Difference between revisions of "Aufgaben:Exercise 3.5: Eye Opening with Pseudoternary Coding"

From LNTwww
 
(6 intermediate revisions by 2 users not shown)
Line 3: Line 3:
 
}}
 
}}
  
[[File:P_ID1421__Dig_A_3_5.png|right|frame|Eye diagrams with AMI and duobinary code]]
+
[[File:EN_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:
+
Three digital transmission systems are considered,  each with the following matching properties:
 
* NRZ rectangular pulses with amplitude  $s_0 = 2 \, {\rm V}$,
 
* NRZ rectangular pulses with amplitude  $s_0 = 2 \, {\rm V}$,
* Coaxial cable with characteristic cable attenuation  $a_* = 40 \, {\rm dB}$,
+
 
 +
* coaxial cable with characteristic cable attenuation  $a_* = 40 \, {\rm dB}$,
 +
 
 
* AWGN noise with noise power density  $N_0$,
 
* 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  $T_{\rm D} = 0$.
+
* 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:
+
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:
+
⇒   $\text{System A}$  uses a binary bipolar redundancy-free transmission signal. The following descriptive variables are known:
* Basic pulse values  $g_0 = 1.56 \, {\rm V}$, $g_1 = g_{\rm –1} = 0.22 \, {\rm V}$, $g_2 = g_{\rm –2} = \, \text{ ...} \, \approx 0$
+
* 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}
 
:$$\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}
Line 24: Line 28:
 
  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:
+
⇒   $\text{System B}$  uses AMI coding:
 
*Here the outer symbols  $"+1"$  or  $"–1"$  occur only in isolation.
 
*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{ ...} $"  among others, are not possible,
+
*In the case of three consecutive symbols, the sequences  "$\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{ ...} $".
 
* in contrast to the sequence  "$\hspace{-0.1cm}\text{ ...}  \, , \, +1, \, –1, \, +1, \, \text{ ...} $".
  
  
$\text{System C}$  uses the duobinary code:
+
⇒   $\text{System C}$  uses the duobinary code:
*Here the alternating sequence  "$\hspace{-0.1cm} \text{ ...}  \, , \, –1, \, +1, \, –1, \, \text{ ...}  $"  is excluded by the code, which has a favorable effect on the eye opening.
+
*Here the alternating sequence  "$\hspace{-0.1cm} \text{ ...}  \, , \, –1, \, +1, \, –1, \, \text{ ...}  $"  is excluded by the code,  which has a favorable effect on the eye opening.
  
  
  
  
 
+
Notes:  
 
 
''Notes:''
 
 
*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission|"Intersymbol Interference for Multi-Level Transmission"]].
 
*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission|"Intersymbol Interference for Multi-Level Transmission"]].
 
   
 
   
Line 47: Line 49:
 
===Questions===
 
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Calculate the half eye opening for the '''AMI code'''.
+
{Calculate the half eye opening for the&nbsp; '''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.
 
{Calculate the worst-case signal-to-noise ratio for this system.
Line 60: Line 62:
 
$E_2 \ = \ $ { 0.667 3% } $\ {\rm V}$
 
$E_2 \ = \ $ { 0.667 3% } $\ {\rm V}$
  
{Calculate the half eye opening at the '''duobinary code'''.
+
{Calculate the half eye opening at the&nbsp; '''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% } $\ {\rm V}$
Line 71: Line 73:
 
===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Since the symbol rate is not changed in the AMI code compared to the redundancy-free binary system, the basic pulse values remain unchanged:
+
'''(1)'''&nbsp; Since the symbol rate is not changed in the AMI code compared to the redundancy-free binary system,&nbsp; the basic detection pulse values remain unchanged:
 
:$$g_0 = 1.56 \, {\rm V}, \ g_1 = g_{\rm &ndash;1} = 0.22 \, {\rm V}, \ g_2 = g_{\rm &ndash;2} \approx 0.$$
 
:$$g_0 = 1.56 \, {\rm V}, \ g_1 = g_{\rm &ndash;1} = 0.22 \, {\rm V}, \ g_2 = g_{\rm &ndash;2} \approx 0.$$
  
In pseudo ternary coding, there are always two eye openings:
+
In pseudo-ternary coding,&nbsp; 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:
 
*The upper boundary line of the upper eye results in the AMI code as in the redundancy-free binary system:
:$$d_{\rm top}= g_0 - 2 \cdot g_1 \hspace{0.2cm}{\rm (associated} \hspace{0.1cm}{\rm
+
:$$d_{\rm top}= g_0 - 2 \cdot g_1 \hspace{0.2cm}\text{(associated sequence: } "\hspace{-0.1cm}-1, +1, -1\hspace{-0.1cm}")
sequence:}-1, +1, -1{\rm )}
 
