underline{\it u}_i = \left ( u_i^{(1)}, u_i^{(2)}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, u_i^{(k)}\right )\hspace{0.05cm},\hspace{0.5cm} \underline{\it x}_i = \left ( x_i^{(1)}, x_i^{(2)}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, x_i^{(n)}\right )\hspace{0.05cm}.\]

Considering the sequences

\[\underline{\it u} = \big( \underline{\it u}_1\hspace{0.05cm}, \underline{\it u}_2\hspace{0.05cm}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline{\it u}_i\hspace{0.05cm}, \hspace{0.05cm}\text{...} \hspace{0.1cm} \big)\hspace{0.05cm},\hspace{0.5cm} \underline{\it x} = \big( \underline{\it x}_1\hspace{0.05cm}, \underline{\it x}_2\hspace{0.05cm}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline{\it x}_i\hspace{0.05cm}, \hspace{0.05cm}\text{...} \hspace{0.1cm} \big)\hspace{0.05cm},\]

starting at $i = 1$ and extending in time to infinity, this relation can be expressed by the matrix equation $\underline{x} = \underline{u} \cdot \mathbf{G}$. Here, holds for the generator matrix:

\[{ \boldsymbol{\rm G}}=\begin{pmatrix} { \boldsymbol{\rm G}}_0 & { \boldsymbol{\rm G}}_1 & { \boldsymbol{\rm G}}_2 & \cdots & { \boldsymbol{\rm G}}_m & & & \\ & { \boldsymbol{\rm G}}_0 & { \boldsymbol{\rm G}}_1 & { \boldsymbol{\rm G}}_2 & \cdots & { \boldsymbol{\rm G}}_m & &\\ & & { \boldsymbol{\rm G}}_0 & { \boldsymbol{\rm G}}_1 & { \boldsymbol{\rm G}}_2 & \cdots & { \boldsymbol{\rm G}}_m &\\ & & & \cdots & \cdots & & & \cdots \end{pmatrix}\hspace{0.05cm}.\]

From this equation one immediately recognizes the memory $m$ of the convolutional code.

The parameters $k$ and $n$ are not directly readable.

However, they are determined by the number of rows and columns of the partial matrices $\mathbf{G}_l$.

$\text{Example 2:}$ With the two matrices $\mathbf{G}_0$ and $\mathbf{G}_1$ – see $\text{Example 1}$ – the matrix sketched on the right $\mathbf{G}$ is obtained.

Generator matrix of a convolutional code

It should be noted:

The generator matrix $\mathbf{G}$ actually extends downwards and to the right to infinity. Explicitly shown, however, are only eight rows and twelve columns.

For the temporal information sequence $\underline{u} = (0, 1, 1, 0, 0, 0, 1, 1)$ the drawn matrix part is sufficient. The code sequence is then:

$$\underline{x} = (0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0).$$

On the basis of the label colors, the $n = 3$ code word strings can be read.

We got the same result $($in a different way$)$ in the $\text{Example 4}$ at the end of the last chapter:

$$\underline{\it x}^{(1)} = (0\hspace{0.05cm}, 0\hspace{0.05cm}, 1\hspace{0.05cm}, 1) \hspace{0.05cm},$$

$$\underline{\it x}^{(2)} = (1\hspace{0.05cm}, 0\hspace{0.05cm},1\hspace{0.05cm}, 1) \hspace{0.05cm},$$

$$ \underline{\it x}^{(3)} = (1\hspace{0.05cm}, 1\hspace{0.05cm}, 1\hspace{0.05cm}, 0) \hspace{0.05cm}.$$

Generator matrix for convolutional encoder of rate $1/n$

We now consider the special case $k = 1$,

on the one hand for reasons of simplest possible representation,

but also because convolutional encoders of rate $1/n$ have great importance for practice.

Convolutional encoder
$(k = 1, \ n = 2, \ m = 1)$

Convolutional encoder with $k = 1, \ n = 2, \ m = 1$

From the adjacent sketch can be derived:

$${ \boldsymbol{\rm G}}_0=\begin{pmatrix} 1 & 1 \end{pmatrix}\hspace{0.05cm},\hspace{0.3cm} { \boldsymbol{\rm G}}_1=\begin{pmatrix} 0 & 1 \end{pmatrix}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}$$

Thus, the resulting generator matrix is:

$${ \boldsymbol{\rm G}}=\begin{pmatrix} 11 & 01 & 00 & 00 & 00 & \cdots & \\ 00 & 11 & 01 & 00 & 00 & \cdots & \\ 00 & 00 & 11 & 01 & 00 & \cdots & \\ 00 & 00 & 00 & 11 & 01 & \cdots & \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{pmatrix}\hspace{0.05cm}.$$

For the input sequence $\underline{u} = (1, 0, 1, 1)$, the code sequence starts with $\underline{x} = (1, 1, 0, 1, 1, 1, 1, 0, \ \text{...})$.

