Aufgaben:Exercise 4.11Z: Error Probability with QAM - Revision history

Guenter at 06:43, 18 April 2022

2022-04-18T06:43:36Z

Guenter at 06:37, 18 April 2022

2022-04-18T06:37:41Z

Reed at 16:38, 19 March 2022

2022-03-19T16:38:25Z

Reed: /* Solution */

2022-03-19T16:38:11Z

Solution

Reed: /* Questions */

2022-03-19T16:31:04Z

Questions

Reed at 16:27, 19 March 2022

2022-03-19T16:27:35Z

Reed at 16:26, 19 March 2022

2022-03-19T16:26:28Z

Reed at 16:25, 19 March 2022

2022-03-19T16:25:48Z

Guenter at 13:41, 19 March 2022

2022-03-19T13:41:09Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.11Z: Fehlerwahrscheinlichkeit bei QAM to Aufgaben:Exercise 4.11Z: Error Probability with QAM

2022-03-19T13:36:25Z

Guenter moved page Aufgabe 4.11Z: Fehlerwahrscheinlichkeit bei QAM to Exercise 4.11Z: Error Probability with QAM

@@ Line 59: / Line 59: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; With the values given, for &nbsp;''Binary Phase Shift Keying''&nbsp; (BPSK), one gets:
+'''(1)'''&nbsp; With the values given,&nbsp; for &nbsp;"Binary Phase Shift Keying"&nbsp; (BPSK),&nbsp; one gets:
 :$$E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} = \frac{1}{2}\cdot (2\,{\rm V})^2 \cdot 1\,{\rm &micro; s} = 2 \cdot 10^{-6}\,{\rm V^2s}
 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}  {E_{\rm B}}/{N_0} = \frac {2 \cdot 10^{-6}\,{\rm V^2s}}{0.25 \cdot 10^{-6}\,{\rm V^2/Hz}} = 8$$
 :$$ \Rightarrow \hspace{0.3cm} p_\text{B, BPSK} = {\rm Q}\left ( \sqrt{16} \right ) = {\rm Q}\left ( 4 \right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{8}\right )\hspace{0.05cm}.$$
-*Based on the given &nbsp; $x$–values in the table, it is convenient to use the first equation in this subtask:
+*Based on the given &nbsp; $x$–values in the table,&nbsp; it is convenient to use the first equation in this subtask:
 :$$p_\text{B, BPSK} = {\rm Q}\left ( 4 \right ) \hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4} }\hspace{0.05cm}.$$
-'''(2)'''&nbsp; With twice the bit duration, the energy is also twice as large:&nbsp; $E_{\rm B} = 4 · 10^{–6} \ \rm  V^2s$  ⇒  $E_{\rm B}/N_0 = 16$.
+'''(2)'''&nbsp; With twice the bit duration,&nbsp; the energy is also twice as large:&nbsp; $E_{\rm B} = 4 · 10^{–6} \ \rm  V^2s$  ⇒  $E_{\rm B}/N_0 = 16$.
 *It follows that:
 :$$p_\text{B, BPSK} = {\rm Q}\left ( \sqrt{32} \right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{16}\right ) ={1}/{2}\cdot {\rm erfc}\left ( 4\right ) \hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$
-*For pragmatic reasons, the last column of the table was used here.
+*For pragmatic reasons,&nbsp; the last column of the table was used here.
-'''(3)'''&nbsp; Substituting the equations given for the 4-QAM into each other, we get the same result as for the BPSK:
+'''(3)'''&nbsp; Substituting the equations given for the 4-QAM into each other,&nbsp; we get the same result as for the BPSK:
 :$$p_{\rm B, \hspace{0.05cm}4-QAM} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) \equiv p_\text{B, BPSK}.$$
-*Also, since the energy per bit has not changed from subtask&nbsp; '''(2)'''&nbsp;,  the same error probability will arise:
+*Also,&nbsp; since the energy per bit has not changed from subtask&nbsp; '''(2)''',&nbsp;  the same error probability will arise:
