Aufgaben:Exercise 3.4Z: Equivalent Convolution Codes?

• Return to book

Non-systematic and systematic convolutional encoder

The top figure shows a convolutional encoder described by the following equations:

$$x_i^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i}^{(1)} + u_{i-1}^{(1)}+ u_{i-1}^{(2)} \hspace{0.05cm},$$

$$x_i^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i}^{(2)} + u_{i-1}^{(2)} \hspace{0.05cm},$$

$$x_i^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i}^{(2)}\hspace{0.05cm}.$$

We are looking for the transfer function matrices

• $\mathbf{G}(D)$  of this non-systematic code,  and

• $\mathbf{G}_{\rm sys}(D)$  of the equivalent systematic code.

The matrix  $\mathbf{G}_{\rm sys}(D)$  is obtained in the following way:

• One splits off from the  $k × n$  matrix  $\mathbf{G}(D)$  in front a square matrix  $\mathbf{T}(D)$  with  $k$  rows and  $k$  columns.  The remainder is denoted by  $\mathbf{Q}(D)$.

• Calculate the inverse matrix  $\mathbf{T}^{-1}(D)$  of   $\mathbf{T}(D)$.  From this calculate 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  $k × k$  identity matrix  $\mathbf{I}_k$,  the transfer function matrix of the equivalent systematic code can be written in the desired form:

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

• The circuit below will certainly generate a systematic code with the same parameters  $k$  and  $n$.

In subtask  (5)  it has to be clarified whether this is indeed the  "equivalent systematic code".  That is,  whether in fact for the two circuits exactly the same quantity   $\{ \hspace{0.1cm} \underline{x} \hspace{0.1cm}\}$   of code sequences results when all possible information sequences   $\{ \hspace{0.1cm} \underline{u} \hspace{0.1cm} \}$   are taken into account.

Hints:

• This exercise belongs to the chapter  "Algebraic and Polynomial Description".
• Reference is made in particular to the sections 

• "Transfer Function Matrix"  and 
• "Equivalent systematic convolutional code".

Questions

1

What are the parameters of the encoder shown above?

$k \hspace{0.25cm} = \ $

$n \hspace{0.22cm} = \ $

$m \hspace{0.10cm} = \ $

$ν \hspace{0.28cm} = \ $

$R \hspace{0.18cm} = \ $

2

What is the form of the transfer function matrix  $\mathbf{G}(D)$?

	The first row of  $\mathbf{G}(D)$  is  $(1 + D, \, 0, \, 0)$.
	The first row of  $\mathbf{G}(D)$  is  $(1 + D^2, \, 0, \, D^2)$.
	The second row of  $\mathbf{G}(D)$  is  $(D, \, 1 + D, \, 1)$.
	The third row of  $\mathbf{G}(D)$  is  $(D, \, 1 + D, \, 1)$.

3

Specify   $\mathbf{T}(D)$  and   $\mathbf{T}^{-1}(D)$.  What is the determinant?

	$\det {\mathbf{T}(D)} = 1$,
	$\det {\mathbf{T}(D)} = D$,
	$\det {\mathbf{T}(D)} = 1 + D^2$.

4

What is true for the equivalent systematic transfer function matrix?

	The first row of  $\mathbf{G}_{\rm sys}(D)$  is  $(1, \, 0, \, 0)$.
	The second row of  $\mathbf{G}_{\rm sys}(D)$  is  $(0, \, 1, \, 1 + D)$.
	The second row of  $\mathbf{G}_{\rm sys}(D)$  is  $(0, \, 1, \, 1/(1 + D))$.

5

Are the two given circuits actually equivalent?

	YES.
	NO.

Solution

Solution

(1)  Here  $\underline{k = 2}$  and  $\underline{n = 3}$   ⇒   Rate  $\underline{R = 2/3}$.

• The memory order  $\underline{m = 1}$  $($number of memory elements per input$)$.

• The influence length is equal to the sum of all memory elements   ⇒   $\underline{\nu = 2}$.

(2)  The information bit  $u_i^{(1)}$  affects only the first output  $x_i^{(1)}$,  while  $u_i^{(2)}$  is used for  $x_i^{(2)}$  and  $x_i^{(3)}$.

