Aufgaben:Exercise 3.5: Eye Opening with Pseudoternary Coding

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Eye diagrams with AMI and duobinary code

Three digital transmission systems are considered,  each with the following matching properties:

• NRZ rectangular pulses with amplitude  $s_0 = 2 \, {\rm V}$,

• coaxial cable with characteristic cable attenuation  $a_* = 40 \, {\rm dB}$,

• AWGN noise with noise power density  $N_0$,

• receiver filter  $H_{\rm E}(f) = 1/H_{\rm K}(f) \cdot H_{\rm G}(f) $ consisting of an ideal channel equalizer  $H_{\rm K}(f)^{-1}$  and a Gaussian low-pass filter  $H_{\rm G}(f)$  with normalized cutoff frequency  $f_{\rm G} \cdot T \approx 0.5$,

• threshold decision with optimal decision thresholds and optimal detection time  $T_{\rm D} = 0$.

The system variants to be investigated in the exercise differ only in terms of the line code:

⇒   $\text{System A}$  uses a binary bipolar redundancy-free transmission signal. The following descriptive variables are known:

• Basic detection pulse values  $g_0 = 1.56 \, {\rm V}$, $g_1 = g_{\rm –1} = 0.22 \, {\rm V}$, $g_2 = g_{\rm –2} = \, \text{ ...} \, \approx 0$

$$\Rightarrow \hspace{0.3cm}{\ddot{o}(T_{\rm D})}/{ 2} = g_{0} -g_{1}-g_{-1} = 1.12\,{\rm V} \hspace{0.05cm}.$$

• Noise rms value  $\sigma_d \approx 0.2 \, {\rm V}$

$$\Rightarrow \hspace{0.3cm}\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2\big]^2}{ \sigma_d^2}\approx 31.36\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \approx 15\,{\rm dB}\hspace{0.05cm}.$$

⇒   $\text{System B}$  uses AMI coding:

• Here the outer symbols  $"+1"$  or  $"–1"$  occur only in isolation.
• In the case of three consecutive symbols, the sequences  "$\hspace{-0.1cm}\text{ ...} \, , \, +1, \, +1, \, +1, \,\text{ ...}$"  and  "$\hspace{-0.1cm}\text{ ...} \, , \, +1, \, 0, \, +1, \, \text{ ...} $"  among others,  are not possible,
• in contrast to the sequence  "$\hspace{-0.1cm}\text{ ...} \, , \, +1, \, –1, \, +1, \, \text{ ...} $".

⇒   $\text{System C}$  uses the duobinary code:

• Here the alternating sequence  "$\hspace{-0.1cm} \text{ ...} \, , \, –1, \, +1, \, –1, \, \text{ ...} $"  is excluded by the code,  which has a favorable effect on the eye opening.

Notes:

• The exercise belongs to the chapter  "Intersymbol Interference for Multi-Level Transmission".

• Not all of the numerical values given here are necessary to solve this exercise.

Questions

1

Calculate the half eye opening for the  AMI code.

$\text{System B:}\hspace{0.4cm} \ddot{o}(T_{\rm D})/2 \ = \ $ $\ {\rm V}$

2

Calculate the worst-case signal-to-noise ratio for this system.

$\text{System B:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ $\ {\rm dB}$

3

How must the thresholds  $E_1$  and  $E_2$  be chosen so that the result just calculated is correct?

$E_1 \ \hspace{0.05cm} = \ $ $\ {\rm V}$

$E_2 \ = \ $ $\ {\rm V}$

4

Calculate the half eye opening at the  duobinary code.

$\text{System C:}\hspace{0.4cm} \ddot{o}(T_{\rm D})/2 \ = \ $ $\ {\rm V}$

5

Calculate the worst-case signal-to-noise ratio for duobinary coding.