 :$$p_{\rm B, \hspace{0.05cm}4-QAM}= {\rm Q}\left ( \sqrt{32} \right ) = {1}/{2}\cdot {\rm erfc}\left ( 4\right ) \hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Only <u>Answer 1</u> is correct:
+'''(4)'''&nbsp; Only&nbsp; <u>Answer 1</u>&nbsp; is correct:
-*Of course the error probability of error is the same in the two branches.  Why would it not be?
+*Of course the error probability is the same in the two branches.&nbsp;  Why would it not be?
 *This would no longer be true with a phase offset between the transmitter and receiver though.
-*However, the distance of the useful samples from the threshold is&nbsp; $g_0$&nbsp; here and thus smaller than the envelope &nbsp; $s_0$&nbsp; of the entire 4-QAM by a factor of $\sqrt{2}$&nbsp;.
+*However,&nbsp; the distance of the useful samples from the threshold is&nbsp; $g_0$&nbsp; here and thus smaller than the envelope &nbsp; $s_0$&nbsp; of the entire 4-QAM by a factor of $\sqrt{2}$&nbsp;.
-*However, if the inphase branch (or the quadrature branch) is considered as a stand-alone BPSK, the noise power is also half that of BPSK because of the lower symbol rate.  Therefore, the error probability remains the same.
+*However,&nbsp; if the inphase branch&nbsp; (or the quadrature branch)&nbsp; is considered as a stand-alone BPSK,&nbsp; the noise power is also half that of BPSK because of the lower symbol rate.&nbsp;  Therefore,&nbsp; the error probability remains the same.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1721__Mod_Z_4_10.png|right|frame|Table with two different Gaussian error functions]]
+[[File:P_ID1721__Mod_Z_4_10.png|right|frame|Table with two different&nbsp; "Gaussian error functions"]]
 We now make the following assumptions:
 * binary bipolar amplitude coefficients &nbsp;$a_ν ∈ \{±1\}$,
-* rectangular fundamental transmission pulse with amplitude &nbsp;$s_0$&nbsp; and bit time &nbsp;$T_{\rm B}$,
+* rectangular basic transmission pulse with amplitude &nbsp;$s_0$&nbsp; and bit duration &nbsp;$T_{\rm B}$,
 * AWGN noise with noise power density  &nbsp;$N_0$,
 * a receiver according to the matched-filter principle,
@@ Line 12: / Line 12: @@
-As has been shown several times, the bit error probability of binary phase modulation &nbsp; $\rm (BPSK)$&nbsp; under these conditions can be calculated using the following equations:
+The bit error probability of binary phase modulation &nbsp; $\rm (BPSK)$&nbsp; under these conditions can be calculated using the following equations:
 :$$ p_{\rm B, \hspace{0.05cm}BPSK} = {\rm Q}\left ({s_0}/{\sigma_d } \right ), \hspace{0.2cm} E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/{T_{\rm B} }$$
 :$$\Rightarrow \hspace{0.3cm} p_{\rm B, \hspace{0.05cm}BPSK} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$
-The corresponding equations of&nbsp; $\rm 4–QAM$&nbsp;  are:
+*The corresponding equations of&nbsp; $\rm 4–QAM$&nbsp;  are:
 :$$ p_{\rm B, \hspace{0.05cm}4-QAM} = {\rm Q}\left ( {g_0}/{\sigma_d } \right ), \hspace{0.2cm}g_{0} = {s_0}/{\sqrt{2}}, \hspace{0.2cm}E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/({2 \cdot T_{\rm B} }).$$