• Thus,  for the zeroth  "partial matrix"  is obtained:

$${ \boldsymbol{\rm G}}_0 = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 1 \end{pmatrix} \hspace{0.05cm}. $$

• The delayed inputs affect as follows:

• $u_{i–1}^{(1)}$  affects  $x_i^{(1)}$,

• $u_{i–1}^{(2)}$  affects  $x_i^{(1)}$  and  $x_i^{(2)}$:

• Thus,  the partial matrix  $\mathbf{G}_1$  and the transfer function matrix  $\mathbf{G}(D)$:

$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 & 0 & 0\\ 1 & 1 & 0 \end{pmatrix} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} { \boldsymbol{\rm G}}(D) = { \boldsymbol{\rm G}}_0 + { \boldsymbol{\rm G}}_1 \cdot D = \begin{pmatrix} 1+D & 0 & 0\\ D & 1+D & 1 \end{pmatrix} \hspace{0.05cm}. $$

• Therefore the  proposed solutions 1 and 3  are correct.

• Answer 2 cannot be correct,  because no element with  $D^2$  can occur in the transfer function matrix when  $m = 1$.

• $\mathbf{G}(D)$ is moreover a $2 × 3$ matrix; there is no third row.

(3)  Splitting  $\mathbf{G}(D)$  gives the  $2 × 2$  matrix.

$${ \boldsymbol{\rm T}}(D) = \begin{pmatrix} 1+D & 0 \\ D & 1+D \end{pmatrix} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm det}\hspace{0.1cm}{ \boldsymbol{\rm T}}(D) = (1+D) \cdot (1+D) = 1+D^2 $$

$$\Rightarrow \hspace{0.3cm}{ \boldsymbol{\rm T}}^{-1}(D) = \frac{1}{1+D^2} \cdot \begin{pmatrix} 1+D & 0 \\ D & 1+D \end{pmatrix} \hspace{0.05cm}. $$

• The correct solution is  solution 3.  For control:

$${ \boldsymbol{\rm T}}(D) \cdot { \boldsymbol{\rm T}}^{-1}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac{1}{1+D^2} \cdot \begin{pmatrix} 1+D & 0 \\ D & 1+D \end{pmatrix} \cdot \begin{pmatrix} 1+D & 0 \\ D & 1+D \end{pmatrix} =$$

$$ \ = \ \hspace{-0.15cm} ... \hspace{0.1cm}= \frac{1}{1+D^2} \cdot \begin{pmatrix} 1+D^2 & 0 \\ 0 & 1+D^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\hspace{0.05cm}. $$

(4)  According to the data sheet applies:

$${ \boldsymbol{\rm P}}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} { \boldsymbol{\rm T}}^{-1}(D) \cdot { \boldsymbol{\rm Q}}(D) = \frac{1}{1+D^2} \cdot \begin{pmatrix} 1+D & 0 \\ D & 1+D \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix} =$$

$$\ = \ \hspace{-0.15cm} \frac{1}{1+D^2} \cdot \begin{pmatrix} (1+D)\cdot 0 + 0 \cdot 1 \\ D\cdot 0 + (1+D)\cdot 1 \end{pmatrix} = \frac{1}{1+D^2} \cdot \begin{pmatrix} 0 \\ 1+D \end{pmatrix} = \begin{pmatrix} 0 \\ 1/(1+D) \end{pmatrix} $$

$$\Rightarrow \hspace{0.3cm} {\boldsymbol{\rm G}}_{\rm sys}(D) = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 1/(1+D) \end{pmatrix}\hspace{0.05cm}. $$

• The correct solution is therefore the  proposals 1 and 3.

(5)  Correct is  YES.   The lower circuit on the data sheet is identified by the equations  $x_i^{(1)} = u_i^{(1)}$,  $x_i^{(2)} = u_i^{(2)}$,  and

$$x_i^{(3)}= x_{i-1}^{(3)} + u_i^{(2)} \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm} X^{(3)}(D)= X^{(3)}(D) \cdot D +U^{(2)}(D)$$

$$\Rightarrow \hspace{0.3cm} G(D) = \frac {X^{(3)}(D)}{U^{(2)}(D)} = \frac {1}{1+D} \hspace{0.05cm}.$$

• This corresponds exactly to the last element of  $\mathbf{G}_{\rm sys}(D)$  from subtask  (4).