Channel Coding/Code Description with State and Trellis Diagram

1. Channel Coding
2. Convolutional Codes and Their Decoding
3. Code Description with State and Trellis Diagram

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State definition for a memory register

A convolutional encoder can also be understood as an automaton with a finite number of states. The number of states results from the number of memory elements   ⇒   memory  $m$  thereby to  $2^m$.

Convolutional encoder with  $k = 1, \ n = 2, \ m = 2$

In this chapter we mostly start from the drawn convolutional encoder, which is characterized by the following parameters:

• $k = 1, \ n = 2$   ⇒   Coderate $R = 1/2$,
  

• Memory  $m = 2$   ⇒   influence length $\nu = 3$,
  

• Transfer function matrix in octal form  $(7, 5)$:

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

The code sequence at time  $i$   ⇒   $\underline{x}_i = (x_i^{(1)}, \ x_i^{(2)})$  depends not only on the information bit  $u_i$  but also on the content  $(u_{i-1}, \ u_{i-2})$  of the memory.

• There are  $2^m = 4$  possibilities for this, namely the states  $S_0, \ S_1, \ S_2$  and  $S_3$.
• Let the register state  $S_{\mu}$  be defined by the index:

\[\mu = u_{i-1} + 2 \cdot u_{i-2}\hspace{0.05cm}, \hspace{0.5cm}{\rm allgemein\hspace{-0.1cm}:}\hspace{0.15cm} \mu = \sum_{l = 1}^{m} \hspace{0.1cm}2\hspace{0.03cm}^{l-1} \cdot u_{i-l} \hspace{0.05cm}.\]

To avoid confusion, we distinguish in the following by upper or lower case letters:

• the possible states  $S_{\mu}$  with the indices  $0 ≤ \mu ≤ 2^m \,-1$,
  

• the current states  $s_i$  at the times.  $i = 1, \ 2, \ 3, \ \text{...}$
  

$\text{Example 1:}$  The graph shows for the convolutional encoder sketched above

• the information sequence  $\underline{u} = (u_1,u_2, \text{...} \ ) $   ⇒   information bits  $u_i$,

• the current states at the times  $i$   ⇒   $s_i ∈ \{S_0, \ S_1, \ S_2, \ S_3\}$,
  

• the 2 bit sequences  $\underline{x}_i = (x_i^{(1)}, \ x_i^{(2)})$   ⇒   encoding the information bit $u_i$.
  

To illustrate the states  $S_{\mu}$

For example, you can see from this illustration (the colors are to facilitate the transition to later graphics):

• At time  $i = 5$  $u_{i-1} = u_4 = 0$,  $u_{i-2} = u_3 = 1$ holds. That is, the automaton is in state  $s_5 = S_2$. With the information bit  $u_i = u_5 = 0$  the code sequence  $\underline{x}_5 = (11)$ is obtained.
  

• The state for time  $i = 6$  results from  $u_{i-1} = u_5 = 0$  and  $u_{i-2} = u_4 = 0$  to  $s_6 = S_0$. Because of  $u_6 = 0$  now  $\underline{x}_6 = (00)$  is output and the new state  $s_7$  is again  $S_0$.
  

• At time  $i = 12$  the code sequence  $\underline{x}_{12} = (11)$  is output because of  $u_{12} = 0$  as at time  $i = 5$  and one passes from state  $s_{12} = S_2$  to state  $s_{13} = S_0$ .
  

• In contrast, at time  $i = 9$  the code sequence  $\underline{x}_{9} = (00)$  is output and followed by  $s_9 = S_2$  $s_{10} = S_1$. The same is true for  $i = 15$. In both cases, the current information bit is  $u_i = 1$.

$\text{Conclusion:}$  From this example it can be seen that the code sequence  $\underline{x}_i$  at time  $i$  alone depends on

• from the current state  $s_i = S_{\mu} \ (0 ≤ \mu ≤ 2^m -1)$, as well as
  
• from the current information bit  $u_i$.
  
  

Similarly, the successor state  $s_{i+1}$  is determined by  $s_i$  and  $u_i$  alone.

These properties are considered in the so-called state transition diagram  on the next page

.

Representation in the state transition diagram

The graph shows the  state transition diagram for our "standard encoder".

• This provides all information about the  $(n = 2, \ k = 1, \ m = 2)$–convolutional encoder in compact form.
• The color scheme is aligned with the  "sequential representation"  on the previous page.
• For the sake of completeness, the derivation table is also given again.
  