-Here it is taken into account that - in order to achieve the same transmission energy per bit as with BPSK - one must reduce the pulse amplitude&nbsp;$g_0$&nbsp; of the square-wave impulses in the two sub-branches of 4-QAM by a factor of &nbsp;$\sqrt{2}$&nbsp;.  The envelope is then equal to &nbsp;$s_0$ for both systems.
+*Here it is taken into account that &ndash; in order to achieve the same transmission energy per bit as with BPSK &ndash; one must reduce the  amplitude&nbsp;$g_0$&nbsp; of the rectangular pulses in the two sub-branches of 4-QAM by a factor of &nbsp;$\sqrt{2}$&nbsp;.
 Hints:
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation|"Quadrature Amplitude Modulation"]].
+*Reference is also made to the page&nbsp; [[Modulation_Methods/Linear_Digital_Modulation#Error_probabilities_-_a_brief_overview|"Error probabilities &ndash; a brief overview"]]&nbsp; in the previous chapter.
 * Always assume the following numerical values: &nbsp; $s_0 = 2\,{\rm V}, \hspace{0.05cm} N_0 = 0.25 \cdot 10^{-6}\,{\rm V^2/Hz}\hspace{0.05cm}.$
-*The bit time is &nbsp;$T_{\rm B} = 1 \ \rm &micro; s$&nbsp; (question 1) and &nbsp;$T_{\rm B} = 2 \ \rm &micro; s$&nbsp; (from question 2 onwards).
+*The bit duration is &nbsp;$T_{\rm B} = 1 \ \rm &micro; s$&nbsp; (question 1)&nbsp; and &nbsp;$T_{\rm B} = 2 \ \rm &micro; s$&nbsp; (from question 2 onwards).
-*In the table, the two common Gaussian error functions &nbsp;${\rm Q}(x)$&nbsp;  and &nbsp;$1/2 \cdot {\rm erfc}(x)$&nbsp; are given.
+*In the table,&nbsp; the two common Gaussian error functions &nbsp;${\rm Q}(x)$&nbsp;  and &nbsp;$1/2 \cdot {\rm erfc}(x)$&nbsp; are given.
-*Energies are to be given in &nbsp;$\rm V^2s$&nbsp; ; thus, they refer to the reference resistance &nbsp;$R = 1 \ \rm \Omega$.
+*Energies are to be given in &nbsp;$\rm V^2s$;&nbsp;  thus, they refer&nbsp; to the reference resistance &nbsp;$R = 1 \ \rm \Omega$.
@@ Line 41: / Line 38: @@
 <quiz display=simple>
-{What error probability &nbsp;$p_\text{B, BPSK}$&nbsp; results for&nbsp; ''Binary Phase Shift Keying''&nbsp; (BPSK) when $T_{\rm B} = 1 \ \rm &micro; s$?
+{What error probability &nbsp;$p_\text{B, BPSK}$&nbsp; results for&nbsp; '''BPSK'''&nbsp; when $T_{\rm B} = 1 \ \rm &micro; s$?
 |type="{}"}
 $p_\text{B, BPSK} \ = \ $ { 0.317 3% } $\ \rm 10^{-4}$
-{What error probability &nbsp;$p_\text{B, BPSK}$&nbsp; results for&nbsp; ''Binary Phase Shift Keying''&nbsp; (BPSK) when $T_{\rm B} = 2 \ \rm &micro; s$?
+{What error probability &nbsp;$p_\text{B, BPSK}$&nbsp; results for&nbsp;  '''BPSK'''&nbsp; when $T_{\rm B} = 2 \ \rm &micro; s$?
 |type="{}"}
 $p_\text{B, BPSK} \ = \ $ { 0.771 3% } $\ \rm 10^{-8}$
-{What error probability &nbsp;$p_\text{B, 4-QAM}$&nbsp; is obtained for 4–QAM when &nbsp;$E_{\rm B} = 4 · 10^{–6} \ \rm V^2s$?
+{What error probability &nbsp;$p_\text{B, 4-QAM}$&nbsp; is obtained for&nbsp; '''4-QAM'''&nbsp; when &nbsp;$E_{\rm B} = 4 · 10^{–6} \ \rm V^2s$?
 |type="{}"}
 $p_\text{B, 4-QAM} \ = \ $ { 0.771 3% } $\ \rm 10^{-8}$
@@ Line 56: / Line 53: @@
 |type="[]"}
 + The same result is obtained as for the entire 4-QAM.
-- The spacing of the useful samples is equal to &nbsp;$s_0$ as in BPSK.
+- The distance of the noisless samples from the threshold&nbsp; $E=0$&nbsp; is the same &nbsp;$(s_0)$&nbsp;  as in BPSK.
 - The same result is obtained for the noise power as in BPSK.
 </quiz>