This result is equal to the sum of rows 1, 3 and 4 of the generator matrix.

Convolutional encoder $(k = 1, \ n = 2, \ m = 2)$

Convolutional encoder with $k = 1, \ n = 2, \ m = 2$

Due to the memory order $m = 2$ there are three submatrices here:

\[{ \boldsymbol{\rm G}}_0=\begin{pmatrix} 1 & 1 \end{pmatrix}\hspace{0.05cm},\hspace{0.3cm} { \boldsymbol{\rm G}}_1=\begin{pmatrix} 1 & 0 \end{pmatrix}\hspace{0.05cm},\hspace{0.3cm} { \boldsymbol{\rm G}}_2=\begin{pmatrix} 1 & 1 \end{pmatrix}\]

Thus, the resulting generator matrix is now:

\[ { \boldsymbol{\rm G}}=\begin{pmatrix} 11 & 10 & 11 & 00 & 00 & 00 & \cdots & \\ 00 & 11 & 10 & 11 & 00 & 00 & \cdots & \\ 00 & 00 & 11 & 10 & 11 & 00 & \cdots & \\ 00 & 00 & 00 & 11 & 10 & 11 & \cdots & \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{pmatrix}\hspace{0.05cm}.\]

Here the input sequence $\underline{u} = (1, 0, 1, 1)$ leads to the code sequence $\underline{x} = (1, 1, 1, 0, 0, 0, 0, 1, \ \text{...})$.

Convolutional encoder $(k = 1, \ n = 3, \ m = 3)$

Convolutional encoder with $k = 1, \ n = 3, \ m = 3$

Because of $m = 3$ there are now four partial matrices of the respective dimension $1 × 3$:

\[{ \boldsymbol{\rm G}}_0=\begin{pmatrix} 1 & 1 & 0 \end{pmatrix}\hspace{0.05cm},\hspace{0.3cm} { \boldsymbol{\rm G}}_1=\begin{pmatrix} 0 & 0 & 1 \end{pmatrix}\hspace{0.05cm},\hspace{0.3cm} { \boldsymbol{\rm G}}_2=\begin{pmatrix} 0 & 0 & 1 \end{pmatrix}\hspace{0.05cm},\hspace{0.3cm} { \boldsymbol{\rm G}}_3=\begin{pmatrix} 0 & 1 & 1 \end{pmatrix}\hspace{0.05cm}.\]

Thus, the resulting generator matrix is:

\[{ \boldsymbol{\rm G}}=\begin{pmatrix} 110 & 001 & 001 & 011 & 000 & 000 & 000 & \cdots & \\ 000 & 110 & 001 & 001 & 011 & 000 & 000 & \cdots & \\ 000 & 000 & 110 & 001 & 001 & 011 & 000 & \cdots & \\ 000 & 000 & 000 & 110 & 001 & 001 & 011 & \cdots & \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{pmatrix}\hspace{0.05cm}.\]

One obtains for $\underline{u} = (1, 0, 1, 1)$ the code sequence $\underline{x} = (1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, \ \text{...})$.

GF(2) description forms of a digital filter

Digital filter in ${\rm GF}(2)$ of order $m$

In the chapter "Basics of Convolutional Coding" it was already pointed out,

that a rate $1/n$ convolutional encoder can be realized by several digital filters,
where the filters operate in parallel with the same input sequence $\underline{u}$ .

Before we elaborate on this statement, we shall first mention the properties of a digital filter for the Galois field ${\rm GF(2)}$.

The graph is to be interpreted as follows:

The filter has impulse response $\underline{g} = (g_0,\ g_1,\ g_2, \ \text{...} \ ,\ g_m)$.

For all filter coefficients $($with indices $0 ≤ l ≤ m)$ holds: $g_l ∈ {\rm GF}(2) = \{0, 1\}$.

