Aufgaben:Exercise 4.10: Union Bound - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]
	[[de:4.~~10_Union_Bound~~]]		[[de:Aufgaben:Aufgabe 4.10: Union Bound]]

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	*The actual  (average)  error probability is generally given as follows:		*The actual  (average)  error probability is generally given as follows:
	:$${\rm Pr}({ \cal E}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sum\limits_{i = 0 }^{M-1} {\rm Pr}(m_i) \cdot {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{0.05cm},$$		:$${\rm Pr}({ \cal E}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sum\limits_{i = 0 }^{M-1} {\rm Pr}(m_i) \cdot {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{0.05cm},$$
	:$$ {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr} \left [ \bigcup_{k \ne i} { \cal E}_{ik}\right ] \hspace{0.05cm},\hspace{0.2cm}{ \rm where}\hspace{0.2cm}		:$$ {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr} \left [ \bigcup_{k \ne i} { \cal E}_{ik}\right ] \hspace{0.05cm},\hspace{0.2cm}{ \rm where}\hspace{0.2cm}{ \cal E}_{ik}\text{:} \ \ \boldsymbol{ r }{\rm \hspace{0.15cm}is \hspace{0.15cm}closer \hspace{0.15cm}to \hspace{0.15cm}}\boldsymbol{ s }_k {\rm \hspace{0.15cm}than \hspace{0.15cm}to \hspace{0.15cm}the \hspace{0.15cm}setpoint \hspace{0.15cm}}\boldsymbol{ s }_i\hspace{0.05cm}.$$
	{ \cal E}_{ik}\text{:} \ \ \boldsymbol{ r }{\rm \hspace{0.15cm}is \hspace{0.15cm}closer \hspace{0.15cm}to \hspace{0.15cm}}\boldsymbol{ s }_k {\rm \hspace{0.15cm}than \hspace{0.15cm}to \hspace{0.15cm}the \hspace{0.15cm}setpoint \hspace{0.15cm}}\boldsymbol{ s }_i
	\hspace{0.05cm}.$$

	*The simpler  "Union Bound"  $\rm (UB)$  provides an upper bound on the falsification probability assuming that message  $m_i$  $($or signal  $\boldsymbol{s}_i)$  was sent:		*The simpler  "Union Bound"  $\rm (UB)$  provides an upper bound on the falsification probability assuming that message  $m_i$  $($or signal  $\boldsymbol{s}_i)$  was sent:
	:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ \ge \ \hspace{-0.1cm} {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} \boldsymbol{ s }_i ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i )\hspace{0.05cm},\ $$		:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ \ge \ \hspace{-0.1cm} {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} \boldsymbol{ s }_i ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i )\hspace{0.05cm},\ $$
	:$$ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ = \ \hspace{-0.2cm}\sum\limits_{k = 0 ,\hspace{0.1cm} k \ne i}^{M-1}\hspace{-0.1cm}		:$$ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ = \ \hspace{-0.2cm}\sum\limits_{k = 0 ,\hspace{0.1cm} k \ne i}^{M-1}\hspace{-0.1cm}{\rm Pr}({ \cal E}_{ik}) = \hspace{-0.1cm}\sum\limits_{k = 0, \hspace{0.1cm} k \ne i}^{M-1}\hspace{-0.1cm}{\rm Q} \left ( \frac{d_{ik}/2}{\sigma_n} \right )\hspace{0.05cm}. $$
	{\rm Pr}({ \cal E}_{ik}) = \hspace{-0.1cm}\sum\limits_{k = 0, \hspace{0.1cm} k \ne i}^{M-1}\hspace{-0.1cm}{\rm Q} \left ( \frac{d_{ik}/2}{\sigma_n} \right )\hspace{0.05cm}. $$

	The following abbreviations are used:		The following abbreviations are used:
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	*Thus,  the signal space coordinates are also pure numerical values without unit:		*Thus,  the signal space coordinates are also pure numerical values without unit:
	:$$\boldsymbol{ s }_1 = (-1, \hspace{0.1cm}+1)\hspace{0.05cm}, \hspace{0.2cm}		:$$\boldsymbol{ s }_1 = (-1, \hspace{0.1cm}+1)\hspace{0.05cm}, \hspace{0.2cm}\boldsymbol{ s }_2 = (+1, \hspace{0.1cm}+1)\hspace{0.05cm}.$$
	\boldsymbol{ s }_2 = (+1, \hspace{0.1cm}+1)\hspace{0.05cm}.$$

