Exercise 3.4: Systematic Convolution Codes

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 a "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 pages

"Transfer Function Matrix" and

"Equivalent systematic convolutional code".

Questions

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 considered here $\rm A$ 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 $k = 1$ 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 the encoder $\rm A$.
Let us now consider the encoder $\rm C$. Here the second matrix element of $\mathbf{G}(D)$ holds:

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

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

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

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

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