@@ Line 86: / Line 86: @@
-'''(4)'''&nbsp; Only<u>Answer 1</u> is correct:
+'''(4)'''&nbsp; Only <u>Answer 1</u> is correct:
 *Of course the error probability of error is the same in the two branches.  Why would it not be?
 *This would no longer be true with a phase offset between the transmitter and receiver though.

@@ Line 62: / Line 62: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Mit den vorgegebenen Werten erhält man für &nbsp;''Binary Phase Shift Keying''&nbsp; (BPSK):
+'''(1)'''&nbsp; With the values given, for &nbsp;''Binary Phase Shift Keying''&nbsp; (BPSK), one gets:
 :$$E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} = \frac{1}{2}\cdot (2\,{\rm V})^2 \cdot 1\,{\rm &micro; s} = 2 \cdot 10^{-6}\,{\rm V^2s}
 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}  {E_{\rm B}}/{N_0} = \frac {2 \cdot 10^{-6}\,{\rm V^2s}}{0.25 \cdot 10^{-6}\,{\rm V^2/Hz}} = 8$$
 :$$ \Rightarrow \hspace{0.3cm} p_\text{B, BPSK} = {\rm Q}\left ( \sqrt{16} \right ) = {\rm Q}\left ( 4 \right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{8}\right )\hspace{0.05cm}.$$
-*Aufgrund der gegebenen&nbsp; $x$–Werte in der Tabelle ist bei dieser Teilaufgabe zweckmäßigerweise die erste Gleichung anzuwenden:
+*Based on the given &nbsp; $x$–values in the table, it is convenient to use the first equation in this subtask:
 :$$p_\text{B, BPSK} = {\rm Q}\left ( 4 \right ) \hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4} }\hspace{0.05cm}.$$
-'''(2)'''&nbsp; Bei doppelter Bitdauer ist auch die Energie doppelt so groß:&nbsp; $E_{\rm B} = 4 · 10^{–6} \ \rm  V^2s$  ⇒  $E_{\rm B}/N_0 = 16$.
+'''(2)'''&nbsp; With twice the bit duration, the energy is also twice as large:&nbsp; $E_{\rm B} = 4 · 10^{–6} \ \rm  V^2s$  ⇒  $E_{\rm B}/N_0 = 16$.
-*Daraus folgt:
+*It follows that:
 :$$p_\text{B, BPSK} = {\rm Q}\left ( \sqrt{32} \right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{16}\right ) ={1}/{2}\cdot {\rm erfc}\left ( 4\right ) \hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$
-*Aus pragmatischen Gründen wurde hier die letzte Spalte der Tabelle benutzt.
+*For pragmatic reasons, the last column of the table was used here.
-'''(3)'''&nbsp; Setzt man die für die 4–QAM gegebenen Gleichungen ineinander ein, so kommt man zum gleichen Ergebnis wie bei der BPSK:
+'''(3)'''&nbsp; Substituting the equations given for the 4-QAM into each other, we get the same result as for the BPSK:
 :$$p_{\rm B, \hspace{0.05cm}4-QAM} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) \equiv p_\text{B, BPSK}.$$
-*Da sich auch die Energie pro Bit gegenüber der Teilaufgabe&nbsp; '''(2)'''&nbsp; nicht geändert hat, wird sich auch die gleiche Fehlerwahrscheinlichkeit einstellen:
+*Also, since the energy per bit has not changed from subtask&nbsp; '''(2)'''&nbsp;,  the same error probability will arise:
 :$$p_{\rm B, \hspace{0.05cm}4-QAM}= {\rm Q}\left ( \sqrt{32} \right ) = {1}/{2}\cdot {\rm erfc}\left ( 4\right ) \hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Richtig ist nur der <u>erste Lösungsvorschlag</u>:
+'''(4)'''&nbsp; Only<u>Answer 1</u> is correct:
-*Die Fehlerwahrscheinlichkeit ist natürlich in den beiden Zweigen gleich groß.&nbsp; Warum auch nicht?
+*Of course the error probability of error is the same in the two branches.  Why would it not be?
-*Das würde allerdings bei einem Phasenversatz zwischen Sender und Empfänger nicht mehr gelten.
+*This would no longer be true with a phase offset between the transmitter and receiver though.
-*Der Abstand der Nutzabtastwerte von der Schwelle ist hier allerdings&nbsp; $g_0$&nbsp; und damit um den Faktor&nbsp; $\sqrt{2}$&nbsp; kleiner als die Hüllkurve&nbsp; $s_0$&nbsp; der gesamten 4–QAM.
+*However, the distance of the useful samples from the threshold is&nbsp; $g_0$&nbsp; here and thus smaller than the envelope &nbsp; $s_0$&nbsp; of the entire 4-QAM by a factor of $\sqrt{2}$&nbsp;.
-*Betrachtet man den Inphase–Zweig (oder den Quadratur–Zweig) als eine eigenständige BPSK, so ist aber auch die Rauschleistung wegen der geringeren Symbolrate nur halb so groß wie bei der BPSK.&nbsp; Deshalb bleibt die Fehlerwahrscheinlichkeit gleich.
+*However, if the inphase branch (or the quadrature branch) is considered as a stand-alone BPSK, the noise power is also half that of BPSK because of the lower symbol rate.  Therefore, the error probability remains the same.
 {{ML-Fuß}}