The individual symbols $u_i$ of the input sequence $\underline{u}$ are also binary: $u_i ∈ \{0, 1\}$.

Thus, for the output symbol at times $i ≥ 1$ with addition and multiplication in ${\rm GF(2)}$:

\[x_i = \sum_{l = 0}^{m} g_l \cdot u_{i-l} \hspace{0.05cm}.\]

This corresponds to the $($discrete time$)$ »$\rm convolution$«, denoted by an asterisk. This can be used to write for the entire output sequence:

\[\underline{x} = \underline{u} * \underline{g}\hspace{0.05cm}.\]

Major difference compared to the chapter »$\text{Digital Filters}$« in the book "Theory of Stochastic Signals" is the modulo-2 addition $(1 + 1 = 0)$ instead of the conventional addition $(1 + 1 = 2)$.

$\text{Example 3:}$ The impulse response of the shown third order digital filter is: $\underline{g} = (1, 0, 1, 1)$.

Digital filter with impulse response $(1, 0, 1, 1)$

Let the input sequence of this filter be unlimited in time: $\underline{u} = (1, 1, 0, 0, 0, \ \text{ ...})$.

This gives the $($infinite$)$ initial sequence $\underline{x}$ in the binary Galois field ⇒ ${\rm GF(2)}$:

\[\underline{x} = (\hspace{0.05cm}1,\hspace{0.05cm} 1,\hspace{0.05cm} 0,\hspace{0.05cm} 0,\hspace{0.05cm} 0, \hspace{0.05cm} \text{ ...} \hspace{0.05cm}) * (\hspace{0.05cm}1,\hspace{0.05cm} 0,\hspace{0.05cm} 1,\hspace{0.05cm} 1\hspace{0.05cm})\]

\[\Rightarrow \hspace{0.3cm} \underline{x} =(\hspace{0.05cm}1,\hspace{0.05cm} 0,\hspace{0.05cm} 1,\hspace{0.05cm} 1,\hspace{0.05cm} 0, \hspace{0.05cm}0,\hspace{0.05cm} \text{ ...} \hspace{0.05cm}) \oplus (\hspace{0.05cm}0,\hspace{0.05cm}\hspace{0.05cm}1,\hspace{0.05cm} 0,\hspace{0.05cm} 1,\hspace{0.05cm} 1,\hspace{0.05cm}0, \hspace{0.05cm} \hspace{0.05cm} \text{ ...}\hspace{0.05cm}) = (\hspace{0.05cm}1,\hspace{0.05cm}\hspace{0.05cm}1,\hspace{0.05cm} 1,\hspace{0.05cm} 0,\hspace{0.05cm} 1,\hspace{0.05cm} 0, \hspace{0.05cm} \text{ ...} \hspace{0.05cm}) \hspace{0.05cm}.\]

In the conventional convolution $($for real numbers$)$, on the other hand, the result would have been:

\[\underline{x}= (\hspace{0.05cm}1,\hspace{0.05cm}\hspace{0.05cm}1,\hspace{0.05cm} 1,\hspace{0.05cm} 2,\hspace{0.05cm} 1,\hspace{0.05cm} 0, \text{ ...} \hspace{0.05cm}) \hspace{0.05cm}.\]

However, discrete time signals can also be represented by polynomials with respect to a dummy variable.

$\text{Definition:}$ The »D–transform« belonging to the discrete time signal $\underline{x} = (x_0, x_1, x_2, \ \text{...}) $ reads:

\[X(D) = x_0 + x_1 \cdot D + x_2 \cdot D^2 + \hspace{0.05cm}\text{...}\hspace{0.05cm}= \sum_{i = 0}^{\infty} x_i \cdot D\hspace{0.05cm}^i \hspace{0.05cm}.\]

For this particular transformation to an image area, we also use the following notation, where "D" stands for "delay operator":

\[\underline{x} = (x_0, x_1, x_2,\hspace{0.05cm}...\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad X(D) = \sum_{i = 0}^{\infty} x_i \cdot D\hspace{0.05cm}^i \hspace{0.05cm}.\]

Note: In the literature, sometimes $x(D)$ is used instead of $X(D)$. However, we write in our learning tutorial all image domain functions ⇒ "spectral domain functions" with capital letters,&nbsp: for example the Fourier transform, the Laplace transform and the D–transform:

\[x(t) \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}}\!\!\!-\!\!\bullet\hspace{0.15cm} X(f)\hspace{0.05cm},\hspace{0.4cm} x(t) \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\bullet\hspace{0.15cm} X(p) \hspace{0.05cm},\hspace{0.4cm} \underline{x} \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm} X(D) \hspace{0.05cm}.\]

We now apply the D–transform also

to the information sequence $\underline{u}$, and

the impulse response $\underline{g}$.

