Difference between revisions of "Aufgaben:Exercise 4.06Z: Signal Space Constellations"
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[[File:EN_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: | + | The (mean) symbol error probability of an optimal binary system is: |
:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )\hspace{0.05cm}.$$ | :$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )\hspace{0.05cm}.$$ | ||
It should be noted here: | It should be noted here: | ||
− | * ${\rm Q}(x)$ denotes the complementary Gaussian error function (definition and 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 \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2} | \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | * $d$ specifies the distance between the two transmitted signal points $s_0$ and $s_1$ in vector space: | + | * The parameter $d$ specifies the distance between the two transmitted signal points $s_0$ and $s_1$ in vector space: |
:$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} \hspace{0.05cm}.$$ | :$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} \hspace{0.05cm}.$$ | ||
− | * $\sigma_n^2$ is the variance of the AWGN noise after the detector, which | + | * $\sigma_n^2$ 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 $\sigma_n^2 = N_0/2$. |
− | The graphic shows three different signal space constellations, namely | + | The graphic shows three different signal space constellations, namely |
− | * Variant $\rm A$: $s_0 = (+1, \, +5), \hspace{0.4cm} s_1 = (+4, \, +1)$, | + | :* Variant $\rm A$: $s_0 = (+1, \, +5), \hspace{0.4cm} s_1 = (+4, \, +1)$, |
− | * Variant $\rm B$: $s_0 = (-1.5, \, +2), \, s_1 = (+1.5, \, -2)$, | + | :* Variant $\rm B$: $s_0 = (-1.5, \, +2), \, s_1 = (+1.5, \, -2)$, |
− | * Variant $\rm C$: $s_0 = (-2.5, \, 0), \hspace{0.45cm} s_1 = (+2.5, \, 0)$. | + | :* Variant $\rm C$: $s_0 = (-2.5, \, 0), \hspace{0.45cm} s_1 = (+2.5, \, 0)$. |
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+ | Notes: | ||
+ | * The chapter belongs to the chapter [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]]. | ||
+ | |||
+ | * For numeric calculations, the energy $E = 1$ can be set for simplification. | ||
− | |||
− | |||
− | |||
− | |||
* Unless otherwise specified, equally probable symbols can be assumed: | * Unless otherwise specified, equally probable symbols can be assumed: | ||
:$${\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) = 0.5\hspace{0.05cm}.$$ | :$${\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_0) = {\rm Pr}(\boldsymbol{ s } = \boldsymbol{ s }_1) = 0.5\hspace{0.05cm}.$$ | ||
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===Questions=== | ===Questions=== | ||
<quiz display=simple> | <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 (in any case) be fulfilled so that the given error probability equation is valid? |
|type="[]"} | |type="[]"} | ||
+ Additive white Gaussian noise with variance $\sigma_n^2$. | + Additive white Gaussian noise with variance $\sigma_n^2$. | ||
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+ All variants show the same error behavior. | + All variants show the same error behavior. | ||
− | {Give the error probability for variant $\rm A$ with $\sigma_n^2 = E$. You can calculate ${\rm Q}(x)$ according to the approximation. | + | {Give the error probability for variant $\rm A$ with $\sigma_n^2 = E$. You can calculate ${\rm Q}(x)$ according to the approximation. |
|type="{}"} | |type="{}"} | ||
$p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$ | $p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$ | ||
− | {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 probability for variant $\rm C$ with equally probable symbols? | + | {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? |
|type="{}"} | |type="{}"} | ||
$p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$ | $p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$ | ||
− | {What is the error probability for variant $\rm B$ under the same conditions? | + | {What is the error probability for variant $\rm B$ under the same conditions? |
|type="{}"} | |type="{}"} | ||
$p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$ | $p_{\rm S} \ = \ $ { 0.7 3% } $\ \%$ | ||
− | {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 | + | {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 $\rm C$? |
|type="{}"} | |type="{}"} | ||
$E_{\rm S} \ = \ $ { 21.5 3% } $\ \cdot 10^{\rm –6} \ \rm Ws$ | $E_{\rm S} \ = \ $ { 21.5 3% } $\ \cdot 10^{\rm –6} \ \rm Ws$ | ||
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' The <u>first three prerequisites</u> must be met in any case: | + | '''(1)''' The <u>first three prerequisites</u> must be met in any case: |