	*The signal space point  $\boldsymbol{s}_0$  in configuration  $\rm A$  is located such that  $\boldsymbol{s}_0$,  $\boldsymbol{s}_1$,  $\boldsymbol{s}_2$  describe an equilateral triangle.		*The signal space point  $\boldsymbol{s}_0$  in configuration  $\rm A$  is located such that  $\boldsymbol{s}_0$,  $\boldsymbol{s}_1$,  $\boldsymbol{s}_2$  describe an equilateral triangle.
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	'''(3)'''  Here,  the falsification probabilities differ for each symbol.		'''(3)'''  Here,  the falsification probabilities differ for each symbol.
	*If  $\boldsymbol{s}_0$  was sent,  with $d_{01} = d_{02} = 2^{0.5}$  and  $\sigma = 0.5$:		*If  $\boldsymbol{s}_0$  was sent,  with $d_{01} = d_{02} = 2^{0.5}$  and  $\sigma = 0.5$:
	:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} = 2 \cdot {\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )		:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} = 2 \cdot {\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )= 2 \cdot {\rm Q}(\sqrt{2}) = 2 \cdot 0.079 = 0.158 \hspace{0.05cm}. $$
	= 2 \cdot {\rm Q}(\sqrt{2}) = 2 \cdot 0.079 = 0.158 \hspace{0.05cm}. $$

	*In contrast,  the other two conditional probabilities are smaller:		*In contrast,  the other two conditional probabilities are smaller:
	:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} = p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}		:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} = p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Q} \left ( \frac{{2}/2}{0.5} \right )+{\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )= {\rm Q}(2) +{\rm Q}(\sqrt{2}) = 0.023 + 0.079 = 0.102 \hspace{0.05cm}.$$
	{\rm Q} \left ( \frac{{2}/2}{0.5} \right )+{\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )= {\rm Q}(2) +{\rm Q}(\sqrt{2}) = 0.023 + 0.079 = 0.102 \hspace{0.05cm}.$$

	*By averaging,  we obtain for the Union Bound considering the different distances:		*By averaging,  we obtain for the Union Bound considering the different distances:
	:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{3} \cdot \left [ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} + p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} +p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2}\right ]=		:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{3} \cdot \left [ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} + p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} +p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2}\right ]={1}/{3} \cdot \left [ 2 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot ({\rm Q}({2}) + {\rm Q}(\sqrt{2})) \right ] ={1}/{3} \cdot \left [ 4 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot {\rm Q}({2}) \right ] $$
	{1}/{3} \cdot \left [ 2 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot ({\rm Q}({2}) + {\rm Q}(\sqrt{2})) \right ] =		:$$ \Rightarrow \hspace{0.3cm} p_{\rm UB} ={1}/{3} \cdot \big [ 4 \cdot 0.079+ 2 \cdot 0.023 \big ] \hspace{0.1cm}\hspace{0.12cm}\underline {\approx 12.1\% } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm UB}\ge {\rm Pr}({ \cal E})\hspace{0.05cm}.$$
	{1}/{3} \cdot \left [ 4 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot {\rm Q}({2}) \right ] $$
	:$$ \Rightarrow \hspace{0.3cm} p_{\rm UB} =
	{1}/{3} \cdot \big [ 4 \cdot 0.079+ 2 \cdot 0.023 \big ] \hspace{0.1cm}\hspace{0.12cm}\underline {\approx 12.1\% } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm UB}\ge {\rm Pr}({ \cal E})\hspace{0.05cm}.$$


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	'''(4)'''  This configuration is described by the following equations:		'''(4)'''  This configuration is described by the following equations:
	:$$d_{01} = d_{02} = \sqrt{2^2 + 1^2}= \sqrt{5} \approx 2.24\hspace{0.2cm}, d_{12} = 2$$		:$$d_{01} = d_{02} = \sqrt{2^2 + 1^2}= \sqrt{5} \approx 2.24\hspace{0.2cm}, d_{12} = 2$$
	:$$\Rightarrow \hspace{0.3cm} p_{\rm UB} =		:$$\Rightarrow \hspace{0.3cm} p_{\rm UB} ={1}/{3} \cdot \big [ 4 \cdot {\rm Q}(\sqrt{5})+ 2 \cdot {\rm Q}({2}) \big ] = {1}/{3} \cdot \big [ 4 \cdot 0.013+ 2 \cdot 0.023 \big ]\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 3.2\%} \hspace{0.05cm}. $$
	{1}/{3} \cdot \big [ 4 \cdot {\rm Q}(\sqrt{5})+ 2 \cdot {\rm Q}({2}) \big ] = {1}/{3} \cdot \big [ 4 \cdot 0.013+ 2 \cdot 0.023 \big ]\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 3.2\%} \hspace{0.05cm}. $$


	'''(5)'''  Let it be:		'''(5)'''  Let it be:
	:$$p_{\rm UB} = 2 \cdot {\rm Q}\left ( {1}/{\sigma_n} \right ) = 0.032 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$p_{\rm UB} = 2 \cdot {\rm Q}\left ( {1}/{\sigma_n} \right ) = 0.032 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Q}\left ( {1}/{\sigma_n} \right ) = 0.016\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {1}/{\sigma_n} \approx 2.14\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\hspace{0.1cm}\hspace{0.15cm}\sigma_n \hspace{0.15cm}\underline {\approx 0.467}\hspace{0.05cm}. $$
	{\rm Q}\left ( {1}/{\sigma_n} \right ) = 0.016\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {1}/{\sigma_n} \approx 2.14
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\hspace{0.1cm}\hspace{0.15cm}\sigma_n \hspace{0.15cm}\underline {\approx 0.467}\hspace{0.05cm}. $$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]
			[[de:4.10_Union_Bound]]

Guenter at 15:13, 6 September 2022

2022-09-06T15:13:37Z

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}

	[[File:~~P_ID2043__Dig_A_4_10~~.png\|right\|frame\|Signal space constellations with $N=2$  and  $M=3$ ~~'''KORREKTUR: Konfig.'''~~]]		[[File:EN_Dig_A_4_10.png\|right\|frame\|Signal space constellations with $N=2$  and  $M=3$]]
	The so-called  "Union Bound"  gives an upper bound for the error probability of a non-binary transmission system  $(M > 2)$.		The so-called  "Union Bound"  gives an upper bound for the error probability of a non-binary transmission system  $(M > 2)$.