@@ Line 41: / Line 41: @@
 <quiz display=simple>
-{Welche Fehlerwahrscheinlichkeit &nbsp;$p_\text{B, BPSK}$&nbsp; ergibt sich für&nbsp; ''Binary Phase Shift Keying''&nbsp; (BPSK) mit $T_{\rm B} = 1 \ \rm &micro; s$?
+{What error probability &nbsp;$p_\text{B, BPSK}$&nbsp; results for&nbsp; ''Binary Phase Shift Keying''&nbsp; (BPSK) when $T_{\rm B} = 1 \ \rm &micro; s$?
 |type="{}"}
 $p_\text{B, BPSK} \ = \ $ { 0.317 3% } $\ \rm 10^{-4}$
-{Welche Fehlerwahrscheinlichkeit &nbsp;$p_\text{B, BPSK}$&nbsp; ergibt sich für &nbsp; ''Binary Phase Shift Keying''&nbsp; (BPSK) mit $T_{\rm B} = 2 \ \rm &micro; s$?
+{What error probability &nbsp;$p_\text{B, BPSK}$&nbsp; results for&nbsp; ''Binary Phase Shift Keying''&nbsp; (BPSK) when $T_{\rm B} = 2 \ \rm &micro; s$?
 |type="{}"}
 $p_\text{B, BPSK} \ = \ $ { 0.771 3% } $\ \rm 10^{-8}$
-{Welche Fehlerwahrscheinlichkeit &nbsp;$p_\text{B, 4-QAM}$&nbsp; erhält man für die 4–QAM mit &nbsp;$E_{\rm B} = 4 · 10^{–6} \ \rm V^2s$?
+{What error probability &nbsp;$p_\text{B, 4-QAM}$&nbsp; is obtained for 4–QAM when &nbsp;$E_{\rm B} = 4 · 10^{–6} \ \rm V^2s$?
 |type="{}"}
 $p_\text{B, 4-QAM} \ = \ $ { 0.771 3% } $\ \rm 10^{-8}$
-{Was trifft zu, wenn man nur einen Zweig&nbsp; $\rm (I$&nbsp; oder&nbsp; $\rm Q)$&nbsp; der 4–QAM betrachtet?
+{Which statements apply if we consider only one branch &nbsp; $\rm (I$&nbsp; or&nbsp; $\rm Q)$&nbsp; of the 4–QAM?
 |type="[]"}
-+ Es ergibt sich das gleiche Ergebnis wie für die gesamte 4–QAM.
++ The same result is obtained as for the entire 4-QAM.
-- Der Abstand der Nutzabtastwerte ist wie bei der BPSK gleich &nbsp;$s_0$.
+- The spacing of the useful samples is equal to &nbsp;$s_0$ as in BPSK.
-- Es ergibt sich die gleiche Rauschleistung wie bei der BPSK.
+- The same result is obtained for the noise power as in BPSK.
 </quiz>

