Aufgaben:Exercise 4.15: Optimal Signal Space Allocation - Revision history

Guenter at 15:06, 1 October 2022

2022-10-01T15:06:29Z

Hwang at 19:45, 1 September 2022

2022-09-01T19:45:00Z

Guenter at 16:03, 23 August 2022

2022-08-23T16:03:46Z

Guenter at 15:39, 23 August 2022

2022-08-23T15:39:57Z

Hwang at 20:48, 15 August 2022

2022-08-15T20:48:39Z

Hwang at 20:03, 15 August 2022

2022-08-15T20:03:45Z

Guenter at 11:59, 13 July 2022

2022-07-13T11:59:38Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.15: Optimale Signalraumbelegung to Aufgaben:Exercise 4.15: Optimal Signal Space Allocation

2022-07-13T11:53:41Z

Guenter moved page Aufgabe 4.15: Optimale Signalraumbelegung to Exercise 4.15: Optimal Signal Space Allocation

Javier: Text replacement - "”" to """

2021-05-28T14:36:04Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:31:27Z

Text replacement - "„" to """

@@ Line 83: / Line 83: @@
 <br clear=all>
 '''(2)'''&nbsp; <u>All statements are true</u>:
-[[File:P_ID2074__Dig_A_4_15c.png|right|frame|To calculate the minimum distance]]
+[[File:P_ID2074__Dig_A_4_15c.png|right|frame|To calculate minimum distance]]
 *In the drawn example on the front page with&nbsp; $R = R_{\rm max}$,&nbsp; the distance between two neighboring blue points is exactly the same as the distance between a red (outer) and a blue (inner) point.

