Exercise 3.6Z: Transition Diagram at 3 States

State transition diagram for $m = 3$ $($incomplete$)$

In the state transition diagram of an encoder with memory $m$ there are $2^m$ states. Therefore, the diagram shown with eight states describes a convolutional encoder with memory $m = 3$.

Usually the states are denoted by $S_0, \ \text{...} \ , \ S_{\mu}, \ \text{...} \ , \ S_7$, where the index $\mu$ is determined from the occupancy of the shift register $($contents from left to right: $u_{i-1}, u_{i-2}, u_{i-3})$ :

$$\mu = \sum_{l = 1}^{m} \hspace{0.1cm}2\hspace{0.03cm}^{l-1} \cdot u_{i-l} \hspace{0.05cm}.$$

The state $S_0$ therefore results for the shift register content "$000$", the state $S_1$ for "$100$" and the state $S_7$ for "$111$".

However, in the above graphic, for the states $S_0, \, \text{...} \, , \, S_7$ only placeholder names $\mathbf{A}, \, \text{...} \, , \, \mathbf{H}$ are used. In the subtasks (1) and (2) you should clarify which placeholder stands for which state.

For convolutional encoders of rate $1/n$, which will be exclusively considered here, two arrows depart from each state $S_{\mu}$ ,

a red one for the current information bit $u_i = 0$ and

a blue one for $u_i = 1$.

This is another reason why the state transition diagram shown is not complete. It is to be mentioned furthermore:

At each state also two arrows arrive, whereby these can be absolutely of the same color.

Next to the arrows there are usually the $n$ code bits. This was also omitted here.

Hints:

The exercise belongs to the chapter "Code Description with State and Trellis Diagram".

In $\text{Exercise 3.7Z}$ two convolutional codes with memory $m = 3$ are examined, both of which can be described by the transition diagram analyzed here.

Please include the appropriate index $\mu$ for all questions.

Reference is made in particular to the sections

Questions

${\rm state} \ \mathbf{A} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm state} \ \mathbf{F} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm state} \ \mathbf{B} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm state} \ \mathbf{C} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm state} \ \mathbf{D} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm state} \ \mathbf{E} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm state} \ \mathbf{G} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm state} \ \mathbf{H} \ ⇒ \ {\rm index} \ {\mu} \ = \ $

${\rm From \ {\it S}_{\rm 1} \ to \ the\ state \ with \ index \ {\mu}} \ = \ $

${\rm From \ {\it S}_{\rm 3} \ to \ the\ state \ with \ index \ {\mu}} \ = \ $

${\rm From \ {\it S}_{\rm 5} \ to \ the\ state \ with \ index \ {\mu}} \ = \ $

${\rm From \ {\it S}_{\rm 7} \ to \ the\ state \ with \ index \ {\mu}} \ = \ $

Solution

Relationship between placeholders and states

State transition diagram with $2^3$ states

(1) The placeholder $\mathbf{A}$ represents the state $S_0$ ⇒ $u_{i-1} = 0, \ u_{i-2} = 0, \ u_{i-3} = 0$.

This is the only state $S_{\mu}$ where one remains in the same state $S_{\mu}$ by the info-bit $u_i = 0$ (red arrow).

From the state $S_7$ ⇒ $u_{i-1} = 1, \ u_{i-2} = 1, \ u_{i-3} = 1$ one comes with $u_i = 1$ (blue arrow) also again to the state $S_7$.

Thus, for $\mathbf{A}$ the index $\underline{\mu = 0}$ and for $\mathbf{F}$ the index $\underline{\mu = 7}$ had to be entered.

(2) Starting from the state $\mathbf{A} = S_0$, one arrives at the following states according to the initial graph in a clockwise direction with the red arrows $(u_i = 0)$ or the blue arrows $(u_i = 1)$:

$$u_{i–3} = 0, \ u_{i–2} = 0, \ u_{i–1} = 0, \ u_i = 1 ⇒ s_{i+1} = \mathbf{B} = S_1,$$

$$u_{i–3} = 0, \ u_{i–2} = 0, \ u_{i–1} = 1, \ u_i = 0 ⇒ s_{i+1} = \mathbf{C} = S_2,$$

$$u_{i–3} = 0, \ u_{i–2} = 1, \ u_{i–1} = 0, \ u_i = 1 ⇒ s_{i+1} = \mathbf{D} = S_5,$$

$$u_{i–3} = 1, \ u_{i–2} = 0, \ u_{i–1} = 1, \ u_i = 1 ⇒ s_{i+1} = \mathbf{E} = S_3,$$

$$u_{i–3} = 0, \ u_{i–2} = 1, \ u_{i–1} = 1, \ u_i = 1 ⇒ s_{i+1} = \mathbf{F} = S_7,$$

$$u_{i–3} = 1, \ u_{i–2} = 1, \ u_{i–1} = 1, \ u_i = 0 ⇒ s_{i+1} = \mathbf{G} = S_6,$$

$$u_{i–3} = 1, \ u_{i–2} = 1, \ u_{i–1} = 0, \ u_i = 0 ⇒ s_{i+1} = \mathbf{H} = S_4,$$

$$u_{i–3} = 1, \ u_{i–2} = 0, \ u_{i–1} = 0, \ u_i = 0 ⇒ s_{i+1} = \mathbf{A} = S_0.$$

So the indices $\mu$ are to be entered in the order 1, 2, 5, 3, 6, 4.

The graphic shows the connection between the placeholders and the states $S_{\mu}$.

(3) From state $S_1$ ⇒ $u_{i–1} = 1, \ u_{i–2} = 0, \ u_{i–3} = 0$ one arrives with $u_i = 0$ (red arrow) at state $S_2$. On the other hand, with $u_i = 1$ (blue arrow) one ends up at the state $S_3$ ⇒ $u_{i–1} = 1, \ u_{i–2} = 1, \ u_{i–3} = 0$.

The adjacent graphic shows the state transition diagram with all transitions. From this it can be read:

From state $S_3$ one comes with $u_i = 0$ to state $S_6$.

From state $S_5$ one comes with $u_i = 0$ to the state $S_2$.

From the state $S_7$ one comes with $u_i = 0$ to the state $S_6$.

Thus, the indices are to be entered in the order 3, 6, 2, 6.