Aufgaben:Exercise 3.4: Systematic Convolution Codes - Revision history

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	[[Category:Channel Coding: Exercises\|^3.2 Polynomial Description^]]		[[Category:Channel Coding: Exercises\|^3.2 Polynomial Description^]]
	[[de:Aufgaben:3.~~4_Systematische_Faltungscodes~~]]		[[de:Aufgaben:Aufgabe 3.4: Systematische Faltungscodes]]

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	*The transfer function matrix of such a code is:		*The transfer function matrix of such a code is:
	:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*The encoder  $\rm A$  shown in the upper graph is certainly not systematic,  since for this  $G^{(1)}(D) ≠ 1$  holds.  To derive the matrix  $\mathbf{G}(D)$  we refer to an  [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1. 2Fn-convolution_encoders\|$\text{earlier example}$]],  where for our standard rate–1/2  encoder with memory  $m = 2$  the transfer function matrix was determined:		*The encoder  $\rm A$  shown in the upper graph is certainly not systematic,  since for this  $G^{(1)}(D) ≠ 1$  holds.  To derive the matrix  $\mathbf{G}(D)$  we refer to an  [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1. 2Fn-convolution_encoders\|$\text{earlier example}$]],  where for our standard rate–1/2  encoder with memory  $m = 2$  the transfer function matrix was determined:
	:$${\boldsymbol{\rm G}}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big ) = \big ( \hspace{0.05cm} 1 + D + D^2\hspace{0.05cm} , \hspace{0.2cm} 1 + D^2 \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big ) = \big ( \hspace{0.05cm} 1 + D + D^2\hspace{0.05cm} , \hspace{0.2cm} 1 + D^2 \hspace{0.05cm}\big )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	:The encoder  $\rm A$  differs from this example only by swapping the two outputs.		:The encoder  $\rm A$  differs from this example only by swapping the two outputs.
	*If the transfer function matrix of a code is		*If the transfer function matrix of a code is
	:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )\hspace{0.05cm},$$
	\hspace{0.05cm},$$

	:then the equivalent systematic representation of this rate–1/2  convolutional code holds in general:		:then the equivalent systematic representation of this rate–1/2  convolutional code holds in general:
	:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {G^{(2)}(D)}/{G^{(1)}(D)} \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {G^{(2)}(D)}/{G^{(1)}(D)} \hspace{0.05cm}\big )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	*The desired result is then with the  $k × k$  identity matrix  $\mathbf{I}_k$:		*The desired result is then with the  $k × k$  identity matrix  $\mathbf{I}_k$:
	:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm I}}_k\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm T}}^{-1}(D) \cdot {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm I}}_k\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm T}}^{-1}(D) \cdot {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*We assume here the  $\mathbf{G}(D)$  matrix for the coder  $\rm A$.  Because of  $k = 1$  here both  $\mathbf{T}(D)$  and $\mathbf{Q}(D)$  have dimension  $1 × 1$,  so strictly speaking they are not matrices at all:		*We assume here the  $\mathbf{G}(D)$  matrix for the coder  $\rm A$.  Because of  $k = 1$  here both  $\mathbf{T}(D)$  and $\mathbf{Q}(D)$  have dimension  $1 × 1$,  so strictly speaking they are not matrices at all:
	:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm T}}(D)\hspace{0.05cm} ; \hspace{0.2cm} {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm T}}(D)\hspace{0.05cm} ; \hspace{0.2cm} {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\boldsymbol{\rm T}}(D) = \big ( 1+D^2\big )\hspace{0.05cm} , \hspace{0.2cm}{\boldsymbol{\rm Q}}(D) = \big ( 1+D+D^2\big )\hspace{0.05cm} .$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\boldsymbol{\rm T}}(D) = \big ( 1+D^2\big )\hspace{0.05cm} , \hspace{0.2cm}
	{\boldsymbol{\rm Q}}(D) = \big ( 1+D+D^2\big )\hspace{0.05cm} .$$

