Difference between revisions of "Aufgaben:Exercise 3.3: Code Sequence Calculation via U(D) and G(D)"

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{{quiz-Header|Buchseite=Kanalcodierung/Algebraische und polynomische Beschreibung}}
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{{quiz-Header|Buchseite=Channel_Coding/Algebraic_and_Polynomial_Description}}
  
−
[[File:P_ID2627__KC_A_3_3_v1.png|right|frame|Betrachtete Generatormatrix]]
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[[File:P_ID2627__KC_A_3_3_v1.png|right|frame|Considered generator matrix]]
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Nebenstehend ist für den betrachteten Faltungscode der linke obere Ausschnitt der Generatormatrix $\mathbf{G}$ dargestellt. Daraus sollen unter der Randbedingung $m ≤ 2$ die Teilmatrizen $\mathbf{G}_l$ extrahiert werden, womit dann die Übergangsfunktionsmatrix entsprechend folgender Gleichung zusammengestellt werden kann:
+
The upper left section of the generator matrix  $\mathbf{G}$  is shown for the considered convolutional code.  
−
:$${\boldsymbol{\rm G}}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \sum_{l = 0}^{m} {\boldsymbol{\rm G}}_l \cdot D\hspace{0.03cm}^l
+
*From this the partial matrices  $\mathbf{G}_l$  are to be extracted under the boundary condition  $m ≤ 2$,  with which then the transfer function matrix can be composed according to the following equation:
−
=$$
+
:$${\boldsymbol{\rm G}}(D) =  \sum_{l = 0}^{m} {\boldsymbol{\rm G}}_l \cdot D\hspace{0.03cm}^l
−
:$$ \ = \ \hspace{-0.15cm} {\boldsymbol{\rm G}}_0 + {\boldsymbol{\rm G}}_1 \cdot D + ... \hspace{0.05cm}+ {\boldsymbol{\rm G}}_m \cdot D\hspace{0.03cm}^m
+
= {\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}.$$
 
  \hspace{0.02cm}.$$
  
−
Gesucht werden die $n$ Codesequenzen $\underline{x}^{(1)}, \ \underline{x}^{(2)}, \ ... \ , \underline{x}^{(n)}$, wobei von der Informationssequenz
+
*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:
−
:$$\underline{u} =  (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}, ... \hspace{0.05cm})  $$
+
:$$\underline{u} =  (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm}\hspace{0.05cm})  $$
  
−
auszugehen ist. Diese Sequenz ist dabei in $k$ Teilsequenzen $\underline{u}^{(1)}, \ \underline{u}^{(2)}, \ ... \ , \ \underline{u}^{(k)}$ aufzuspalten. Aus deren $D$–Transformierten
+
*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}^{(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)} $$
 
  {U}^{(k)}(D) \hspace{0.15cm}\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\circ\hspace{0.15cm} \underline{u}^{(k)} $$
  
−
wird dann der Vektor $\underline{U}(D) = (U^{(1)}(D), \ ... \ , \ U^{(k)}(D))$ gebildet. Dann gilt für den Codesequenzvektor in $D$–Darstellung:
+
:the vector   $\underline{U}(D) = (U^{(1)}(D), \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ U^{(k)}(D))$   is formed.
−
:$$\underline{X}(D) = \left (\hspace{0.05cm} {X}^{(1)}(D)\hspace{0.05cm}, \hspace{0.05cm} ... \hspace{0.12cm}, \hspace{0.05cm} {X}^{(k)}(D)\hspace{0.05cm}\right ) = \underline{U}(D) \cdot {\boldsymbol{\rm G}}(D)\hspace{0.05cm}.$$
+
*Then applies to the code sequence vector in  D–representation:
 +
:$$\underline{X}(D) = \left (\hspace{0.05cm} {X}^{(1)}(D)\hspace{0.05cm}, \hspace{0.05cm} \text{...} \hspace{0.05cm}, \hspace{0.05cm} {X}^{(k)}(D)\hspace{0.05cm}\right ) = \underline{U}(D) \cdot {\boldsymbol{\rm G}}(D)\hspace{0.05cm}.$$
  
−
''Hinweis:''
 
−
* Die Aufgabe gehört zum Themengebiet des Kapitels [[Kanalcodierung/Algebraische_und_polynomische_Beschreibung| Algebraische und polynomische Beschreibung]].
 
