Aufgaben:Exercise 4.13: Four-level QAM - Revision history

Guenter at 12:03, 21 August 2022

2022-08-21T12:03:05Z

Guenter at 11:39, 21 August 2022

2022-08-21T11:39:00Z

Hwang at 11:20, 12 August 2022

2022-08-12T11:20:55Z

Hwang at 09:59, 12 August 2022

2022-08-12T09:59:57Z

Hwang at 09:46, 12 August 2022

2022-08-12T09:46:09Z

Hwang at 09:33, 12 August 2022

2022-08-12T09:33:05Z

Guenter at 11:57, 13 July 2022

2022-07-13T11:57:54Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.13: Vierstufige QAM to Aufgaben:Exercise 4.13: Four-level QAM

2022-07-13T11:51:16Z

Guenter moved page Aufgabe 4.13: Vierstufige QAM to Exercise 4.13: Four-level QAM

Javier: Text replacement - "„" to """

2021-05-28T14:41:55Z

Text replacement - "„" to """

Javier: Text replacement - "”" to """

2021-05-28T14:38:11Z

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@@ Line 20: / Line 20: @@
-''Notes:''
+Notes:
 * The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The "Union Bound" is an upper bound for the mean symbol error probability. For the latter holds:
+'''(1)'''&nbsp; The&nbsp; "Union Bound"&nbsp; is an upper bound for the average symbol error probability.&nbsp; For the latter holds:
 :$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm Pr}({\cal{E}}) =  {\rm Pr}( {\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted})= {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted})
   \hspace{0.05cm}.$$
-*In contrast, for the (improved) "Union Bound" in the present example:
+*In contrast,&nbsp; for the&nbsp; (improved)&nbsp; "Union Bound"&nbsp; in the present example:
 :$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}    {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted}) +{\rm Pr}( \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted}) = 2p = \underline{0.2}
   \hspace{0.05cm}.$$
-'''(2)'''&nbsp; The two probabilities that make up the "Union Bound" additive can be interpreted geometrically as follows:
+'''(2)'''&nbsp; The two probabilities that make up the&nbsp; "Union Bound"&nbsp; additive can be interpreted geometrically as follows:
-* ${\rm Pr}(\boldsymbol{s}_{\rm B} \cup \boldsymbol{s}_{\rm C} | \boldsymbol{s}_{\rm A})$ iis the probability that the receiving point is located in the left half-plane <br> &#8658; &nbsp; the AWGN noise component $n_1$ is negative and greater in magnitude than $\sqrt {E}$.
+* ${\rm Pr}(\boldsymbol{s}_{\rm B} \cup \boldsymbol{s}_{\rm C} | \boldsymbol{s}_{\rm A})$&nbsp; is the probability that the receiving point is located in the left half-plane <br> &#8658; &nbsp; the AWGN noise component&nbsp; $n_1$&nbsp; is negative and greater in magnitude than&nbsp; $\sqrt {E}$.
 Thus, the "Union Bound" considers the third quadrant twice. It is relatively easy to compensate for this error here:
 :$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  p_{\rm UB} -  {\rm Pr}( \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted}) = 2 p -  {\rm Pr}\left [ ( n_1 < -\sqrt{E})\cap ( n_2 < -\sqrt{E})\right ] = 2p - p^2 = \underline{0.19}
   \hspace{0.05cm}.$$
-Here it is considered that the noise components $n_1$ and $n_2$ are independent of each other.