Due to the time limit of $\underline{g}$ the upper summation limit at $G(D)$ results in $i = m$:

\[\underline{u} = (u_0, u_1, u_2,\hspace{0.05cm}\text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad U(D) = \sum_{i = 0}^{\infty} u_i \cdot D\hspace{0.05cm}^i \hspace{0.05cm},\]

\[\underline{g} = (g_0, g_1, \hspace{0.05cm}\text{...}\hspace{0.05cm}, g_m) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad G(D) = \sum_{i = 0}^{m} g_i \cdot D\hspace{0.05cm}^i \hspace{0.05cm}.\]

$\text{Theorem:}$ As with all spectral transformations, the multiplication applies to the D–transform in the image domain, since the $($discrete$)$ time functionss $\underline{u}$ and $\underline{g}$ are interconnected by the »convolution«:

\[\underline{x} = \underline{u} * \underline{g} \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad X(D) = U(D) \cdot G(D) \hspace{0.05cm}.\]

The $($rather simple$)$ $\rm proof$ of this important result can be found in the specification for $\text{Exercise 3.3Z}$.

As in »$\text{system theory}$« commonly, the D–transform $G(D)$ of the impulse response $\underline{g}$ is also called "transfer function".

Impulse response $(1, 0, 1, 1)$ of a digital filter

$\text{Example 4:}$ We consider again the discrete time signals

\[\underline{u} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad U(D) = 1+ D \hspace{0.05cm},\]

\[\underline{g} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad G(D) = 1+ D^2 + D^3 \hspace{0.05cm}.\]

As in $\text{Example 3}$ $($on this page above$)$, you get also on this solution path:

\[X(D) = U(D) \cdot G(D) = (1+D) \cdot (1+ D^2 + D^3) \]

\[\Rightarrow \hspace{0.3cm} X(D) = 1+ D^2 + D^3 +D + D^3 + D^4 = 1+ D + D^2 + D^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{x} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}, \text{...} \hspace{0.05cm}) \hspace{0.05cm}.\]

Multiplication by the "delay operator" $D$ in the image domain corresponds to a shift of one place to the right in the time domain:

\[W(D) = D \cdot X(D) \quad \bullet\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\circ\quad \underline{w} = (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}, \text{...} \hspace{0.05cm}) \hspace{0.05cm}.\]

Application of the D–transform to rate $1/n$ convolution encoders

We now apply the results of the last page to a convolutional encoder, restricting ourselves for the moment to the special case $k = 1$.

Such a $(n, \ k = 1)$ convolutional code can be realized with $n$ digital filters operating in parallel on the same information sequence $\underline{u}$.

The graph shows the arrangement for the code parameter $n = 2$ ⇒ code rate $R = 1/2$.

Two filters working in parallel, each with order $m$

The following equations apply equally to both filters, setting $j = 1$ for the upper filter and $j = 2$ for the lower filter:

The impulse responses of the two filters result in

\[\underline{g}^{(j)} = (g_0^{(j)}, g_1^{(j)}, \hspace{0.05cm}\text{...}\hspace{0.05cm}, g_m^{(j)}\hspace{0.01cm}) \hspace{0.05cm},\hspace{0.2cm}{\rm with }\hspace{0.15cm} j \in \{1,2\}\hspace{0.05cm}.\]

The two output sequences are as follows, considering that both filters operate on the same input sequence $\underline{u} = (u_0, u_1, u_2, \hspace{0.05cm} \text{...})$ :

\[\underline{x}^{(j)} = (x_0^{(j)}, x_1^{(j)}, x_2^{(j)}, \hspace{0.05cm}\text{...}\hspace{0.05cm}) = \underline{u} \cdot \underline{g}^{(j)} \hspace{0.05cm},\hspace{0.2cm}{\rm with }\hspace{0.15cm} j \in \{1,2\}\hspace{0.05cm}.\]