State transition diagram  $\rm A$  for  $n = 2, \ k = 1, \ m = 2$.

The  $2^{m+k}$  transitions are labeled "$u_i \ | \ \underline{x}_i$". For example, it can be read:

• The information bit  $u_i = 0$  (indicated by a red label) takes you from state  $s_i = S_1$  to state  $s_{i+1} = S_2$  and the two code bits are  $x_i^{(1)} = 1, \ x_i^{(2)} = 0$.
  

• With the information bit  $u_i = 1$  (blue label) in the state  $s_i = S_1$  on the other hand, the code bits result in  $x_i^{(1)} = 0, \ x_i^{(2)} = 1$, and one arrives at the new state  $s_{i+1} = S_3$.
  
  

Let us now consider another systematic code, also with the characteristics  $n = 2, \ k = 1, \ m = 2$. This is the  "equivalent systematic representation"  of the upper coder. This is also referred to as  Recursive Systematic Convolutional Encoder .

State transition diagram  $\rm B$  for  $n = 2, \ k = 1, \ m = 2$.

Compared to the upper state transition diagram, one can see the following differences:

• Since the earlier information bits  $u_{i-1}$  and  $u_{i-2}$  are not stored, here the states  $s_i = S_{\mu}$  to the processed sizes  $w_{i-1}$  and  $w_{i-2}$, where  $w_i = u_i + w_{i-1} + w_{i-2}$  holds.
  

• Since this code is also systematic, again  $x_i^{(1)} = u_i$ holds. The derivation of the second code bits  $x_i^{(2)}$  can be found in the  "Exercise 3.5". It is a recursive filter, as described in the  "Filter structure with fractional–rational transfer function"  section.
  
  

$\text{Conclusion:}$  Image comparison shows that a similar state transition diagram is obtained for both encoders:

• The structure of the state transition diagram is determined solely by the parameters  $m$  and  $k$ .

• The number of states is  $2^{m \hspace{0.05cm}\cdot \hspace{0.05cm}k}$.
  

• Arrows go from each state  $2^k$ .
  

• You also get from each state  $s_i ∈ \{S_0, \ S_1, \ S_2, \ S_3\}$  to the same successor states  $s_{i+1}$.
  

• A difference exists solely with respect to the code sequences output, where in both cases  $\underline{x}_i ∈ \{00, \ 01, \ 10, \ 11\}$  applies.

Representation in the trellis diagram

One gets from the state transition diagram to the so-called  trellis diagram  (short:  Trellis) by additionally considering a time component   ⇒   run variable  $i$ . The following graph contrasts both diagrams for our "standard encoder"  $(n = 2, \ k = 1, \ m = 2)$ .

State transition diagram vs. trellis diagram  $(n = 2, \ k = 1, \ m = 2)$

The bottom illustration has a resemblance to a garden trellis – assuming some imagination. Further is to be recognized on the basis of this diagram:

• Since all memory registers are preallocated with zeros, the trellis always starts from the state  $s_1 = S_0$. At the next time  $(i = 2)$  only the two (start–)states  $S_0$  and  $S_1$  are possible.
  

• From time  $i = m + 1 = 3$  the trellis has exactly the same appearance for each transition from  $s_i$  to  $s_{i+1}$ . Each state  $S_{\mu}$  is connected to a successor state by a red arrow  $(u_i = 0)$  and a blue arrow  $(u_i = 1)$ .
  

• Versus a  code tree that grows exponentially with each time step  $i$  – see for example section  "Error sizes and total error sizes"  in the book "Digital Signal Transmission" – here the number of nodes (i.e. possible states) is limited to  $2^m$ .
  

• This pleasing property of any trellis diagram is also used by  "Viterbi algorithm"  for efficient maximum likelihood decoding of convolutional codes.
  
  

The following example is to show that there is a  $1\text{:}1$ mapping between the string of code sequences  $\underline{x}_i$  and the paths through the trellis diagram.

• Also, the information sequence  $\underline{u}$  can be read from the selected trellis path based on the colors of the individual branches.
• A red branch stands for the information bit  $u_i = 0$, a blue one for  $u_i = 1$.
  

$\text{Example 2:}$  Im  $\text{"Example 1"}$  was created for our rate– $1/2$ standard encoder with memory  $m = 2$  and the information sequence  $\underline{u} = (1, 1, 1, 0, 0, 0, 1, 0, 1, \ \text{. ..})$  the code sequence  $\underline{x}$  is derived, which is given again here for  $i ≤ 9$ .

