Difference between revisions of "Aufgaben:Exercise 4.06Z: Signal Space Constellations"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Approximation der Fehlerwahrscheinlichkeit}}
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}
  
[[File:P_ID2016__Dig_Z_4_6.png|right|frame|Drei Signalraumkonstellationen]]
+
[[File:EN_Dig_Z_4_6.png|right|frame|Three signal space constellations]]
Die (mittlere) Fehlerwahrscheinlichkeit eines optimalen Binärsystems lautet:
+
The  (mean)  symbol error probability of an optimal binary system is:
 
:$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =  {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )\hspace{0.05cm}.$$
 
:$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =  {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )\hspace{0.05cm}.$$
  
Hierzu ist anzumerken:
+
It should be noted here:
* ${\rm Q}(x)$ bezeichnet die komplementäre Gaußsche Fehlerfunktion (Definition und Approximation):
+
* ${\rm Q}(x)$  denotes the complementary Gaussian error function  (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  
 
:$${\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 \\
+
\approx  \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}
\hspace{-0.1cm} \ \approx \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}
 
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
  
* $d$ gibt den Abstand der beiden Sendesignalpunkte $s_0$ und $s_1$ im vorgegebenen Vektorraum an:
+
* The parameter  $d$  specifies the distance between the two transmitted signal points  $s_0$  and  $s_1$  in vector space:
 
:$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} \hspace{0.05cm}.$$
 
:$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} \hspace{0.05cm}.$$
  
* $\sigma_n^2$ ist die Varianz des AWGN–Rauschens nach dem Detektor, der zum Beispiel als Matched–Filter realisiert sein kann. Es gelte $\sigma_n^2 = N_0/2$.
+
* $\sigma_n^2$&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; $\sigma_n^2 = N_0/2$.
  
  
Durch die Grafik sind drei unterschiedliche Signalraumkonstellationen gegeben, nämlich
+
The graphic shows three different signal space constellations,&nbsp; namely
* Variante <i>A</i>: $s_0 = (+1, \ \, +5), \hspace{0.5cm} s_1 = (+4, \ \, +1)$,
 
* Variante <i>B</i>: $s_0 = (&ndash;1, \ \, +2), \hspace{0.35cm} s_1 = (+1.5, \ \, &ndash;2)$,
 
* Variante <i>C</i>: $s_0 = (&ndash;2.5, \ \, 0), \hspace{0.4cm} s_1 = (+2.5, \ \, 0)$.
 
  
Die jeweils mittlere Energie pro Symbol ($E_{\rm S$) kann nach folgender Gleichung berechnet werden:
+
:* 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 $\rm C$: &nbsp; $s_0 = (-2.5, \,  0), \hspace{0.45cm} s_1 = (+2.5, \,  0)$.
 +
 
 +
 
 +
The mean energy per symbol &nbsp;$(E_{\rm S})$&nbsp; can be calculated as follows:
 
:$$E_{\rm S}  = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \cdot  ||  \boldsymbol{ s }_0||^2 +  
 
:$$E_{\rm S}  = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \cdot  ||  \boldsymbol{ s }_0||^2 +  
 
  {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) \cdot  ||  \boldsymbol{ s }_1||^2\hspace{0.05cm}.$$
 
  {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) \cdot  ||  \boldsymbol{ s }_1||^2\hspace{0.05cm}.$$
  
''Hinweise:''
+
 
* Die Aufgabe gehört zum Themengebiet von Kapitel [[Digitalsignal%C3%BCbertragung/Approximation_der_Fehlerwahrscheinlichkeit| Approximation der Fehlerwahrscheinlichkeit]].
+
 
* Wenn bei einer Teilaufgabe keine anderslautende Angabe gemacht ist, so kann von gleichwahrscheinlichen Symbolen ausgegangen werden:
+
 
 +
Notes:
 +
* The chapter belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
 +
 +
* For numeric calculations, the energy&nbsp; $E = 1$&nbsp; can be set for simplification.
 +
 
 +
* Unless otherwise specified, equally probable symbols can be assumed:
 
:$${\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) = 0.5\hspace{0.05cm}.$$
 
:$${\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) = 0.5\hspace{0.05cm}.$$
 +
  
# Die Normierungsenergie $E$ ist hier stillschweigend zu $1$ gesetzt.
 
  
  
  
===Fragebogen===
+
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice
+
{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; $\sigma_n^2$.
 +
+ Optimal binary receiver.
 +
+ Decision boundary in the middle between the symbols.
 +
- Equally likely symbols&nbsp; $s_0$&nbsp; and&nbsp; $s_1$.
 +
 
 +
{Which statement applies to the error probability with&nbsp; $\sigma_n^2 = E$?
 
|type="[]"}
 
|type="[]"}
+ correct
+
- Variant &nbsp;$\rm A$&nbsp; has the lowest error probability.
- false
+
- Variant &nbsp;$\rm B$&nbsp; has the lowest error probability.
 +
- Variant &nbsp;$\rm C$&nbsp; has the lowest error probability.
 +
+ All variants show the same error behavior.
  