@@ Line 69: / Line 69: @@
 {{ML-Kopf}}
 [[File:P_ID2073__Dig_A_4_15a.png|right|frame|Special cases of&nbsp; "8–QAM"]]
-'''(1)'''&nbsp; Because of&nbsp; $M = 8$ &nbsp; &#8658; &nbsp; $b = 3$,&nbsp; the averae signal energy per bit is&nbsp; $E_{\rm B} = E_{\rm S}/3$,&nbsp; where the average signal energy per symbol&nbsp; $(E_{\rm S})$&nbsp; is to be calculated as the mean square distance of the signal space points from the origin.&nbsp; With&nbsp; $r = 1$&nbsp; one obtains:
+'''(1)'''&nbsp; Because of&nbsp; $M = 8$ &nbsp; &#8658; &nbsp; $b = 3$,&nbsp; the average signal energy per bit is&nbsp; $E_{\rm B} = E_{\rm S}/3$,&nbsp; where the average signal energy per symbol&nbsp; $(E_{\rm S})$&nbsp; is to be calculated as the mean square distance of the signal space points from the origin.&nbsp; With&nbsp; $r = 1$&nbsp; one obtains:
 :$$E_{\rm S} = {1}/{8  } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 +  R^2})/{2  }$$
 :$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 +  R^2})/{6}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2069__Dig_A_4_15.png|right|frame|Considered 8–QAM]]
+[[File:P_ID2069__Dig_A_4_15.png|right|frame|Considered&nbsp;  "8–QAM"]]
 A signal space constellation with&nbsp; $M = 8$&nbsp; signal space points is considered here:
 * Four points lie on a circle with radius&nbsp; $r = 1$.
@@ Line 68: / Line 68: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Because of $M = 8$ &nbsp;&#8658;&nbsp; $b = 3$, the mean signal energy per bit is $E_{\rm B} = E_{\rm S}/3$, where the mean signal energy per symbol ($E_{\rm S}$) is to be calculated as the mean square distance of the signal space points from the origin. With $r = 1$ one obtains:
+[[File:P_ID2073__Dig_A_4_15a.png|right|frame|Special cases of&nbsp; "8–QAM"]]
-[[File:P_ID2073__Dig_A_4_15a.png|right|frame|Special cases of 8–QAM]]
+'''(1)'''&nbsp; Because of&nbsp; $M = 8$ &nbsp; &#8658; &nbsp; $b = 3$,&nbsp; the averae signal energy per bit is&nbsp; $E_{\rm B} = E_{\rm S}/3$,&nbsp; where the average signal energy per symbol&nbsp; $(E_{\rm S})$&nbsp; is to be calculated as the mean square distance of the signal space points from the origin.&nbsp; With&nbsp; $r = 1$&nbsp; one obtains:
-:$$E_{\rm S} = {1}/{8  } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 +  R^2})/{2  }
+:$$E_{\rm S} = {1}/{8  } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 +  R^2})/{2  }$$
- \hspace{0.3cm}\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 +  R^2})/{6}
+:$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 +  R^2})/{6}
    \hspace{0.05cm}.$$
 In particular:
-* For $R = 1$, there is an 8&ndash;PSK &nbsp; &rArr; &nbsp; $E_{\rm S} = 1$ and $E_{\rm B} \ \underline {= 0.333}$ (see left graph).
+* For&nbsp; $R = 1$,&nbsp; there is an&nbsp; "8&ndash;PSK" &nbsp; &rArr; &nbsp; $E_{\rm S} = 1$&nbsp; and&nbsp; $E_{\rm B} \ \underline {= 0.333}$&nbsp; (see left graph).
 * The right graph is valid for $R = \sqrt{2}$. In this case, $E_{\rm B} \ \underline {= 0.5}$.
 Note that these energies actually still have to be multiplied by the normalization energy $E$.
-'''(3)'''&nbsp; The graphic illustrates the geometric calculation. With "Pythagoras" one obtains:
+'''(3)'''&nbsp; The graphic illustrates the geometric calculation.&nbsp; With&nbsp; "Pythagoras"&nbsp; one obtains:
-:$$d_{\rm min}^2 =(R/\sqrt{2})^2 +  (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2  \hspace{0.3cm}
+:$$d_{\rm min}^2 =(R/\sqrt{2})^2 +  (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2$$
-\Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}
+:$$  \Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}
    \hspace{0.05cm}.$$
-In particular, for $R = 1$ (8&ndash;PSK):
+*In particular,&nbsp; for $R = 1$&nbsp; ("8&ndash;PSK"):
 :$$d_{\rm min} = \sqrt{ 2 - \sqrt{2} }  \hspace{0.1cm} \underline{= 0.765} \hspace{0.1cm} (= 2 \cdot \sin (22.5^{\circ}) )
    \hspace{0.05cm}.$$
-In contrast, for $\underline {R = \sqrt{2}}$ corresponding to the right graph for subtask '''(1)''', the minimum distance is $d_{\rm min} \ \underline {= 1}$.
+*In contrast,&nbsp; for&nbsp; $\underline {R = \sqrt{2}}$&nbsp; corresponding to the right graph for subtask&nbsp; '''(1)''',&nbsp; the minimum distance is&nbsp; $d_{\rm min} \ \underline {= 1}$.
 '''(4)'''&nbsp; Using the results of '''(1)''' and '''(3)''', we obtain in general or for $R = 1$ (8&ndash;PSK):
 :$$\eta = \frac{ d_{\rm min}^2}{ 4 \cdot E_{\rm B}} = \frac{ 1 - \sqrt{2} \cdot R + R^2}{ 4 \cdot (1 +  R^2)/6}
   = \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 +  R^2}\hspace{0.3cm}
@@ Line 108: / Line 113: @@
 '''(5)'''&nbsp; For $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$, the following value is obtained:
 :$$\eta =  \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 +  R^2} = 3/2 \cdot \left [ 1 - \frac{  \sqrt{2} \cdot R }{ 1 +  R^2}\right ]\hspace{0.05cm},$$
 :$$\sqrt{2} \cdot R = \sqrt{3}- 1\hspace{0.05cm},\hspace{0.2cm} 1 +  R^2 = 3 - \sqrt{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
   \eta =  3/2 \cdot \left [ 1 - \frac{  \sqrt{3}- 1 }{ 3 - \sqrt{3}}\right ]\hspace{0.1cm} \underline{\approx  0.634}\hspace{0.05cm}.$$
-For $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$ exactly the same value results.
+*For&nbsp; $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$&nbsp; exactly the same value results.
-The previously considered special cases $R = 1$ (8&ndash;PSK, left graph in the first subtask) and $R = \sqrt{2}$ (right graph) have a noticeably smaller $\eta$ with $\eta = 0.439$ and $\eta = 0.5$, respectively (compared to $\eta = 0.634$).
+#The&nbsp; (always desired)&nbsp; maximum of the power efficiency&nbsp; $\eta$&nbsp; results e.g. for&nbsp; $R = R_{\rm max}$ &ndash; i.e. for the signal space constellation in the information section.
 {{ML-Fuß}}

