Difference between revisions of "Aufgaben:Exercise 3.5: Eye Opening with Pseudoternary Coding"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission |
}} | }} | ||
− | [[File:P_ID1421__Dig_A_3_5.png|right|frame| | + | [[File:P_ID1421__Dig_A_3_5.png|right|frame|Eye diagrams with AMI and duobinary code]] |
− | + | Three message transmission systems are considered, each with the following matching properties: | |
− | * NRZ | + | * NRZ rectangular pulses with amplitude $s_0 = 2 \, {\rm V}$, |
− | * | + | * Coaxial cable with characteristic cable attenuation $a_* = 40 \, {\rm dB}$, |
− | * AWGN | + | * 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 transmission code: | |
− | + | $\text{System A}$ uses a binary bipolar redundancy-free transmission signal. The following descriptive variables are known: | |
− | * | + | * Basic pulse values $g_0 = 1.56 \, {\rm V}$, $g_1 = g_{\rm –1} = 0.22 \, {\rm V}$, $g_2 = g_{\rm –2} = \, \text{ ...} \, \approx 0$ |
:$$\Rightarrow \hspace{0.3cm}{\ddot{o}(T_{\rm D})}/{ 2} = g_{0} | :$$\Rightarrow \hspace{0.3cm}{\ddot{o}(T_{\rm D})}/{ 2} = g_{0} | ||
-g_{1}-g_{-1} = 1.12\,{\rm V} | -g_{1}-g_{-1} = 1.12\,{\rm V} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | * | + | * Noise rms value $\sigma_d \approx 0.2 \, {\rm V}$ |
:$$\Rightarrow \hspace{0.3cm}\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2\big]^2}{ | :$$\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} | \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}.$$ | 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \approx 15\,{\rm dB}\hspace{0.05cm}.$$ | ||
− | + | $\text{System B}$ 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. |
Line 38: | Line 38: | ||
− | '' | + | ''Notes:'' |
− | * | + | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission|Intersymbol Interference for Multi-Level Transmission]]. |
− | * | + | * Not all of the numerical values given here are necessary to solve this exercise. |
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Calculate the half eye opening for the '''AMI code'''. |
|type="{}"} | |type="{}"} | ||
$\text{System B:}\hspace{0.4cm} \ddot{o}(T_{\rm D})/2$ = { 0.45 3% } $\ {\rm V}$ | $\text{System B:}\hspace{0.4cm} \ddot{o}(T_{\rm D})/2$ = { 0.45 3% } $\ {\rm V}$ | ||
− | { | + | {Calculate the worst-case signal-to-noise ratio for this system. |
|type="{}"} | |type="{}"} | ||
$\text{System B:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 7 3% } $\ {\rm dB}$ | $\text{System B:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 7 3% } $\ {\rm dB}$ | ||
− | { | + | {How must the thresholds $E_1$ and $E_2$ be chosen so that the result just calculated is correct? |
|type="{}"} | |type="{}"} | ||
$E_1 \ \hspace{0.05cm} = \ ${ -0.69--0.65 } $\ {\rm V}$ | $E_1 \ \hspace{0.05cm} = \ ${ -0.69--0.65 } $\ {\rm V}$ | ||
$E_2 \ = \ $ { 0.667 3% } $\ {\rm V}$ | $E_2 \ = \ $ { 0.667 3% } $\ {\rm V}$ | ||
− | { | + | {Calculate the half eye opening at the '''duobinary code'''. |
|type="{}"} | |type="{}"} | ||
$\text{System C:}\hspace{0.4cm} \ddot{o}(T_{\rm D})/2 \ = \ $ { 0.67 3% } $\ {\rm V}$ | $\text{System C:}\hspace{0.4cm} \ddot{o}(T_{\rm D})/2 \ = \ $ { 0.67 3% } $\ {\rm V}$ | ||
− | { | + | {Calculate the worst-case signal-to-noise ratio for duobinary coding. |
|type="{}"} | |type="{}"} | ||
$\text{System C:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 10.5 3% } $\ {\rm dB}$ | $\text{System C:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 10.5 3% } $\ {\rm dB}$ | ||
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Since the symbol rate is not changed in the AMI code compared to the redundancy-free binary system, the basic pulse values remain unchanged: |
:$$g_0 = 1.56 \, {\rm V}, \ g_1 = g_{\rm –1} = 0.22 \, {\rm V}, \ g_2 = g_{\rm –2} \approx 0.$$ | :$$g_0 = 1.56 \, {\rm V}, \ g_1 = g_{\rm –1} = 0.22 \, {\rm V}, \ g_2 = g_{\rm –2} \approx 0.$$ | ||
− | + | 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: |