	*For the two elements of the systematic transfer function matrix,  we obtain:		*For the two elements of the systematic transfer function matrix,  we obtain:
	:$$G^{(1)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm T}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = 1 C,\hspace{0.8cm}		:$$G^{(1)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm T}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = 1 C,\hspace{0.8cm}G^{(2)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm Q}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = \frac{1+D+D^2}{1+D^2}$$
	G^{(2)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm Q}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = \frac{1+D+D^2}{1+D^2}$$		:$$\Rightarrow \hspace{0.3cm}{\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.1cm} \frac{1+D+D^2}{1+D^2} \hspace{0.05cm}\big )\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}{\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.1cm} \frac{1+D+D^2}{1+D^2} \hspace{0.05cm}\big )
	\hspace{0.05cm}.$$

	So the  <u>last proposed solution</u>  is correct:		So the  <u>last proposed solution</u>  is correct:
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	'''(3)'''  The generator function matrix of encoder  $\rm B$  is:		'''(3)'''  The generator function matrix of encoder  $\rm B$  is:
	:$${\boldsymbol{\rm G}}_{\rm B}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {1+D+D^2} \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}_{\rm B}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {1+D+D^2} \hspace{0.05cm}\big )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*So this encoder is not equivalent to encoder  $\rm A$.		*So this encoder is not equivalent to encoder  $\rm A$.

	*Let us now consider the encoder  $\rm C$.  Here the second matrix element of  $\mathbf{G}(D)$:		*Let us now consider the encoder  $\rm C$.  Here the second matrix element of  $\mathbf{G}(D)$:
	:$$w_i \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_i + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}		:$$w_i \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_i + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{U}(D) = W(D) \cdot (1 + D^2 ) \hspace{0.05cm},\hspace{0.8cm}x_i^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} w_i + w_{i-1} + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{X}^{(2)}(D) = W(D) \cdot (1 +D + D^2 )$$
	{U}(D) = W(D) \cdot (1 + D^2 ) \hspace{0.05cm},\hspace{0.8cm}
	x_i^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} w_i + w_{i-1} + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}
	{X}^{(2)}(D) = W(D) \cdot (1 +D + D^2 )$$
	:$$\Rightarrow \hspace{0.3cm} G^{(2)}(D) = \frac{{X}^{(2)}(D)}{{U}(D)} = \frac{1+D+D^2}{1+D^2}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} G^{(2)}(D) = \frac{{X}^{(2)}(D)}{{U}(D)} = \frac{1+D+D^2}{1+D^2}\hspace{0.05cm}.$$

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	[[Category:Channel Coding: Exercises\|^3.2 Polynomial Description^]]		[[Category:Channel Coding: Exercises\|^3.2 Polynomial Description^]]
			[[de:Aufgaben:3.4_Systematische_Faltungscodes]]

Guenter at 17:34, 10 November 2022

2022-11-10T17:34:36Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct is the <u>proposed solution 1</u>:		'''(1)'''  Correct is the  <u>proposed solution 1</u>:
	*Proposition 2 would result if the two outputs were swapped, that is, for the [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1.2Fn-convolution_encoders\|"~~Rate~~ 1/2 standard code"]] mostly considered in the theory section.		*Proposition 2 would result if the two outputs were swapped,  that is,  for the [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1.2Fn-convolution_encoders\|"rate–1/2 standard code"]] mostly considered in the theory section.

	*Proposition 3 applies to a systematic code ⇒ $\underline{x}^{(1)} = \underline{u}$. However, the coder ~~considered here~~  $\rm A$  does not exhibit this property.		*Proposition 3 applies to a systematic code   ⇒   $\underline{x}^{(1)} = \underline{u}$.  However,  the coder $\rm A$  considered here does not exhibit this property.