@@ Line 30: / Line 30: @@
 ''Hints:''
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation|Quadrature Amplitude Modulation]].
-*Reference is also made to the page&nbsp; [[Modulation_Methods/Lineare_digitale_Modulation#Fehlerwahrscheinlichkeiten_-_ein_kurzer_.C3.9Cberblick|Fehlerwahrscheinlichkeiten &ndash; ein kurzer Überblick]]&nbsp; in the previous chapter.
+*Reference is also made to the page&nbsp; [[Modulation_Methods/Linear_Digital_Modulation#Error_probabilities_-_a_brief_overview|Error probabilities &ndash; a brief overview]]&nbsp; in the previous chapter.
 * Always assume the following numerical values: &nbsp; $s_0 = 2\,{\rm V}, \hspace{0.05cm} N_0 = 0.25 \cdot 10^{-6}\,{\rm V^2/Hz}\hspace{0.05cm}.$
 *The bit time is &nbsp;$T_{\rm B} = 1 \ \rm &micro; s$&nbsp; (question 1) and &nbsp;$T_{\rm B} = 2 \ \rm &micro; s$&nbsp; (from question 2 onwards).

@@ Line 4: / Line 4: @@
 [[File:P_ID1721__Mod_Z_4_10.png|right|frame|Tabelle zweier unterschiedlicher Gaußscher Fehlerfunktionen]]
-Wir gehen von den folgenden Voraussetzungen aus:
+We now make the following assumptions:
-* binäre bipolare Amplitudenkoeffizienten &nbsp;$a_ν ∈ \{±1\}$,
+* binary bipolar amplitude coefficients &nbsp;$a_ν ∈ \{±1\}$,
-* rechteckförmiger Sendegrundimpuls mit Amplitude &nbsp;$s_0$&nbsp; und Bitdauer &nbsp;$T_{\rm B}$,
+* rectangular fundamental transmission pulse with amplitude &nbsp;$s_0$&nbsp; and bit time &nbsp;$T_{\rm B}$,
-* AWGN–Rauschen mit der Rauschleistungsdichte &nbsp;$N_0$,
+* AWGN noise with noise power density  &nbsp;$N_0$,
-* Empfänger gemäß dem Matched–Filter–Prinzip,
+* a receiver according to the matched-filter principle,
-* bestmögliche Demodulation und Detektion.
+* the best possible demodulation and detection.
-Wie schon mehrfach gezeigt wurde, kann man die Bitfehlerwahrscheinlichkeit der binären Phasenmodulation&nbsp; $\rm (BPSK)$&nbsp; bei diesen Randbedingungen mit den folgenden Gleichungen berechnen:
+As has been shown several times, the bit error probability of binary phase modulation &nbsp; $\rm (BPSK)$&nbsp; under these conditions can be calculated using the following equations:
 :$$ p_{\rm B, \hspace{0.05cm}BPSK} = {\rm Q}\left ({s_0}/{\sigma_d } \right ), \hspace{0.2cm} E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/{T_{\rm B} }$$
 :$$\Rightarrow \hspace{0.3cm} p_{\rm B, \hspace{0.05cm}BPSK} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$
-Die entsprechenden Gleichungen der&nbsp; $\rm 4–QAM$&nbsp;  lauten:
+The corresponding equations of&nbsp; $\rm 4–QAM$&nbsp;  are:
 :$$ p_{\rm B, \hspace{0.05cm}4-QAM} = {\rm Q}\left ( {g_0}/{\sigma_d } \right ), \hspace{0.2cm}g_{0} = {s_0}/{\sqrt{2}}, \hspace{0.2cm}E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/({2 \cdot T_{\rm B} }).$$
