Aufgaben:Exercise 4.06Z: Signal Space Constellations - Revision history

Guenter at 16:18, 27 July 2022

2022-07-27T16:18:21Z

Guenter at 15:57, 27 July 2022

2022-07-27T15:57:25Z

Guenter at 15:13, 8 July 2022

2022-07-08T15:13:36Z

Hwang at 13:55, 5 July 2022

2022-07-05T13:55:07Z

Hwang at 13:39, 5 July 2022

2022-07-05T13:39:10Z

Show changes

Hwang at 13:04, 5 July 2022

2022-07-05T13:04:15Z

Guenter at 10:08, 23 June 2022

2022-06-23T10:08:59Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.06Z: Signalraumkonstellationen to Aufgaben:Exercise 4.06Z: Signal Space Constellations

2022-06-23T09:57:53Z

Guenter moved page Aufgabe 4.06Z: Signalraumkonstellationen to Exercise 4.06Z: Signal Space Constellations

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

2021-03-23T12:49:05Z

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

Guenter at 15:35, 13 March 2019

2019-03-13T15:35:54Z

@@ Line 79: / Line 79: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The <u>first three prerequisites</u> must be met in any case:
+'''(1)'''&nbsp; The&nbsp; <u>first three prerequisites</u>&nbsp; must be met in any case:
 *The equation then applies independently of the occurrence probabilities.
-*In the case of ${\rm Pr}(\boldsymbol{s} = \boldsymbol{s}_0) &ne; {\rm Pr}(\boldsymbol{s} = \boldsymbol{s}_1)$, a lower error probability can be achieved by shifting the decision threshold.
+*In the case of&nbsp; ${\rm Pr}(\boldsymbol{s} = \boldsymbol{s}_0) &ne; {\rm Pr}(\boldsymbol{s} = \boldsymbol{s}_1)$,&nbsp; a lower error probability can be achieved by shifting the decision threshold.
-'''(2)'''&nbsp; The noise rms value $\sigma_n$ and thus also the signal energy $E = \sigma_n^2$ are the same for all three considered variants. The same applies to the distance of the signal space points. For variant &nbsp;$\rm A$,&nbsp; for example, the following applies:
+'''(2)'''&nbsp; The noise rms value&nbsp; $\sigma_n$&nbsp; and thus also the signal energy&nbsp; $E = \sigma_n^2$&nbsp; are the same for all three considered variants.&nbsp;
 :$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} = \sqrt{ E \cdot (4-1)^2 + E \cdot (1-5)^2} = 5 \cdot \sqrt{E}\hspace{0.05cm}.$$
-Due to the shifting of the coordinate system, the distance between $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$ does not change (variant &nbsp;$\rm B$), and the same distance results in variant &nbsp;$\rm C$&nbsp; (after rotation).
+*Due to the shifting of the coordinate system,&nbsp; the distance between&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$&nbsp; does not change (variant &nbsp;$\rm B$),&nbsp; the same distance results in variant &nbsp;$\rm C$&nbsp; (after rotation).
-<u>Solution 4</u> is correct:
+*<u>Solution 4</u> is correct:
-*By rotating the coordinate system, one can always get by with a basis function $(N = 1)$ for a binary system $(M = 2)$.
+#By rotating the coordinate system,&nbsp; one can always get by with one basis function&nbsp; $(N = 1)$&nbsp; for a binary system&nbsp; $(M = 2)$.
-*Since the two-dimensional noise is circularly symmetric &nbsp; &#8658; &nbsp; equal standard deviation $\sigma_n$ in all directions, the noise term can also be described one-dimensionally as in the chapter [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|"Error Probability for Baseband Transmission"]].
+#Since the two-dimensional noise is circularly symmetric &nbsp; &#8658; &nbsp; equal standard deviation&nbsp; $\sigma_n$&nbsp; in all directions,&nbsp; <br>the noise term can also be described one-dimensionally as in the chapter&nbsp; [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|"Error Probability for Baseband Transmission"]].
-'''(3)'''&nbsp; For all variants considered here, i.e., also for variant &nbsp;$\rm A$, the following holds:
+'''(3)'''&nbsp; For all variants considered here,&nbsp; i.e.,&nbsp; also for variant &nbsp;$\rm A$,&nbsp; the following holds:
 :$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )=   {\rm Q} \left ( \frac{5/2 \cdot \sqrt{E}}{\sigma_n} \right )
