Aufgaben:Exercise 3.4Z: Eye Opening and Level Number - Revision history

Guenter at 15:09, 20 June 2022

2022-06-20T15:09:53Z

Guenter at 14:46, 20 June 2022

2022-06-20T14:46:34Z

Guenter: Guenter moved page Aufgaben:Exercise 3.4Z: Eye Opening and Step Number to Aufgaben:Exercise 3.4Z: Eye Opening and Level Number

2022-06-20T12:57:40Z

Guenter moved page Exercise 3.4Z: Eye Opening and Step Number to Exercise 3.4Z: Eye Opening and Level Number

Guenter: Guenter moved page Aufgaben:Exercise 3.4Z: Eye Opening and Number of Steps to Aufgaben:Exercise 3.4Z: Eye Opening and Step Number

2022-06-20T12:51:41Z

Guenter moved page Exercise 3.4Z: Eye Opening and Number of Steps to Exercise 3.4Z: Eye Opening and Step Number

Hwang at 15:58, 16 May 2022

2022-05-16T15:58:37Z

Guenter at 15:07, 7 May 2022

2022-05-07T15:07:09Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.4Z: Augenöffnung und Stufenzahl to Aufgaben:Exercise 3.4Z: Eye Opening and Number of Steps

2022-05-07T15:03:14Z

Guenter moved page Aufgabe 3.4Z: Augenöffnung und Stufenzahl to Exercise 3.4Z: Eye Opening and Number of Steps

Hwang at 13:01, 4 May 2022

2022-05-04T13:01:50Z

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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:49:59Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