+Here it is considered that the noise components&nbsp; $n_1$&nbsp; and&nbsp; $n_2$&nbsp; are independent of each other.
-'''(3)'''&nbsp; As proved in subtask '''(2)''', the following are valid for the individual falsification probabilities:
+'''(3)'''&nbsp; As proved in subtask&nbsp; '''(2)''',&nbsp; the following are valid for the individual falsification probabilities:
-* Quadrant 2: ${\rm Pr}(\boldsymbol{s}_{\rm B} \  {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.09$,
+* Quadrant 2:&nbsp; ${\rm Pr}(\boldsymbol{s}_{\rm B} \  {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.09$,
-* Quadrant 3: ${\rm Pr}(\boldsymbol{s}_{\rm C} \ {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.01$,
+* Quadrant 3:&nbsp; ${\rm Pr}(\boldsymbol{s}_{\rm C} \ {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.01$,
-* Quadrant 4: ${\rm Pr}(\boldsymbol{s}_{\rm D} \ {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.09$.
+* Quadrant 4:&nbsp; ${\rm Pr}(\boldsymbol{s}_{\rm D} \ {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.09$.
-Thus, for the mean bit error probability we obtain:
+Thus,&nbsp; for the average bit error probability we obtain:
 :$$p_{\rm B} = { 1}/{ 2} \cdot \big [ 1 \cdot 0.09 + 2 \cdot 0.01 + 1 \cdot 0.09\big ]= \underline{0.1} = p
   \hspace{0.05cm}.$$
-*It is taken into account that quadrant 2 and quadrant 4 each lead to only one bit error, while quadrant 3 leads to two.
+*It is taken into account that&nbsp; "quadrant 2"&nbsp; and&nbsp; "quadrant 4"&nbsp; each lead to only one bit error,&nbsp; while&nbsp; "quadrant 3"&nbsp; leads to two.
 *The factor $1/2$ considers again that in each case one symbol contains two binary characters (bits).
-'''(4)'''&nbsp; According to the solution to subtask '''(2)''', the bit error probability is equal to the probability that the two noise components exceed certain limits:
+'''(4)'''&nbsp; According to the solution to subtask&nbsp; '''(2)''',&nbsp; the bit error probability is equal to the probability that the two noise components exceed certain limits:
 :$$p_{\rm B}  = {\rm Pr}( n_1 < -\sqrt{E}) = {\rm Pr}( n_2 < -\sqrt{E})
   \hspace{0.05cm}.$$
-*For the AWGN channel, this probability with variance is $\sigma_n^2 = N_0/2$:
+*For the AWGN channel,&nbsp; this probability with variance&nbsp; $\sigma_n^2 = N_0/2$&nbsp; is:
 :$$p_{\rm B} = {\rm Q} \left ( { { \sqrt{E}}/{ \sigma_n} }\right ) =  {\rm Q} \left ( \sqrt{ { {2E}}/{ N_0} }\right )
   \hspace{0.05cm}.$$
-*The mean energy per symbol can be most easily determined by averaging over the squared distances of the signal space points from the origin. This yields $E_{\rm S} = 2E$.
+*The average energy per symbol can be most easily determined by averaging over the squared distances of the signal space points from the origin.&nbsp; <br>This yields&nbsp; $E_{\rm S} = 2E$.
 *The average energy per bit is half: $E_{\rm B} = E_{\rm S}/2 = E$. From this follows:
 :$$p_{\rm B}  =  {\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )
   \hspace{0.05cm}.$$
-*The <u>second solution</u> is therefore correct.
+*The&nbsp; <u>second solution</u>&nbsp; is therefore correct.