{Input-Box Frage
+
{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="{}"}
 
|type="{}"}
$xyz$ = { 5.4 3% } $ab$
+
$p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$
 +
 
 +
{It is assumed that&nbsp; $N_0 = 2 \cdot 10^{\rm &ndash;6} \ {\rm W/Hz}$, &nbsp; $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$. &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; &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 variant &nbsp;$\rm C$?&nbsp;
 +
|type="{}"}
 +
$E_{\rm S} \ = \ $ { 21.5 3% } $\ \cdot 10^{\rm &ndash;6} \ \rm Ws$
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;  
+
'''(1)'''&nbsp; The&nbsp; <u>first three prerequisites</u>&nbsp; must be met in any case:
'''(2)'''&nbsp;  
+
*The equation then applies independently of the occurrence probabilities.
'''(3)'''&nbsp;  
+
*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.
'''(4)'''&nbsp;  
+
 
'''(5)'''&nbsp;  
+
 
 +
 
 +
'''(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;  
 +
*The same applies to the distance of the signal space points.&nbsp; For variant &nbsp;$\rm A$,&nbsp; for example,&nbsp; the following applies:
 +
:$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} = \sqrt{ E \cdot (4-1)^2 + E \cdot (1-5)^2} = 5 \cdot \sqrt{E}\hspace{0.05cm}.$$
 +
 
 +
*Due to the shifting of the coordinate system,&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:
 +
#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&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,&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
 +
:$$p_{\rm S}  = \frac{1}{\sqrt{2\pi} \cdot 2.5} \cdot {\rm e}^{-2.5^2/2} \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.7 \%}\hspace{0.05cm}.$$
 +
 
 +
 
 +
'''(4)'''&nbsp; For variant &nbsp;$\rm C$,&nbsp; the average energy per symbol is given by:
 +
:$$E_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \cdot  (-2.5 \cdot \sqrt{E})^2 +
 +
{\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) \cdot  (+ 2.5 \cdot \sqrt{E})^2 =
 +
\left [ {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) + {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \right ] \cdot 6.25 \cdot E = 6.25 \cdot E$$
 +
:$$\Rightarrow \hspace{0.3cm} E = \frac {E_{\rm S}}{6.25} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sqrt{E}= \frac {\sqrt{E_{\rm S}}}{2.5}
 +
\hspace{0.05cm}.$$
 +
 
 +
*Substituting this result into the equation found in&nbsp; '''(3)''',&nbsp; we obtain with&nbsp; $\sigma_n^2 = N_0/2$:
 +
:$$p_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm Q} \left ( \frac{2.5 \cdot \sqrt{E}}{\sigma_n} \right )=  {\rm Q} \left ( \frac{ \sqrt{E_{\rm S}}}{\sigma_n} \right )
 +
=  {\rm Q} \left ( \frac{ \sqrt{2 \cdot E_{\rm S}}}{N_0} \right ) ={\rm Q} \left ( \sqrt{\frac{ 2 \cdot 6.25 \cdot 10^{-6}\,{\rm Ws}}{2 \cdot 10^{-6}\,{\rm W/Hz}}} \right )
 +
={\rm Q}(2.5) \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.7 \%}\hspace{0.05cm}. $$
 +
 
 +
 
 +
'''(5)'''&nbsp; Rotating the coordinate system does not change the energy ratios.
 +
*Therefore,&nbsp; $p_{\rm S} \ \underline {\approx 0.7\%}$&nbsp; is obtained again.
 +
 
 +
 
 +
 
 +
'''(6)'''&nbsp; In variant &nbsp;$\rm A$,&nbsp; the average energy per symbol is
 +
:$$E_{\rm S}  = {1}/{2} \cdot    \left [ (1^2 + 5^2) \cdot E + (4^2 + 1^2) \cdot E \right ] = 21.5 \cdot E
 +
\hspace{0.05cm}. $$
 +
 
