Aufgaben:Exercise 3.1Z: Convolution Codes of Rate 1/2

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Convolutional codes of rate  $1/2$

The graphic shows two convolutional encoders of rate  $R = 1/2$: 

• At the input there is the information sequence  $\underline {u} = (u_1, u_2, \ \text{...} \ , u_i, \ \text{...})$. 

• From this,  modulo-2 operations generate the two sequences

$$\underline{\it x}^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \big( \hspace{0.05cm}x_1^{(1)}\hspace{0.05cm},\hspace{0.05cm} x_2^{(1)}\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm},\hspace{0.05cm} x_i^{(1)} \hspace{0.05cm},\text{...} \hspace{0.05cm} \big )\hspace{0.05cm},$$

$$\underline{\it x}^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \big( \hspace{0.05cm}x_1^{(2)}\hspace{0.05cm},\hspace{0.05cm} x_2^{(2)}\hspace{0.05cm},\hspace{0.05cm}\text{...} \hspace{0.05cm},\hspace{0.05cm} x_i^{(2)} \hspace{0.05cm}, \text{...} \hspace{0.05cm} \big )$$

where  $x_i^{(j)}$  with  $j = 1$  resp.  $j = 2$  may depend except from  $u_i$  also from the previous information bits  $u_{i-1}, \ \text{...} \ , u_{i-m}$.

• One refers  $m$  as the  "memory"  and  $\nu = m + 1$  as the  "influence length"  of the code  $($or of the encoder$)$.

• The considered encoders  $\rm A$  and  $\rm B$  differ with respect to these quantities.

Hints:

• The exercise refers to the chapter  "Basics of Convolutional Coding".

• Not shown in the diagram is the multiplexing of the two subsequences  $\underline {x}^{(1)}$  and  $\underline {x}^{(2)}$  to the resulting code sequence 

$$\underline {x} = (x_1^{(1)}, x_1^{(2)}, x_2^{(1)}, x_2^{(2)}, \ \text{...}).$$

• In subtasks  (3)  to  (5)  you are to determine the start of the sequences  $\underline {x}^{(1)}, \underline{x}^{(2)}$  and  $\underline{x}$  assuming the information sequence 

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

Questions

1

In which code parameters do encoder  $\rm A$  and encoder  $\rm B$ differ?

	$k$:     Number of information bits processed per coding step,
	$n$:     Number of code bits output per coding step,
	$m$:   memory order of the code,
	$\nu$:     influence length of the code.

2

Which encoder exhibits the memory  $m = 2$?

	Encoder  $\rm A$,
	encoder  $\rm B$.

3

What is the partial code sequence  $\underline {x}^{(1)}$  of encoder  $\rm B$  for  $\underline {u} = (1, 0, 1, 1, 0, 0, \ \text{...})$?

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

4

What is the partial code sequence  $\underline {x}^{(2)}$  of encoder  $\rm B$  for  $\underline {u} = (1, 0, 1, 1, 0, 0, \ \text{...})$

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

5

How does the entire code sequence  $\underline {x}$  start from encoder  $\rm B$  after multiplexing?

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

Solution

Solution

(1)  For both encoders,  $k = 1$  and  $n = 2$.

• The memory  $m$  and the influence length  $\nu$  are different   ⇒   Answers 3 and 4.

(2)  The shift register of encoder  $\rm A$  does contain two memory cells.  However,  since  $x_i^{(1)} = u_i$  and  $x_i^{(2)} = u_i + u_{i-1}$  is influenced only by the immediately preceding bit  $u_{i-1}$  besides the current information bit  $u_i$,

Equivalent encoder representations

• the memory is  $m = 1$,  and

• the influence length is  $\nu = m + 1 = 2$.

The graphic shows the two encoders in another representation,  whereby the  "memory cells"  are highlighted in yellow.

• For the encoder  $\rm A$  one recognizes only one memory ⇒ $m = 1$.

• In contrast,  for the encoder  $\rm B$  actually  $m = 2$  and  $\nu = 3$.

• Thus,  the  proposed solution 2  is correct.

(3)  For the upper output of encoder  $\rm B$  applies in general:

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

• Considering the preassignment ($(u_0 = u_{-1} = 0)$,  we obtain with the above data:

$$x_1^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{1} + u_{0}+ u_{-1} = 1+0+0 = 1 \hspace{0.05cm},\hspace{1cm}x_2^{(1)} = u_{2} + u_{1}+ u_{0} = 0+1+0 = 1\hspace{0.05cm},$$

$$x_3^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{3} + u_{2}+ u_{1} \hspace{0.25cm}= 1+0+1 = 0 \hspace{0.05cm},\hspace{1cm}x_4^{(1)} = u_{4} + u_{3}+ u_{2} = 1+1+0 = 0\hspace{0.05cm},$$

$$x_5^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{5} + u_{4}+ u_{3} \hspace{0.25cm}= 0+1+1 = 0 \hspace{0.05cm},\hspace{1cm}x_6^{(1)} = u_{6} + u_{5}+ u_{4} = 0+0+1 = 1\hspace{0.05cm},$$

$$x_7^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} x_8^{(1)} = \text{...} \hspace{0.05cm}= 0 \hspace{0.05cm}.$$

• The  proposed solution 1  is correct.  The second solution suggestion   ⇒   $\underline {x}^{(1)} = \underline {u}$  would only be valid for a systematic code  $($which is not present here$)$.

(4)  Analogous to subtask  (3)  ⇒   $x_i^{(2)} = u_i + u_{i–2}$:

$$x_1^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1+0 = 1 \hspace{0.05cm},\hspace{0.2cm}x_2^{(2)} = 0+0 = 0\hspace{0.05cm}, \hspace{0.2cm}x_3^{(3)} = 1+1 = 0\hspace{0.05cm},\hspace{0.2cm}x_4^{(2)} = 1+0 = 1 \hspace{0.05cm},$$

$$x_5^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0+1 = 1\hspace{0.05cm}, \hspace{0.2cm}x_6^{(2)} = 0+1 = 1\hspace{0.05cm},\hspace{0.2cm} x_7^{(2)} = x_8^{(2)} = \text{...} \hspace{0.05cm}= 0 \hspace{0.05cm}.$$

• The correct solution is the  proposed solution 2.

(5)  For the  $($entire$)$  code sequence,  one can formally write:

$$\underline{\it x} = \big( \hspace{0.05cm}\underline{\it x}_1\hspace{0.05cm}, \hspace{0.05cm} \underline{\it x}_2\hspace{0.05cm}, \hspace{0.05cm}\text{...}\hspace{0.05cm} \underline{\it x}_i \hspace{0.05cm}, \text{...} \hspace{0.05cm} \big )\hspace{0.05cm}, \hspace{0.3cm} \underline{\it x}_i = \big( x_i^{(1)}\hspace{0.05cm}, x_i^{(2)} \big) \hspace{0.4cm}\Rightarrow \hspace{0.4cm} \underline{\it x} = \big( \hspace{0.05cm}x_1^{(1)}\hspace{0.01cm},\hspace{0.05cm} x_2^{(1)}\hspace{0.01cm},\hspace{0.05cm} x_1^{(2)}\hspace{0.01cm},\hspace{0.05cm} x_2^{(2)}\hspace{0.01cm}, \hspace{0.05cm} \text{...} \hspace{0.05cm} \big )\hspace{0.05cm}. $$

• A comparison with the solutions of subtasks  (3)  and  (4)  shows the correctness of the  proposed solution 1.