For the D–transform of the output sequences:

\[X^{(j)}(D) = U(D) \cdot G^{(j)}(D) \hspace{0.05cm},\hspace{0.2cm}{\rm with }\hspace{0.15cm} j \in \{1,2\}\hspace{0.05cm}.\]

In order to represent this fact more compactly, we now define the following vectorial quantities of a convolutional code of rate $1/n$:

$\text{Definition:}$ The »D– transfer functions« of the $n$ parallel arranged digital filters are combined in the vector $\underline{G}(D)$:

\[\underline{G}(D) = \left ( G^{(1)}(D), G^{(2)}(D), \hspace{0.05cm}\text{...}\hspace{0.1cm}, G^{(n)} (D) \right )\hspace{0.05cm}.\]

The vector $\underline{X}(D)$ contains the D–transform of $n$ code sequences $\underline{x}^{(1)}, \underline{x}^{(2)}, \ \text{...} \ , \underline{x}^{(n)}$:

\[\underline{X}(D) = \left ( X^{(1)}(D), X^{(2)}(D), \hspace{0.05cm}\text{...}\hspace{0.1cm}, X^{(n)} (D) \right )\hspace{0.05cm}.\]

This gives the following vector equation:

\[\underline{X}(D) = U(D) \cdot \underline{G}(D)\hspace{0.05cm}.\]

$U(D)$ is not a vector quantity here because of the code parameter $k = 1$.

$\text{Example 5:}$ We consider the convolutional encoder with code parameters $n = 2, \ k = 1, \ m = 2$. For this one holds:

Convolutional encoder $(n = 2, \ k = 1,\ m = 2)$

\[\underline{g}^{(1)} =(\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad G(D) = 1+ D + D^2 \hspace{0.05cm},\]

\[\underline{g}^{(2)}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad G(D) = 1+ D^2 \]

\[\Rightarrow \hspace{0.3cm} \underline{G}(D) = \big ( 1+ D + D^2 \hspace{0.05cm}, \hspace{0.1cm}1+ D^2 \big )\hspace{0.05cm}.\]

Let the information sequence be $\underline{u} = (1, 0, 1, 1)$ ⇒ D–transform $U(D) = 1 + D^2 + D^3$. This gives:

\[\underline{X}(D) = \left ( X^{(1)}(D),\hspace{0.1cm} X^{(2)}(D) \right ) = U(D) \cdot \underline{G}(D) \hspace{0.05cm}, \hspace{0.2cm}\]

where

\[{X}^{(1)}(D) = (1+ D^2 + D^3) \cdot (1+ D + D^2)=1+ D + D^2 + D^2 + D^3 + D^4 + D^3 + D^4 + D^5 = 1+ D + D^5\]

\[\Rightarrow \hspace{0.3cm} \underline{x}^{(1)} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} \text{...} \hspace{0.05cm} \hspace{0.05cm}) \hspace{0.05cm},\]

\[{X}^{(2)}(D) = (1+ D^2 + D^3) \cdot (1+ D^2)=1+ D^2 + D^2 + D^4 + D^3 + D^5 = 1+ D^3 + D^4 + D^5\]

\[\Rightarrow \underline{x}^{(2)} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} \text{...} \hspace{0.05cm} \hspace{0.05cm}) \hspace{0.05cm}.\]

We got the same result in $\text{Exercise 3.1Z}$ on other way. After multiplexing the two strands, you get again:

$$\underline{x} = (11, 10, 00, 01, 01, 11, 00, 00, \hspace{0.05cm} \text{...} \hspace{0.05cm}).$$

Transfer Function Matrix

We have seen that a convolutional code of rate $1/n$ can be most compactly described as a vector equation in the D–transformed domain:

General $(n, \ k)$ convolutional encoder

$$\underline{X}(D) = U(D) \cdot \underline{G}(D).$$

Now we extend the result to convolutional encoders with more than one input ⇒ $k ≥ 2$ $($see graph$)$.