 *The same result is also obtained if the 4&ndash;QAM is considered as two orthogonal (i.e.: not interfering with each other) BPSK systems over the same channel as in the chapter [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver|"Structure of the Optimal Receiver"]] of the book "Modulation methods".
 {{ML-Fuß}}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2066__Dig_A_4_13.png|right|frame|Signal space constellation of the 4–QAM]]
+[[File:P_ID2066__Dig_A_4_13.png|right|frame|Signal space constellation&nbsp; "4–QAM"]]
-We now consider a quadrature amplitude modulation with&nbsp; $M = 4$&nbsp; symbols and the (normalized) signal space points
+We now consider a quadrature amplitude modulation with&nbsp; $M = 4$&nbsp; symbols and the&nbsp; (normalized)&nbsp; signal space points
 :$$\boldsymbol{ s}_{\rm A} = (+1, +1)\hspace{0.05cm},\hspace{0.2cm}\boldsymbol{ s}_{\rm B} = (-1, +1)\hspace{0.05cm},\hspace{0.2cm} \boldsymbol{ s}_{\rm C} = (-1, -1)\hspace{0.05cm},\hspace{0.2cm}\boldsymbol{ s}_{\rm D} = (+1, -1)
   \hspace{0.05cm}.$$
-The symbols are equally probable. Thus, averaging can be omitted to calculate the mean symbol error probability.
+The symbols are equally probable.&nbsp; Thus,&nbsp; averaging can be omitted to calculate the average symbol error probability.
 For example:
@@ Line 13: / Line 13: @@
   \hspace{0.05cm}.$$
-The assignment of the symbols to <i>bit-duples</i>&nbsp; can also be taken from the graphic (red labels). Gray coding is assumed here.
+The assignment of the symbols to&nbsp; "bit-duples"&nbsp; can also be taken from the graphic&nbsp; (red labels).&nbsp; Gray coding is assumed here.
@@ Line 21: / Line 21: @@
 ''Notes:''
 * The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
 * Reference is made in particular to the section&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"Quadrature amplitude modulation"]]&nbsp; (QAM).
-* For subtask '''(4)''', the (discrete-time) AWGN channel with variance&nbsp; $\sigma_n^2 = N_0/2$&nbsp; is assumed.
+* Reference is made in particular to the section&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"Quadrature amplitude modulation"]]&nbsp; $\rm (QAM)$.
 * For the probability that a symbol is falsified horizontally or vertically by the noise signal&nbsp; $n$,&nbsp; with the complementary Gaussian error function &nbsp;$\rm Q(x)$ holds:
 :$$p = {\rm Pr}( n < -x_0) = {\rm Pr}( n > + x_0) = {\rm Q}(x_0 / \sigma_n)
@@ Line 33: / Line 36: @@
 ===Questions===
 <quiz display=simple>
-{Give the "Union Bound" as the upper bound for the symbol error probability&nbsp; $p_{\rm S}$ &nbsp; $(p_{\rm UB} &#8805; p_{\rm S})$. Let&nbsp; $p = 0.1$.
+{Give the&nbsp; "Union Bound"&nbsp; as the upper bound for the symbol error probability&nbsp; $p_{\rm S}$ &nbsp; $(p_{\rm UB} &#8805; p_{\rm S})$.&nbsp; Let&nbsp; $p = 0.1$.
 |type="{}"}
 $p_{\rm UB}\ = \ $ { 0.2 3% }
@@ Line 41: / Line 44: @@
 $p_{\rm S}\ = \ $ { 0.19 3% }
-{What is the bit error probability for Gray coding&nbsp; $p_{\rm B}$?
+{What is the bit error probability&nbsp; $p_{\rm B}$&nbsp; for Gray coding?
 |type="{}"}
 $p_{\rm B}\ = \ $ { 0.1 3% }