 +
*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.
 +
 
 +
*Thus,&nbsp; with&nbsp; $\sigma_n^2 = N_0/2$,&nbsp; we obtain the governing equation:
 +
:$$p_{\rm S}  = {\rm Q} \left ( \frac{ 2.5 \cdot \sqrt{E}}{\sqrt{N_0/2}} \right ) 
 +
={\rm Q}(2.5)\approx 0.7 \cdot 10^{-2} \hspace{0.3cm}
 +
\Rightarrow \hspace{0.3cm} \sqrt{\frac {2E}{N_0}} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac {E}{N_0} = 0.5
 +
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac {E_{\rm S}}{21.5 \cdot N_0} = 0.5$$
 +
:$$\Rightarrow \hspace{0.3cm} {E_{\rm S}} = 0.5 \cdot {21.5 \cdot N_0} \hspace{0.1cm} \hspace{0.15cm}\underline { = 21.5 \cdot 10^{-6}\,{\rm Ws}}\hspace{0.05cm}.$$
 +
 
 +
*This means:&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:
 +
#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 $E_{\rm S} = 6.25 \cdot 10^{\rm &ndash;6} \ \rm Ws$,&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.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
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[[Category:Aufgaben zu Digitalsignalübertragung|^4.3 BER-Approximation^]]
+
[[Category:Digital Signal Transmission: Exercises|^4.3 BER Approximation^]]

Latest revision as of 17:18, 27 July 2022

Three signal space constellations

The  (mean)  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:

  • ${\rm Q}(x)$  denotes the complementary Gaussian error function  (definition and approximation):
$${\rm Q}(x) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} {\rm e}^{-u^2/2} \,{\rm d} u \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2} \hspace{0.05cm}.$$
  • The parameter  $d$  specifies the distance between the two transmitted signal points  $s_0$  and  $s_1$  in vector space:
$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} \hspace{0.05cm}.$$
  • $\sigma_n^2$  is the variance of the AWGN noise after the detector, which e.g. can be implemented as a matched filter.
    It is assumed that  $\sigma_n^2 = N_0/2$.


The graphic shows three different signal space constellations,  namely

  • Variant $\rm A$:   $s_0 = (+1, \, +5), \hspace{0.4cm} s_1 = (+4, \, +1)$,
  • Variant $\rm B$:   $s_0 = (-1.5, \, +2), \, s_1 = (+1.5, \, -2)$,
  • Variant $\rm C$:   $s_0 = (-2.5, \, 0), \hspace{0.45cm} s_1 = (+2.5, \, 0)$.


The mean energy per symbol  $(E_{\rm S})$  can be calculated as follows:

$$E_{\rm S} = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \cdot || \boldsymbol{ s }_0||^2 + {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) \cdot || \boldsymbol{ s }_1||^2\hspace{0.05cm}.$$



Notes:

  • For numeric calculations, the energy  $E = 1$  can be set for simplification.
  • Unless otherwise specified, equally probable symbols can be assumed:
$${\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) = 0.5\hspace{0.05cm}.$$



Questions

1

Which prerequisites must absolutely  (in any case)  be fulfilled so that the given error probability equation is valid?

Additive white Gaussian noise with variance  $\sigma_n^2$.
Optimal binary receiver.
Decision boundary in the middle between the symbols.
Equally likely symbols  $s_0$  and  $s_1$.

2

Which statement applies to the error probability with  $\sigma_n^2 = E$?

Variant  $\rm A$  has the lowest error probability.
Variant  $\rm B$  has the lowest error probability.
Variant  $\rm C$  has the lowest error probability.
All variants show the same error behavior.

3

Give the error probability for variant  $\rm A$  with  $\sigma_n^2 = E$.   You can calculate  ${\rm Q}(x)$  according to the approximation.

$p_{\rm S} \ = \ $

$\ \%$

4

It is assumed that  $N_0 = 2 \cdot 10^{\rm –6} \ {\rm W/Hz}$,   $E_{\rm S} = 6.25 \cdot 10^{\rm –6} \ \rm Ws$.   What is the error probability for variant $\rm C$  with equally probable symbols?

$p_{\rm S} \ = \ $

$\ \%$

5

What is the error probability for variant   $\rm B$  under the same conditions?

$p_{\rm S} \ = \ $

$\ \%$

6

How large should the average energy per symbol  $(E_{\rm S})$  be chosen for variant $\rm A$  in order to obtain the same error probability as for variant  $\rm C$? 