@@ Line 5: / Line 5: @@
 A signal space constellation with&nbsp; $M = 8$&nbsp; signal space points is considered here:
 * Four points lie on a circle with radius&nbsp; $r = 1$.
 * Four further points lie offset by&nbsp; $45^\circ$&nbsp; on a second circle with radius&nbsp; $R$, where the following shall hold:
 :$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}=  \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518
   \hspace{0.05cm},\hspace{0.2cm}
 R_{\rm max}=  \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$
-Let the two axes (basis functions) be normalized respectively and denoted&nbsp; $I$&nbsp; and&nbsp; $Q$&nbsp; for simplicity. For further simplification,&nbsp; $E = 1$&nbsp; can be set.
+Let the two axes&nbsp; ("basis functions")&nbsp; be normalized respectively and denoted&nbsp; $I$&nbsp; and&nbsp; $Q$&nbsp; for simplicity.&nbsp; For further simplification,&nbsp; $E = 1$&nbsp; can be set.
-In the question section, we speak of blue and red points. According to the diagram, the blue points lie on the circle with radius&nbsp; $r = 1$, the red points on the circle with radius&nbsp; $R$. The case&nbsp; $R = R_{\rm max}$ is drawn.
+In the question section,&nbsp; we speak of&nbsp; "blue"&nbsp; and&nbsp; "red"&nbsp; points.&nbsp; According to the diagram,&nbsp; the blue points lie on the circle with radius&nbsp; $r = 1$,&nbsp; the red points on the circle with radius&nbsp; $R$.&nbsp; The case&nbsp; $R = R_{\rm max}$ is drawn.
 The system parameter&nbsp; $R$&nbsp; is to be determined in this exercise in such a way that the quotient
 :$$\eta = \frac{ (d_{\rm min}/2)^2}{ E_{\rm B}} $$
-becomes maximum. $\eta$&nbsp; is a measure for the quality of a modulation alphabet at given transmission energy per bit ("Power Efficiency"). It is calculated from
+becomes maximum.&nbsp; $\eta$&nbsp; is a measure for the quality of a modulation alphabet at given transmission energy per bit&nbsp; ("power efficiency").&nbsp; It is calculated from
-* the minimum distance&nbsp; $d_{\rm min}$, and
+* the minimum distance&nbsp; $d_{\rm min}$,&nbsp; and
-''Notes:''
-* The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
+Notes:
-* Reference is made in particular to the sections&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"Quadrature amplitude modulation"]]&nbsp; and&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#M.E2.80.93level_amplitude_shift_keying_.28M.E2.80.93ASK.29|"Multilevel phase modulation"]].
+* The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
@@ Line 37: / Line 40: @@
 ===Questions===
 <quiz display=simple>
-{Calculate the average energy&nbsp; $E_{\rm B}$&nbsp; per bit depending on&nbsp; $R$, in particular for&nbsp; $R = 1$&nbsp; and&nbsp; $R = \sqrt{2}$.
+{Calculate the average energy&nbsp; $E_{\rm B}$&nbsp; per bit depending on&nbsp; $R$,&nbsp; in particular for&nbsp; $R = 1$&nbsp; and&nbsp; $R = \sqrt{2}$.
 |type="{}"}
 $R = 1 \text{:} \hspace{0.55cm} E_{\rm B}\  = \ $ { 0.333 3% }
@@ Line 46: / Line 49: @@
 + For&nbsp; $R < R_{\rm min}$,&nbsp; the minimum distance occurs between two red points.
 + For&nbsp; $R > R_{\rm max}$,&nbsp; the minimum distance occurs between two blue points.
-+ For&nbsp; $R_{\rm min} &#8804; R &#8804; R_{\rm max}$,&nbsp; the minimum distance occurs between "red" and "blue".
++ For&nbsp; $R_{\rm min} &#8804; R &#8804; R_{\rm max}$,&nbsp; the minimum distance occurs between&nbsp; "red"&nbsp; and&nbsp; "blue".
-{Calculate the minimum distance depending on&nbsp; $R$, in particular for
+{Calculate the minimum distance depending on&nbsp; $R$,&nbsp; in particular for
 |type="{}"}
 $R = 1 \text{:} \hspace{0.55cm} d_{\rm min}\ = \ $ { 0.765 3% }
 $R = \sqrt{2} \text{:} \hspace{0.2cm} d_{\rm min}\ = \ $ { 1 3% }
-{Give the power efficiency&nbsp; $\eta$&nbsp; in general terms. What&nbsp; $\eta$&nbsp; results for&nbsp; $R = 1$?
+{Give the power efficiency&nbsp; $\eta$&nbsp; in general terms.&nbsp; What&nbsp; $\eta$&nbsp; results for&nbsp; $R = 1$?
 |type="{}"}
 $\eta\ = \ $ { 0.439 3% }
-{What power efficiency values result for&nbsp; $R = R_{\rm min}$&nbsp; and&nbsp; $R = R_{\rm max}$? Interpretation.
+{What power efficiency values result for&nbsp; $R = R_{\rm min}$&nbsp; and&nbsp; $R = R_{\rm max}$?&nbsp; Interpretation.
 |type="{}"}
 $R = R_{\rm min} \text{:} \hspace{0.35cm} \eta\ = \ ${ 0.634 3% }

