Exercise 3.4Z: Eye Opening and Level Number

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Binary and quaternary
eye diagrams

In this exercise,  a redundancy-free binary system and a redundancy-free quaternary system are compared with respect to vertical eye opening.  The same boundary conditions apply to the two transmission systems:

• The basic transmission pulse  $g_s(t)$  is NRZ rectangular in each case and has the height  $s_0 = 1 \, {\rm V}$.

• The (equivalent) bit rate is  $R_{\rm B} = 100 \, {\rm Mbit/s}$.

• The AWGN noise has the  (one-sided)  noise power density  $N_0$.

• Let the receiver filter be a Gaussian low-pass filter with cutoff frequency  $f_{\rm G} = 30 \, {\rm MHz}$:

$$H_{\rm G}(f) = {\rm e}^{{- \pi \cdot f^2}/{(2f_{\rm G})^2}}\hspace{0.05cm}.$$

• The decision thresholds are optimal. The detection time is  $T_{\rm D} = 0$.

For the half-eye opening of an M-level transmission system,  the following holds in general:

$${\ddot{o}(T_{\rm D})}/{ 2} = \frac{g_0}{ M-1} - \sum_{\nu = 1}^{\infty} |g_\nu | - \sum_{\nu = 1}^{\infty} |g_{-\nu} |\hspace{0.05cm}.$$

• Here,  $g_0 = g_d(t = 0)$  is the  "main value"  of the basic detection pulse  $g_d(t) = g_s(t) * h_{\rm G}(t)$. 

• The second term describes the trailers  ("postcursors")  $g_{\rm \nu} = g_d(t = \nu T)$. 

• The last term describes the  "precursors"  $g_{\rm -\nu} = g_d(t = -\nu T)$.

Note that in the present configuration with Gaussian low-pass

• all the basic detection pulse values  $\text{...} \, g_{\rm -1}, \, g_0, \, g_1, \, \text{...}$  are positive,

• the (infinite) sum  $\text{...} \, + \, g_{\rm -1} + g_0 + g_1\,\text{...}$  gives the constant value  $s_0$, 

• the main value can be calculated with the complementary Gaussian error function  ${\rm Q}(x)$: 

$$g_0 = s_0 \cdot\big [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\big] \hspace{0.05cm}.$$

The graph shows the  (noiseless)  eye diagrams of the binary and quaternary systems and, in red, the corresponding basic detection pulses  $g_d(t)$:

• The optimal decision thresholds  $E$  $($for $M = 2)$  and  $E_1$,  $E_2$,  $E_3$  $($for $M = 4)$  are also drawn.

• In subtask  (7)  these are to be determined numerically.

Notes:

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

• For the complementary Gaussian error function applies:

$${\rm Q}(0.25) = 0.4013,\hspace{0.2cm}{\rm Q}(0.50) = 0.3085,\hspace{0.2cm}{\rm Q}(0.75) = 0.2266,\hspace{0.2cm}{\rm Q}(1.00) = 0.1587,$$

$${\rm Q}(1.25) = 0.1057,\hspace{0.2cm}{\rm Q}(1.50) = 0.0668,\hspace{0.2cm}{\rm Q}(1.75) = 0.0401,\hspace{0.2cm}{\rm Q}(2.00) = 0.0228.$$

Questions

1

What is the symbol duration  $T$  for the binary and the quaternary system?

$M = 2\text{:}\hspace{0.4cm} T \ = \ $ $\ {\rm ns}$

$M = 4\text{:}\hspace{0.4cm} T \ = \ $ $\ {\rm ns}$

2

Calculate the main value  $g_0$  for the binary system.

$M = 2\text{:}\hspace{0.4cm} g_0\ = \ $ $\ {\rm V}$

3

Calculate the main value  $g_0$  for the quaternary system.

$M = 4\text{:}\hspace{0.4cm} g_0\ = \ $ $\ {\rm V}$

4

Which equations are valid considering the Gaussian low-pass?

	$\ddot{o}(T_{\rm D})/2 = M \cdot g_0/(M - 1) - s_0,$
	$\ddot{o}(T_{\rm D})/2 = M \cdot s_0/(M - 1) - g_0,$
	$\ddot{o}(T_{\rm D})/2 = s_0/(M - 1) \cdot \big [1 - 2 \cdot M \cdot {\rm Q}(\sqrt{2\pi} \cdot {\rm log_2} \, (M) \cdot f_{\rm G}/R_{\rm B}) \big ].$

5

What is the eye opening for the binary system?

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

6

What is the eye opening for the quaternary system?

$M = 4\text{:}\hspace{0.4cm} \ddot{o}(T_{\rm D})\ = \ $ $\ {\rm V}$

7

Determine the optimal thresholds of the quaternary system.  Enter the lower threshold  $E_1$  as a control.