Trellis diagram of a sample sequence

The trellis diagram is drawn below. You can see:

• The selected path through the trellis is obtained by lining up red and blue arrows representing the possible information bits  $u_i \in \{ 0, \ 1\}$ . This statement is true for any rate $1/n$ convolutional code. For a code with  $k > 1$  there would be  $2^k$  different colored arrows.
  

• For a rate–$1/n$–convolutional code, the arrows are labeled with the code words  $\underline{x}_i = (x_i^{(1)}, \ \text{...} \ , \ x_i^{(n)})$  which are derived from the information bit  $u_i$  and the current register states  $s_i$  . For  $n = 2$  there are only four possible code words, namely  $00, \ 01, \ 10$  and  $11$.
  

• In the vertical direction, one recognizes from the trellis the register states  $S_{\mu}$. Given a rate $k/n$ convolutional code with memory order  $m$  there are  $2^{k \hspace{0.05cm}\cdot \hspace{0.05cm}m}$  different states. In the present example  $(k = 1, \ m = 2)$  these are the states  $S_0, \ S_1, \ S_2$ and $S_3$.

Definition of the free distance

An important parameter of linear block codes is the  "minimum Hamming distance"  between any two codewords  $\underline{x}$  and  $\underline{x}\hspace{0.05cm}'$:

\[d_{\rm min}(\mathcal{C}) = \min_{\substack{\underline{x},\hspace{0.05cm}\underline{x}' \hspace{0.05cm}\in \hspace{0.05cm} \mathcal{C} \\ {\underline{x}} \hspace{0.05cm}\ne \hspace{0.05cm} \underline{x}\hspace{0.03cm}'}}\hspace{0.1cm}d_{\rm H}(\underline{x}, \hspace{0.05cm}\underline{x}\hspace{0.03cm}')\hspace{0.05cm}.\]

• Due to linearity, the zero word  $\underline{0}$ also belongs to each block code. Thus  $d_{\rm min}$  is identical to the minimum  "Hamming weight"  $w_{\rm H}(\underline{x})$  of a codeword  $\underline{x} ≠ \underline{0}$.
  

• For convolutional codes, however, the description of the distance ratios proves to be much more complex, since a convolutional code consists of infinitely long and infinitely many code sequences. Therefore we define here instead of the minimum Hamming distance:

$\text{Definition:}$ 

• The  free distance  $d_{\rm F}$  of a convolutional code is equal to the number of bits in which any two code sequences  $\underline{x}$  and  $\underline{x}\hspace{0.03cm}'$  differ (at least).
• In other words,   The free distance is equal to the minimum  Hamming distance  between any two paths  through the trellis.

Since convolutional codes are also linear, we can also use the infinite zero sequence as a reference sequence:   $\underline{x}\hspace{0.03cm}' = \underline{0}$. Thus, the free distance  $d_{\rm F}$  is equal to the minimum Hamming weight (number of ones) of a code sequence  $\underline{x} ≠ \underline{0}$.

$\text{Example 3:}$  We consider the null sequence  $\underline{0}$  (marked with white filled circles) as well as two other code sequences  $\underline{x}$  and  $\underline{x}\hspace{0.03cm}'$  (with yellow and dark background, respectively) in our standard trellis and characterize these sequences based on the state sequences:

\[\underline{0} =\left ( S_0 \rightarrow S_0 \rightarrow S_0\rightarrow S_0\rightarrow S_0\rightarrow \hspace{0.05cm}\text{...} \hspace{0.05cm}\right)= \left ( 00, 00, 00, 00, 00,\hspace{0.05cm} \text{...} \hspace{0.05cm}\right) \hspace{0.05cm},\] \[\underline{x} =\left ( S_0 \rightarrow S_1 \rightarrow S_2\rightarrow S_0\rightarrow S_0\rightarrow \hspace{0.05cm}\text{...} \hspace{0.05cm}\right)= \left ( 11, 10, 11, 00, 00,\hspace{0.05cm} \text{...} \hspace{0.05cm}\right) \hspace{0.05cm},\] \[\underline{x}\hspace{0.03cm}' = \left ( S_0 \rightarrow S_1 \rightarrow S_3\rightarrow S_2\rightarrow S_0\rightarrow \hspace{0.05cm}\text{...} \hspace{0.05cm}\right)= \left ( 11, 01, 01, 11, 00,\hspace{0.05cm} \text{...} \hspace{0.05cm}\right) \hspace{0.05cm}.\]

On the definition of free distance

One can see from these plots:

• The free distance  $d_{\rm F} = 5$  is equal to the Hamming–weight  $w_{\rm H}(\underline{x})$. No sequence other than the one highlighted in yellow differs from  $\underline{0}$  by less than five bits. For example  $w_{\rm H}(\underline{x}') = 6$.
  

