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  $S_0, \, \text{...} \, , \, S_7$  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 

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

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

2

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

${\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} \ = \ $

3

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

${\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

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.