 =   {\rm Q}(2.5)\hspace{0.05cm}.$$
-With the given approximation we obtain
+*With the given approximation we obtain
 :$$p_{\rm S}  = \frac{1}{\sqrt{2\pi} \cdot 2.5} \cdot {\rm e}^{-2.5^2/2} \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.7 \%}\hspace{0.05cm}.$$
@@ Line 111: / Line 112: @@
   \hspace{0.05cm}.$$
-Substituting this result into the equation found in (3), we obtain with $\sigma_n^2 = N_0/2$:
+*Substituting this result into the equation found in&nbsp; '''(3)''',&nbsp; we obtain with&nbsp; $\sigma_n^2 = N_0/2$:
 :$$p_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}   {\rm Q} \left ( \frac{2.5 \cdot \sqrt{E}}{\sigma_n} \right )=   {\rm Q} \left ( \frac{ \sqrt{E_{\rm S}}}{\sigma_n} \right )
 =   {\rm Q} \left ( \frac{ \sqrt{2 \cdot E_{\rm S}}}{N_0} \right ) ={\rm Q} \left ( \sqrt{\frac{ 2 \cdot 6.25 \cdot 10^{-6}\,{\rm Ws}}{2 \cdot 10^{-6}\,{\rm W/Hz}}} \right )
@@ Line 117: / Line 118: @@
-'''(5)'''&nbsp; Rotating the coordinate system does not change the energy ratios. Therefore, $p_{\rm S} \ \underline {\approx 0.7\%}$ is obtained again.
+'''(5)'''&nbsp; Rotating the coordinate system does not change the energy ratios.
@@ Line 125: / Line 127: @@
 \hspace{0.05cm}. $$
-The distance from the threshold, which should be midway between $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$ for equally probable symbols, is $d/2 = 2.5 \cdot E^{\rm 1/2}$, as in the other variants. Thus, with $\sigma_n^2 = N_0/2$, we obtain the governing equation:
+*The distance from the threshold,&nbsp; which should be midway between&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$&nbsp; for equally probable symbols,&nbsp; is&nbsp; $d/2 = 2.5 \cdot E^{\rm 1/2}$,&nbsp; as in the other variants.
 :$$p_{\rm S}  = {\rm Q} \left ( \frac{ 2.5 \cdot \sqrt{E}}{\sqrt{N_0/2}} \right )
 ={\rm Q}(2.5)\approx 0.7 \cdot 10^{-2} \hspace{0.3cm}
@@ Line 132: / Line 136: @@
 :$$\Rightarrow \hspace{0.3cm} {E_{\rm S}} = 0.5 \cdot {21.5 \cdot N_0} \hspace{0.1cm} \hspace{0.15cm}\underline { = 21.5 \cdot 10^{-6}\,{\rm Ws}}\hspace{0.05cm}.$$
-This means: For variant &nbsp;$\rm A$,&nbsp; compared to the other two symbols, a mean symbol energy $E_{\rm S}$ larger by a factor of $3.44$is required to achieve the same error probability $p_{\rm S} = 0.7%$.
+*This means:&nbsp; For variant &nbsp;$\rm A$,&nbsp; compared to the other two variants,&nbsp; a mean symbol energy&nbsp; $E_{\rm S}$&nbsp; larger by a factor of&nbsp; $3.44$&nbsp; is required to achieve the same&nbsp; $p_{\rm S} = 0.7\%$.&nbsp; Because:
-*That means: This signal space constellation is very unfavorable. It results in a very large $E_{\rm S}$ without increasing the distance $d$ at the same time.
+#This signal space constellation is very unfavorable.&nbsp; It results in a very large&nbsp; $E_{\rm S}$&nbsp; without increasing the distance&nbsp; $d$&nbsp; at the same time.
-*With $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$, on the other hand, $p_{\rm S} = {\rm Q}(2.5/3.44^{\rm 1/2}) \approx {\rm Q}(1.35) \approx 9\%$ would result.
+#With $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$,&nbsp; on the other hand,&nbsp; $p_{\rm S} = {\rm Q}(2.5/3.44^{\rm 1/2}) \approx {\rm Q}(1.35) \approx 9\%$&nbsp; would result.
-*That means: &nbsp; The error probability would be larger by more than one power of ten.
+#That means: &nbsp; The error probability would be larger by more than one power of ten.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:EN_Dig_Z_4_6.png|right|frame|Three signal space constellations]]
-The (mean) error probability of an optimal binary system is:
+The&nbsp; (mean)&nbsp; symbol error probability of an optimal binary system is:
 :$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )\hspace{0.05cm}.$$
 It should be noted here:
-* ${\rm Q}(x)$&nbsp; denotes the complementary Gaussian error function (definition and approximation):
+* ${\rm Q}(x)$&nbsp; denotes the complementary Gaussian error function&nbsp; (definition and approximation):