$E_{\rm S} \ = \ $

$\ \cdot 10^{\rm –6} \ \rm Ws$


Solution

(1)  The  first three prerequisites  must be met in any case:

  • The equation then applies independently of the occurrence probabilities.
  • In the case of  ${\rm Pr}(\boldsymbol{s} = \boldsymbol{s}_0) ≠ {\rm Pr}(\boldsymbol{s} = \boldsymbol{s}_1)$,  a lower error probability can be achieved by shifting the decision threshold.


(2)  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  $\rm A$,  for example,  the following applies:
$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} = \sqrt{ E \cdot (4-1)^2 + E \cdot (1-5)^2} = 5 \cdot \sqrt{E}\hspace{0.05cm}.$$
  • Due to the shifting of the coordinate system,  the distance between  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$  does not change (variant  $\rm B$),  the same distance results in variant  $\rm C$  (after rotation).
  • Solution 4 is correct:
  1. By rotating the coordinate system,  one can always get by with one basis function  $(N = 1)$  for a binary system  $(M = 2)$.
  2. Since the two-dimensional noise is circularly symmetric   ⇒   equal standard deviation  $\sigma_n$  in all directions, 
    the noise term can also be described one-dimensionally as in the chapter  "Error Probability for Baseband Transmission".


(3)  For all variants considered here,  i.e.,  also for variant  $\rm A$,  the following holds:

$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )= {\rm Q} \left ( \frac{5/2 \cdot \sqrt{E}}{\sigma_n} \right ) = {\rm Q}(2.5)\hspace{0.05cm}.$$
  • 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)  For variant  $\rm C$,  the average energy per symbol is given by:

$$E_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \cdot (-2.5 \cdot \sqrt{E})^2 + {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) \cdot (+ 2.5 \cdot \sqrt{E})^2 = \left [ {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) + {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) \right ] \cdot 6.25 \cdot E = 6.25 \cdot E$$
$$\Rightarrow \hspace{0.3cm} E = \frac {E_{\rm S}}{6.25} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sqrt{E}= \frac {\sqrt{E_{\rm S}}}{2.5} \hspace{0.05cm}.$$
  • Substituting this result into the equation found in  (3),  we obtain with  $\sigma_n^2 = N_0/2$:
$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Q} \left ( \frac{2.5 \cdot \sqrt{E}}{\sigma_n} \right )= {\rm Q} \left ( \frac{ \sqrt{E_{\rm S}}}{\sigma_n} \right ) = {\rm Q} \left ( \frac{ \sqrt{2 \cdot E_{\rm S}}}{N_0} \right ) ={\rm Q} \left ( \sqrt{\frac{ 2 \cdot 6.25 \cdot 10^{-6}\,{\rm Ws}}{2 \cdot 10^{-6}\,{\rm W/Hz}}} \right ) ={\rm Q}(2.5) \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.7 \%}\hspace{0.05cm}. $$


(5)  Rotating the coordinate system does not change the energy ratios.

  • Therefore,  $p_{\rm S} \ \underline {\approx 0.7\%}$  is obtained again.


(6)  In variant  $\rm A$,  the average energy per symbol is

$$E_{\rm S} = {1}/{2} \cdot \left [ (1^2 + 5^2) \cdot E + (4^2 + 1^2) \cdot E \right ] = 21.5 \cdot E \hspace{0.05cm}. $$
  • 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:
$$p_{\rm S} = {\rm Q} \left ( \frac{ 2.5 \cdot \sqrt{E}}{\sqrt{N_0/2}} \right ) ={\rm Q}(2.5)\approx 0.7 \cdot 10^{-2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sqrt{\frac {2E}{N_0}} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac {E}{N_0} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac {E_{\rm S}}{21.5 \cdot N_0} = 0.5$$
$$\Rightarrow \hspace{0.3cm} {E_{\rm S}} = 0.5 \cdot {21.5 \cdot N_0} \hspace{0.1cm} \hspace{0.15cm}\underline { = 21.5 \cdot 10^{-6}\,{\rm Ws}}\hspace{0.05cm}.$$
  • This means:  For variant  $\rm A$,  compared to the other two variants,  a mean symbol energy  $E_{\rm S}$  larger by a factor of  $3.44$  is required to achieve the same  $p_{\rm S} = 0.7\%$.  Because:
  1. 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.
  2. With $E_{\rm S} = 6.25 \cdot 10^{\rm –6} \ \rm Ws$,  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.
  3. That means:   The error probability would be larger by more than one power of ten.