@@ Line 37: / Line 37: @@
 ===Questions===
 <quiz display=simple>
-{Berechnen Sie die mittlere Energie&nbsp; $E_{\rm B}$&nbsp; pro Bit abhängig von&nbsp; $R$, insbesondere für&nbsp; $R = 1$&nbsp; und&nbsp; $R = \sqrt{2}$.
+{Calculate the average energy&nbsp; $E_{\rm B}$&nbsp; per bit depending on&nbsp; $R$, in particular for&nbsp; $R = 1$&nbsp; and&nbsp; $R = \sqrt{2}$.
 |type="{}"}
 $R = 1 \text{:} \hspace{0.55cm} E_{\rm B}\  = \ $ { 0.333 3% }
 $R = \sqrt{2} \text{:} \hspace{0.2cm} E_{\rm B}\ = \ $ { 0.5 3% }
-{Welche Aussagen gelten für den minimalen Abstand zweier Signalraumpunkte?
+{Which statements are true for the minimum distance between two signal space points?
 |type="[]"}
-+ Für&nbsp; $R < R_{\rm min}$&nbsp; tritt die minimale Distanz zwischen zwei roten Punkten auf.
++ For&nbsp; $R < R_{\rm min}$,&nbsp; the minimum distance occurs between two red points.
-+ Für&nbsp; $R > R_{\rm max}$&nbsp; tritt die minimale Distanz zwischen zwei blauen Punkten auf.
++ For&nbsp; $R > R_{\rm max}$,&nbsp; the minimum distance occurs between two blue points.
-+ Für&nbsp; $R_{\rm min} &#8804; R &#8804; R_{\rm max}$&nbsp; tritt die minimale Distanz zwischen "Rot" und "Blau" auf.
++ For&nbsp; $R_{\rm min} &#8804; R &#8804; R_{\rm max}$,&nbsp; the minimum distance occurs between "red" and "blue".
-{Berechnen Sie die minimale Distanz abhängig von&nbsp; $R$, insbesondere für
+{Calculate the minimum distance depending on&nbsp; $R$, in particular for
 |type="{}"}
 $R = 1 \text{:} \hspace{0.55cm} d_{\rm min}\ = \ $ { 0.765 3% }
 $R = \sqrt{2} \text{:} \hspace{0.2cm} d_{\rm min}\ = \ $ { 1 3% }
-{Geben Sie die Leistungseffizienz&nbsp; $\eta$&nbsp; allgemein an. Welches&nbsp; $\eta$&nbsp; ergibt sich für&nbsp; $R = 1$?
+{Give the power efficiency&nbsp; $\eta$&nbsp; in general terms. What&nbsp; $\eta$&nbsp; results for&nbsp; $R = 1$?
 |type="{}"}
 $\eta\ = \ $ { 0.439 3% }
-{Welche Leistungseffizienzwerte ergeben sich für&nbsp; $R = R_{\rm min}$&nbsp; und&nbsp; $R = R_{\rm max}$? Interpretation.
+{What power efficiency values result for&nbsp; $R = R_{\rm min}$&nbsp; and&nbsp; $R = R_{\rm max}$? Interpretation.
 |type="{}"}
 $R = R_{\rm min} \text{:} \hspace{0.35cm} \eta\ = \ ${ 0.634 3% }
@@ Line 63: / Line 63: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Wegen $M = 8$ &nbsp;&#8658;&nbsp; $b = 3$ gilt für die mittlere Signalenergie pro Bit $E_{\rm B} = E_{\rm S}/3$, wobei die mittlere Signalenergie pro Symbol ($E_{\rm S}$) als der mittlere quadratische Abstand der Signalraumpunkte vom Ursprung zu berechnen ist. Mit $r = 1$ erhält man:
+'''(1)'''&nbsp; Because of $M = 8$ &nbsp;&#8658;&nbsp; $b = 3$, the mean signal energy per bit is $E_{\rm B} = E_{\rm S}/3$, where the mean signal energy per symbol ($E_{\rm S}$) is to be calculated as the mean square distance of the signal space points from the origin. With $r = 1$ one obtains:
-[[File:P_ID2073__Dig_A_4_15a.png|right|frame|Sonderfälle der 8–QAM]]