Guenter at 15:24, 23 February 2019

2019-02-23T15:24:57Z

@@ Line 91: / Line 91: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; In the binary system, the bit duration is equal to the reciprocal of the equivalent bit rate:
+'''(1)'''&nbsp; In the binary system,&nbsp; 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:
+*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)'''&nbsp; According to the given equation, the following holds for the binary system:
 :$$g_0 \ = \ s_0   \cdot\left [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot
    T  \right)\right]= 1\,{\rm V}   \cdot\left [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot 30\,{\rm MHz} \cdot
@@ Line 106: / Line 107: @@
-'''(3)'''&nbsp; Due to the double symbol duration, the same cutoff frequency for $M = 4$:
+'''(3)'''&nbsp; Due to the double symbol duration,&nbsp; with the same cutoff frequency for&nbsp; $M = 4$:
 :$$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}}
@@ Line 112: / Line 113: @@
-'''(4)'''&nbsp; Extending the given equation by $&plusmn;g_0$, we obtain:
+'''(4)'''&nbsp; Extending the given equation by&nbsp; $\pm g_0$,&nbsp; 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.
+*In the case of the Gaussian low-pass filter,&nbsp; the magnitude formation can be omitted.
-*The sum over all detection pulse values is equal to $s_0$.
+*The sum over all detection pulse values is equal to&nbsp; $s_0$.
-The <u>first, but also the last solution</u> is correct:
+The&nbsp; <u>first, but also the last solution</u>&nbsp; is correct:
 :$${\ddot{o}(T_{\rm D})}/{ 2} \ = \ \frac{M}{ M-1} \cdot g_0 - s_0
   = \frac{M}{ M-1} \cdot s_0 \cdot\left [ 1- 2 \cdot {\rm Q} \left( \sqrt{2\pi} \cdot f_{\rm G} \cdot
@@ Line 128: / Line 129: @@
    T  \right)\right]
    \hspace{0.05cm}.$$
-Using the relation $T = {\rm log_2} \,(M)/R_{\rm B}$, we arrive at the third proposed solution, which is also applicable.
+Using the relation&nbsp; $T = {\rm log_2} \,(M)/R_{\rm B}$,&nbsp; we arrive at the third proposed solution,&nbsp; which is also applicable.
-'''(5)'''&nbsp; Using the results from '''(2)''' and '''(4)''' and $M = 2$, one obtains:
+'''(5)'''&nbsp; Using the results from&nbsp; '''(2)'''&nbsp; and&nbsp; '''(4)''',&nbsp; one obtains with&nbsp; $M = 2$:
 :$${\ddot{o}(T_{\rm D})} = 2 \cdot (2 \cdot g_0 -  s_0) = 2 \cdot (2 \cdot 0.547\,{\rm V} -  1\,{\rm V}) \hspace{0.15cm}\underline {= 0.188\,{\rm V}}
 \hspace{0.05cm}.$$
-'''(6)'''&nbsp; On the other hand, with $g_0 = 0.867 \, {\rm V}$, $s_0 = 1 \, {\rm V}$ and $M = 4$, we get:
+'''(6)'''&nbsp; On the other hand,&nbsp; with $g_0 = 0.867 \, {\rm V}$, $s_0 = 1 \, {\rm V}$&nbsp; and&nbsp; $M = 4$,&nbsp; 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)'''&nbsp; 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).
+'''(7)'''&nbsp; According to subtask&nbsp; '''(3)''',&nbsp; $g_0 = 0.867 \, {\rm V}$&nbsp; and correspondingly $g_{\rm VN} = 0.133 \, {\rm V}$&nbsp; (sum of all precursors and trailers).
-*The eye opening is $\ddot{o} = 0.312 \, {\rm V}$.
+*The eye opening is&nbsp; $\ddot{o} = 0.312 \, {\rm V}$.
-*From the sketch on the information section, we can see that the upper boundary of the upper eye has the following value (for $T_{\rm D} = 0$):
+*From the sketch on the information section,&nbsp; we can see that the upper boundary&nbsp; $($German:&nbsp; "obere Grenzlinie" &nbsp; &rArr; &nbsp; "o"$)$&nbsp; of the upper eye has the following value&nbsp; $($for&nbsp; $T_{\rm D} = 0)$:
 :$$o = s_0 - 2 \cdot g_{\rm VN}= g_0 -  g_{\rm VN}=  0.867\,{\rm V} -  0.133\,{\rm V} = 0.734\,{\rm V}
 \hspace{0.05cm}.$$
-*The lower limit is at:
+*The lower limit&nbsp; $($German:&nbsp; "untere Grenzlinie" &nbsp; &rArr; &nbsp; "u"$)$&nbsp; is at:
 :$$u = o -{\ddot{o}} =  0.734\,{\rm V} -   0.312\,{\rm V} = 0.422\,{\rm V}
 \hspace{0.05cm}.$$
@@ Line 156: / Line 157: @@
 \hspace{0.05cm}.$$
-*The sought threshold (for the lower eye) is $E_1 \, \underline {= \, &ndash;0.578 \, V}$.
