Aufgaben:Exercise 3.6Z: Transition Diagram at 3 States

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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  $S_0, \, \text{...} \, , \, S_7$  placeholder names  $\mathbf{A}, \, \text{...} \, , \, \mathbf{H}$  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  $S_{\mu}$ ,

• one 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  "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 
  • "State definition for a memory register"  as well as.
  • "Representation in the state transition diagram".

Questions

1

For which states  $S_{\mu}$  do the placeholders  $\mathbf{A}$  and  $\mathbf{F}$ stand?

${\rm Zustand} \ \mathbf{A} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

${\rm Zustand} \ \mathbf{F} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

2

Also name the mappings of the other placeholders to the indexes.

${\rm Zustand} \ \mathbf{B} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

${\rm Zustand} \ \mathbf{C} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

${\rm Zustand} \ \mathbf{D} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

${\rm Zustand} \ \mathbf{E} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

${\rm Zustand} \ \mathbf{G} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

${\rm Zustand} \ \mathbf{H} \ ⇒ \ {\rm Index} \ {\mu} \ = \ $

3

To which state  $S_{\mu}$  does the second arrow in each case go?

${\rm Von \ {\it S}_{\rm 1} \ zum \ Zustand \ mit \ Index \ {\mu}} \ = \ $

${\rm Von \ {\it S}_{\rm 3} \ zum \ Zustand \ mit \ Index \ {\mu}} \ = \ $

${\rm Von \ {\it S}_{\rm 5} \ zum \ Zustand \ mit \ Index \ {\mu}} \ = \ $

${\rm Von \ {\it S}_{\rm 7} \ zum \ Zustand \ mit \ Index \ {\mu}} \ = \ $

Solution

Solution

Relationship between placeholders and 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 infobit $u_i = 0$ (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 $S_3$ ⇒ $u_{i–1} = 1, \ u_{i–2} = 1, \ u_{i–3} = 0$.

State transition diagram with $2^3$ states

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 the 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.