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
  
*In contrast, for the lower boundary line of the upper eye:
+
*In contrast,&nbsp; for the lower boundary line of the upper eye:
:$$d_{\rm bottom}= g_1 \hspace{0.2cm}{\rm (associated} \hspace{0.1cm}{\rm
+
:$$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}.$$
sequence:}\hspace{0.2cm}0, \hspace{0.05cm}0, +1\hspace{0.2cm}{\rm bzw.}\hspace{0.2cm}+1, \hspace{0.05cm}0, \hspace{0.05cm}0{\rm )}\hspace{0.05cm}.$$
 
  
Thus, for the half eye opening, the following holds true:
+
Thus,&nbsp; for the half eye opening,&nbsp; 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 {=
 
:$${\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}.$$
Line 93: Line 93:
  
  
'''(2)'''&nbsp; 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:
+
'''(2)'''&nbsp; In terms of noise,&nbsp; there is no difference between the three systems since the same symbol rate is always present.&nbsp; 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 loss compared to the redundancy-free binary system is thus almost&nbsp; $8 \, {\rm dB}$.
*The reason for this serious loss of signal-to-noise ratio is that with the AMI code, despite $37\%$ redundancy, the symbol sequence &nbsp;$\text{ ...} , \, &ndash;1, \, +1, \, &ndash;1, \text{ ...} $&nbsp; which is particularly unfavorable with respect to intersymbol interference, is not excluded.
+
 +
*The reason for this serious loss of signal-to-noise ratio is that with the AMI code,&nbsp; despite $37\%$&nbsp; redundancy,&nbsp; the symbol sequence &nbsp;"$\text{ ...} , \, &ndash;1, \, +1, \, &ndash;1, \text{ ...} $"&nbsp; is not excluded,&nbsp; which is particularly unfavorable with respect to intersymbol interference.
  
  
'''(3)'''&nbsp; The threshold $E_2$ must be in the middle between $d_{\rm top}$ and $d_{\rm bottom}$:
+
 
 +
'''(3)'''&nbsp; The threshold&nbsp; $E_2$&nbsp; must be in the middle between&nbsp; $d_{\rm top}$&nbsp; and&nbsp; $d_{\rm bottom}$:
 
:$$E_2= {1}/{2} \cdot (d_{\rm top} + d_{\rm bottom}) = {1}/{2} \cdot (g_0 -  g_1 ) \hspace{0.15cm}\underline {=
 
:$$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: $E_1 \, \underline {= \, &ndash;0.67 {\rm V}}$.
+
*The threshold value $E_1$ is symmetrical to this:&nbsp; $E_1 \, \underline {= \, &ndash;0.67 {\rm V}}$.
  
  
'''(4)'''&nbsp; We again assume the same basic pulse values.
+
 
*The worst-case sequence with respect to the upper boundary line of the upper eye is &nbsp;$\text{ ...} , 0, \, +1, \, 0, \text{ ...} $,  
+
'''(4)'''&nbsp; We again assume the same basic detection  pulse values.
*while the lower boundary line is defined by &nbsp;$\text{ ...} ,  0, \, 0, \, +1, \text{ ...} $&nbsp; or &nbsp;$\text{ ...} , +1, \, 0, \, 0, \text{ ...} $&nbsp; respectively.
+
*The worst-case sequence with respect to the upper boundary line of the upper eye is &nbsp;"$\text{ ...} , 0, \, +1, \, 0, \text{ ...} $",  
 +
*while the lower boundary line is defined by &nbsp;"$\text{ ...} ,  0, \, 0, \, +1, \text{ ...} $"&nbsp; or &nbsp;"$\text{ ...} , +1, \, 0, \, 0, \text{ ...} $"&nbsp; respectively.
 +
 
 
*From this follows:
 
*From this follows:
 
:$$d_{\rm top}= g_0, \hspace{0.2cm} d_{\rm bottom} = g_1 \hspace{0.3cm}\Rightarrow
 
:$$d_{\rm top}= g_0, \hspace{0.2cm} d_{\rm bottom} = g_1 \hspace{0.3cm}\Rightarrow
Line 118: Line 122:
  
  
'''(5)'''&nbsp; Using the result from '''(4)''', we obtain analogous to subtask '''(2)''':
+
'''(5)'''&nbsp; Using the result from&nbsp; '''(4)''',&nbsp; we obtain analogous to subtask&nbsp; '''(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 129: Line 133:
 
  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.  
+
*It should be noted that the same cutoff frequency&nbsp; $f_{\rm G} \cdot T = 0.5$&nbsp; 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.
+
 +
*If the cutoff frequency is optimized,&nbsp; 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 Eye Opening with Multilevel Systems^]]
 
[[Category:Digital Signal Transmission: Exercises|^3.4 Eye Opening with Multilevel Systems^]]

Latest revision as of 15:45, 28 June 2022

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:

  • 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

(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.