-Hierbei ist berücksichtigt, dass man – um die gleiche Sendeenergie pro Bit wie bei der BPSK zu erreichen – die Impulsamplitude &nbsp;$g_0$&nbsp; der Rechteckimpulse in den beiden Teilzweigen der 4–QAM um den Faktor &nbsp;$\sqrt{2}$&nbsp; herabsetzen muss.&nbsp; Die Hüllkurve ist dann bei beiden Systemen gleich &nbsp;$s_0$.
+Here it is taken into account that - in order to achieve the same transmission energy per bit as with BPSK - one must reduce the pulse amplitude&nbsp;$g_0$&nbsp; of the square-wave impulses in the two sub-branches of 4-QAM by a factor of &nbsp;$\sqrt{2}$&nbsp;.  The envelope is then equal to &nbsp;$s_0$ for both systems.
@@ Line 28: / Line 28: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Quadratur%E2%80%93Amplitudenmodulation|Quadratur&ndash;Amplitudenmodulation]].
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation|Quadrature Amplitude Modulation]].
-*Bezug genommen wird aber auch auf die Seite&nbsp; [[Modulation_Methods/Lineare_digitale_Modulation#Fehlerwahrscheinlichkeiten_-_ein_kurzer_.C3.9Cberblick|Fehlerwahrscheinlichkeiten &ndash; ein kurzer Überblick]]&nbsp; im vorherigen Kapitel.
+*Reference is also made to the page&nbsp; [[Modulation_Methods/Lineare_digitale_Modulation#Fehlerwahrscheinlichkeiten_-_ein_kurzer_.C3.9Cberblick|Fehlerwahrscheinlichkeiten &ndash; ein kurzer Überblick]]&nbsp; in the previous chapter.
-* Gehen Sie stets von den folgenden Zahlenwerten aus: &nbsp; $s_0 = 2\,{\rm V}, \hspace{0.05cm} N_0 = 0.25 \cdot 10^{-6}\,{\rm V^2/Hz}\hspace{0.05cm}.$
+* Always assume the following numerical values: &nbsp; $s_0 = 2\,{\rm V}, \hspace{0.05cm} N_0 = 0.25 \cdot 10^{-6}\,{\rm V^2/Hz}\hspace{0.05cm}.$
-*Die Bitdauer beträgt &nbsp;$T_{\rm B} = 1 \ \rm &micro; s$&nbsp; (Teilaufgabe 1) bzw. &nbsp;$T_{\rm B} = 2 \ \rm &micro; s$&nbsp; (ab Teilaufgabe 2).
+*The bit time is &nbsp;$T_{\rm B} = 1 \ \rm &micro; s$&nbsp; (question 1) and &nbsp;$T_{\rm B} = 2 \ \rm &micro; s$&nbsp; (from question 2 onwards).
-*In der Tabelle sind die beiden gebräuchlichen Gaußschen Fehlerfunktionen &nbsp;${\rm Q}(x)$&nbsp;  und &nbsp;$1/2 \cdot {\rm erfc}(x)$&nbsp;  angegeben.
+*In the table, the two common Gaussian error functions &nbsp;${\rm Q}(x)$&nbsp;  and &nbsp;$1/2 \cdot {\rm erfc}(x)$&nbsp; are given.
-*Energien sind in  &nbsp;$\rm V^2s$&nbsp;  anzugeben; sie beziehen sich somit auf den Bezugswiderstand &nbsp;$R = 1 \ \rm \Omega$.
+*Energies are to be given in &nbsp;$\rm V^2s$&nbsp; ; thus, they refer to the reference resistance &nbsp;$R = 1 \ \rm \Omega$.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
@@ Line 60: / Line 60: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Mit den vorgegebenen Werten erhält man für &nbsp;''Binary Phase Shift Keying''&nbsp; (BPSK):

← Older revision		Revision as of 13:41, 19 March 2022
Line 96:		Line 96:


−	[[Category:Modulation Methods: Exercises\|^4.3 ~~Quadratur–Amplitudenmodulation~~^]]	+	[[Category:Modulation Methods: Exercises\|^4.3 Quadrature Amplitude Modulation^]]

← Older revision	Revision as of 13:36, 19 March 2022
(No difference)