Aufgaben:Exercise 4.12: Calculations for the 16-QAM - Revision history

Hwang at 19:06, 1 September 2022

2022-09-01T19:06:15Z

Guenter at 15:55, 20 August 2022

2022-08-20T15:55:48Z

Guenter at 15:22, 20 August 2022

2022-08-20T15:22:57Z

Guenter at 14:50, 20 August 2022

2022-08-20T14:50:52Z

Hwang at 09:31, 12 August 2022

2022-08-12T09:31:56Z

Hwang at 14:23, 19 July 2022

2022-07-19T14:23:07Z

Hwang at 14:14, 19 July 2022

2022-07-19T14:14:44Z

Guenter at 11:57, 13 July 2022

2022-07-13T11:57:05Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.12: Berechnungen zur 16-QAM to Aufgaben:Exercise 4.12: Calculations for the 16-QAM

2022-07-13T11:50:14Z

Guenter moved page Aufgabe 4.12: Berechnungen zur 16-QAM to Exercise 4.12: Calculations for the 16-QAM

Javier: Text replacement - "„" to """

2021-05-28T14:36:36Z

Text replacement - "„" to """

@@ Line 71: / Line 71: @@
 [[File:P_ID2063__Dig_A_4_12c.png|right|frame|Illustration:&nbsp; 16–QAM error probability]]
-'''(3)'''&nbsp; The&nbsp; ">Union Bound"&nbsp; is an upper bound on the symbol error probability.
+'''(3)'''&nbsp; The&nbsp; "Union Bound"&nbsp; is an upper bound on the symbol error probability.
 *It only takes into account the transition to adjacent decision regions due to AWGN noise.

@@ Line 22: / Line 22: @@
 # For numerical calculations,&nbsp; use&nbsp; $E = 1 \ \rm mWs$&nbsp; and &nbsp;$p = 0.4\%$.
 #The AWGN noise power density&nbsp; $N_0$&nbsp; can be calculated approximately from these values:
-:$$p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right ) = 0.004 \hspace{0.1cm}\Rightarrow\hspace{0.1cm}
+::$$p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right ) = 0.004 \hspace{0.1cm}\Rightarrow\hspace{0.1cm}
   { 2E}{ N_0} \approx 2.65^2 \approx 7 \hspace{0.1cm}\Rightarrow\hspace{0.1cm} N_0 = { E}/{ 3.5}\approx 1.4 \cdot 10^{-4}\,{\rm W/Hz}
   \hspace{0.05cm}.$$

