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

{{quiz-Header|Buchseite=Channel_Coding/Algebraic_and_Polynomial_Description}}

[[File:P_ID2625__KC_Z_3_2_neu.png|right|frame|Convolutional encoder with parameters&nbsp; $k = 1, \ n = 3, \ m = 3$.]]

The presented convolutional encoder is defined by the parameters&nbsp; 
*$k = 1$&nbsp; $($only one information sequence&nbsp; $\underline{u})$,&nbsp; and&nbsp;

* $n = 3$&nbsp; $($three code sequences&nbsp;  $\underline{x}^{(1)}, \ \underline{x}^{(2)}, \ \underline{x}^{(3)}).$&nbsp; 


From the number of memory cells,&nbsp; the memory&nbsp; $m = 3$. 

With the information bit&nbsp; $u_i$&nbsp; to the coding step&nbsp; $i$,&nbsp; 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,&nbsp; partial matrices &nbsp; $\mathbf{G}_l$ &nbsp; can be derived,&nbsp; as described on the page&nbsp; [[Channel_Coding/Algebraic_and_Polynomial_Description#Division_of_the_generator_matrix_into_partial_matrices|"Division of the generator matrix into partial matrices"]]&nbsp;. 

*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 &nbsp; $\underline{x} = (x_1^{(1)}, \ x_1^{(2)}, \ x_1^{(3)}, \ x_2^{(1)}, \ x_2^{(2)}, \ x_2^{(3)}, \ \text{...})$ &nbsp; holds:
:$$\underline{x} = \underline{u} \cdot { \boldsymbol{\rm G}}  \hspace{0.05cm}.$$





Hints:
*This exercise belongs to the chapter&nbsp; [[Channel_Coding/Algebraic_and_Polynomial_Description| "Algebraic and Polynomial Description"]].

*Reference is made in particular to the section&nbsp; [[Channel_Coding/Algebraic_and_Polynomial_Description#Division_of_the_generator_matrix_into_partial_matrices|"Division of the generator matrix into partial matrices"]]. 
 



===Questions===
<quiz display=simple>
{From how many partial matrices&nbsp; $\mathbf{G}_l$&nbsp; is the matrix&nbsp; $\mathbf{G}$&nbsp; composed?
|type="{}"}
${\rm Number \ of \ partial \ matrices} \ = \ ${ 4 }

{What dimension do the partial matrices&nbsp; $\mathbf{G}_l$ have?
|type="{}"}
${\rm Rows \ of \ the \ partial \ matrices} \hspace{0.425cm} = \ ${ 1 } 
${\rm Columns \ of \ the \ partial \ matrices} \ = \ ${ 3 } 

{Which statements are correct?
|type="[]"}
+ It holds&nbsp; $\mathbf{G}_0 = (1, 1, 1)$.
+ It holds&nbsp; $\mathbf{G}_ 1 = (1, 1, 0)$.
+ It holds&nbsp; $\mathbf{G}_2 = (0, 1, 1)$.
+ It holds&nbsp; $\mathbf{G}_3 = (1, 1, 0)$.

{Create the generator matrix&nbsp; $\mathbf{G}$&nbsp; with five rows and fifteen columns.&nbsp; What code sequence results for&nbsp; $\underline{u} = (1, 0, 1, 1, 0)$?
|type="()"}
- It holds&nbsp; $\underline{x} = (1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, \text{...}).$
+ It holds&nbsp; $\underline{x} = (1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, \text{...}).$
- It holds&nbsp; $\underline{x} = (0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, \text{...}).$
</quiz>


===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; For the index&nbsp; $l$&nbsp; of the partial matrices:&nbsp; $0 &#8804; l &#8804; m$. 
*The coder under consideration has memory&nbsp; $m = 3$.
 
*Thereby&nbsp; <u>four partial matrices</u>&nbsp; are to be considered.



'''(2)'''&nbsp; Each partial matrix&nbsp; $\mathbf{G}_l$&nbsp; consists of 
*<u>one row</u> &nbsp; &#8658; &nbsp; $k = 1$,&nbsp; and
 
*<u>three columns</u> &nbsp; &#8658; &nbsp; $n = 3$.



'''(3)'''&nbsp; <u>All statements</u> are correct: 
*Since the current information bit&nbsp; $u_i$&nbsp; affects all three outputs&nbsp; $x_i^{(1)}, \ x_i^{(2)}$&nbsp; and&nbsp; $x_i^{(3)}$ &nbsp; &rArr; &nbsp;  $\mathbf{G}_0 = (1, 1, 1)$.
 
*In contrast,&nbsp; $\mathbf{G}_3 = (1, 1, 0)$&nbsp; states that only the first two inputs are affected by&nbsp; $u_{i-3}$,&nbsp; but not $x_i^{(3)}$.



'''(4)'''&nbsp; Correct is the&nbsp; <u>proposed solution 2</u>:
[[File:EN_KC_Z_3_2_d_v2.png|right|frame|Generator matrix&nbsp; $\mathbf{G}$]]
*The searched generator matrix&nbsp; $\mathbf{G}$&nbsp; is shown on the right,&nbsp; where the four partial matrices&nbsp; $\mathbf{G}_0, \ ... , \mathbf{G}_3$&nbsp; 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 &nbsp; $\underline{x}$ &nbsp; 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.&nbsp; For example,&nbsp; 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,&nbsp; this result can be verified:
:$${x}_1^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  u_1 + u_{-1} = 1+ (0) = 1 \hspace{0.05cm},$$
:$${x}_2^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  u_2 + u_{0} = 0+ (0) = 0 \hspace{0.05cm},$$
:$${x}_3^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  u_3 + u_{1} = 1+1 = 0 \hspace{0.05cm},$$
:$${x}_4^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  u_4 + u_{2} = 1+0 = 1 \hspace{0.05cm},$$
:$${x}_5^{(3)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  u_5 + u_{3} = 0+ 1 = 1 \hspace{0.05cm}.$$

<u>Notes:</u> 
*The memory preallocation with zeros is taken into account here:&nbsp; $u_0 = u_{&ndash;1} = 0$.

*If,&nbsp; as assumed here,&nbsp; the information sequence is limited to four bits,&nbsp; then&nbsp; "ones"&nbsp; can occur in the code sequence up to the position&nbsp; $(4 + m) \cdot n = 21$.
{{ML-Fuß}}



[[Category:Channel Coding: Exercises|^3.2 Polynomial Description^]]