In order to map a convolutional code of rate $k/n$ in the D&ndashdomain, the dimension of the above vector equation must be increased with respect to input and transfer function:

\[\underline{X}(D) = \underline{U}(D) \cdot { \boldsymbol{\rm G}}(D)\hspace{0.05cm}.\]

This requires the following measures:

From the scalar function $U(D)$ we get the vector

$$\underline{U}(D) = (U^{(1)}(D), \ U^{(2)}(D), \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ U^{(k)}(D)).$$

From the vector $\underline{G}(D)$ we get the $k × n$ transfer function matrix $($or "polynomial generator matrix"$)$ $\mathbf{G}(D)$ :

\[{\boldsymbol{\rm G}}(D)=\begin{pmatrix} G_1^{(1)}(D) & G_1^{(2)}(D) & \hspace{0.05cm} \text{...} \hspace{0.05cm} & G_1^{(n)}(D)\\ G_2^{(1)}(D) & G_2^{(2)}(D) & \hspace{0.05cm} \text{...} \hspace{0.05cm} & G_2^{(n)}(D)\\ \vdots & \vdots & & \vdots\\ G_k^{(1)}(D) & G_k^{(2)}(D) & \hspace{0.05cm} \text{...} \hspace{0.05cm} & G_k^{(n)}(D) \end{pmatrix}\hspace{0.05cm}.\]

Each of the $k \cdot n$ matrix elements $G_i^{(j)}(D)$ with $1 ≤ i ≤ k,\ 1 ≤ j ≤ n$ is a polynomial over the dummy variable $D$ in the Galois field ${\rm GF}(2)$, maximal of degree $m$, where $m$ denotes the memory.

For the above transfer function matrix, using the $\text{partial matrices}$ $\mathbf{G}_0, \ \text{...} \ , \mathbf{G}_m$ also be written $($as index we use again $l)$:

\[{\boldsymbol{\rm G}}(D) = \sum_{l = 0}^{m} {\boldsymbol{\rm G}}_l \cdot D\hspace{0.03cm}^l = {\boldsymbol{\rm G}}_0 + {\boldsymbol{\rm G}}_1 \cdot D + {\boldsymbol{\rm G}}_2 \cdot D^2 + \hspace{0.05cm} \text{...} \hspace{0.05cm}+ {\boldsymbol{\rm G}}_m \cdot D\hspace{0.03cm}^m \hspace{0.05cm}.\]

$\text{Example 6:}$ We consider the $(n = 3, \ k = 2, \ m = 1)$ convolutional encoder whose partial matrices have already been determined in the $\text{Example 1}$ as follows:

Convolutional encoder with $k = 2, \ n = 3, \ m = 1$

\[{ \boldsymbol{\rm G} }_0 = \begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.5cm} { \boldsymbol{\rm G} }_1 = \begin{pmatrix} 1 & 1 & 1\\ 1 & 0 & 0 \end{pmatrix}\hspace{0.05cm}.\]

Because of $m = 1$ no partial matrices exist for $l ≥ 2$. Thus the transfer function matrix is:

\[{\boldsymbol{\rm G} }(D) = {\boldsymbol{\rm G} }_0 + {\boldsymbol{\rm G} }_1 \cdot D = \begin{pmatrix} 1+D & D & 1+D\\ D & 1 & 1 \end{pmatrix} \hspace{0.05cm}.\]

Let the $($time limited$)$ information sequence be $\underline{u} = (0, 1, 1, 0, 0, 0, 1, 1)$, from which the two input sequences are as follows:

\[\underline{u}^{(1)} = (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad {U}^{(1)}(D) = D + D^3 \hspace{0.05cm},\]

\[\underline{u}^{(2)} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad {U}^{(2)}(D) = 1 + D^3 \hspace{0.05cm}.\]

From this follows for the vector of the D–transform at the encoder output:

\[\underline{X}(D) = \big (\hspace{0.05cm} {X}^{(1)}(D)\hspace{0.05cm}, \hspace{0.05cm} {X}^{(2)}(D)\hspace{0.05cm}, \hspace{0.05cm} {X}^{(3)}(D)\hspace{0.05cm}\big ) = \underline{U}(D) \cdot {\boldsymbol{\rm G} }(D) \begin{pmatrix} D+D^3 & 1+D^3 \end{pmatrix} \cdot \begin{pmatrix} 1+D & D & 1+D\\ D & 1 & 1 \end{pmatrix}\hspace{0.05cm}.\]