$M = 4\text{:}\hspace{0.4cm} E_1\ = \ $ $\ {\rm V}$

Solution

Solution

(1)  In the binary system,  the bit duration is equal to the reciprocal of the equivalent bit rate:

$$T = \frac{1}{R_{\rm B}}= \frac{1}{100\,{\rm Mbit/s}}\hspace{0.15cm}\underline {= 10\,{\rm ns}}\hspace{0.05cm}.$$

• The symbol duration of the quaternary system is twice as large:

$$T = \frac{{\rm log_2}\hspace{0.1cm}4}{R_{\rm B}}\hspace{0.15cm}\underline {= 20\,{\rm ns}}\hspace{0.05cm}.$$

(2)  According to the given equation,  the following holds for the binary system:

$$g_0 \ = \ s_0 \cdot\left [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\right]= 1\,{\rm V} \cdot\left [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot 30\,{\rm MHz} \cdot 10\,{\rm ns} \right)\right]$$

$$\Rightarrow \hspace{0.3cm} g_0 \ \approx \ 1\,{\rm V} \cdot\left [ 1- 2 \cdot {\rm Q} \left( 0.75 \right)\right] = 1\,{\rm V} \cdot\left [ 1- 2 \cdot 0.2266 \right]\hspace{0.15cm}\underline { = 0.547\,{\rm V}} \hspace{0.05cm}.$$

(3)  Due to the double symbol duration,  with the same cutoff frequency for  $M = 4$:

$$g_0 \ = 1\,{\rm V} \cdot\left [ 1- 2 \cdot {\rm Q} \left( 1.5 \right)\right] = 1\,{\rm V} \cdot\left [ 1- 2 \cdot 0.0668 \right] \hspace{0.15cm}\underline {= 0.867\,{\rm V}} \hspace{0.05cm}.$$

(4)  Extending the given equation by  $\pm g_0$,  we obtain:

$${\ddot{o}(T_{\rm D})}/{ 2} = \frac{g_0}{ M-1} + g_0 - g_0 - \sum_{\nu = 1}^{\infty} g_\nu - \sum_{\nu = 1}^{\infty} g_{-\nu} = \frac{M}{ M-1} \cdot g_0 - s_0 \hspace{0.05cm}.$$

Here is taken into account:

• In the case of the Gaussian low-pass filter,  the magnitude formation can be omitted.
• The sum over all detection pulse values is equal to  $s_0$.

The  first, but also the last solution  is correct:

$${\ddot{o}(T_{\rm D})}/{ 2} \ = \ \frac{M}{ M-1} \cdot g_0 - s_0 = \frac{M}{ M-1} \cdot s_0 \cdot\left [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\right]- s_0$$

$$\Rightarrow \hspace{0.3cm} {\ddot{o}(T_{\rm D})}/{ 2} \ = \ \frac{s_0}{ M-1} \cdot \left [ 1- 2 \cdot M \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot T \right)\right] \hspace{0.05cm}.$$

Using the relation  $T = {\rm log_2} \,(M)/R_{\rm B}$,  we arrive at the third proposed solution,  which is also applicable.

(5)  Using the results from  (2)  and  (4),  one obtains with  $M = 2$:

$${\ddot{o}(T_{\rm D})} = 2 \cdot (2 \cdot g_0 - s_0) = 2 \cdot (2 \cdot 0.547\,{\rm V} - 1\,{\rm V}) \hspace{0.15cm}\underline {= 0.188\,{\rm V}} \hspace{0.05cm}.$$

(6)  On the other hand,  with $g_0 = 0.867 \, {\rm V}$, $s_0 = 1 \, {\rm V}$  and  $M = 4$,  we get:

$${\ddot{o}(T_{\rm D})} = 2 \cdot ({4}/{3} \cdot 0.867\,{\rm V} - 1\,{\rm V}) \hspace{0.15cm}\underline {= 0.312\,{\rm V}} \hspace{0.05cm}.$$

(7)  According to subtask  (3),  $g_0 = 0.867 \, {\rm V}$  and correspondingly $g_{\rm VN} = 0.133 \, {\rm V}$  (sum of all precursors and trailers).

• The eye opening is  $\ddot{o} = 0.312 \, {\rm V}$.

• From the sketch on the information section,  we can see that the upper boundary  $($German:  "obere Grenzlinie"   ⇒   "o"$)$  of the upper eye has the following value  $($for  $T_{\rm D} = 0)$:

$$o = s_0 - 2 \cdot g_{\rm VN}= g_0 - g_{\rm VN}= 0.867\,{\rm V} - 0.133\,{\rm V} = 0.734\,{\rm V} \hspace{0.05cm}.$$

• The lower limit  $($German:  "untere Grenzlinie"   ⇒   "u"$)$  is at:

$$u = o -{\ddot{o}} = 0.734\,{\rm V} - 0.312\,{\rm V} = 0.422\,{\rm V} \hspace{0.05cm}.$$

• From this follows for the optimal decision threshold of the upper eye:

$$E_3 = \frac{o + u}{2} = \frac{0.734\,{\rm V} + 0.422\,{\rm V}}{2} { = 0.578\,{\rm V}} \hspace{0.05cm}.$$

• The sought threshold  (for the lower eye)  is $E_1 \, \underline {= \, –0.578 \, V}$.

• The average decision threshold is  $E_2 = 0$  for symmetry reasons.