+[[File:P_ID2073__Dig_A_4_15a.png|right|frame|Special cases of 8–QAM]]
 :$$E_{\rm S} = {1}/{8  } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 +  R^2})/{2  }
   \hspace{0.3cm}\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 +  R^2})/{6}
    \hspace{0.05cm}.$$
-Insbesondere gilt:
+In particular:
-* Für $R = 1$ ergibt sich eine 8&ndash;PSK &nbsp; &rArr; &nbsp; $E_{\rm S} = 1$ und $E_{\rm B} \ \underline {= 0.333}$ (siehe linke Grafik).
+* For $R = 1$, there is an 8&ndash;PSK &nbsp; &rArr; &nbsp; $E_{\rm S} = 1$ and $E_{\rm B} \ \underline {= 0.333}$ (see left graph).
-* Die rechte Grafik gilt für $R = \sqrt{2}$. In diesem Fall ist $E_{\rm B} \ \underline {= 0.5}$.
+* The right graph is valid for $R = \sqrt{2}$. In this case, $E_{\rm B} \ \underline {= 0.5}$.
-Anzumerken ist, dass diese Energien eigentlich noch mit der Normierungsenergie $E$ zu multiplizieren sind.
+Note that these energies actually still have to be multiplied by the normalization energy $E$.
-'''(2)'''&nbsp; <u>Alle Aussagen treffen zu</u>:
+'''(2)'''&nbsp; <u>All statements are true</u>:
-*Im gezeichneten Beispiel auf dem Angabenblatt mit $R = R_{\rm max}$ ist der Abstand zwischen zwei benachbarten blauen Punkten genau so groß wie der Abstand zwischen einem roten (äußeren) und einem blauen (inneren) Punkt.
+*In the drawn example on the information section with $R = R_{\rm max}$, the distance between two neighboring blue points is exactly the same as the distance between a red (outer) and a blue (inner) point.
-[[File:P_ID2074__Dig_A_4_15c.png|right|frame|Zur Berechnung der minimalen Distanz]]
+[[File:P_ID2074__Dig_A_4_15c.png|right|frame|To calculate the minimum distance]]
-*Für $R > R_{\rm max}$ ist der Abstand zwischen zwei blauen Punkten am geringsten.
+*For $R > R_{\rm max}$, the distance between two blue points is the smallest.
-*Für $R < R_{\rm min}$ tritt der minimale Abstand zwischen zwei roten Punkten auf.
+*For $R < R_{\rm min}$, the minimum distance occurs between two red points.
-'''(3)'''&nbsp; Die Grafik verdeutlicht die geometrische Berechnung. Mit "Pythagoras" erhält man:
+'''(3)'''&nbsp; The graphic illustrates the geometric calculation. With "Pythagoras" one obtains:
 :$$d_{\rm min}^2 =(R/\sqrt{2})^2 +  (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2  \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}
    \hspace{0.05cm}.$$
-Insbesondere gilt für $R = 1$ (8&ndash;PSK):
+In particular, for $R = 1$ (8&ndash;PSK):
 :$$d_{\rm min} = \sqrt{ 2 - \sqrt{2} }  \hspace{0.1cm} \underline{= 0.765} \hspace{0.1cm} (= 2 \cdot \sin (22.5^{\circ}) )
    \hspace{0.05cm}.$$
-Dagegen ist für $\underline {R = \sqrt{2}}$ entsprechend der rechten Grafik zur Teilaufgabe '''(1)''' die minimale Distanz $d_{\rm min} \ \underline {= 1}$.
+In contrast, for $\underline {R = \sqrt{2}}$ corresponding to the right graph for subtask '''(1)''', the minimum distance is $d_{\rm min} \ \underline {= 1}$.