@@ Line 54: / Line 54: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The quotient $E_{\rm S}/E$ is obtained as the mean square distance of the $M = 16$ signal space points $\boldsymbol{s}_i$ from the origin.
+'''(1)'''&nbsp; The quotient&nbsp; $E_{\rm S}/E$&nbsp; is obtained as the mean square distance of the&nbsp; $M = 16$&nbsp; signal space points&nbsp; $\boldsymbol{s}_i$&nbsp; from the origin.
 *With the given signal space constellation of the 16&ndash;QAM we obtain:
 :$$E_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} { E}/{ 16} \cdot \left [ 4 \cdot (1^2 + 1^2) + 8 \cdot (1^2 + 3^2) + 4 \cdot (3^2 + 3^2)\right ]={ E}/{ 16} \cdot \left [ 4 \cdot 2 + 8 \cdot 10 + 4 \cdot 18\right ] = 10 \cdot E = \underline{10 \ {\rm mWs}}
   \hspace{0.05cm}.$$
-*The same result is obtained with the equation given in the [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]
+*The same result is obtained with the equation given in the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]:
 :$$E_{\rm S} = \frac{ 2 \cdot (M-1)}{ 3 } \cdot  E = \frac{ 2 \cdot 15}{ 3 } \cdot  E = 10 E
   \hspace{0.05cm}.$$
-'''(2)'''&nbsp; Each individual symbol represents four binary symbols. Thus, the average energy per bit is
+'''(2)'''&nbsp; Each individual symbol represents four binary symbols.&nbsp; Thus,&nbsp; the average energy per bit is
 :$$E_{\rm B} = \frac{ E_{\rm S}}{ {\rm log_2} \hspace{0.05cm}(M)} = 2.5 \cdot  E = \underline{2.5 \ {\rm mWs}}
   \hspace{0.05cm}.$$
-[[File:P_ID2063__Dig_A_4_12c.png|right|frame|Illustration of the 16–QAM error probability]]
+[[File:P_ID2063__Dig_A_4_12c.png|right|frame|Illustration:&nbsp; 16–QAM error probability]]
-'''(3)'''&nbsp; The <i>Union Bound</i> is an upper bound on the symbol error probability.
+'''(3)'''&nbsp; The&nbsp; ">Union Bound"&nbsp; is an upper bound on the symbol error probability.
 *It only takes into account the transition to adjacent decision regions due to AWGN noise.
 *From the graph, it can be seen that the corner symbols (filled in yellow) can only be biased towards two other symbols and the remaining edge symbols (filled in green) can be biased in three directions.
-*The "worst case" are the four inner symbols (with blue filling) with four falsification possibilities each. From this follows:
+*From the graph,&nbsp; it can be seen that the corner symbols&nbsp; (filled in yellow)&nbsp; can only be biased towards two other symbols and the remaining edge symbols&nbsp; (filled in green)&nbsp; can be biased in three directions.
 :$$p_{\rm S} = {\rm Pr}({\cal{E}}) \le 4 \cdot p = \underline{1.6\%}= p_{\rm UB}
   \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Counting the blue arrows in the above graph, we get
+'''(4)'''&nbsp; Counting the blue arrows in the above graph,&nbsp; we get
 :$$4 \cdot 2 + 8 \cdot 3 + 4 \cdot 4 = 48.$$
-*Thus, the mean symbol error probability is equal to
+*Thus,&nbsp; the average symbol error probability is equal to
 :$$p_{\rm S} = { E}/{ 16}   \cdot 48 p = 3p = \underline{1.2\%}
   \hspace{0.05cm}.$$
-*The same result is obtained with the equation given in the [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]
+*The same result is obtained with the equation given in the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]:
 :$$p_{\rm S} = 4p \cdot \left [ 1 - { 1}/{ \sqrt{M}} \right ] = 4p \cdot \left [ 1 - { 1}/{ 4} \right ] = 3p
   \hspace{0.05cm}.$$
@@ Line 91: / Line 96: @@
 '''(5)'''&nbsp; With Gray coding according to the red labeling in the graph, each symbol error causes exactly one bit error.
-*But since with each symbol $M = 4$ binary symbols are transmitted, is
+'''(5)'''&nbsp; With Gray coding according to the red labeling in the graph,&nbsp; each symbol error causes exactly one bit error.