• In this example, the free distance  $d_{\rm F} = 5$  also results as the Hamming distance between the sequences  $\underline{x}$  and  $\underline{x}\hspace{0.03cm}'$, which differ in exactly five bit positions.

The larger the free distance  $d_{\rm F}$  is, the better is the "asymptotic behavior" of the convolutional encoder.

• For the exact error probability calculation, however, one needs the exact knowledge of the  "weight enumerator function"  just as for the linear block codes.
• This is given for the convolutional codes in detail only in the chapter  "Distance Characteristics and Error Probability Barriers" .

Optimal convolutional codes of rate  $1/2$

At the outset, just a few remarks:

• The free distance  $d_{\rm F}$  increases with increasing memory  $m$  provided that one uses appropriate polynomials for the transfer function matrix  $\mathbf{G}(D)$ .
• In the table, for rate $1/2$ convolutional codes, the  $n = 2$  polynomials are given together with the  $d_{\rm F}$ value.
• Of greater practical importance is the industry standard code  with  $m = 6$  ⇒   $64$  states and the free distance  $d_{\rm F} = 10$  (in the table this is highlighted in red).

The following example shows the effects of using unfavorable polynomials.

$\text{Example 4:}$  In  $\text{Example 3}$  it was shown that our standard convolutional encoder  $(n = 2, \ k = 1, \ m = 2)$  has the free distance  $d_{\rm F} = 5$  . This is based on the transfer function matrix  $\mathbf{G}(D) = (1 + D + D^2, \ 1 + D^2)$, as shown in the model shown (without red deletion).

State transition diagram  $\rm C$  for  $n = 2, \ k = 1, \ m = 2$.

Using  $\mathbf{G}(D) = (1 + D, \ 1 + D^2)$   ⇒   red change, the state transition diagram changes from the  "State transition diagram $\rm A$"  little at first glance. However, the effects are serious:

• The free distance is now no longer  $d_{\rm F} = 5$, but only  $d_{\rm F} = 3$, where the code sequence  $\underline{x} = (11, \ 01, \ 00, \ 00, \ text{. ..})$  with only three ones belonging to the information sequence  $\underline{u} = \underline{1} = (1, \ 1, \ 1, \ 1, \ \text{...})$ .
  

• That is:  By only three transmission errors at positions 1, 2, and 4, one corrupts the one sequence  $(\underline{1})$  into the zero sequence  $(\underline{0})$  and thus produces infinitely many decoding errors in state  $S_3$.
  

• A convolutional encoder with these properties is called "catastrophic". The reason for this extremely unfavorable behavior is that here the two polynomials  $1 + D$  and  $1 + D^2$  are not divisor-remote.

Terminated convolutional codes

In the theoretical description of convolutional codes, one always assumes information sequences  $\underline{u}$  and code sequences  $\underline{x}$  that are infinite in length by definition.

• In praktischen Anwendungen – zum Beispiel  GSM  und  UMTS  – verwendet man dagegen stets eine Informationssequenz endlicher Länge $L$.
• Bei einem Rate–$1/n$–Faltungscode hat dann die Codesequenz mindestens die Länge  $n \cdot L$.
  

Terminierter Faltungscode der Rate  $R = 128/260$

The graph without the gray background on the right shows the trellis of our standard rate $1/2$ convolutional code at binary input sequence  $\underline{u}$  of finite length  $L = 128$. Thus the code sequence  $\underline{x}$  has length  $2 \cdot L = 256$.

However, due to the undefined final state, a complete maximum likelihood decoding of the transmitted sequence is not possible. Since it is not known which of the states  $S_0, \ \text{...} \ , \ S_3$  would occur for  $i > L$ , the error probability will be (somewhat) larger than in the limiting case  $L → ∞$.

To prevent this, the convolutional code is terminated as shown in the diagram above:

• You add  $m = 2$  zeros to the  $L = 128$  information bits   ⇒   $L' = 130$.

