Aufgaben:Exercise 1.4: Entropy Approximations for the AMI Code

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Binary source signal (top) and
ternary encoder signal (bottom)

The graph above shows the binary source signal  $q(t)$, which can also be described by the symbol sequence  $\langle q_\nu \rangle$  with  $q_\nu \in \{ {\rm L}, {\rm H} \}$ .  In the entire task,  $p_{\rm L} = p_{\rm H} =0.5$ applies.

The coded signal  $c(t)$  and the corresponding symbol sequence  $\langle c_\nu \rangle$  with  $c_\nu \in \{{\rm P}, {\rm N}, {\rm M} \}$  results from the AMI coding (Alternate Mark Inversion) according to the following rule:

• The binary symbol  $\rm L$  ⇒  Low is always represented by the ternary symbol  $\rm N$  ⇒  Null .
• The binary symbol  $\rm H$  ⇒  High  is also coded deterministically but alternately (hence the name „AMI”) by the symbols  $\rm P$  ⇒  Plus  and  $\rm M$  ⇒  Minus .

In this task, the entropy approximations for the AMI–coded signal are to be calculated:

• Approximation  $H_1$  refers only to the symbol probabilities  $p_{\rm P}$,  $p_{\rm N}$  and  $p_{\rm M}$.

• The  $k$–th entropy approximation  $(k = 2, 3, \text{...} \ )$  can be determined according to the following equation:

$$H_k = \frac{1}{k} \cdot \sum_{i=1}^{3^k} p_i^{(k)} \cdot {\rm log}_2\hspace{0.1cm}\frac {1}{p_i^{(k)}} \hspace{0.5cm}({\rm Einheit\hspace{-0.1cm}: \hspace{0.1cm}bit/Symbol})

\hspace{0.05cm}.$$

Here,  $p_i^{(k)}$  the  $i$–th composite probability of a  $k$–tuple.

Hints:

• This task belongs to the chapter  Sources with Memory.
• Reference is made in particular to the page  The Entropy of the AMI-code.
• In task  Aufgabe 1.4Z  the actual entropy of the code symbol sequence  $\langle c_\nu \rangle$  is calculated to  $H = 1 \; \rm bit/symbol$ .
• The following relations between the units are to be expected:   $H \le$ ...$ \le H_3 \le H_2 \le H_1 \le H_0

\hspace{0.05cm}.$

Questions

1 What is the decision content of the AMI code?

$H_0 \ = \ $ $\ \rm bit/symbol$

2 Calculate the first entropy approximation of the AMI code.

$H_1 \ = \ $ $\ \rm bit/symbol$

3 What is the entropy approximation  $H_2$, based on two-tuples?

$H_2 \ = \ $ $\ \rm bit/symbol$

4 What is the value of the entropy approximation  $H_3$, based on three-tuples?

$H_3 \ = \ $ $\ \rm bit/symbol$

5 Which statements apply to the entropy approximation  $H_4$?

	It must be averaged over  $3^4 = 81$  summands.
	$1 \; {\rm bit/symbol} < H_4 < H_3$ is valid.
	After a long calculation you get  $H_4 = 1.333 \; {\rm bit/symbol}$.

Solution

Solution

(1)  The symbol size is  $M = 3$.  This gives the decision content with the  binary logarithm  to the base $2$   ⇒   $\log_2$ bzw $\rm ld$:

$$H_0 = {\rm log}_2\hspace{0.1cm} M = {\rm log}_2\hspace{0.1cm} (3) \hspace{0.15cm} \underline { = 1.585 \,{\rm bit/symbol}}

\hspace{0.05cm}.$$

(2)  The first-order entropy approximation takes into account only the symbol probabilities  $p_{\rm P}$,  $p_{\rm N}$  and  $p_{\rm M}$  and not the statistical bonds within the code sequence  $\langle c_\nu \rangle$.  Thus one obtains:

$$p_{\rm N} = p_{\rm L} = 1/2\hspace{0.05cm},\hspace{0.2cm}p_{\rm P} = p_{\rm M} = p_{\rm H}/2 = 1/4 \hspace{0.3cm}

\Rightarrow\hspace{0.3cm} H_1 = \frac{1}{2} \cdot {\rm log}_2\hspace{0.1cm} (2) +

2 \cdot \frac{1}{4} \cdot {\rm log}_2\hspace{0.1cm}(4)  \hspace{0.15cm} \underline {= 1.5 \,{\rm bit/Symbol}} 
\hspace{0.05cm}.$$

(3)  First, the  $M^2 = 9$  composite probabilities of two-tuples have to be determined here, in the following marked by the first two code symbols  $c_1$  and  $c_2$:

• Since in the AMI code neither  $\rm P$  can follow  $\rm P$  nor  $\rm M$  can follow  $\rm M$ ,  $p_{\rm PP} = p_{\rm MM} =0$.
• For the composite probabilities under the condition  $c_2 = \rm N$  applies:

$$p_{\rm NN} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}( c_1 = \mathbf{N}) \cdot {\rm Pr}(c_2 = \mathbf{N}\hspace{0.05cm} | c_1 = \mathbf{N}) = 1/2 \cdot 1/2 = 1/4 \hspace{0.05cm},$$

$$ p_{\rm MN} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}( c_1 = \mathbf{M}) \cdot {\rm Pr}(c_2 = \mathbf{N}\hspace{0.05cm} | c_1 = \mathbf{M}) = 1/4 \cdot 1/2 = 1/8 \hspace{0.05cm},$$

$$ p_{\rm PN} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}( c_1 = \mathbf{P}) \cdot {\rm Pr}(c_2 = \mathbf{N}\hspace{0.05cm} | c_1 = \mathbf{P}) = 1/4 \cdot 1/2 = 1/8

\hspace{0.05cm}.$$

• The composite probabilities of the two-tuples  $\rm PM$  and  $\rm MP$  are:

$$p_{\rm PM} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}( c_1 = \mathbf{P}) \cdot {\rm Pr}(c_2 = \mathbf{M}\hspace{0.05cm} | c_1 = \mathbf{P}) = 1/4 \cdot 1/2 = 1/8 \hspace{0.05cm},$$

$$ p_{\rm MP} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}( c_1 = \mathbf{M}) \cdot {\rm Pr}(c_2 = \mathbf{P}\hspace{0.05cm} | c_1 = \mathbf{M}) = 1/4 \cdot 1/2 = 1/8 \hspace{0.05cm}.$$

• For the remaining probabilities, it must also be taken into account whether the binary symbol  $\rm H$  with  $\rm P$  or with  $\rm M$  last time  ⇒  further factor  $1/2$:

$$p_{\rm NM} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}( c_1 = \mathbf{N}) \cdot {\rm Pr}(c_2 = \mathbf{M}\hspace{0.05cm} | c_1 = \mathbf{N}) = 1/2 \cdot 1/2 \cdot 1/2= 1/8 \hspace{0.05cm},$$

$$ p_{\rm NP} \hspace{0.1cm} = \hspace{0.1cm} {\rm Pr}( c_1 = \mathbf{N}) \cdot {\rm Pr}(c_2 = \mathbf{P}\hspace{0.05cm} | c_1 = \mathbf{N}) = 1/2 \cdot 1/2 \cdot 1/2 = 1/8 \hspace{0.05cm}.$$

Thus the entropy  $H_2'$  of a two-tuple or its entropy  $H_2$  per code symbol:

$$H_2\hspace{0.01cm}' = \frac{1}{4} \cdot {\rm log}_2\hspace{0.1cm} (4) +

6 \cdot \frac{1}{8} \cdot {\rm log}_2\hspace{0.1cm}(8)  \hspace{0.15cm} {= 2.75 \,{\rm bit/two-tuple}}\hspace{0.3cm}

\Rightarrow\hspace{0.3cm} H_2 = \frac{H_2\hspace{0.01cm}'}{2} \hspace{0.15cm} \underline {= 1.375 \,{\rm bit/Symbol}}

\hspace{0.05cm}.$$

(4)  The calculation of  $H_3$  is similar to the last subtask for  $H_2$, except that now  $3^3 = 27$  composite probabilities must be determined:

$$p_{\rm NNN} = 1/8\hspace{0.4cm}{\rm (nur \hspace{0.15cm}einmal)}

\hspace{0.05cm},$$

$$p_{\rm NMM} = p_{\rm NPP} = p_{\rm MNM} = ... = 0 \hspace{0.4cm}{\rm (ingesamt \hspace{0.15cm}12)}

\hspace{0.05cm},$$

$$p_{\rm NNM} = p_{\rm NNP} = p_{\rm PMP} = ... = 1/16 \hspace{0.4cm}{\rm (ingesamt \hspace{0.15cm}14)}$$

$$\Rightarrow\hspace{0.3cm} H_3 = \frac{1}{3} \cdot \left [

\frac{1}{8} \cdot {\rm log}_2\hspace{0.1cm} (8) + 
14 \cdot \frac{1}{16} \cdot {\rm log}_2\hspace{0.1cm}(16)
\right ]  \hspace{0.15cm} \underline {= 1.292 \,{\rm bit/symbol}} 
\hspace{0.05cm}.$$

(5)  Correct are the proposed solutions 1 and 2.

• On the other hand, statement 3 is wrong, because  $H_4$  must in any case be smaller than  $H_3 = 1.292 \; \rm bit/symbol$.