This results in the following code sequences in the three strands:

\[{X}^{(1)}(D) = (D + D^3) \cdot (1+D) + (1 + D^3) \cdot D =D + D^2 + D^3 + D^4 + D + D^4 = D^2 + D^3\]

\[\Rightarrow \hspace{0.3cm} \underline{x}^{(1)} = (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm}) \hspace{0.05cm},\]

\[{X}^{(2)}(D)= (D + D^3) \cdot D + (1 + D^3) \cdot 1 = D^2 + D^4 + 1 + D^3 = 1+D^2 + D^3 + D^4\]

\[\Rightarrow \hspace{0.3cm}\underline{x}^{(2)} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\hspace{0.05cm} \text{...} \hspace{0.05cm}) \hspace{0.05cm},\]

\[{X}^{(3)}(D)=(D + D^3) \cdot (1 + D) + (1 + D^3) \cdot 1 = D + D^2 + D^3+ D^4 + 1 + D^3 = 1+ D + D^2 + D^4\]

\[\Rightarrow \hspace{0.3cm}\underline{x}^{(3)} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\hspace{0.05cm} \text{...} \hspace{0.05cm}) \hspace{0.05cm}.\]

We have already obtained the same results in other ways in previous examples:

in $\text{Example 4}$ the chapter "Basics of Convolutional Coding",

in $\text{Example 2}$ of the current chapter.

Systematic convolutional codes

Polynomial representation using the transfer function matrix $\mathbf{G}(D)$ provides insight into the structure of a convolutional code.

Systematic convolutional code with $k = 3, \ n = 4$

For example, this $k × n$ matrix is used to recognize whether it is a $\text{systematic code}$.

This refers to a code where the code sequences $\underline{x}^{(1)}, \ \text{...} \ , \ \underline{x}^{(k)}$ are identical with the information sequences $\underline{u}^{(1)}, \ \text{...} \ , \ \underline{u}^{(k)}$.

The graph shows an example of a systematic $(n = 4, \ k = 3)$ convolutional code.

A systematic $(n, k)$ convolutional code exists whenever the transfer function matrix $($with $k$ rows and $n$ columns$)$ has the following appearance:

\[{\boldsymbol{\rm G}}(D) = {\boldsymbol{\rm G}}_{\rm sys}(D) = \left [ \hspace{0.05cm} {\boldsymbol{\rm I}}_k\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm P}}(D) \hspace{0.05cm}\right ] \hspace{0.05cm}.\]

The following nomenclature is used:

$\mathbf{I}_k$ denotes a diagonal unit matrix of dimension $k × k$.
$\mathbf{P}(D)$ is a $k × (n -k)$ matrix, where each matrix element describes a polynomial in $D$.

$\text{Example 7:}$ A systematic convolutional code with $n = 3, \ k = 2, \ m = 2$ might have the following transfer function matrix:

\[{\boldsymbol{\rm G} }_{\rm sys}(D) = \begin{pmatrix} 1 & 0 & 1+D^2\\ 0 & 1 & 1+D \end{pmatrix}\hspace{0.05cm}.\]

In contrast, other systematic convolutional codes with equal $n$ and equal $k$ differ only by the two matrix elements in the last column.

Equivalent systematic convolutional code

For every $(n, \ k)$ convolutional code with matrix $\mathbf{G}(D)$ there is an "equivalent systematic code" whose D–matrix we denote by $\mathbf{G}_{\rm sys}(D)$.

To get from the transfer function matrix $\mathbf{G}(D)$ to the matrix $\mathbf{G}_{\rm sys}(D)$ of the equivalent systematic convolutional code, proceed as follows according to the diagram:

Subdivision of $\mathbf{G}(D)$ into $\mathbf{T}(D)$ and $\mathbf{Q}(D)$

Divide the $k × n$ matrix $\mathbf{G}(D)$ into a square matrix $\mathbf{T}(D)$ with $k$ rows and $k$ columns and denote the remainder by $\mathbf{Q}(D)$.

Then calculate the inverse matrix $\mathbf{T}^{-1}(D)$ to $\mathbf{T}(D)$ and from this the matrix for the equivalent systematic code:

\[{\boldsymbol{\rm G}}_{\rm sys}(D)= {\boldsymbol{\rm T}}^{-1}(D) \cdot {\boldsymbol{\rm G}}(D) \hspace{0.05cm}.\]

Since the product $\mathbf{T}^{-1}(D) \cdot \mathbf{T}(D)$ yields the $k × k$ unit matrix $\mathbf{I}_k$
⇒ the transfer function matrix of the equivalent systematic code can be written in the desired form:

\[{\boldsymbol{\rm G}}_{\rm sys}(D) = \bigg [ \hspace{0.05cm} {\boldsymbol{\rm I}}_k\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm P}}(D) \hspace{0.05cm}\bigg ] \hspace{0.5cm}{\rm with}\hspace{0.5cm} {\boldsymbol{\rm P}}(D)= {\boldsymbol{\rm T}}^{-1}(D) \cdot {\boldsymbol{\rm Q}}(D) \hspace{0.05cm}. \hspace{0.05cm}\]