@@ Line 10: / Line 10: @@
 For example:
-:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent})
+:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted})
   \hspace{0.05cm}.$$
@@ Line 55: / Line 55: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The "Union Bound" is an upper bound for the mean symbol error probability. For the latter holds:
-:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm Pr}({\cal{E}}) =  {\rm Pr}( {\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent})= {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent})
+:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm Pr}({\cal{E}}) =  {\rm Pr}( {\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted})= {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted})
   \hspace{0.05cm}.$$
 *In contrast, for the (improved) "Union Bound" in the present example:
-:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}    {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent}) +{\rm Pr}( \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent}) = 2p = \underline{0.2}
+:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}    {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted}) +{\rm Pr}( \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted}) = 2p = \underline{0.2}
   \hspace{0.05cm}.$$
@@ Line 69: / Line 69: @@
 Thus, the "Union Bound" considers the third quadrant twice. It is relatively easy to compensate for this error here:
-:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  p_{\rm UB} -  {\rm Pr}( \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent}) = 2 p -  {\rm Pr}\left [ ( n_1 < -\sqrt{E})\cap ( n_2 < -\sqrt{E})\right ] = 2p - p^2 = \underline{0.19}
+:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  p_{\rm UB} -  {\rm Pr}( \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm transmitted}) = 2 p -  {\rm Pr}\left [ ( n_1 < -\sqrt{E})\cap ( n_2 < -\sqrt{E})\right ] = 2p - p^2 = \underline{0.19}
   \hspace{0.05cm}.$$
@@ Line 75: / Line 75: @@
-'''(3)'''&nbsp; Wie in der Teilaufgabe '''(2)''' nachgewiesen wurde, gelten für die einzelnen Verfälschungswahrscheinlichkeiten:
+'''(3)'''&nbsp; As proved in subtask '''(2)''', the following are valid for the individual falsification probabilities:
-* Quadrant 2: ${\rm Pr}(\boldsymbol{s}_{\rm B} \  {\rm empfangen} \ | \ \boldsymbol{s}_{\rm A} \ {\rm gesendet}) = 0.09$,
+* Quadrant 2: ${\rm Pr}(\boldsymbol{s}_{\rm B} \  {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.09$,
-* Quadrant 3: ${\rm Pr}(\boldsymbol{s}_{\rm C} \ {\rm empfangen} \ | \ \boldsymbol{s}_{\rm A} \ {\rm gesendet}) = 0.01$,
+* Quadrant 3: ${\rm Pr}(\boldsymbol{s}_{\rm C} \ {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.01$,
-* Quadrant 4: ${\rm Pr}(\boldsymbol{s}_{\rm D} \ {\rm empfangen} \ | \ \boldsymbol{s}_{\rm A} \ {\rm gesendet}) = 0.09$.
+* Quadrant 4: ${\rm Pr}(\boldsymbol{s}_{\rm D} \ {\rm received} \ | \ \boldsymbol{s}_{\rm A} \ {\rm transmitted}) = 0.09$.