 :$${\rm Q}(x) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}   \frac{1}{\sqrt{2\pi}}  \int_{x}^{\infty} {\rm e}^{-u^2/2} \,{\rm d} u
 \approx  \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}
 \hspace{0.05cm}.$$
-* $d$&nbsp; specifies the distance between the two transmitted signal points&nbsp; $s_0$&nbsp; and &nbsp;$s_1$&nbsp; in vector space:
+* The parameter&nbsp; $d$&nbsp; specifies the distance between the two transmitted signal points&nbsp; $s_0$&nbsp; and &nbsp;$s_1$&nbsp; in vector space:
 :$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} \hspace{0.05cm}.$$
-* $\sigma_n^2$&nbsp; is the variance of the AWGN noise after the detector, which, for example, can be implemented as a matched filter. <br>It is assumed that&nbsp; $\sigma_n^2 = N_0/2$.
+* $\sigma_n^2$&nbsp; is the variance of the AWGN noise after the detector, which e.g. can be implemented as a matched filter. <br>It is assumed that&nbsp; $\sigma_n^2 = N_0/2$.
-The graphic shows three different signal space constellations, namely
+The graphic shows three different signal space constellations,&nbsp; namely
-* Variant $\rm A$: &nbsp; $s_0 = (+1, \,  +5), \hspace{0.4cm} s_1 = (+4, \,  +1)$,
+:* Variant $\rm A$: &nbsp; $s_0 = (+1, \,  +5), \hspace{0.4cm} s_1 = (+4, \,  +1)$,
-* Variant $\rm B$: &nbsp; $s_0 = (-1.5, \,  +2), \, s_1 = (+1.5, \,  -2)$,
+:* Variant $\rm B$: &nbsp; $s_0 = (-1.5, \,  +2), \, s_1 = (+1.5, \,  -2)$,
-* Variant $\rm C$: &nbsp; $s_0 = (-2.5, \,  0), \hspace{0.45cm} s_1 = (+2.5, \,  0)$.
+:* Variant $\rm C$: &nbsp; $s_0 = (-2.5, \,  0), \hspace{0.45cm} s_1 = (+2.5, \,  0)$.
@@ Line 32: / Line 32: @@
 * Unless otherwise specified, equally probable symbols can be assumed:
 :$${\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) = 0.5\hspace{0.05cm}.$$
@@ Line 46: / Line 46: @@
 ===Questions===
 <quiz display=simple>
-{Which prerequisites must absolutely (in any case) be fulfilled so that the given error probability equation is valid?
+{Which prerequisites must absolutely&nbsp; (in any case)&nbsp; be fulfilled so that the given error probability equation is valid?
 |type="[]"}
 + Additive white Gaussian noise with variance&nbsp; $\sigma_n^2$.
@@ Line 60: / Line 60: @@
 + All variants show the same error behavior.
-{Give the error probability for variant &nbsp;$\rm A$&nbsp; with &nbsp;$\sigma_n^2 = E$.&nbsp; You can calculate&nbsp; ${\rm Q}(x)$&nbsp; according to the approximation.
+{Give the error probability for variant &nbsp;$\rm A$&nbsp; with &nbsp;$\sigma_n^2 = E$. &nbsp; You can calculate&nbsp; ${\rm Q}(x)$&nbsp; according to the approximation.
 |type="{}"}
 $p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$
-{It is assumed that&nbsp; $N_0 = 2 \cdot 10^{\rm &ndash;6} \ {\rm W/Hz}$,&nbsp; $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$. What is the probability for variant&nbsp;$\rm C$&nbsp; with equally probable symbols?
+{It is assumed that&nbsp; $N_0 = 2 \cdot 10^{\rm &ndash;6} \ {\rm W/Hz}$, &nbsp; $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$. &nbsp; What is the error probability for variant&nbsp;$\rm C$&nbsp; with equally probable symbols?
 |type="{}"}
 $p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$
-{What is the error probability for variant &nbsp;$\rm B$ under the same conditions?
+{What is the error probability for variant&nbsp; &nbsp;$\rm B$&nbsp; under the same conditions?
 |type="{}"}
 $p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$
-{How large should the average energy per symbol &nbsp;$(E_{\rm S})$&nbsp; be chosen for variant&nbsp;$\rm A$&nbsp; in order to obtain the same error probability as for system &nbsp;$\rm C$?&nbsp;
+{How large should the average energy per symbol &nbsp;$(E_{\rm S})$&nbsp; be chosen for variant&nbsp;$\rm A$&nbsp; in order to obtain the same error probability as for variant &nbsp;$\rm C$?&nbsp;
 |type="{}"}
 $E_{\rm S} \ = \ $ { 21.5 3% } $\ \cdot 10^{\rm &ndash;6} \ \rm Ws$

