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 codeword $\underline{x}$ 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 information word), respectively. $n$ (bit count of a codeword) 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 equal to $0$.

This definition will now be illustrated by an example.

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

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

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

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

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

Because of the memory order $m = 1$ this encoder is fully characterized by the two 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: $\hspace{1.1cm}u_i^{(1)}$ 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

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

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

From the 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 $\mathbf{G}_l$ .

Generator matrix of a convolutional code

$\text{Example 2:}$ 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.

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 $n = 3$ codeword 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:

$$\underline{\it x}^{(1)} = (0\hspace{0.05cm}, 0\hspace{0.05cm}, 1\hspace{0.05cm}, 1) \hspace{0.05cm},\hspace{0.5cm} \underline{\it x}^{(2)} = (1\hspace{0.05cm}, 0\hspace{0.05cm},1\hspace{0.05cm}, 1) \hspace{0.05cm},\hspace{0.5cm} \underline{\it x}^{(3)} = (1\hspace{0.05cm}, 1\hspace{0.05cm}, 1\hspace{0.05cm}, 0) \hspace{0.05cm}.$$

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

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

From the adjacent sketch can be derived:

\[{ \boldsymbol{\rm G}}_0=\begin{pmatrix} 1 & 1 \end{pmatrix}\hspace{0.05cm},\hspace{0.3cm} { \boldsymbol{\rm G}}_1=\begin{pmatrix} 0 & 1 \end{pmatrix}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} { \boldsymbol{\rm G}}=\begin{pmatrix} 11 & 01 & 00 & 00 & 00 & \cdots & \\ 00 & 11 & 01 & 00 & 00 & \cdots & \\ 00 & 00 & 11 & 01 & 00 & \cdots & \\ 00 & 00 & 00 & 11 & 01 & \cdots & \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \end{pmatrix}\hspace{0.05cm}.\]

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

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:

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

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

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:

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

and 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) "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 "Digital Filters" in the book "Theory of Stochastic Signals" is the modulo 2 addition $(1 + 1 = 0)$ instead of the conventional addition $(1 + 1 = 2)$.

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

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

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 $\underline{x} = (x_0, x_1, x_2, \ \text{...})$ belonging to the discrete time signal $D$ transform 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 all image domain functions with capital letters in our learning tutorial, for example the Fourier–, the Laplace 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, multiplication applies to the $D$ transform in the image domain, since the (discrete) time signals $\underline{u}$ and $\underline{g}$ are linked by the convolution :

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

One – as in the "System theory" commonly – also the $D$ transform $G(D)$ of the impulse response $\underline{g}$ as transfer function. The (rather simple) $\rm proof$ of this important result can be found in the specification for "Exercise 3.3Z".

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

$\text{Example 4:}$ 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 the $\text{Example 3}$ (on this page above), you also get 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 $D$ in the image domain corresponds to a shift of one place to the right in the time domain, which is why $D$ is called the delay operator:

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

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

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

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

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

For the $D$ transform of the output sequences:

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

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:

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

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

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

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

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

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

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

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 the "Exercise 3.1Z" 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

General $(n, \ k)$ convolutional encoder

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

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 $D$ domain, 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 $\underline{G}(D)$ 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 memory.

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

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

Faltungscoder mit $k = 2, \ n = 3, \ m = 1$

$\text{Example 6:}$ 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:

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

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 $D$& transforms 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.

For example, this $k × n$ matrix is used to recognize whether it is a "systematic code" .
This refers to a code where the code sequences $\underline{x}^{(1)}, \ \text{...} \ , \ \underline{x}^{(k)}$ with the information sequences $\underline{u}^{(1)}, \ \text{...} \ , \ \underline{u}^{(k)}$ are identical.

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

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

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 $D$ .

$\text{Example 7:}$ For example, 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 $\mathbf{G}_{\rm sys}(D)$.

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

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:

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 to $\mathbf{T}(D)$ $\mathbf{T}^{-1}(D)$ 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 $\mathbf{T}^{-1}(D) \cdot \mathbf{T}(D)$ 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:

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

Convolutional encoder of rate $2/3$

$\text{Example 8:}$ The coder of rate $2/3$ considered more often in the last pages is not systematic because, for example $\underline{x}^{(1)} ≠ \underline{u}^{(1)}, \ \underline{x}^{(2)} ≠ \underline{u}^{(2)}$ holds (see adjacent encoder circuit).

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 $\mathbf{I}_2$, and 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

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

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:

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

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

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

The coefficients $a_0, \ \text{...} \ , \ a_m$ describe the forward branch.
The coefficients $b_1, \ \text{...} \ , \ b_m$ form a feedback branch.
All coefficients are binary, so $1$ (continuous connection) or $0$ (missing connection).