−
* Der hier zugrunde liegende Codierer ist identisch mit dem von [[Aufgaben:3.2_G%E2%80%93Matrix_eines_Faltungscoders| Aufgabe A3.2]].
 
−
* Nachdem auch von der gleichen Informationssequenz $\underline{u}$ ausgegangen wird, muss sich hier die gleiche Codesequenz $\underline{x}$ ergeben wie in Aufgabe A3.2, siehe [[Aufgaben:3.2_G%E2%80%93Matrix_eines_Faltungscoders| Musterlösung]].
 
−
* Die Lösungswege beider Aufgaben unterscheiden sich allerdings grundlegend.
 
  
  
  
−
===Fragebogen===
+
 
 +
Hints:
 +
* This exercise belongs to the chapter  [[Channel_Coding/Algebraic_and_Polynomial_Description| "Algebraic and Polynomial Description"]].
 +
 
 +
* The underlying encoder here is identical to that of  [[Aufgaben:Exercise_3.2:_G-matrix_of_a_Convolutional_Encoder|$\text{Exercise 3.2}$]].
 +
 
 +
* Since also  $\underline{u}$  remains  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===
 
<quiz display=simple>
 
<quiz display=simple>
−
{Multiple-Choice
+
{What are the code parameters? &nbsp; <u>Hint:</u> &nbsp; For the memory applies:&nbsp; $m &#8804; 2$.
 +
|type="{}"}
 +
$n \hspace{0.25cm} = \ ${ 4 }
 +
$k \hspace{0.28cm} = \ ${ 3 }
 +
$m \hspace{0.13cm} = \ ${ 2 }
 +
 
 +
{Which statements are true for the transfer function matrix&nbsp; $\mathbf{G}(D)$?
 
|type="[]"}
 
|type="[]"}
−
+ correct
+
+ The&nbsp; $\mathbf{G}(D)$&nbsp; element in row 1,&nbsp; column&nbsp; 1 is&nbsp; "$1$".
−
- false
+
+ The&nbsp; $\mathbf{G}(D)$&nbsp; element in row 2,&nbsp; column 2 is&nbsp; "$1 + D$".
 +
+ The&nbsp; $\mathbf{G}(D)$&nbsp; element in row 3,&nbsp; column 3 is&nbsp; "$1 + D^2$".
 +
 
 +
{Which statements are true for the&nbsp; D&ndash;transforms of the input sequences?
 +
|type="[]"}
 +
- $U^{(1)}(D) = 1$,
 +
+ $U^{(2)}(D) = 1 + D$,
 +
- $U^{(3)}(D) = D^2$.
 +
 
 +
{What are the first three bits of the code sequence&nbsp; $\underline{x}^{(1)}$?
 +
|type="()"}
 +
+ $\underline{x}^{(1)} = (0, \, 1, \, 1, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$,
 +
- $\underline{x}^{(1)} = (1, \, 0, \, 0, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$,
 +
- $\underline{x}^{(1)} = (0, \, 0, \, 1, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$.
 +
 
 +
{What are the first three bits of the code sequence&nbsp; $\underline{x}^{(2)}$?
 +
|type="()"}
 +
- $\underline{x}^{(2)} = (0, \, 1, \, 1, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$,
 +
+ $\underline{x}^{(2)} = (1, \, 0, \, 0, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$,
 +
- $\underline{x}^{(2)} = (0, \, 0, \, 1, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$.
  
−
{Input-Box Frage
+
{What are the first three bits of the code sequence&nbsp; $\underline{x}^{(3)}$?
−
|type="{}"}
+
|type="()"}
−
$xyz \ = \ ${ 5.4 3% } $ab$
+
- $\underline{x}^{(3)} = (0, \, 1, \, 1, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$,
 +
- $\underline{x}^{(3)} = (1, \, 0, \, 0, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$,
 +
+ $\underline{x}^{(3)} = (0, \, 0, \, 1, \, \hspace{0.05cm} \text{...} \hspace{0.05cm})$.
 