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}
-[[File:P_ID2016__Dig_Z_4_6.png|right|frame|Three signal space constellations]]
+[[File:EN_Dig_Z_4_6.png|right|frame|Three signal space constellations]]
 The (mean) error probability of an optimal binary system is:
 :$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )\hspace{0.05cm}.$$

@@ Line 91: / Line 91: @@
 <u>Solution 4</u> is correct:
-*Durch eine Drehung des Koordinatensystems kann man bei einem Binärsystem $(M = 2)$ stets mit einer Basisfunktion $(N = 1)$ auskommen.
+*By rotating the coordinate system, one can always get by with a basis function $(N = 1)$ for a binary system $(M = 2)$.
-*Da das zweidimensionale Rauschen zirkulär symmetrisch ist &nbsp; &#8658; &nbsp; gleiche Streuung $\sigma_n$ in alle Richtungen, kann auch der Rauschterm wie im Kapitel [[Digitalsignal%C3%BCbertragung/Fehlerwahrscheinlichkeit_bei_Basisband%C3%BCbertragung|Fehlerwahrscheinlichkeit bei Basisbandübertragung]] eindimensional beschrieben werden.
+*Since the two-dimensional noise is circularly symmetric &nbsp; &#8658; &nbsp; equal standard deviation $\sigma_n$ in all directions, the noise term can also be described one-dimensionally as in the chapter [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|"Error Probability for Baseband Transmission"]].
-'''(3)'''&nbsp; Für alle hier betrachteten Varianten gilt, also auch für die Variante &nbsp;$\rm A$:
+'''(3)'''&nbsp; For all variants considered here, i.e., also for variant &nbsp;$\rm A$, the following holds:
 :$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )=   {\rm Q} \left ( \frac{5/2 \cdot \sqrt{E}}{\sigma_n} \right )
 =   {\rm Q}(2.5)\hspace{0.05cm}.$$
-Mit der angegebenen Näherung erhält man
+With the given approximation we obtain
 :$$p_{\rm S}  = \frac{1}{\sqrt{2\pi} \cdot 2.5} \cdot {\rm e}^{-2.5^2/2} \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.7 \%}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Bei der Variante &nbsp;$\rm C$&nbsp; ergibt sich für die mittlere Energie pro Symbol:
+'''(4)'''&nbsp; For variant &nbsp;$\rm C$,&nbsp; the average energy per symbol is given by:
 :$$E_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \cdot  (-2.5 \cdot \sqrt{E})^2 +
   {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) \cdot  (+ 2.5 \cdot \sqrt{E})^2 =
@@ Line 111: / Line 111: @@
   \hspace{0.05cm}.$$
-Setzt man dieses Ergebnis in die unter (3) gefundene Gleichung ein, so erhält man mit $\sigma_n^2 = N_0/2$:
+Substituting this result into the equation found in (3), we obtain with $\sigma_n^2 = N_0/2$:
 :$$p_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}   {\rm Q} \left ( \frac{2.5 \cdot \sqrt{E}}{\sigma_n} \right )=   {\rm Q} \left ( \frac{ \sqrt{E_{\rm S}}}{\sigma_n} \right )
 =   {\rm Q} \left ( \frac{ \sqrt{2 \cdot E_{\rm S}}}{N_0} \right ) ={\rm Q} \left ( \sqrt{\frac{ 2 \cdot 6.25 \cdot 10^{-6}\,{\rm Ws}}{2 \cdot 10^{-6}\,{\rm W/Hz}}} \right )
@@ Line 117: / Line 117: @@
-'''(5)'''&nbsp; Durch Drehung des Koordinatensystems ändert sich nichts an den Energieverhältnissen. Deshalb erhält man wieder $p_{\rm S} \ \underline {\approx 0.7\%}$.
+'''(5)'''&nbsp; Rotating the coordinate system does not change the energy ratios. Therefore, $p_{\rm S} \ \underline {\approx 0.7\%}$ is obtained again.
-'''(6)'''&nbsp; Bei der Variante &nbsp;$\rm A$&nbsp; ist die mittlere Energie pro Symbol
+'''(6)'''&nbsp; In variant &nbsp;$\rm A$,&nbsp; the average energy per symbol is
 :$$E_{\rm S}  = {1}/{2} \cdot    \left [ (1^2 + 5^2) \cdot E + (4^2 + 1^2) \cdot E \right ] = 21.5 \cdot E
 \hspace{0.05cm}. $$
-Der Abstand von der Schwelle, die bei gleichwahrscheinlichen Symbolen in der Mitte zwischen $\boldsymbol{s}_0$ und $\boldsymbol{s}_1$ liegen sollte, ist wie bei den anderen Varianten $d/2 = 2.5 \cdot E^{\rm 1/2}$. Mit $\sigma_n^2 = N_0/2$ erhält man somit die Bestimmungsgleichung:
+The distance from the threshold, which should be midway between $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$ for equally probable symbols, is $d/2 = 2.5 \cdot E^{\rm 1/2}$, as in the other variants. Thus, with $\sigma_n^2 = N_0/2$, we obtain the governing equation:
 :$$p_{\rm S}  = {\rm Q} \left ( \frac{ 2.5 \cdot \sqrt{E}}{\sqrt{N_0/2}} \right )
 ={\rm Q}(2.5)\approx 0.7 \cdot 10^{-2} \hspace{0.3cm}
@@ Line 132: / Line 132: @@
 :$$\Rightarrow \hspace{0.3cm} {E_{\rm S}} = 0.5 \cdot {21.5 \cdot N_0} \hspace{0.1cm} \hspace{0.15cm}\underline { = 21.5 \cdot 10^{-6}\,{\rm Ws}}\hspace{0.05cm}.$$
-Das bedeutet: Bei der Variante &nbsp;$\rm A$&nbsp; ist gegenüber den beiden anderen Symbolen eine um den Faktor $3.44$ größere mittlere Symbolenergie $E_{\rm S}$ erforderlich, um die gleiche Fehlerwahrscheinlichkeit $p_{\rm S} = 0.7%$ zu erzielen.
+This means: For variant &nbsp;$\rm A$,&nbsp; compared to the other two symbols, a mean symbol energy $E_{\rm S}$ larger by a factor of $3.44$is required to achieve the same error probability $p_{\rm S} = 0.7%$.
-*Das heißt: Diese Signalraumkonstellation ist sehr ungünstig. Es ergibt sich ein sehr großes $E_{\rm S}$, ohne dass gleichzeitig der Abstand $d$ vergrößert wird.
+*That means: This signal space constellation is very unfavorable. It results in a very large $E_{\rm S}$ without increasing the distance $d$ at the same time.
-*Mit $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$ würde sich dagegen $p_{\rm S} = {\rm Q}(2.5/3.44^{\rm 1/2}) \approx {\rm Q}(1.35) \approx 9\%$ ergeben.
+*With $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$, on the other hand, $p_{\rm S} = {\rm Q}(2.5/3.44^{\rm 1/2}) \approx {\rm Q}(1.35) \approx 9\%$ would result.
-*Das heißt: &nbsp; Die Fehlerwahrscheinlichkeit würde um mehr als eine Zehnerpotenz größer.
+*That means: &nbsp; The error probability would be larger by more than one power of ten.
 {{ML-Fuß}}