	'''(2)'''  To go from a nonsystematic $(n, \ k)$ code with matrix $\mathbf{G}(D)$ to the equivalent systematic code   ⇒   matrix $\mathbf{G}_{\rm sys}(D)$, ~~<br>~~ one must generally split $\mathbf{G}(D)$ into a $k × k$ matrix $\mathbf{T}(D)$ and a remainder matrix $\mathbf{Q}(D)$.		'''(2)'''  To go from a nonsystematic  $(n, \ k)$  code with matrix  $\mathbf{G}(D)$  to the equivalent systematic code   ⇒   matrix  $\mathbf{G}_{\rm sys}(D)$,  one must generally split  $\mathbf{G}(D)$  into a  $k × k$  matrix  $\mathbf{T}(D)$  and a remainder matrix  $\mathbf{Q}(D)$.

	*The desired result is then with the $k × k$ identity matrix $\mathbf{I}_k$:		*The desired result is then with the  $k × k$  identity matrix  $\mathbf{I}_k$:
	:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm I}}_k\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm T}}^{-1}(D) \cdot {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm I}}_k\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm T}}^{-1}(D) \cdot {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*We assume here the $\mathbf{G}(D)$ matrix for the coder  $\rm A$ .		*We assume here the  $\mathbf{G}(D)$  matrix for the coder  $\rm A$.  Because of  $k = 1$  here both  $\mathbf{T}(D)$  and $\mathbf{Q}(D)$  have dimension  $1 × 1$,  so strictly speaking they are not matrices at all:
	*Because $k = 1$ here both $\mathbf{T}(D)$ and $\mathbf{Q}(D)$ have dimension $1 × 1$, so strictly speaking they are not matrices at all:
	:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm T}}(D)\hspace{0.05cm} ; \hspace{0.2cm} {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm T}}(D)\hspace{0.05cm} ; \hspace{0.2cm} {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\boldsymbol{\rm T}}(D) = \big ( 1+D^2\big )\hspace{0.05cm} , \hspace{0.2cm}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\boldsymbol{\rm T}}(D) = \big ( 1+D^2\big )\hspace{0.05cm} , \hspace{0.2cm}
	{\boldsymbol{\rm Q}}(D) = \big ( 1+D+D^2\big )\hspace{0.05cm} .$$		{\boldsymbol{\rm Q}}(D) = \big ( 1+D+D^2\big )\hspace{0.05cm} .$$

	*For the two elements of the systematic transfer function matrix, we obtain:		*For the two elements of the systematic transfer function matrix,  we obtain:
	:$$G^{(1)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm T}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = 1 C,\hspace{0.8cm}		:$$G^{(1)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm T}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = 1 C,\hspace{0.8cm}
	G^{(2)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm Q}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = \frac{1+D+D^2}{1+D^2}$$		G^{(2)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm Q}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = \frac{1+D+D^2}{1+D^2}$$
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	So the <u>last proposed solution</u> is correct:		So the  <u>last proposed solution</u>  is correct:
	*Proposed solution 1 does not describe a systematic code.		*Proposed solution 1 does not describe a systematic code.
	*A code according to solution suggestion 2 is systematic, but not equivalent to the coder  $\rm A$  according to the given circuit and transfer function matrix $\mathbf{G}(D)$.
			*A code according to solution suggestion 2 is systematic,  but not equivalent to the coder  $\rm A$  according to the given circuit and transfer function matrix  $\mathbf{G}(D)$.