$\text{Example 8:}$ The coder of rate $2/3$ considered often in the last sections is not systematic because e.g. $\underline{x}^{(1)} ≠ \underline{u}^{(1)}, \ \underline{x}^{(2)} ≠ \underline{u}^{(2)}$ holds $($see adjacent graphic$)$.

Convolutional encoder of rate $2/3$

⇒ However, this can also be seen from the transfer function matrix:

\[{\boldsymbol{\rm G} }(D) = \big [ \hspace{0.05cm} {\boldsymbol{\rm T} }(D)\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm Q} }(D) \hspace{0.05cm}\big ]\]

\[\Rightarrow \hspace{0.3cm} {\boldsymbol{\rm T} }(D) = \begin{pmatrix} 1+D & D\\ D & 1 \end{pmatrix}\hspace{0.05cm},\hspace{0.2cm} {\boldsymbol{\rm Q} }(D) = \begin{pmatrix} 1+D \\ 1 \end{pmatrix}\hspace{0.05cm}.\]

The determinant of $\mathbf{T}(D)$ results in $(1 + D) \cdot 1 + D \cdot D = 1 + D + D^2$ and is nonzero.

Thus, for the inverse of $\mathbf{T}(D)$ can be written $($swapping the diagonal elements!$)$:

\[{\boldsymbol{\rm T} }^{-1}(D) = \frac{1}{1+D+D^2} \cdot \begin{pmatrix} 1 & D\\ D & 1+D \end{pmatrix}\hspace{0.05cm}.\]

The product $\mathbf{T}(D) \cdot \mathbf{T}^{–1}(D)$ gives the unit matrix $\mathbf{I}_2$ ⇒ for the third column of $\mathbf{G}_{\rm sys}(D)$ holds:

\[{\boldsymbol{\rm P} }(D)= {\boldsymbol{\rm T} }^{-1}(D) \cdot {\boldsymbol{\rm Q} }(D) = \frac{1}{1+D+D^2} \cdot \begin{pmatrix} 1 & D\\ D & 1+D \end{pmatrix}\cdot \begin{pmatrix} 1+D\\ 1 \end{pmatrix} \]

\[\Rightarrow \hspace{0.3cm} {\boldsymbol{\rm P} }(D) = \frac{1}{1+D+D^2} \cdot \begin{pmatrix} (1+D) + D \\ D \cdot (1+D) + (1+D) \end{pmatrix} = \frac{1}{1+D+D^2} \cdot \begin{pmatrix} 1 \\ 1+D^2 \end{pmatrix} \]

\[\Rightarrow \hspace{0.2cm}{\boldsymbol{\rm G} }_{\rm sys}(D) = \begin{pmatrix} 1 & 0 & \frac{1}{1+D+D^2}\\ 0 & 1 &\frac{1+D^2}{1+D+D^2} \end{pmatrix}\hspace{0.05cm}. \]

It remains to be clarified what the filter of such a fractional–rational transfer function looks like.

Filter structure with fractional–rational transfer function

If a transfer function has the form $G(D) = A(D)/B(D)$, the associated filter is called »recursive«. Given a recursive convolutional encoder with memory $m$, the two polynomials $A(D)$ and $B(D)$ can be written in general terms:

Recursive filter for realization of $G(D) = A(D)/B(D)$

\[A(D) = \sum_{l = 0}^{m} a_l \cdot D\hspace{0.05cm}^l = a_0 + a_1 \cdot D + a_2 \cdot D^2 +\ \text{...} \ \hspace{0.05cm} + a_m \cdot D\hspace{0.05cm}^m \hspace{0.05cm},\]

\[B(D) = 1 + \sum_{l = 1}^{m} b_l \cdot D\hspace{0.05cm}^l = 1 + b_1 \cdot D + b_2 \cdot D^2 + \ \text{...} \ \hspace{0.05cm} + b_m \cdot D\hspace{0.05cm}^m \hspace{0.05cm}.\]

The graphic shows the corresponding filter structure in the so–called "Controller Canonical Form":