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Approximation der Fehlerwahrscheinlichkeit}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}
-[[File:P_ID2016__Dig_Z_4_6.png|right|frame|Drei Signalraumkonstellationen]]
+[[File:P_ID2016__Dig_Z_4_6.png|right|frame|Three signal space constellations]]
 Die (mittlere) Fehlerwahrscheinlichkeit eines optimalen Binärsystems lautet:
 :$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )\hspace{0.05cm}.$$
@@ Line 35: / Line 35: @@
 ''Hinweise:''
-* Die Aufgabe gehört zum  Kapitel&nbsp; [[Digitalsignal%C3%BCbertragung/Approximation_der_Fehlerwahrscheinlichkeit| Approximation der Fehlerwahrscheinlichkeit]].
+* Die Aufgabe gehört zum  Kapitel&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
 * Für numerische Berechnungen kann zur Vereinfachung die Energie&nbsp; $E = 1$&nbsp; gesetzt werden.
 * Wenn keine anderslautende Angabe gemacht ist, so kann von gleichwahrscheinlichen Symbolen ausgegangen werden:

← Older revision		Revision as of 10:08, 23 June 2022
Line 141:		Line 141:


−	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]

← Older revision		Revision as of 12:49, 23 March 2021
Line 141:		Line 141:


−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.3 BER-Approximation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]