 :$$p_{\rm B} = \frac{ p_{\rm S}}{ {\rm log_2} \hspace{0.05cm}(M)}
 = \frac{ 1.2\%}{ 4}  = \underline{0.3\%}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2062__Dig_A_4_12.png|right|frame|Signal space constellation of 16–QAM]]
+[[File:P_ID2062__Dig_A_4_12.png|right|frame|Signal space constellation of&nbsp; $\rm 16–QAM$]]
-The graphic shows the signal space constellation of the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"quadrature amplitude modulation"]]&nbsp; with &nbsp;$M = 16$&nbsp; signal space points.
+The graph shows the signal space constellation of&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"quadrature amplitude modulation"]]&nbsp; with &nbsp;$M = 16$&nbsp; signal space points.&nbsp; The following should be calculated for this modulation method:
 *the&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Union_Bound_-_Upper_bound_for_the_error_probability|"Union Bound"]]&nbsp; $p_{\rm UB}$&nbsp; as upper bound,
 * the average bit error probability&nbsp; $p_{\rm B}$&nbsp; with Gray coding.
 Notes:
+# The exercise deals with a partial aspect of the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
-''Notes:''
+#The Gray assignment is given in the graphic&nbsp; $($red lettering$)$.
-* The exercise deals with a partial aspect of the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation]].
+#The probability that the upper left symbol is falsified into one of the neighboring symbols is abbreviated to&nbsp; $p$&nbsp; $($blue arrows in the graph$)$.
-*The Gray assignment is given in the graphic (red lettering).
+#A diagonal falsification &nbsp; &#8658; &nbsp; two bit falsified&nbsp; $($green arrow$)$&nbsp; is excluded.
-* The probability that the upper left symbol is falsified into one of the neighboring symbols is abbreviated to&nbsp; $p$&nbsp; (blue arrows in the graph).
+#For the AWGN channel,&nbsp; with the complementary Gaussian error integral for this auxiliary variable,&nbsp; the following applies: &nbsp; $p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right )\hspace{0.05cm}.$
-* A diagonal falsification &nbsp;&#8658;&nbsp; two bit falsified (green arrow) is excluded.
+# For numerical calculations,&nbsp; use&nbsp; $E = 1 \ \rm mWs$&nbsp; and &nbsp;$p = 0.4\%$.
-* For the AWGN channel, with the complementary Gaussian error integral for this auxiliary variable, the following applies: &nbsp; $p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right )\hspace{0.05cm}.$
+#The AWGN noise power density&nbsp; $N_0$&nbsp; can be calculated approximately from these values:
 :$$p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right ) = 0.004 \hspace{0.1cm}\Rightarrow\hspace{0.1cm}
   { 2E}{ N_0} \approx 2.65^2 \approx 7 \hspace{0.1cm}\Rightarrow\hspace{0.1cm} N_0 = { E}/{ 3.5}\approx 1.4 \cdot 10^{-4}\,{\rm W/Hz}
@@ Line 32: / Line 31: @@
 ===Questions===
 <quiz display=simple>
-{Let&nbsp; $E = 1 \ \rm mWs$. What is the average energy ''per symbol''?
+{Let&nbsp; $E = 1 \ \rm mWs$.&nbsp; What is the&nbsp; "average energy per symbol"?
 |type="{}"}
 $E_{\rm S}\ = \ $ { 10 3% } $\  \rm mWs$
-{What is the average energy ''per bit''?
+{What is the&nbsp; "average energy per bit"?
 |type="{}"}
 $E_{\rm B}\ = \ $ { 2.5 3% } $\ \rm mWs$
-{Give the (improved) "Union Bound"&nbsp; $(p_{\rm UB})$&nbsp; for&nbsp; $p = 0.4\%$.&nbsp;
+{Give the&nbsp; (improved)&nbsp; "Union Bound"&nbsp; $(p_{\rm UB})$&nbsp; for&nbsp; $p = 0.4\%$.&nbsp;
 |type="{}"}
 $p_{\rm UB} \ = \ $ { 1.6 3% } $\ \%$
@@ Line 48: / Line 47: @@
 $p_{\rm S}  \ = \ $ { 1.2 3% } $\ \%$
-{Calculate the actual bit error rate for Gray coding.
+{Calculate the actual bit error probability&nbsp; $p_{\rm B}$&nbsp; for Gray coding.
 |type="{}"}
 $p_{\rm B}  \ = \ $ { 0.3 3% } $\ \%$