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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*So this encoder is not equivalent to ~~the~~ encoder  $\rm A$.		*So this encoder is not equivalent to encoder  $\rm A$.
	*Let us now consider the encoder  $\rm C$. Here the second matrix element of $\mathbf{G}(D)$ ~~holds~~:
			*Let us now consider the encoder  $\rm C$.  Here the second matrix element of  $\mathbf{G}(D)$:
	:$$w_i \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_i + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}		:$$w_i \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_i + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}
	{U}(D) = W(D) \cdot (1 + D^2 ) \hspace{0.05cm},\hspace{0.8cm}		{U}(D) = W(D) \cdot (1 + D^2 ) \hspace{0.05cm},\hspace{0.8cm}
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	:$$\Rightarrow \hspace{0.3cm} G^{(2)}(D) = \frac{{X}^{(2)}(D)}{{U}(D)} = \frac{1+D+D^2}{1+D^2}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} G^{(2)}(D) = \frac{{X}^{(2)}(D)}{{U}(D)} = \frac{1+D+D^2}{1+D^2}\hspace{0.05cm}.$$

	This corresponds exactly to the result of ~~the~~ subtask '''(2)''''  ⇒  <u>Proposed solution 2</u>.		*This corresponds exactly to the result of subtask  '''(2)'''   ⇒   <u>Proposed solution 2</u>.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 17:06, 10 November 2022

2022-11-10T17:06:07Z

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	[[File:EN_KC_A_3_4.png\|right\|frame\|Predefined filter structures]]		[[File:EN_KC_A_3_4.png\|right\|frame\|Predefined filter structures]]
	One speaks of a systematic convolutional code of rate  $R = 1/2$   ⇒   $k = 1, \ n = 2$, if the code bit  $x_i^{(1)}$  is equal to the currently applied information bit  $u_i$~~ ~~.		One speaks of a  "systematic convolutional code"  of rate  $R = 1/2$   ⇒   $k = 1, \ n = 2$,  if the code bit  $x_i^{(1)}$  is equal to the currently applied information bit  $u_i$.

	The transfer function matrix of such a code is:		*The transfer function matrix of such a code is:
	:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The encoder  $\rm A$  shown in the upper graph is certainly not systematic, since for this  $G^{(1)}(D) ≠ 1$  holds. To derive the matrix  $\mathbf{G}(D)$  we refer to a  [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1. 2Fn-convolution_encoders\|"earlier example"]], where for our standard rate 1/2 encoder with memory  $m = 2$  the transfer function matrix was determined:		*The encoder  $\rm A$  shown in the upper graph is certainly not systematic,  since for this  $G^{(1)}(D) ≠ 1$  holds.  To derive the matrix  $\mathbf{G}(D)$  we refer to an  [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1. 2Fn-convolution_encoders\|$\text{earlier example}$]],  where for our standard rate–1/2  encoder with memory  $m = 2$  the transfer function matrix was determined:
	:$${\boldsymbol{\rm G}}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big ) = \big ( \hspace{0.05cm} 1 + D + D^2\hspace{0.05cm} , \hspace{0.2cm} 1 + D^2 \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big ) = \big ( \hspace{0.05cm} 1 + D + D^2\hspace{0.05cm} , \hspace{0.2cm} 1 + D^2 \hspace{0.05cm}\big )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The encoder  $\rm A$  differs from this example only by swapping the two outputs.		:The encoder  $\rm A$  differs from this example only by swapping the two outputs.
	*If the transfer function matrix of a code is		*If the transfer function matrix of a code is
	:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )
	\hspace{0.05cm},$$		\hspace{0.05cm},$$

	then the equivalent systematic representation of this rate 1/2 convolutional code holds in general:		:then the equivalent systematic representation of this rate–1/2  convolutional code holds in general:
	:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {G^{(2)}(D)}/{G^{(1)}(D)} \hspace{0.05cm}\big )		:$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {G^{(2)}(D)}/{G^{(1)}(D)} \hspace{0.05cm}\big )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	In subtask '''(3)''', check which of the systematic arrangements is equivalent to encoder  $\rm A$~~ ~~?		In subtask  '''(3)''',  check which of the systematic arrangements is equivalent to encoder  $\rm A$?
	*Either encoder  $\rm B$,		*Either encoder  $\rm B$,
	*or encoder  $\rm C$
			*or encoder  $\rm C$

	*or both.		*or both.