$\text{System C:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ $\ {\rm dB}$

Solution

Solution

(1)  Since the symbol rate is not changed in the AMI code compared to the redundancy-free binary system,  the basic detection pulse values remain unchanged:

$$g_0 = 1.56 \, {\rm V}, \ g_1 = g_{\rm –1} = 0.22 \, {\rm V}, \ g_2 = g_{\rm –2} \approx 0.$$

In pseudo-ternary coding,  there are always two eye openings:

• The upper boundary line of the upper eye results in the AMI code as in the redundancy-free binary system:

$$d_{\rm top}= g_0 - 2 \cdot g_1 \hspace{0.2cm}\text{(associated sequence: } "\hspace{-0.1cm}-1, +1, -1\hspace{-0.1cm}") \hspace{0.05cm}.$$

• In contrast,  for the lower boundary line of the upper eye:

$$d_{\rm bottom}= g_1 \hspace{0.2cm}\text{(associated sequences: }\ "\hspace{-0.1cm}0,\ 0, +1\hspace{-0.1cm}"\hspace{0.2cm}\text{ and } "\hspace{-0.1cm}+1,\ 0,\ 0\hspace{-0.1cm}")\hspace{0.05cm}.$$

Thus,  for the half eye opening,  the following holds true:

$${\ddot{o}(T_{\rm D})}/{2}= {1}/{2} \cdot (d_{\rm top} - d_{\rm bottom}) = {1}/{2} \cdot g_0 - {3}/{2} \cdot g_1 \hspace{0.15cm}\underline {= 0.45\,{\rm V}}\hspace{0.05cm}.$$

The corresponding equation for the redundancy-free binary system is:  

$${\ddot{o}(T_{\rm D})}/{2}= g_0 - 2 \cdot g_1 \hspace{0.05cm}.$$

(2)  In terms of noise,  there is no difference between the three systems since the same symbol rate is always present.  It follows for the AMI code:

$$\rho_{\rm U} = \frac{(0.45\,{\rm V})^2}{(0.2\,{\rm V})^2} = 5.06 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {\approx 7\,{\rm dB}} \hspace{0.05cm}.$$

• The loss compared to the redundancy-free binary system is thus almost  $8 \, {\rm dB}$.

• The reason for this serious loss of signal-to-noise ratio is that with the AMI code,  despite $37\%$  redundancy,  the symbol sequence  "$\text{ ...} , \, –1, \, +1, \, –1, \text{ ...} $"  is not excluded,  which is particularly unfavorable with respect to intersymbol interference.

(3)  The threshold  $E_2$  must be in the middle between  $d_{\rm top}$  and  $d_{\rm bottom}$:

$$E_2= {1}/{2} \cdot (d_{\rm top} + d_{\rm bottom}) = {1}/{2} \cdot (g_0 - g_1 ) \hspace{0.15cm}\underline {= 0.67\,{\rm V}}\hspace{0.05cm}.$$

• The threshold value $E_1$ is symmetrical to this:  $E_1 \, \underline {= \, –0.67 {\rm V}}$.

(4)  We again assume the same basic detection pulse values.

• The worst-case sequence with respect to the upper boundary line of the upper eye is  "$\text{ ...} , 0, \, +1, \, 0, \text{ ...} $",
• while the lower boundary line is defined by  "$\text{ ...} , 0, \, 0, \, +1, \text{ ...} $"  or  "$\text{ ...} , +1, \, 0, \, 0, \text{ ...} $"  respectively.

• From this follows:

$$d_{\rm top}= g_0, \hspace{0.2cm} d_{\rm bottom} = g_1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\ddot{o}(T_{\rm D})}/{2} = {g_0}/{2} - {g_1}/{2}\hspace{0.15cm}\underline { = 0.667\,{\rm V}} \hspace{0.05cm}.$$

(5)  Using the result from  (4),  we obtain analogous to subtask  (2):

$$\rho_{\rm U} = \frac{(0.67\,{\rm V})^2}{(0.2\,{\rm V})^2} = 11.2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {\approx 10.5\,{\rm dB}} \hspace{0.05cm}.$$

• Prerequisite for this result are thresholds at

$$E_2= {1}/{2} \cdot (g_0 + g_1 ) = 0.89\,{\rm V}, \hspace{0.2cm}E_1 = - 0.89\,{\rm V}\hspace{0.05cm}.$$

• It should be noted that the same cutoff frequency  $f_{\rm G} \cdot T = 0.5$  was always assumed here.

• If the cutoff frequency is optimized,  it may well be that the duobinary code is superior to the redundancy-free binary code if the characteristic cable attenuation is sufficiently large.