-Für die mittlere Bitfehlerwahrscheinlichkeit erhält man somit:
+Thus, for the mean bit error probability we obtain:
 :$$p_{\rm B} = { 1}/{ 2} \cdot \big [ 1 \cdot 0.09 + 2 \cdot 0.01 + 1 \cdot 0.09\big ]= \underline{0.1} = p
   \hspace{0.05cm}.$$
-*Berücksichtigt ist, dass der Quadrant 2 und der Quadrant 4 jeweils nur zu einem Bitfehler führt, der Quadrant 3 dagegen zu zweien.
+*It is taken into account that quadrant 2 and quadrant 4 each lead to only one bit error, while quadrant 3 leads to two.
-*Der Faktor $1/2$ berücksichtigt wieder, dass jeweils ein Symbol zwei Binärzeichen (Bit) beinhaltet.
+*The factor $1/2$ considers again that in each case one symbol contains two binary characters (bits).
-'''(4)'''&nbsp; Die Bitfehlerwahrscheinlichkeit ist nach der Lösung zur Teilaufgabe '''(2)''' gleich der Wahrscheinlichkeit, dass die beiden Rauschkomponenten gewisse Grenzen überschreiten:
+'''(4)'''&nbsp; According to the solution to subtask '''(2)''', the bit error probability is equal to the probability that the two noise components exceed certain limits:
 :$$p_{\rm B}  = {\rm Pr}( n_1 < -\sqrt{E}) = {\rm Pr}( n_2 < -\sqrt{E})
   \hspace{0.05cm}.$$
-*Beim AWGN&ndash;Kanal lautet diese Wahrscheinlichkeit mit der Varianz $\sigma_n^2 = N_0/2$:
+*For the AWGN channel, this probability with variance is $\sigma_n^2 = N_0/2$:
 :$$p_{\rm B} = {\rm Q} \left ( { { \sqrt{E}}/{ \sigma_n} }\right ) =  {\rm Q} \left ( \sqrt{ { {2E}}/{ N_0} }\right )
   \hspace{0.05cm}.$$
-*Die mittlere Energie pro Symbol kann am einfachsten durch Mittelung über die quadratischen Abstände der Signalraumpunkte vom Ursprung bestimmt werden. Daraus ergibt sich $E_{\rm S} = 2E$.
+*The mean energy per symbol can be most easily determined by averaging over the squared distances of the signal space points from the origin. This yields $E_{\rm S} = 2E$.
-*Die mittlere Energie pro Bit ist halb so groß: $E_{\rm B} = E_{\rm S}/2 = E$. Daraus folgt:
+*The average energy per bit is half: $E_{\rm B} = E_{\rm S}/2 = E$. From this follows:
 :$$p_{\rm B}  =  {\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )
   \hspace{0.05cm}.$$
-*Richtig ist also der <u>zweite Lösungsvorschlag</u>.
+*The <u>second solution</u> is therefore correct.
-*Zum gleichen Ergebnis kommt man auch, wenn man die 4&ndash;QAM wie im Kapitel [[Digitalsignal%C3%BCbertragung/Struktur_des_optimalen_Empf%C3%A4ngers| Struktur des optimalen Empfängers]] des Buches "Modulationsverfahren" als zwei orthogonale (das heißt: sich nicht störende) BPSK&ndash;Systeme über den gleichen Kanal betrachtet.
+*The same result is also obtained if the 4&ndash;QAM is considered as two orthogonal (i.e.: not interfering with each other) BPSK systems over the same channel as in the chapter [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver|"Structure of the Optimal Receiver"]] of the book "Modulation methods".
 {{ML-Fuß}}

