Aufgaben:Exercise 3.3: Code Sequence Calculation via U(D) and G(D)

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Considered generator matrix

The upper left section of the generator matrix  $\mathbf{G}$  is shown for the considered convolutional code.

• From this the partial matrices  $\mathbf{G}_l$  are to be extracted under the boundary condition  $m ≤ 2$,  with which then the transfer function matrix can be composed according to the following equation:

$${\boldsymbol{\rm G}}(D) = \sum_{l = 0}^{m} {\boldsymbol{\rm G}}_l \cdot D\hspace{0.03cm}^l = {\boldsymbol{\rm G}}_0 + {\boldsymbol{\rm G}}_1 \cdot D + \ \text{...} \ \hspace{0.05cm}+ {\boldsymbol{\rm G}}_m \cdot D\hspace{0.03cm}^m \hspace{0.02cm}.$$

• Searched are the  $n$  code sequences  $\underline{x}^{(1)}, \ \underline{x}^{(2)}, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ \underline{x}^{(n)}$,  assuming the following information sequence:

$$\underline{u} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm}\hspace{0.05cm}) $$

• This sequence is thereby divided into  $k$  subsequences  $\underline{u}^{(1)}, \ \underline{u}^{(2)}, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ \underline{u}^{(k)}$  to split.

• From their D–transforms

$${U}^{(1)}(D) \hspace{0.15cm}\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\circ\hspace{0.15cm} \underline{u}^{(1)},\hspace{0.25cm} ...\hspace{0.25cm},\hspace{0.05cm} {U}^{(k)}(D) \hspace{0.15cm}\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\circ\hspace{0.15cm} \underline{u}^{(k)} $$

the vector   $\underline{U}(D) = (U^{(1)}(D), \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ U^{(k)}(D))$   is formed.

• Then applies to the code sequence vector in  D–representation:

$$\underline{X}(D) = \left (\hspace{0.05cm} {X}^{(1)}(D)\hspace{0.05cm}, \hspace{0.05cm} \text{...} \hspace{0.05cm}, \hspace{0.05cm} {X}^{(k)}(D)\hspace{0.05cm}\right ) = \underline{U}(D) \cdot {\boldsymbol{\rm G}}(D)\hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  "Algebraic and Polynomial Description".

• The underlying encoder here is identical to that of  $\text{Exercise 3.2}$.

• Since also  $\underline{u}$  remains the same,  the same code sequence  $\underline{x}$  must result here as in Exercise 3.2.  However,  the solution paths of both exercises are fundamentally different.

Questions

1

What are the code parameters?   Hint:   For the memory applies:  $m ≤ 2$.

$n \hspace{0.25cm} = \ $

$k \hspace{0.28cm} = \ $

$m \hspace{0.13cm} = \ $

2

Which statements are true for the transfer function matrix  $\mathbf{G}(D)$?

	The  $\mathbf{G}(D)$  element in row 1,  column  1 is  "$1$".
	The  $\mathbf{G}(D)$  element in row 2,  column 2 is  "$1 + D$".
	The  $\mathbf{G}(D)$  element in row 3,  column 3 is  "$1 + D^2$".

3

Which statements are true for the  D–transforms of the input sequences?

	$U^{(1)}(D) = 1$,
	$U^{(2)}(D) = 1 + D$,
	$U^{(3)}(D) = D^2$.

4

What are the first three bits of the code sequence  $\underline{x}^{(1)}$?

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

5

What are the first three bits of the code sequence  $\underline{x}^{(2)}$?

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

6

What are the first three bits of the code sequence  $\underline{x}^{(3)}$?

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

Solution

Solution

(1)  The generator matrix of a convolutional code has the general form:

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

• From the graph on the information page,  the  $k × n$  partial matrices can be determined:

$${ \boldsymbol{\rm G}}_0=\begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix}\hspace{0.05cm},\hspace{0.2cm} { \boldsymbol{\rm G}}_1=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}\hspace{0.05cm},\hspace{0.2cm} { \boldsymbol{\rm G}}_2=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}\hspace{0.05cm}. $$

• The code parameters are thus:  $\underline{n = 4}$,     $\underline{k = 3}$,     $\underline{m = 2}$.

Hints:

1. The represented part of  $\mathbf{G}$  would have the same appearance for  $m > 2$  as for  $m = 2$.
2. This is why the additional specification  $m ≤ 2$  was necessary.

(2)  All proposed solutions  are correct.  According to the specification sheet

$${\boldsymbol{\rm G}}(D) = {\boldsymbol{\rm G}}_0 + {\boldsymbol{\rm G}}_1 \cdot D + {\boldsymbol{\rm G}}_2 \cdot D^2 = \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1+D & 1+D & 1 \\ 0 & D & 1+D^2 & 1+D^2 \end{pmatrix}\hspace{0.05cm}.$$

(3)  After splitting the information sequence

$$\underline{u} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}, ... \hspace{0.05cm})$$

into the three partial sequences $\underline{u}^{(1)}$,  $\underline{u}^{(2)}$,  $\underline{u}^{(3)}$  and subsequent D–transformation we get:

$$\underline{u}^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad {U}^{(1)}(D) = D + D^2 \hspace{0.05cm},$$

$$\underline{u}^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad {U}^{(2)}(D) = 1+D \hspace{0.05cm},$$

$$\underline{u}^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad {U}^{(3)}(D) = 1 + D^2 \hspace{0.05cm}.$$

• Accordingly,  only the  proposed solution 2  is correct.

(4)  In the first column of  $\mathbf{G}(D)$  there is only one  "$1$"  in row 1,  the other two matrix elements are zero.

• This is a systematic code   ⇒   $\underline{x}^{(1)} = \underline{u}^{(1)} = (0, \, 1, \, 1)$.

• Correct is the   proposed solution 1.

(5)  The D–transform  $X^{(2)}(D)$  is obtained as the vector product

• of the D–transform of the information sequence   ⇒   $\underline{U}(D) = (U^{(1)}(D), \, U^{(2)}(D), \, U^{(3)}(D))$

• and the second column  of $\mathbf{G}(D)$:

$$X^{(2)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} ( D + D^2) \cdot 1 + ( 1+D) \cdot ( 1+D) +( 1+D^2) \cdot D\hspace{0.03cm}=D + D^2 +1 +D +D + D^2 +D + D^3 = 1+D^3 \hspace{0.05cm}.$$

Correct is the  proposed solution 2:   $\underline{x}^{(2)} = (1, \, 0, \, 0)$.  Since we are only interested in the first three bits,  the contribution  $D^3$  is not relevant.

(6)  Analogous to subtask  (5),  we obtain here:

$$X^{(3)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} ( D + D^2) \cdot 0 + ( 1+D) \cdot ( 1+D) +( 1+D^2) \cdot ( 1+D^2)=1 + D + D + D^2 +1 + D^2 + D^2 + D^4 = D^2 + D^4 \hspace{0.05cm}.$$

• This gives  $\underline{x}^{(3)} = (0, \, 0, \, 1)$   ⇒   Proposed solution 3.

• The same result is obtained for  $\underline{x}^{(4)}$.

• After joining all  $n = 4$  subsequences,  one obtains  (of course)  the same result as in  $\text{Exercise 3.2}$:

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