+*The sought threshold&nbsp; (for the lower eye)&nbsp; is $E_1 \, \underline {= \, &ndash;0.578 \, V}$.
 *The average decision threshold is $E_2 = 0$ for symmetry reasons.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1420__Dig_Z_3_4.png|right|frame|Binary and quaternary eye diagrams]]
+[[File:P_ID1420__Dig_Z_3_4.png|right|frame|Binary and quaternary <br>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:
+In this exercise,&nbsp; a redundancy-free binary system and a redundancy-free quaternary system are compared with respect to vertical eye opening.&nbsp; The same boundary conditions apply to the two transmission systems:
 * The basic transmission pulse &nbsp;$g_s(t)$&nbsp; is NRZ rectangular in each case and has the height &nbsp;$s_0 = 1 \, {\rm V}$.
 * The (equivalent) bit rate is &nbsp;$R_{\rm B} = 100 \, {\rm Mbit/s}$.
-* The AWGN noise has the noise power density &nbsp;$N_0$.
+* The (equivalent) bit rate is &nbsp;$R_{\rm B} = 100 \, {\rm Mbit/s}$.
 * Let the receiver filter be a Gaussian low-pass filter with cutoff frequency &nbsp;$f_{\rm G} = 30 \, {\rm MHz}$:
 :$$H_{\rm G}(f) = {\rm e}^{{- \pi \cdot f^2}/{(2f_{\rm G})^2}}\hspace{0.05cm}.$$
@@ Line 13: / Line 16: @@
-For the half-eye opening of an M-level transmission system, the following holds in general:
+For the half-eye opening of an M-level transmission system,&nbsp; 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, &nbsp;$g_0 = g_d(t = 0)$&nbsp; is the main value of the basic transmitter pulse &nbsp;$g_d(t) = g_s(t) * h_{\rm G}(t)$. The second term describes the trailer &nbsp;$g_{\rm \nu} = g_d(t = \nu T)$&nbsp; and the last term the precursor &nbsp;$g_{\rm -\nu} = g_d(t = -\nu T)$.
+*Here, &nbsp;$g_0 = g_d(t = 0)$&nbsp; is the&nbsp; "main value"&nbsp; of the basic detection pulse &nbsp;$g_d(t) = g_s(t) * h_{\rm G}(t)$.&nbsp;
 Note that in the present configuration with Gaussian low-pass
-* all the basic transmitter pulse values &nbsp;$\text{...} \, g_{\rm -1}, \, g_0, \, g_1, \, \text{...}$&nbsp; are positive,
+* all the basic detection pulse values &nbsp;$\text{...} \, g_{\rm -1}, \, g_0, \, g_1, \, \text{...}$&nbsp; are positive,
 * the (infinite) sum &nbsp;$\text{...} \, + \, g_{\rm -1} + g_0 + g_1\,\text{...}$&nbsp; gives the constant value &nbsp;$s_0$,&nbsp;
 * the main value can be calculated with the complementary Gaussian error function &nbsp;${\rm Q}(x)$:&nbsp;
 :$$g_0 = s_0
@@ Line 28: / Line 38: @@
     \hspace{0.05cm}.$$
-The graph shows the eye diagrams of the binary and quaternary systems and, in red, the corresponding basic transmitter pulses &nbsp;$g_d(t)$:
+The graph shows the&nbsp; (noiseless)&nbsp;  eye diagrams of the binary and quaternary systems and, in red, the corresponding basic detection pulses &nbsp;$g_d(t)$:
-*The optimal decision thresholds &nbsp;$E$&nbsp; $($for $M = 2)$&nbsp; and &nbsp;$E_1$, &nbsp;$E_2$, &nbsp;$E_3$ $($for $M = 4)$ are also drawn.
+*The optimal decision thresholds &nbsp;$E$&nbsp; $($for $M = 2)$&nbsp; and &nbsp;$E_1$, &nbsp;$E_2$, &nbsp;$E_3$&nbsp; $($for $M = 4)$&nbsp; are also drawn.
-''Notes:''
+Notes:
 *The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission|"Intersymbol Interference for Multi-Level Transmission"]].
@@ Line 48: / Line 57: @@
 ===Questions===
 <quiz display=simple>
-{What is the symbol duration &nbsp;$T$&nbsp; for the binary or the quaternary system?
+{What is the symbol duration &nbsp;$T$&nbsp; for the binary and the quaternary system?
 |type="{}"}
 $M = 2\text{:}\hspace{0.4cm} T \ = \ $ { 10 3% } $\ {\rm ns}$
@@ Line 75: / Line 84: @@
 $M = 4\text{:}\hspace{0.4cm} \ddot{o}(T_{\rm D})\ = \ $ { 0.312 3% } $\ {\rm V}$
-{Determine the optimal thresholds of the quaternary system. Enter the lower threshold &nbsp;$E_1$&nbsp; as a control.
+{Determine the optimal thresholds of the quaternary system.&nbsp; Enter the lower threshold &nbsp;$E_1$&nbsp; as a control.
 |type="{}"}
 $M = 4\text{:}\hspace{0.4cm} E_1\ = \ $ { -0.595--0.561 } $\ {\rm V}$