@@ Line 55: / Line 55: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Der Quotient $E_{\rm S}/E$ ergibt sich als der mittlere quadratische Abstand der $M = 16$ Signalraumpunkte $\boldsymbol{s}_i$ vom Ursprung.
+'''(1)'''&nbsp; The quotient $E_{\rm S}/E$ is obtained as the mean square distance of the $M = 16$ signal space points $\boldsymbol{s}_i$ from the origin.
-*Mit der gegebenen Signalraumkonstellation der 16&ndash;QAM erhält man:
+*With the given signal space constellation of the 16&ndash;QAM we obtain:
 :$$E_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} { E}/{ 16} \cdot \left [ 4 \cdot (1^2 + 1^2) + 8 \cdot (1^2 + 3^2) + 4 \cdot (3^2 + 3^2)\right ]={ E}/{ 16} \cdot \left [ 4 \cdot 2 + 8 \cdot 10 + 4 \cdot 18\right ] = 10 \cdot E = \underline{10 \ {\rm mWs}}
   \hspace{0.05cm}.$$
-*Zum gleichen Ergebnis kommt man mit der im [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation| Theorieteil]] angegebenen Gleichung
+*The same result is obtained with the equation given in the [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]
 :$$E_{\rm S} = \frac{ 2 \cdot (M-1)}{ 3 } \cdot  E = \frac{ 2 \cdot 15}{ 3 } \cdot  E = 10 E
   \hspace{0.05cm}.$$
-'''(2)'''&nbsp; Jedes einzelne Symbol stellt vier Binärsymbole dar. Damit ist die mittlere Energie pro Bit.
+'''(2)'''&nbsp; Each individual symbol represents four binary symbols. Thus, the average energy per bit is
 :$$E_{\rm B} = \frac{ E_{\rm S}}{ {\rm log_2} \hspace{0.05cm}(M)} = 2.5 \cdot  E = \underline{2.5 \ {\rm mWs}}
   \hspace{0.05cm}.$$
-[[File:P_ID2063__Dig_A_4_12c.png|right|frame|Zur Verdeutlichung der 16–QAM–Fehlerwahrscheinlichkeit]]
+[[File:P_ID2063__Dig_A_4_12c.png|right|frame|Illustration of the 16–QAM error probability]]
-'''(3)'''&nbsp; Die <i>Union Bound</i> ist eine obere Schranke für die Symbolfehlerwahrscheinlichkeit.
+'''(3)'''&nbsp; The <i>Union Bound</i> is an upper bound on the symbol error probability.
-*Sie berücksichtigt nur den Übergang zu benachbarten Entscheidungsregionen aufgrund von AWGN&ndash;Rauschen.
+*It only takes into account the transition to adjacent decision regions due to AWGN noise.
-*Aus der Grafik geht hervor, dass die Ecksymbole (gelb gefüllt) nur zu zwei anderen Symbolen hin verfälscht werden können und die restlichen Randsymbole (grüne Füllung) in drei Richtungen.
+*From the graph, it can be seen that the corner symbols (filled in yellow) can only be biased towards two other symbols and the remaining edge symbols (filled in green) can be biased in three directions.
-*Der "worst case" sind die vier inneren Symbole (mit blauer Füllung) mit jeweils vier Verfälschungsmöglichkeiten. Daraus folgt:
+*The "worst case" are the four inner symbols (with blue filling) with four falsification possibilities each. From this follows:
 :$$p_{\rm S} = {\rm Pr}({\cal{E}}) \le 4 \cdot p = \underline{1.6\%}= p_{\rm UB}
   \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Zählt man die blauen Pfeile in obiger Grafik, so kommt man auf
+'''(4)'''&nbsp; Counting the blue arrows in the above graph, we get
 :$$4 \cdot 2 + 8 \cdot 3 + 4 \cdot 4 = 48.$$
-*Die mittlere Symbolfehlerwahrscheinlichkeit ist somit gleich
+*Thus, the mean symbol error probability is equal to
 :$$p_{\rm S} = { E}/{ 16}   \cdot 48 p = 3p = \underline{1.2\%}
   \hspace{0.05cm}.$$
-*Zum gleichen Ergebnis kommt man mit der im [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation| Theorieteil]] angegebenen Gleichung
+*The same result is obtained with the equation given in the [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]
 :$$p_{\rm S} = 4p \cdot \left [ 1 - { 1}/{ \sqrt{M}} \right ] = 4p \cdot \left [ 1 - { 1}/{ 4} \right ] = 3p
   \hspace{0.05cm}.$$
-*Beide Gleichungen gelten nur dann exakt, wenn man wie hier diagonale Verfälschungen ausschließt.
+*Both equations hold exactly only if one excludes diagonal falsifications as here.
-'''(5)'''&nbsp; Bei Graycodierung entsprechend der roten Beschriftung in der Grafik bewirkt jeder Symbolfehler genau einen Bitfehler.
+'''(5)'''&nbsp; With Gray coding according to the red labeling in the graph, each symbol error causes exactly one bit error.
-*Da aber mit jedem Symbol $M = 4$ Binärsymbole übertragen werden, ist
+*But since with each symbol $M = 4$ binary symbols are transmitted, is
 :$$p_{\rm B} = \frac{ p_{\rm S}}{ {\rm log_2} \hspace{0.05cm}(M)}
 = \frac{ 1.2\%}{ 4}  = \underline{0.3\%}

