Aufgaben:Exercise 3.4: Systematic Convolution Codes

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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  $u_i$.

• The transfer function matrix of such a code is:

$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )\hspace{0.05cm}.$$

• 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  $\text{earlier example}$,  where for our standard rate–1/2  encoder with memory  $m = 2$  the transfer function matrix was determined:

$${\boldsymbol{\rm G}}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big ) = \big ( \hspace{0.05cm} 1 + D + D^2\hspace{0.05cm} , \hspace{0.2cm} 1 + D^2 \hspace{0.05cm}\big )\hspace{0.05cm}.$$

The encoder  $\rm A$  differs from this example only by swapping the two outputs.

• If the transfer function matrix of a code is

$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} G^{(1)}(D)\hspace{0.05cm} , \hspace{0.2cm} G^{(2)}(D) \hspace{0.05cm}\big )\hspace{0.05cm},$$

then the equivalent systematic representation of this rate–1/2  convolutional code holds in general:

$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {G^{(2)}(D)}/{G^{(1)}(D)} \hspace{0.05cm}\big )\hspace{0.05cm}.$$

In subtask  (3),  check which of the systematic arrangements is equivalent to encoder  $\rm A$?

• Either encoder  $\rm B$,

• or encoder  $\rm C$ 

• or both.

Hints:

• This exercise belongs to the chapter  "Algebraic and Polynomial Description".

• Reference is made in particular to the sections 

• "Transfer Function Matrix"  and 
• "Equivalent systematic convolutional code".

Questions

1 What is the transfer function matrix of  $\rm A$?

	$\mathbf{G}(D) = (1 + D^2, \ 1 + D + D^2)$,
	$\mathbf{G}(D) = (1 + D + D^2, \ 1 + D^2)$,
	$\mathbf{G}(D) = (1, \ 1 + D + D^2)$.

2 What is the equivalent systematic transfer function matrix?

	$\mathbf{G}_{\rm sys}(D) = (1 + D + D^2, \ 1 + D^2)$,
	$\mathbf{G}_{\rm sys}(D) = (1, \ 1 + D + D^2)$,
	$\mathbf{G}_{\rm sys}(D) = (1, \ (1 + D + D^2)/(1 + D^2))$.

3 Which encoder is equivalent to  $\rm A$  and systematic?

	Encoder  $\rm B$ ,
	encoder  $\rm C$ .

Solution

Solution

(1)  Correct is the  proposed solution 1:

• Proposition 2 would result if the two outputs were swapped,  that is,  for the "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 $\rm A$  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  $\mathbf{G}_{\rm sys}(D)$,  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  $\mathbf{I}_k$:

$${\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm I}}_k\hspace{0.05cm} ; \hspace{0.1cm} {\boldsymbol{\rm T}}^{-1}(D) \cdot {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )\hspace{0.05cm}.$$

• 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:

$${\boldsymbol{\rm G}}(D) = \big ( \hspace{0.05cm} {\boldsymbol{\rm T}}(D)\hspace{0.05cm} ; \hspace{0.2cm} {\boldsymbol{\rm Q}}(D)\hspace{0.05cm}\big )\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\boldsymbol{\rm T}}(D) = \big ( 1+D^2\big )\hspace{0.05cm} , \hspace{0.2cm}{\boldsymbol{\rm Q}}(D) = \big ( 1+D+D^2\big )\hspace{0.05cm} .$$

• For the two elements of the systematic transfer function matrix,  we obtain:

$$G^{(1)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm T}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = 1 C,\hspace{0.8cm}G^{(2)}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\boldsymbol{\rm Q}}(D) \cdot {\boldsymbol{\rm T}}^{-1}(D) = \frac{1+D+D^2}{1+D^2}$$

$$\Rightarrow \hspace{0.3cm}{\boldsymbol{\rm G}}_{\rm sys}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.1cm} \frac{1+D+D^2}{1+D^2} \hspace{0.05cm}\big )\hspace{0.05cm}.$$

So the  last proposed solution  is correct:

• 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  $\mathbf{G}(D)$.

(3)  The generator function matrix of encoder  $\rm B$  is:

$${\boldsymbol{\rm G}}_{\rm B}(D) = \big ( \hspace{0.05cm} 1\hspace{0.05cm} , \hspace{0.2cm} {1+D+D^2} \hspace{0.05cm}\big )\hspace{0.05cm}.$$

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

• Let us now consider the encoder  $\rm C$.  Here the second matrix element of  $\mathbf{G}(D)$:

$$w_i \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_i + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{U}(D) = W(D) \cdot (1 + D^2 ) \hspace{0.05cm},\hspace{0.8cm}x_i^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} w_i + w_{i-1} + w_{i-2} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{X}^{(2)}(D) = W(D) \cdot (1 +D + D^2 )$$

$$\Rightarrow \hspace{0.3cm} G^{(2)}(D) = \frac{{X}^{(2)}(D)}{{U}(D)} = \frac{1+D+D^2}{1+D^2}\hspace{0.05cm}.$$

• This corresponds exactly to the result of subtask  (2)   ⇒   Proposed solution 2.