@@ Line 37: / Line 37: @@
 ''Notes:''
-*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission|Intersymbol Interference for Multi-Level Transmission]].
+*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission|"Intersymbol Interference for Multi-Level Transmission"]].
 *For the complementary Gaussian error function applies:

← Older revision		Revision as of 15:07, 7 May 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^3.4 ~~Auge bei mehrstufigen Systemen~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.4 Eye Opening with Multilevel Systems^]]

← Older revision		Revision as of 12:49, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^3.4 Auge bei mehrstufigen Systemen^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.4 Auge bei mehrstufigen Systemen^]]

← Older revision	Revision as of 12:57, 20 June 2022
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← Older revision	Revision as of 12:51, 20 June 2022
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@@ Line 95: / Line 95: @@
    = 1\,{\rm V}   \cdot\left [ 1- 2 \cdot 0.2266 \right]\hspace{0.15cm}\underline { =  0.547\,{\rm V}}
     \hspace{0.05cm}.$$
 '''(3)'''&nbsp; Aufgrund der doppelten Symboldauer ergibt sich bei gleicher Grenzfrequenz für $M = 4$:
@@ Line 100: / Line 101: @@
    = 1\,{\rm V}   \cdot\left [ 1- 2 \cdot 0.0668 \right] \hspace{0.15cm}\underline {=  0.867\,{\rm V}}
     \hspace{0.05cm}.$$
 '''(4)'''&nbsp; Erweitert man die angegebene Gleichung um $&plusmn;g_0$, so erhält man:
@@ Line 105: / Line 107: @@
   = \frac{M}{ M-1} \cdot g_0 - s_0 \hspace{0.05cm}.$$
-Hierbei ist berücksichtigt, dass beim Gaußtiefpass auf die Betragsbildung verzichtet werden kann und zum zweiten, dass die Summe über alle Detektionsimpulswerte gleich $s_0$ ist.
+Hierbei ist berücksichtigt:
 Richtig ist also der <u>erste, aber auch der letzte Lösungsvorschlag</u>:
@@ Line 117: / Line 122: @@
-'''(5)'''&nbsp; Mit den Ergebnissen aus (2) und (4) sowie $M = 2$ erhält man:
+'''(5)'''&nbsp; Mit den Ergebnissen aus '''(2)''' und '''(4)''' sowie $M = 2$ erhält man:
 :$${\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}.$$
 '''(6)'''&nbsp; Mit $g_0 = 0.867 \, {\rm V}$, $s_0 = 1 \, {\rm V}$ und $M = 4$ ergibt sich dagegen:
@@ Line 126: / Line 132: @@
-'''(7)'''&nbsp; Entsprechend Teilaufgabe (3) ist $g_0 = 0.867 \, {\rm V}$ und dementsprechend $g_{\rm VN} = 0.133 \, {\rm V}$ (Summe aller Vor&ndash; und Nachläufer). Die Augenöffnung beträgt $\ddot{o} = 0.312 \, {\rm V}$.
+'''(7)'''&nbsp; Entsprechend Teilaufgabe '''(3)''' ist $g_0 = 0.867 \, {\rm V}$ und dementsprechend $g_{\rm VN} = 0.133 \, {\rm V}$ (Summe aller Vor&ndash; und Nachläufer).
-Aus der Skizze auf der Angabenseite erkennt man, dass die obere Begrenzung des oberen Auges folgenden Wert besitzt (für $T_{\rm D} = 0$):
+*Aus der Skizze auf der Angabenseite erkennt man, dass die obere Begrenzung des oberen Auges folgenden Wert besitzt (für $T_{\rm D} = 0$):
 :$$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}.$$
-Die untere Begrenzung liegt bei:
+*Die untere Begrenzung liegt bei:
 :$$u = o -{\ddot{o}} =  0.734\,{\rm V} -   0.312\,{\rm V} = 0.422\,{\rm V}
 \hspace{0.05cm}.$$
-Daraus folgt für die optimale Entscheiderschwelle des oberen Auges:
+*Daraus folgt für die optimale Entscheiderschwelle des oberen Auges:
 :$$E_3 = \frac{o + u}{2} = \frac{0.734\,{\rm V} + 0.422\,{\rm V}}{2}  { =  0.578\,{\rm V}}
 \hspace{0.05cm}.$$
-Der gesuchte Schwellenwert (für das untere Auge) ist $E_1 \, \underline {= \, &ndash;0.578 \, V}$. Die mittlere Entscheiderschwelle liegt aus Symmetriegründen bei $E_2 = 0$.
+*Der gesuchte Schwellenwert (für das untere Auge) ist $E_1 \, \underline {= \, &ndash;0.578 \, V}$.
 {{ML-Fuß}}