</quiz>
 
</quiz>
  
−
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
−
'''(1)'''&nbsp;  
+
'''(1)'''&nbsp; The generator matrix of a convolutional code has the general form:
−
'''(2)'''&nbsp;  
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:$${ \boldsymbol{\rm G}}=\begin{pmatrix}
−
'''(3)'''&nbsp;  
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{ \boldsymbol{\rm G}}_0 & { \boldsymbol{\rm G}}_1 & { \boldsymbol{\rm G}}_2 & \cdots  & { \boldsymbol{\rm G}}_m & & & \\
−
'''(4)'''&nbsp;  
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& { \boldsymbol{\rm G}}_0 & { \boldsymbol{\rm G}}_1 & { \boldsymbol{\rm G}}_2 & \cdots & { \boldsymbol{\rm G}}_m & &\\
−
'''(5)'''&nbsp;  
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&          & { \boldsymbol{\rm G}}_0 & { \boldsymbol{\rm G}}_1 & { \boldsymbol{\rm G}}_2 & \cdots & { \boldsymbol{\rm G}}_m &\\
 +
&          &          & \ddots  & \ddots & & & \ddots
 +
\end{pmatrix}\hspace{0.05cm}.$$
 +
 
 +
*From the graph on the information page,&nbsp; the&nbsp; $k &times; n$&nbsp; 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: &nbsp;$\underline{n = 4}$, &nbsp; &nbsp; $\underline{k = 3}$, &nbsp; &nbsp; $\underline{m = 2}$.
 +
 
 +
 
 +
<u>Hints:</u>
 +
# The represented part of&nbsp; $\mathbf{G}$&nbsp; would have the same appearance for&nbsp; $m > 2$&nbsp; as for&nbsp; $m = 2$.
 +
# This is why the additional specification&nbsp; $m &#8804; 2$&nbsp; was necessary.
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; <u>All proposed solutions</u>&nbsp; are correct.&nbsp; 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)'''&nbsp; 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 $\underline{u}^{(1)}$,&nbsp; $\underline{u}^{(2)}$,&nbsp; $\underline{u}^{(3)}$&nbsp; and subsequent D&ndash;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,&nbsp; only the&nbsp; <u>proposed solution 2</u>&nbsp; is correct.
 +
 
 +
 
 +
 
 +
 
 +
'''(4)'''&nbsp; In the first column of&nbsp; $\mathbf{G}(D)$&nbsp; there is only one&nbsp; "$1$"&nbsp; in row 1,&nbsp; the other two matrix elements are zero.
 +
 
 +
*This is a systematic code &nbsp; &#8658; &nbsp; $\underline{x}^{(1)} = \underline{u}^{(1)} = (0, \, 1, \, 1)$.
 +
 
 +
*Correct is the &nbsp; <u>proposed solution 1</u>.
 +
 
 +
 
 +
 
 +
 
 +
 
 +
'''(5)'''&nbsp; The D&ndash;transform&nbsp; $X^{(2)}(D)$&nbsp; is obtained as the vector product
 +
*of the D&ndash;transform of the information sequence &nbsp; &#8658; &nbsp; $\underline{U}(D) = (U^{(1)}(D), \, U^{(2)}(D), \, U^{(3)}(D))$
 +
 +
*and the second column&nbsp; 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&nbsp; <u>proposed solution 2</u>: &nbsp; $\underline{x}^{(2)} = (1, \, 0, \, 0)$.&nbsp; Since we are only interested in the first three bits,&nbsp; the contribution&nbsp; $D^3$&nbsp; is not relevant.
 +
 
 +
 
 +
 
 +
 
 +
'''(6)'''&nbsp; Analogous to subtask&nbsp; '''(5)''',&nbsp; 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&nbsp; $\underline{x}^{(3)} = (0, \, 0, \, 1)$ &nbsp; &#8658; &nbsp; <u>Proposed solution 3</u>.
 +
 
 +
*The same result is obtained for&nbsp; $\underline{x}^{(4)}$.
 +
 +
*After joining all&nbsp; $n = 4$&nbsp; subsequences,&nbsp; one obtains&nbsp; (of course)&nbsp; the same result as in&nbsp; [[Aufgaben:Exercise_3.2:_G-matrix_of_a_Convolutional_Encoder|$\text{Exercise 3.2}$]]:
 +
:$$\underline{x} =  (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}\text{ ...} \hspace{0.05cm})\hspace{0.05cm}.$$
 +
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
−
[[Category:Aufgaben zu  Kanalcodierung|^3.2 Algebraische und polynomische Beschreibung^]]
+
[[Category:Channel Coding: Exercises|^3.2 Polynomial Description^]]

Latest revision as of 18:08, 10 November 2022

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:

  • Since also  $\underline{u}$  remains 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  $\text{Exercise 3.2}$:
$$\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}\text{ ...} \hspace{0.05cm})\hspace{0.05cm}.$$

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