-'''(4)'''&nbsp; Mit den Ergebnissen von '''(1)''' und '''(3)''' erhält man allgemein bzw. für $R = 1$ (8&ndash;PSK):
+'''(4)'''&nbsp; Using the results of '''(1)''' and '''(3)''', we obtain in general or for $R = 1$ (8&ndash;PSK):
 :$$\eta = \frac{ d_{\rm min}^2}{ 4 \cdot E_{\rm B}} = \frac{ 1 - \sqrt{2} \cdot R + R^2}{ 4 \cdot (1 +  R^2)/6}
   = \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 +  R^2}\hspace{0.3cm}
@@ Line 105: / Line 105: @@
-'''(5)'''&nbsp; Für $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$ ergibt sich folgender Wert:
+'''(5)'''&nbsp; For $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$, the following value is obtained:
 :$$\eta =  \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 +  R^2} = 3/2 \cdot \left [ 1 - \frac{  \sqrt{2} \cdot R }{ 1 +  R^2}\right ]\hspace{0.05cm},$$
 :$$\sqrt{2} \cdot R = \sqrt{3}- 1\hspace{0.05cm},\hspace{0.2cm} 1 +  R^2 = 3 - \sqrt{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
   \eta =  3/2 \cdot \left [ 1 - \frac{  \sqrt{3}- 1 }{ 3 - \sqrt{3}}\right ]\hspace{0.1cm} \underline{\approx  0.634}\hspace{0.05cm}.$$
-Für $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$ ergibt sich genau der gleiche Wert.
+For $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$ exactly the same value results.
-*Das (stets gewünschte) Maximum der Leistungseffizienz $\eta$ ergibt sich beispielsweise für $R = R_{\rm max}$ &ndash; also für die Signalraumkonstellation auf dem Angabenblatt.
+*The (always desired) maximum of the power efficiency $\eta$ results, for example, for $R = R_{\rm max}$ &ndash; i.e. for the signal space constellation in the information section.
-*In diesem Fall sind alle Dreiecke aus zwei benachbarten blauen Punkten und dem dazwischenliegenden roten Punkt gleichseitig.
+*In this case all triangles of two neighboring blue points and the red point in between are equilateral.
-*Auch für $R = R_{\rm min}$ ergeben sich gleichseitige Dreiecke, jetzt aber jeweils gebildet durch zwei rotee und einen blauen Punkt.
+*Also for $R = R_{\rm min}$ there are equilateral triangles, but now each formed by two red and one blue point.
-*In diesem Fall ist zwar die Kantenlänge $d_{\rm min}$ deutlich kleiner, aber gleichzeitig ergibt sich auch ein kleineres $E_{\rm B}$, so dass die Leistungseffizienz $\eta$ den gleichen Wert besitzt.
+*In this case the edge length $d_{\rm min}$ is clearly smaller, but at the same time a smaller $E_{\rm B}$ results, so that the power efficiency $\eta$ has the same value.
-Die vorher betrachteten Sonderfälle $R = 1$ (8&ndash;PSK, linke Grafik bei der ersten Teilaufgabe) und $R = \sqrt{2}$ (rechte Grafik) weisen mit $\eta = 0.439$ bzw. $\eta = 0.5$ (gegenüber $\eta = 0.634$) ein merklich kleineres $\eta$ auf.
+The previously considered special cases $R = 1$ (8&ndash;PSK, left graph in the first subtask) and $R = \sqrt{2}$ (right graph) have a noticeably smaller $\eta$ with $\eta = 0.439$ and $\eta = 0.5$, respectively (compared to $\eta = 0.634$).
 {{ML-Fuß}}