@@ Line 37: / Line 37: @@
 $p_{\rm UB}\ = \ $ { 0.2 3% }
-{Wie groß ist die tatsächliche Symbolfehlerwahrscheinlichkeit&nbsp; $p_{\rm S}$?
+{What is the actual symbol error probability&nbsp; $p_{\rm S}$?
 |type="{}"}
 $p_{\rm S}\ = \ $ { 0.19 3% }
-{Wie groß ist die Bitfehlerwahrscheinlichkeit bei Graycodierung&nbsp; $p_{\rm B}$?
+{What is the bit error probability for Gray coding&nbsp; $p_{\rm B}$?
 |type="{}"}
 $p_{\rm B}\ = \ $ { 0.1 3% }
-{Welcher Zusammenhang besteht zwischen&nbsp; $p_{\rm B}$&nbsp; und &nbsp;$E_{\rm B}/N_0$?
+{What is the relationship between&nbsp; $p_{\rm B}$&nbsp; and &nbsp;$E_{\rm B}/N_0$?
 |type="()"}
 - $p_{\rm B} = {\rm Q}\big [\sqrt {E_{\rm B}/N_0}\big ]$,
@@ Line 52: / Line 52: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die "Union Bound" ist eine obere Schranke für die mittlere Symbolfehlerwahrscheinlichkeit. Für letztere gilt:
+'''(1)'''&nbsp; The "Union Bound" is an upper bound for the mean symbol error probability. For the latter holds:
-:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm Pr}({\cal{E}}) =  {\rm Pr}( {\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm gesendet})= {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm entschieden} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm gesendet})
+:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm Pr}({\cal{E}}) =  {\rm Pr}( {\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent})= {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent})
   \hspace{0.05cm}.$$
-*Dagegen gilt für die (verbesserte) "Union Bound" im vorliegenden Beispiel:
+*In contrast, for the (improved) "Union Bound" in the present example:
-:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}    {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm entschieden} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm gesendet}) +{\rm Pr}( \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D}  \hspace{0.15cm}{\rm entschieden} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm gesendet}) = 2p = \underline{0.2}
+:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}    {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent}) +{\rm Pr}( \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent}) = 2p = \underline{0.2}
   \hspace{0.05cm}.$$
-'''(2)'''&nbsp; Die beiden Wahrscheinlichkeiten, aus der sich die "Union Bound" additiv zusammensetzt, lassen sich geometrisch wie folgt deuten:
+'''(2)'''&nbsp; The two probabilities that make up the "Union Bound" additive can be interpreted geometrically as follows:
-* ${\rm Pr}(\boldsymbol{s}_{\rm B} \cup \boldsymbol{s}_{\rm C} | \boldsymbol{s}_{\rm A})$ ist die Wahrscheinlichkeit, dass der Empfangspunkt in der linken Halbebene liegt <br> &#8658; &nbsp; die AWGN&ndash;Rauschkomponente $n_1$ ist negativ und betragsmäßig größer als $\sqrt {E}$.
+* ${\rm Pr}(\boldsymbol{s}_{\rm B} \cup \boldsymbol{s}_{\rm C} | \boldsymbol{s}_{\rm A})$ iis the probability that the receiving point is located in the left half-plane <br> &#8658; &nbsp; the AWGN noise component $n_1$ is negative and greater in magnitude than $\sqrt {E}$.
-* ${\rm Pr}(\boldsymbol{s}_{\rm C} \cup \boldsymbol{s}_{\rm D} | \boldsymbol{s}_{\rm A})$ ist die Wahrscheinlichkeit, dass der Empfangspunkt in der unteren Halbebene liegt <br> &#8658; &nbsp; die AWGN&ndash;Rauschkomponente $n_2$ ist negativ und betragsmäßig größer als $\sqrt {E}$.
+* ${\rm Pr}(\boldsymbol{s}_{\rm C} \cup \boldsymbol{s}_{\rm D} | \boldsymbol{s}_{\rm A})$ is the probability that the receiving point lies in the lower half-plane <br> &#8658; &nbsp; the AWGN noise component $n_2$ is negative and greater in magnitude than $\sqrt {E}$.
-Die "Union Bound"  berücksichtigt also den dritten Quadranten zweimal. Diesen Fehler kann man hier relativ einfach kompensieren:
+Thus, the "Union Bound" considers the third quadrant twice. It is relatively easy to compensate for this error here:
-:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  p_{\rm UB} -  {\rm Pr}( \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm entschieden} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm gesendet}) = 2 p -  {\rm Pr}\left [ ( n_1 < -\sqrt{E})\cap ( n_2 < -\sqrt{E})\right ] = 2p - p^2 = \underline{0.19}
+:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  p_{\rm UB} -  {\rm Pr}( \boldsymbol{ s}_{\rm C}  \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent}) = 2 p -  {\rm Pr}\left [ ( n_1 < -\sqrt{E})\cap ( n_2 < -\sqrt{E})\right ] = 2p - p^2 = \underline{0.19}
   \hspace{0.05cm}.$$
-Hierbei ist berücksichtigt, dass die Rauschkomponenten $n_1$ und $n_2$ voneinander unabhängig sind.
+Here it is considered that the noise components $n_1$ and $n_2$ are independent of each other.