− | :$$d_{\rm | + | :$$d_{\rm top}= g_0 - 2 \cdot g_1 \hspace{0.2cm}{\rm (associated} \hspace{0.1cm}{\rm |
− | + | sequence:}-1, +1, -1{\rm )} | |
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | * | + | *In contrast, for the lower boundary line of the upper eye: |
− | :$$d_{\rm | + | :$$d_{\rm bottom}= g_1 \hspace{0.2cm}{\rm (associated} \hspace{0.1cm}{\rm |
− | + | sequence:}\hspace{0.2cm}0, \hspace{0.05cm}0, +1\hspace{0.2cm}{\rm bzw.}\hspace{0.2cm}+1, \hspace{0.05cm}0, \hspace{0.05cm}0{\rm )}\hspace{0.05cm}.$$ | |
− | + | Thus, for the half eye opening, the following holds true: | |
− | :$${\ddot{o}(T_{\rm D})}/{2}= {1}/{2} \cdot (d_{\rm | + | :$${\ddot{o}(T_{\rm D})}/{2}= {1}/{2} \cdot (d_{\rm top} - d_{\rm bottom}) = {1}/{2} \cdot g_0 - {3}/{2} \cdot g_1 \hspace{0.15cm}\underline {= |
0.45\,{\rm V}}\hspace{0.05cm}.$$ | 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}.$$ | :$${\ddot{o}(T_{\rm D})}/{2}= g_0 - 2 \cdot g_1 \hspace{0.05cm}.$$ | ||
− | '''(2)''' | + | '''(2)''' In terms of noise, there is no difference between the three systems since the same symbol rate is always present. It follows for the AMI code: |
:$$\rho_{\rm U} = \frac{(0.45\,{\rm V})^2}{(0.2\,{\rm V})^2} = | :$$\rho_{\rm U} = \frac{(0.45\,{\rm V})^2}{(0.2\,{\rm V})^2} = | ||
5.06 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | 5.06 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | ||
Line 99: | Line 99: | ||
lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {\approx 7\,{\rm dB}} \hspace{0.05cm}.$$ | lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {\approx 7\,{\rm dB}} \hspace{0.05cm}.$$ | ||
− | * | + | *The loss compared to the redundancy-free binary system is thus almost $8 \, {\rm dB}$. |
− | * | + | *The reason for this serious loss of signal-to-noise ratio is that with the AMI code, despite $37\%$ redundancy, the symbol sequence $\text{ ...} , \, –1, \, +1, \, –1, \text{ ...} $ which is particularly unfavorable with respect to intersymbol interference, is not excluded. |
− | '''(3)''' | + | '''(3)''' The threshold $E_2$ must be in the middle between $d_{\rm top}$ and $d_{\rm bottom}$: |
− | :$$E_2= {1}/{2} \cdot (d_{\rm | + | :$$E_2= {1}/{2} \cdot (d_{\rm top} + d_{\rm bottom}) = {1}/{2} \cdot (g_0 - g_1 ) \hspace{0.15cm}\underline {= |
0.67\,{\rm V}}\hspace{0.05cm}.$$ | 0.67\,{\rm V}}\hspace{0.05cm}.$$ | ||
− | + | The threshold value $E_1$ is symmetrical to this: $E_1 \, \underline {= \, –0.67 {\rm V}}$. | |
− | '''(4)''' | + | '''(4)''' We again assume the same basic pulse values. |
− | * | + | *The worst-case sequence with respect to the upper boundary line of the upper eye is $\text{ ...} , 0, \, +1, \, 0, \text{ ...} $, |
− | * | + | *while the lower boundary line is defined by $\text{ ...} , 0, \, 0, \, +1, \text{ ...} $ or $\text{ ...} , +1, \, 0, \, 0, \text{ ...} $ respectively. |
− | * | + | *From this follows: |
− | :$$d_{\rm | + | :$$d_{\rm top}= g_0, \hspace{0.2cm} d_{\rm bottom} = g_1 \hspace{0.3cm}\Rightarrow |
\hspace{0.3cm}{\ddot{o}(T_{\rm D})}/{2} = {g_0}/{2} - | \hspace{0.3cm}{\ddot{o}(T_{\rm D})}/{2} = {g_0}/{2} - | ||
{g_1}/{2}\hspace{0.15cm}\underline { = 0.667\,{\rm V}} \hspace{0.05cm}.$$ | {g_1}/{2}\hspace{0.15cm}\underline { = 0.667\,{\rm V}} \hspace{0.05cm}.$$ | ||
− | '''(5)''' | + | '''(5)''' Using the result from '''(4)''', we obtain analogous to subtask '''(2)''': |
:$$\rho_{\rm U} = \frac{(0.67\,{\rm V})^2}{(0.2\,{\rm V})^2} = | :$$\rho_{\rm U} = \frac{(0.67\,{\rm V})^2}{(0.2\,{\rm V})^2} = | ||
11.2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | 11.2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | ||
Line 125: | Line 125: | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | * | + | *Prerequisite for this result are thresholds at |
:$$E_2= {1}/{2} \cdot (g_0 + g_1 ) = | :$$E_2= {1}/{2} \cdot (g_0 + g_1 ) = | ||
0.89\,{\rm V}, \hspace{0.2cm}E_1 = - 0.89\,{\rm V}\hspace{0.05cm}.$$ | 0.89\,{\rm V}, \hspace{0.2cm}E_1 = - 0.89\,{\rm V}\hspace{0.05cm}.$$ | ||
