Aufgaben:Exercise 3.2Z: (3, 1, 3) Convolutional Encoder

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

The presented convolutional encoder is defined by the parameters 

• $k = 1$  $($only one information sequence  $\underline{u})$  and 
• $n = 3$  $($three code sequences $\underline{x}^{(1)}, \ \underline{x}^{(2)}, \ \underline{x}^{(3)})$ 

characterized. From the number of memory cells, the memory  $m = 3$.

With the information bit  $u_i$  to the coding step  $i$  the following code bits are obtained:

$$x_i^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i} + u_{i-1} + u_{i-3}\hspace{0.05cm},$$

$$x_i^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i} + u_{i-1} + u_{i-2} + u_{i-3} \hspace{0.05cm},$$

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

From this, partial matrices  $\mathbf{G}_l$  can be derived, as described on the page  "Division of the generator matrix into partial matrices" .

• For the generator matrix can thus be written:

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

• For the code sequence $\underline{x} = (x_1^{(1)}, \ x_1^{(2)}, \ x_1^{(3)}, \ x_2^{(1)}, \ x_2^{(2)}, \ x_2^{(3)}, \ \text{...})$ holds:

$$\underline{x} = \underline{u} \cdot { \boldsymbol{\rm G}} \hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  "Algebraic and Polynomial Description".
• Reference is made in particular to the page  "Division of the generator matrix into partial matrices".

Questions

1

From how many partial matrices  $\mathbf{G}_l$  is the matrix  $\mathbf{G}$  composed?

${\rm Number \ of \ partial \ matrices} \ = \ $

2

What dimension do the partial matrices  $\mathbf{G}_l$ have?

${\rm Rows \ of \ the \ partial \ matrices} \hspace{0.425cm} = \ $

${\rm Columns \ of \ the \ partial \ matrices} \ = \ $

3

Which statements are correct?

	It holds  $\mathbf{G}_0 = (1, 1, 1)$.
	It holds  $\mathbf{G}_ 1 = (1, 1, 0)$.
	It holds  $\mathbf{G}_2 = (0, 1, 1)$.
	It holds  $\mathbf{G}_3 = (1, 1, 0)$.

4

Create the generator matrix  $\mathbf{G}$  with five rows and fifteen columns.
What code sequence results for  $\underline{u} = (1, 0, 1, 1, 0)$?

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

Solution

Solution

(1)  For the index $l$ of the partial matrices, $0 ≤ l ≤ m$.

• The coder under consideration has memory $m = 3$.
• Thereby four partial matrices are to be considered.

(2)  Each partial matrix $\mathbf{G}_l$ consists of.

• one row  ⇒  $k = 1$, and
• three columns  ⇒  $n = 3$.

(3)  All statements are correct:

• Since the current information bit $u_i$ affects all three outputs $x_i^{(1)}, \ x_i^{(2)}$ and $x_i^{(3)}$, $\mathbf{G}_0 = (1, 1, 1)$.
• In contrast, $\mathbf{G}_3 = (1, 1, 0)$ states that only the first two inputs are affected by $u_{i-3}$, but not $x_i^{(3)}$.

(4)  Correct is the proposed solution 2:

Generator matrix $\mathbf{G}$

• The searched generator matrix $\mathbf{G}$ is shown on the right, where the four partial matrices $\mathbf{G}_0, \ ... , \mathbf{G}_3$ are distinguished by color.
• The following vector equation gives the result corresponding to the second proposed solution 2:

$$\underline{x} = \underline{u} \cdot { \boldsymbol{\rm G}} = (1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1) \cdot { \boldsymbol{\rm G}}. $$

• The code sequence $\underline{x}$ is thereby equal to the modulo 2 sum of the matrix rows 1, 3 and 4.

• The three code sequences of the individual branches are distinguished by color. For example, the following applies to the lower output:

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

Using the equations given above, this result can be verified:

$${x}_1^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_1 + u_{-1} = 1+ (0) = 1 \hspace{0.05cm},$$

$${x}_2^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_2 + u_{0} = 0+ (0) = 0 \hspace{0.05cm},$$

$${x}_3^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_3 + u_{1} = 1+1 = 0 \hspace{0.05cm},$$

$${x}_4^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_4 + u_{2} = 1+0 = 1 \hspace{0.05cm},$$

$${x}_5^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_5 + u_{3} = 0+ 1 = 1 \hspace{0.05cm}.$$

Notes:

• The memory preallocation with zeros is taken into account here: $u_0 = u_{–1} = 0$.
• If, as assumed here, the information sequence is limited to four bits, then ones can occur in the code sequence up to the position $(4 + m) \cdot n = 21$.