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2066__Dig_A_4_13.png|right|frame|Signalraumkonstellation der 4–QAM]]
+[[File:P_ID2066__Dig_A_4_13.png|right|frame|Signal space constellation of the 4–QAM]]
-Wir betrachten nun eine Quadraturamplitudenmodulation mit&nbsp; $M = 4$&nbsp; Symbolen und den (normierten) Signalraumpunkten
+We now consider a quadrature amplitude modulation with&nbsp; $M = 4$&nbsp; symbols and the (normalized) signal space points
 :$$\boldsymbol{ s}_{\rm A} = (+1, +1)\hspace{0.05cm},\hspace{0.2cm}\boldsymbol{ s}_{\rm B} = (-1, +1)\hspace{0.05cm},\hspace{0.2cm} \boldsymbol{ s}_{\rm C} = (-1, -1)\hspace{0.05cm},\hspace{0.2cm}\boldsymbol{ s}_{\rm D} = (+1, -1)
   \hspace{0.05cm}.$$
-Die Symbole sind gleichwahrscheinlich. Damit kann man zur Berechnung der mittleren Symbolfehlerwahrscheinlichkeit auf die Mittelung verzichten.
+The symbols are equally probable. Thus, averaging can be omitted to calculate the mean symbol error probability.
-Beispielsweise gilt:
+For example:
-:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm entschieden} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm gesendet})
+:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}( \boldsymbol{ s}_{\rm B} \cup \boldsymbol{ s}_{\rm C} \cup \boldsymbol{ s}_{\rm D} \hspace{0.15cm}{\rm decided} \hspace{0.05cm}|\hspace{0.05cm} \boldsymbol{ s}_{\rm A}\hspace{0.15cm} {\rm sent})
   \hspace{0.05cm}.$$
-Die Zuordnung der Symbole zu <i>Bitdupeln</i>&nbsp; kann ebenfalls der Grafik (rote Beschriftungen) entnommen werden. Hierbei ist die Graycodierung vorausgesetzt.
+The assignment of the symbols to <i>bit-duples</i>&nbsp; can also be taken from the graphic (red labels). Gray coding is assumed here.
@@ Line 20: / Line 20: @@
-''Hinweise:''
+''Notes:''
-* Die Aufgabe gehört zum Kapitel&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
+* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
-* Bezug genommen eird insbesondere auf die Seite&nbsp;  [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation#Quadraturamplitudenmodulation_.28M.E2.80.93QAM.29| Quadraturamplitudenmodulation]]&nbsp; (QAM).
+* Reference is made in particular to the section&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"Quadrature amplitude modulation"]]&nbsp; (QAM).
-* Für die Teilaufgabe '''(4)''' ist der (zeitdiskrete) AWGN&ndash;Kanal mit der Varianz&nbsp; $\sigma_n^2 = N_0/2$&nbsp; vorausgesetzt.
+* For subtask '''(4)''', the (discrete-time) AWGN channel with variance&nbsp; $\sigma_n^2 = N_0/2$&nbsp; is assumed.
-* Für die Wahrscheinlichkeit, dass durch das Rauschsignal&nbsp; $n$&nbsp; ein Symbol horizontal oder vertikal verfälscht wird, gilt mit der komplementären Gaußschen Fehlerfunktion  &nbsp;$\rm Q(x)$:
+* For the probability that a symbol is falsified horizontally or vertically by the noise signal&nbsp; $n$,&nbsp; with the complementary Gaussian error function &nbsp;$\rm Q(x)$ holds:
 :$$p = {\rm Pr}( n < -x_0) = {\rm Pr}( n > + x_0) = {\rm Q}(x_0 / \sigma_n)
   \hspace{0.05cm}.$$
@@ Line 31: / Line 31: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Geben Sie als obere Schranke für die Symbolfehlerwahrscheinlichkeit&nbsp; $p_{\rm S}$ die "Union Bound" an&nbsp; $(p_{\rm UB} &#8805; p_{\rm S})$. Es gelte&nbsp; $p = 0.1$.
+{Give the "Union Bound" as the upper bound for the symbol error probability&nbsp; $p_{\rm S}$ &nbsp; $(p_{\rm UB} &#8805; p_{\rm S})$. Let&nbsp; $p = 0.1$.
 |type="{}"}
 $p_{\rm UB}\ = \ $ { 0.2 3% }

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−	[[Category:Digital Signal Transmission: Exercises\|^4.4 ~~Kohärente~~ Demodulation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]

← Older revision	Revision as of 11:51, 13 July 2022
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