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Trägerfrequenzsysteme mit kohärenter Demodulation}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2062__Dig_A_4_12.png|right|frame|Signalraumkonstellation der 16–QAM]]
+[[File:P_ID2062__Dig_A_4_12.png|right|frame|Signal space constellation of 16–QAM]]
-Die Grafik zeigt die Signalraumkonstellation der&nbsp; [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation#Quadraturamplitudenmodulation_.28M.E2.80.93QAM.29| Quadraturamplitudenmodulation]]&nbsp; mit &nbsp;$M = 16$&nbsp; Signalraumpunkten.
+The graphic shows the signal space constellation of the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"quadrature amplitude modulation"]]&nbsp; with &nbsp;$M = 16$&nbsp; signal space points.
-Für dieses Modulationsverfahren sollen berechnet werden:
+The following should be calculated for this modulation method:
-* die mittlere Energie pro Symbol bzw. pro Bit,
+* the average energy per symbol or per bit,
-* die mittlere Symbolfehlerwahrscheinlichkeit&nbsp; $p_{\rm S}$,
+* the mean symbol error probability&nbsp; $p_{\rm S}$,
-*die&nbsp; [[Digitalsignal%C3%BCbertragung/Approximation_der_Fehlerwahrscheinlichkeit#Union_Bound_-_Obere_Schranke_f.C3.BCr_die_Fehlerwahrscheinlichkeit| Union Bound]]&nbsp; $p_{\rm UB}$&nbsp; als obere Schranke,
+*the&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Union_Bound_-_Upper_bound_for_the_error_probability|"Union Bound"]]&nbsp; $p_{\rm UB}$&nbsp; as upper bound,
-* die mittlere Bitfehlerwahrscheinlichkeit&nbsp; $p_{\rm B}$&nbsp; bei Graycodierung.
+* the average bit error probability&nbsp; $p_{\rm B}$&nbsp; with Gray coding.
@@ Line 15: / Line 15: @@
-''Hinweise:''
+''Notes:''
-* Die Aufgabe behandelt einen Teilaspekt des Kapitels&nbsp; [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation| Trägerfrequenzsysteme mit kohärenter Demodulation]].
+* The exercise deals with a partial aspect of the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation]].
-*Die Gray&ndash;Zuordnung ist in der Grafik angegeben (rote Beschriftung).
+*The Gray assignment is given in the graphic (red lettering).
-* Die Wahrscheinlichkeit, dass das linke obere Symbol in eines der benachbarten Symbole verfälscht wird, wird mit&nbsp; $p$&nbsp; abgekürzt (blaue Pfeile in der Grafik).
+* The probability that the upper left symbol is falsified into one of the neighboring symbols is abbreviated to&nbsp; $p$&nbsp; (blue arrows in the graph).
-* Eine diagonale Verfälschung &nbsp;&#8658;&nbsp; zwei Bit verfälscht (grüner Pfeil) wird ausgeschlossen.
+* A diagonal falsification &nbsp;&#8658;&nbsp; two bit falsified (green arrow) is excluded.
-* Für den AWGN&ndash;Kanal gilt mit dem komplementären Gaußschen Fehlerintegrale für diese Hilfsgröße: &nbsp; $p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right )\hspace{0.05cm}.$
+* For the AWGN channel, with the complementary Gaussian error integral for this auxiliary variable, the following applies: &nbsp; $p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right )\hspace{0.05cm}.$
-* Verwenden Sie für numerische Berechnungen&nbsp; $E = 1 \ \rm mWs$&nbsp; und &nbsp;$p = 0.4\%$.
+* For numerical calculations, use&nbsp; $E = 1 \ \rm mWs$&nbsp; and &nbsp;$p = 0.4\%$.
-*Aus diesen Werten kann die AWGN&ndash;Rauschleistungsdichte&nbsp; $N_0$&nbsp; näherungsweise berechnet werden:
+*The AWGN noise power density&nbsp; $N_0$&nbsp; can be calculated approximately from these values:
 :$$p = {\rm Q} \left ( \sqrt{ { 2E}/{ N_0} }\right ) = 0.004 \hspace{0.1cm}\Rightarrow\hspace{0.1cm}
   { 2E}{ N_0} \approx 2.65^2 \approx 7 \hspace{0.1cm}\Rightarrow\hspace{0.1cm} N_0 = { E}/{ 3.5}\approx 1.4 \cdot 10^{-4}\,{\rm W/Hz}
@@ Line 30: / Line 30: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Es sei&nbsp; $E = 1 \ \rm mWs$. Wie groß ist die mittlere Energie ''pro Symbol''?
+{Let&nbsp; $E = 1 \ \rm mWs$. What is the average energy ''per symbol''?
 |type="{}"}
 $E_{\rm S}\ = \ $ { 10 3% } $\  \rm mWs$

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

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