*The equation then applies independently of the occurrence probabilities. | *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. | + | *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: | + | '''(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: | ||
:$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} = \sqrt{ E \cdot (4-1)^2 + E \cdot (1-5)^2} = 5 \cdot \sqrt{E}\hspace{0.05cm}.$$ | :$$d = \sqrt{ || \boldsymbol{ s }_1 - \boldsymbol{ s }_0||^2} = \sqrt{ E \cdot (4-1)^2 + E \cdot (1-5)^2} = 5 \cdot \sqrt{E}\hspace{0.05cm}.$$ | ||
− | Due to the shifting of the coordinate system, the distance between $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$ does not change (variant $\rm B$), | + | *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). |
− | <u>Solution 4</u> is correct: | + | *<u>Solution 4</u> is correct: |
− | + | #By rotating the coordinate system, one can always get by with one basis function $(N = 1)$ for a binary system $(M = 2)$. | |
− | + | #Since the two-dimensional noise is circularly symmetric ⇒ equal standard deviation $\sigma_n$ in all directions, <br>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)''' For all variants considered here, i.e., also for variant $\rm A$, the following holds: | + | '''(3)''' For all variants considered here, i.e., also for variant $\rm A$, the following holds: |
:$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {\rm Q} \left ( \frac{d/2}{\sigma_n} \right )= {\rm Q} \left ( \frac{5/2 \cdot \sqrt{E}}{\sigma_n} \right ) | :$$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}.$$ | = {\rm Q}(2.5)\hspace{0.05cm}.$$ | ||
− | With the given approximation we obtain | + | *With the given approximation we obtain |
:$$p_{\rm S} = \frac{1}{\sqrt{2\pi} \cdot 2.5} \cdot {\rm e}^{-2.5^2/2} \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.7 \%}\hspace{0.05cm}.$$ | :$$p_{\rm S} = \frac{1}{\sqrt{2\pi} \cdot 2.5} \cdot {\rm e}^{-2.5^2/2} \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.7 \%}\hspace{0.05cm}.$$ | ||
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\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | 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 '''(3)''', we obtain with $\sigma_n^2 = N_0/2$: |
:$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Q} \left ( \frac{2.5 \cdot \sqrt{E}}{\sigma_n} \right )= {\rm Q} \left ( \frac{ \sqrt{E_{\rm S}}}{\sigma_n} \right ) | :$$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} \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 ) | ||
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− | '''(5)''' Rotating the coordinate system does not change the energy ratios. Therefore, $p_{\rm S} \ \underline {\approx 0.7\%}$ is obtained again. | + | '''(5)''' Rotating the coordinate system does not change the energy ratios. |
+ | *Therefore, $p_{\rm S} \ \underline {\approx 0.7\%}$ is obtained again. | ||
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\hspace{0.05cm}. $$ | \hspace{0.05cm}. $$ | ||
− | 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, 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: | ||
:$$p_{\rm S} = {\rm Q} \left ( \frac{ 2.5 \cdot \sqrt{E}}{\sqrt{N_0/2}} \right ) | :$$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} | ={\rm Q}(2.5)\approx 0.7 \cdot 10^{-2} \hspace{0.3cm} | ||
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:$$\Rightarrow \hspace{0.3cm} {E_{\rm S}} = 0.5 \cdot {21.5 \cdot N_0} \hspace{0.1cm} \hspace{0.15cm}\underline { = 21.5 \cdot 10^{-6}\,{\rm Ws}}\hspace{0.05cm}.$$ | :$$\Rightarrow \hspace{0.3cm} {E_{\rm S}} = 0.5 \cdot {21.5 \cdot N_0} \hspace{0.1cm} \hspace{0.15cm}\underline { = 21.5 \cdot 10^{-6}\,{\rm Ws}}\hspace{0.05cm}.$$ | ||
− | This means: For variant $\rm A$, compared to the other two | + | *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: |
− | + | #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. | |
− | + | #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. | |
− | + | #That means: The error probability would be larger by more than one power of ten. | |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Latest revision as of 17:18, 27 July 2022
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:
- The chapter belongs to the chapter "Approximation of the Error Probability".
- 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
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:
- By rotating the coordinate system, one can always get by with one basis function $(N = 1)$ for a binary system $(M = 2)$.
- 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:
- 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.
- 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.
- That means: The error probability would be larger by more than one power of ten.