− | * | + | *It should be noted that the same cutoff frequency $f_{\rm G} \cdot T = 0.5$ was always assumed here. |
− | * | + | *If the cutoff frequency is optimized, it may well be that the duobinary code is superior to the redundancy-free binary code if the characteristic cable attenuation is sufficiently large. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
[[Category:Digital Signal Transmission: Exercises|^3.4 Auge bei mehrstufigen Systemen^]] | [[Category:Digital Signal Transmission: Exercises|^3.4 Auge bei mehrstufigen Systemen^]] |
Revision as of 14:53, 4 May 2022
Three message transmission systems are considered, each with the following matching properties:
- NRZ rectangular pulses with amplitude $s_0 = 2 \, {\rm V}$,
- Coaxial cable with characteristic cable attenuation $a_* = 40 \, {\rm dB}$,
- AWGN noise with noise power density $N_0$,
- Receiver filter $H_{\rm E}(f) = 1/H_{\rm K}(f) \cdot H_{\rm G}(f) $ consisting of an ideal channel equalizer $H_{\rm K}(f)^{-1}$ and a Gaussian low-pass filter $H_{\rm G}(f)$ with normalized cutoff frequency $f_{\rm G} \cdot T \approx 0.5$.
- Threshold decision with optimal decision thresholds and optimal detection time $T_{\rm D} = 0$.
The system variants to be investigated in the exercise differ only in terms of the transmission code:
$\text{System A}$ uses a binary bipolar redundancy-free transmission signal. The following descriptive variables are known:
- Basic pulse values $g_0 = 1.56 \, {\rm V}$, $g_1 = g_{\rm –1} = 0.22 \, {\rm V}$, $g_2 = g_{\rm –2} = \, \text{ ...} \, \approx 0$
- $$\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}.$$
$\text{System B}$ uses AMI coding:
- Here the outer symbols $"+1"$ or $"–1"$ occur only in isolation.
- In the case of three consecutive symbols, the sequences "$\hspace{-0.1cm}\text{ ...} \, , \, +1, \, +1, \, +1, \,\text{ ...}$" and "$\hspace{-0.1cm}\text{ ...} \, , \, +1, \, 0, \, +1, \, \text{ ...} $" among others, are not possible,
- in contrast to the sequence "$\hspace{-0.1cm}\text{ ...} \, , \, +1, \, –1, \, +1, \, \text{ ...} $".
$\text{System C}$ uses the duobinary code:
- Here the alternating sequence "$\hspace{-0.1cm} \text{ ...} \, , \, –1, \, +1, \, –1, \, \text{ ...} $" is excluded by the code, which has a favorable effect on the eye opening.
Notes:
- The exercise belongs to the chapter Intersymbol Interference for Multi-Level Transmission.
- Not all of the numerical values given here are necessary to solve this exercise.
Questions
Solution
- $$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}{\rm (associated} \hspace{0.1cm}{\rm sequence:}-1, +1, -1{\rm )} \hspace{0.05cm}.$$
- In contrast, for the lower boundary line of the upper eye:
- $$d_{\rm bottom}= g_1 \hspace{0.2cm}{\rm (associated} \hspace{0.1cm}{\rm sequence:}\hspace{0.2cm}0, \hspace{0.05cm}0, +1\hspace{0.2cm}{\rm bzw.}\hspace{0.2cm}+1, \hspace{0.05cm}0, \hspace{0.05cm}0{\rm )}\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}.$$
- The loss compared to the redundancy-free binary system is thus almost $8 \, {\rm dB}$.
- The reason for this serious loss of signal-to-noise ratio is that with the AMI code, despite $37\%$ redundancy, the symbol sequence $\text{ ...} , \, –1, \, +1, \, –1, \text{ ...} $ which is particularly unfavorable with respect to intersymbol interference, is not excluded.
(3) The threshold $E_2$ must be in the middle between $d_{\rm top}$ and $d_{\rm bottom}$:
- $$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}.$$
The threshold value $E_1$ is symmetrical to this: $E_1 \, \underline {= \, –0.67 {\rm V}}$.
(4) We again assume the same basic pulse values.
- The worst-case sequence with respect to the upper boundary line of the upper eye is $\text{ ...} , 0, \, +1, \, 0, \text{ ...} $,
- while the lower boundary line is defined by $\text{ ...} , 0, \, 0, \, +1, \text{ ...} $ or $\text{ ...} , +1, \, 0, \, 0, \text{ ...} $ respectively.
- From this follows:
- $$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.