← Older revision		Revision as of 11:59, 13 July 2022
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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 11:53, 13 July 2022
(No difference)

@@ Line 17: / Line 17: @@
 :$$\eta = \frac{ (d_{\rm min}/2)^2}{ E_{\rm B}} $$
-maximal wird. $\eta$&nbsp; ist ein Maß für die Güte eines Modulationsalphabets bei gegebener Sendeenergie pro Bit ("Power Efficiency&rdquo;). Es berechnet sich aus
+maximal wird. $\eta$&nbsp; ist ein Maß für die Güte eines Modulationsalphabets bei gegebener Sendeenergie pro Bit ("Power Efficiency"). Es berechnet sich aus
 * der minimalen Distanz&nbsp; $d_{\rm min}$, und
 * der Bitenergie&nbsp; $E_{\rm B}$.
@@ Line 46: / Line 46: @@
 + Für&nbsp; $R < R_{\rm min}$&nbsp; tritt die minimale Distanz zwischen zwei roten Punkten auf.
 + Für&nbsp; $R > R_{\rm max}$&nbsp; tritt die minimale Distanz zwischen zwei blauen Punkten auf.
-+ Für&nbsp; $R_{\rm min} &#8804; R &#8804; R_{\rm max}$&nbsp; tritt die minimale Distanz zwischen "Rot&rdquo; und "Blau&rdquo; auf.
++ Für&nbsp; $R_{\rm min} &#8804; R &#8804; R_{\rm max}$&nbsp; tritt die minimale Distanz zwischen "Rot" und "Blau" auf.
 {Berechnen Sie die minimale Distanz abhängig von&nbsp; $R$, insbesondere für
@@ Line 86: / Line 86: @@
-'''(3)'''&nbsp; Die Grafik verdeutlicht die geometrische Berechnung. Mit "Pythagoras&rdquo; erhält man:
+'''(3)'''&nbsp; Die Grafik verdeutlicht die geometrische Berechnung. Mit "Pythagoras" erhält man:
 :$$d_{\rm min}^2 =(R/\sqrt{2})^2 +  (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2  \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}