Aufgaben:Exercise 2.4: Dual Code and Gray Code

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Quaternary signals with dual and gray coding

The two shown signals  $s_{1}(t)$  and  $s_{2}(t)$  are two different realizations of a redundancy-free quaternary transmit signal, both derived from the blue drawn source signal  $q(t)$. 

For one of the transmitted signals, the so-called dual code with mapping

$$\mathbf{LL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -s_0, \hspace{0.35cm} \mathbf{LH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -s_0/3,\hspace{0.35cm}

\mathbf{HL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +s_0/3, \hspace{0.35cm} \mathbf{HH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +s_0$$ was used, for the other one a certain form of a gray code. This is characterized by the fact that the binary representation of adjacent amplitude values always differ only in a single bit.

The solution of the exercise should be based on the following assumptions:

• The amplitude levels are  $±3\, \rm V$ and $±1 \, \rm V$.
• The decision thresholds lie in the middle between two adjacent amplitude values, i.e. at  $–2\, \rm V$, $0\, \rm V$  and  $+2\, \rm V$.
• The noise rms value  $\sigma_{d}$  is to be chosen so that the distortion probability from the outer symbol  $(+s_0)$  to the nearest symbol  $(+s_{0}/3)$  is exactly  $p = 1\%$.
• Distortions to non-adjacent symbols can be excluded; in the case of Gaussian perturbations, this simplification is always allowed in practice.

One distinguishes in principle between

• the  symbol error probability  $p_{\rm S}$  (related to the quaternary signal) and
• the  bit error probability  $p_{B}$  (related to the binary source signal).

Notes:

• The exercise is part of the chapter  Basics of Coded Transmission.
• Reference is also made to the chapter  Redundancy-Free Coding.

• For numerical evaluation of the Q–function you can use the interactive applet  Complementary Gaussian Error Functions  benutzen.

Questions

1 Which of the signals  $s_{1}(t)$  or  $s_{2}(t)$  uses a gray coding?

	$s_{1}(t)$  uses a gray coding.
	$s_{2}(t)$  uses a gray coding.

2 Determine the noise rms value from the given condition.

$\sigma_{d} \ = \ $ $\ \rm V$

3 What is the symbol error probability using the gray code?

$p_{\rm S} \ = \ $ $\ \%$

4 What is the bit error probability with the gray code?

$p_{\rm B} \ = \ $ $\ \%$

5 What is the symbol error probability with the dual code?

$p_{\rm S} \ = \ $ $\ \%$

6 What is the bit error probability with the dual code?

$p_{\rm B} \ = \ $ $\ \%$

Solution

Solution

(1)  In the signal $s_{2}(t)$ one recognizes the realization of the dual code indicated at the beginning. On the other hand, in the signal $s_{2}(t)$ a gray code $\Rightarrow$ solution 1 with the following mapping was used:

$$\mathbf{HH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -1, \hspace{0.35cm} \mathbf{HL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} -1/3, \hspace{0.35cm} \mathbf{LL}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +1/3, \hspace{0.35cm} \mathbf{LH}\hspace{0.1cm}\Leftrightarrow \hspace{0.1cm} +1 \hspace{0.05cm}.$$

(2)  Let the probability $p$ that the amplitude value $3 \, \rm V$ falls below the adjacent decision threshold $2\, \rm V$ due to the Gaussian distributed noise with standard deviation $\sigma_{d}$ be $1\, \%$. It follows that:

$$ p = {\rm Q} \left ( \frac{3\,{\rm V} - 2\,{\rm V}} { \sigma_d}\right ) = 1 \%\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {1\,{\rm V} }/ { \sigma_d} \approx 2.33 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} { \sigma_d}\hspace{0.15cm}\underline {\approx 0.43\,{\rm V}}\hspace{0.05cm}.$$

(3)  The two outer symbols are each distorted with probability $p$, the two inner symbols with double probability $(2p)$. By averaging considering equal symbol occurrence probabilities, we obtain

$$p_{\rm S} = 1.5 \cdot p \hspace{0.15cm}\underline { = 1.5 \,\%} \hspace{0.05cm}.$$

(4)  Each symbol error results in exactly one bit error. However, since each quaternary symbol contains exactly two binary symbols, the bit error probability is obtained:

$$p_{\rm B} = {p_{\rm S}}/ { 2}\hspace{0.15cm}\underline { = 0.75 \,\%} \hspace{0.05cm}.$$

(5)  When calculating the symbol error probability $p_{\rm S}$, the mapping used is not taken into account. As in subtask (3), we obtain $p_{\rm S} \hspace{0.15cm}\underline{ = 1.5 \, \%}$.

(6)  The two outer symbols are distorted with $p$ and lead to only one bit error each even with dual code.

• The inner symbols are distorted with $2p$ and now lead to $1.5$ bit errors on average.
• Taking into account the factor $2$ in the denominator – see subtask (2) – we thus obtain for the bit error probability of the dual code:

$$p_{\rm B} = \frac{1} { 4} \cdot \frac{p + 2p \cdot 1.5 + 2p \cdot 1.5 + p} { 2} = p \hspace{0.15cm}\underline { = 1 \,\%} \hspace{0.05cm}.$$