• This results, for example, for the color highlighted path through the trellis:

\[\underline{x}' = (11\hspace{0.05cm},\hspace{0.05cm} 01\hspace{0.05cm},\hspace{0.05cm} 01\hspace{0.05cm},\hspace{0.05cm} 00 \hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.1cm},\hspace{0.05cm} 10\hspace{0.05cm},\hspace{0.05cm}00\hspace{0.05cm},\hspace{0.05cm} 01\hspace{0.05cm},\hspace{0.05cm} 01\hspace{0.05cm},\hspace{0.05cm} 11 \hspace{0.05cm} ) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\underline{u}' = (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.1cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm} ) \hspace{0.05cm}.\]

• The trellis now always ends (i.e. independent of the data) in the defined final state  $S_0$  and one thus achieves the best possible error probability according to maximum likelihood.
  

• However, the improvement in error probability comes at the cost of a lower code rate.
• For $L \gg m$  this loss is only small. In the considered example, instead of  $R = 0.5$  with termination the code rate results in

$$R\hspace{0.05cm}' = 0.5 \cdot L/(L + m) \approx 0.492.$$

Punctured convolutional codes

$\text{Definition:}$ 

• We assume a convolutional code of rate  $R_0 = 1/n_0$  which we call mother code .
• If one deletes some bits from its code sequence according to a given pattern, it is called a  punctured convolutional code with code rate  $R > R_0$.

The puncturing is done using the puncturing matrix  $\mathbf{P}$  with the following properties:

• The row number is  $n_0$, the column number is equal to the puncturing period  $p$, which is determined by the desired code rate.
  

• The  $n_0 \cdot p$  matrix elements  $P_{ij}$  are binary  $(0$  or  $1)$. If  $P_{ij} = 1$  the corresponding code bit is taken, if  $P_{ij} = 0$  punctured.
  

• The rate of the punctured code is equal to the quotient of the puncturing period  $p$  and the number  $N_1$  of ones in the  $\mathbf{P}$ matrix.
  

One can usually find favorably punctured convolutional codes only by means of computer-aided search.

Here, a punctured convolutional code is said to be favorable if it.

• has the same memory order  $m$  as the parent code $($also the total influence length is the same in both cases:   $\nu = m + 1)$,
  

• has as large a free distance as possible  $d_{\rm F}$  which is of course smaller than that of the mother code..
  
  

$\text{Example 5:}$  Starting from our  "Rate–1/2–Standard code"  with parameters  $n_0 = 2$  and  $m = 2$  we obtain with the puncturing matrix

\[{\boldsymbol{\rm P} } = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p = 3\hspace{0.05cm}, \hspace{0.2cm}N_1 = 4\]

a punctured convolutional code of rate  $R = p/N_1 = 3/4$. We consider the following constellation for this purpose:

• Information sequence: $\hspace{2.5cm} \underline{u} = (1, 0, 0, 1, 1, 0, \ \text{...}\hspace{0.03cm})$,
  

• Code sequence without puncturing: $\hspace{0.7cm} \underline{x} = (11, 1 \color{grey}{0}, \color{gray}{1}1, 11, 0\color{gray}{1}, \color{gray}{0}1, \ \text{...}\hspace{0.03cm})$,
  

• Code sequence with puncturing: $\hspace{0.90cm} \underline{x}\hspace{0.05cm}' = (11, 1\_, \_1, 11, 0\_, \_1, \ \text{...}\hspace{0.03cm})$,
  

• Receive sequence without errors: $\hspace{0.88cm} \underline{y} = (11, 1\_, \_1, 11, 0\_, \_1, \ \text{...}\hspace{0.03cm})$,
  

• Modified receive sequence: $\hspace{0.8cm} \underline{y}\hspace{0.05cm}' = (11, 1\rm E, E1, 11, 0E, E1, \ \text{...}\hspace{0.03cm})$.
  

So each punctured bit in the receive sequence  $\underline{y}$  (marked by an underscore) is replaced by an  Erasure  $\rm E$  – see  "Binary Erasure Channel". Such an erasure created by the puncturing  Erasure  is treated by the decoder in the same way as an erasure by the channel.

Of course, the puncturing increases the error probability. This can already be seen from the fact that the free distance decreases to  $d_{\rm F} = 3$  after the puncturing. In contrast, the mother code has the free distance  $d_{\rm F} = 5$.

Puncturing is used, for example, in the so-called RCPC codes (Rate Compatible Punctured Convolutional Codes). More about this in  Aufgabe 3.8.

Exercises for the chapter

Aufgabe 3.6: Zustandsübergangsdiagramm

Aufgabe 3.6Z: Übergangsdiagramm bei 3 Zuständen

Aufgabe 3.7: Vergleich zweier Faltungscodierer

Aufgabe 3.7Z: Welcher Code ist katastrophal?

Aufgabe 3.8: Rate Compatible Punctured Convolutional Codes