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

			* Reference is made in particular to the sections
	~~Hints:~~		:* [[Channel_Coding/Algebraic_and_Polynomial_Description#Transfer_Function_Matrix\|"Transfer Function Matrix"]]  and
	* This exercise belongs to the chapter  [[Channel_Coding/Algebraic_and_Polynomial_Description\| Algebraic and Polynomial Description]].		:* [[Channel_Coding/Algebraic_and_Polynomial_Description#Equivalent_systematic_convolutional_code\|"Equivalent systematic convolutional code"]].
	* Reference is made in particular to the ~~pages~~
	:: [[Channel_Coding/Algebraic_and_Polynomial_Description#Transfer_Function_Matrix\|"Transfer Function Matrix"]]  and
	:: [[Channel_Coding/Algebraic_and_Polynomial_Description#Equivalent_systematic_convolutional_code\|"Equivalent systematic convolutional code"]].


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	{Which encoder is equivalent to  $\rm A$  and systematic?		{Which encoder is equivalent to  $\rm A$  and systematic?
	\|type="()"}		\|type="[]"}
	- Encoder  $\rm B$ ,		- Encoder  $\rm B$ ,
	+ encoder  $\rm C$ .		+ encoder  $\rm C$ .

Guenter at 15:30, 27 October 2022

2022-10-27T15:30:52Z

← Older revision		Revision as of 17:30, 27 October 2022
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	*Proposition 2 would result if the two outputs were swapped, that is, for the [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1.2Fn-convolution_encoders\|"Rate 1/2 standard code"]] mostly considered in the theory section.		*Proposition 2 would result if the two outputs were swapped, that is, for the [[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D-transform_to_rate-1.2Fn-convolution_encoders\|"Rate 1/2 standard code"]] mostly considered in the theory section.

	*Proposition 3 applies to a systematic code ⇒ $\underline{x}^{(1)} = $\underline{u}$. However, the coder considered here  $\rm A$  does not exhibit this property.		*Proposition 3 applies to a systematic code ⇒ $\underline{x}^{(1)} = \underline{u}$. However, the coder considered here  $\rm A$  does not exhibit this property.

Guenter at 15:29, 27 October 2022

2022-10-27T15:29:17Z

← Older revision		Revision as of 17:29, 27 October 2022
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	{{quiz-Header\|Buchseite=Channel_Coding/Algebraic_and_Polynomial_Description}}		{{quiz-Header\|Buchseite=Channel_Coding/Algebraic_and_Polynomial_Description}}

	[[File:~~P_ID2629__KC_A_3_4~~.png\|right\|frame\|Predefined filter structures]]		[[File:EN_KC_A_3_4.png\|right\|frame\|Predefined filter structures]]
	One speaks of a systematic convolutional code of rate  $R = 1/2$   ⇒   $k = 1, \ n = 2$, if the code bit  $x_i^{(1)}$  is equal to the currently applied information bit  $u_i$ .		One speaks of a systematic convolutional code of rate  $R = 1/2$   ⇒   $k = 1, \ n = 2$, if the code bit  $x_i^{(1)}$  is equal to the currently applied information bit  $u_i$ .

Noah at 17:27, 26 September 2022

2022-09-26T17:27:21Z

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Guenter at 09:14, 12 September 2022

2022-09-12T09:14:33Z

← Older revision		Revision as of 11:14, 12 September 2022
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	[[Category:Channel Coding: Exercises\|^3.2 ~~Polynomische Beschreibung~~^]]		[[Category:Channel Coding: Exercises\|^3.2 Polynomial Description^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 3.4: Systematische Faltungscodes to Aufgaben:Exercise 3.4: Systematic Convolution Codes

2022-09-12T09:06:26Z

Guenter moved page Aufgaben:Aufgabe 3.4: Systematische Faltungscodes to Aufgaben:Exercise 3.4: Systematic Convolution Codes

← Older revision	Revision as of 11:06, 12 September 2022
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