Hwang at 12:16, 1 September 2022

2022-09-01T12:16:11Z

← Older revision		Revision as of 14:16, 1 September 2022
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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}

	[[File:P_ID2043__Dig_A_4_10.png\|right\|frame\|Signal space constellations with $N=2$  and  $M=3$]]		[[File:P_ID2043__Dig_A_4_10.png\|right\|frame\|Signal space constellations with $N=2$  and  $M=3$ '''KORREKTUR: Konfig.''']]
	The so-called  "Union Bound"  gives an upper bound for the error probability of a non-binary transmission system  $(M > 2)$.		The so-called  "Union Bound"  gives an upper bound for the error probability of a non-binary transmission system  $(M > 2)$.

Guenter at 13:59, 30 July 2022

2022-07-30T13:59:28Z

Guenter at 13:46, 30 July 2022

2022-07-30T13:46:49Z

← Older revision		Revision as of 15:46, 30 July 2022
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	[[File:P_ID2043__Dig_A_4_10.png\|right\|frame\|Signal space constellations with $N=2$  and  $M=3$]]		[[File:P_ID2043__Dig_A_4_10.png\|right\|frame\|Signal space constellations with $N=2$  and  $M=3$]]
	The so-called "Union Bound" gives an upper bound for the error probability of a non-binary transmission system  $(M > 2)$.		The so-called  "Union Bound"  gives an upper bound for the error probability of a non-binary transmission system  $(M > 2)$.

	The actual (average) error probability is generally given as follows:		*The actual  (average)  error probability is generally given as follows:
	:$${\rm Pr}({ \cal E}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sum\limits_{i = 0 }^{M-1} {\rm Pr}(m_i) \cdot {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{0.05cm},$$		:$${\rm Pr}({ \cal E}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sum\limits_{i = 0 }^{M-1} {\rm Pr}(m_i) \cdot {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{0.05cm},$$
	:$$ {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr} \left [ \bigcup_{k \ne i} { \cal E}_{ik}\right ] \hspace{0.05cm},\hspace{0.2cm}{ \rm where}\hspace{0.2cm}		:$$ {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr} \left [ \bigcup_{k \ne i} { \cal E}_{ik}\right ] \hspace{0.05cm},\hspace{0.2cm}{ \rm where}\hspace{0.2cm}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The simpler ''Union Bound''  provides an upper bound on the falsification probability assuming that message  $m_i$  $($or signal  $\boldsymbol{s}_i)$  was sent:		*The simpler  "Union Bound"  $\rm (UB)$  provides an upper bound on the falsification probability assuming that message  $m_i$  $($or signal  $\boldsymbol{s}_i)$  was sent:
	:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ \ge \ \hspace{-0.1cm} {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} \boldsymbol{ s }_i ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i )\hspace{0.05cm},\ $$		:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ \ge \ \hspace{-0.1cm} {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} \boldsymbol{ s }_i ) = {\rm Pr}({ \cal E}\hspace{0.05cm}\|\hspace{0.05cm} m_i )\hspace{0.05cm},\ $$
	:$$ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ = \ \hspace{-0.2cm}\sum\limits_{k = 0 ,\hspace{0.1cm} k \ne i}^{M-1}\hspace{-0.1cm}		:$$ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \hspace{-0.1cm} \ = \ \hspace{-0.2cm}\sum\limits_{k = 0 ,\hspace{0.1cm} k \ne i}^{M-1}\hspace{-0.1cm}
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	The following abbreviations are used:		The following abbreviations are used:
	* ${\rm Q}(x)$  is the complementary Gaussian error function;		* ${\rm Q}(x)$  is the complementary Gaussian error function;

	* $d_{ik}$  denotes the distance between the signal points  $\boldsymbol{s}_i$  and $\boldsymbol{s}_k$;		* $d_{ik}$  denotes the distance between the signal points  $\boldsymbol{s}_i$  and $\boldsymbol{s}_k$;
	* $\sigma_n$  is the rms value (⇒ root of the variance) of the additive white Gaussian noise.

			* $\sigma_n$  is the rms value  (⇒ root of the variance)  of the additive white Gaussian noise.

	By averaging over all possible signals  $\boldsymbol{s}_i$,  we then arrive at the actual ~~''Union Bound''~~ :
			⇒   By averaging over all possible signals  $\boldsymbol{s}_i$,  we then arrive at the actual  "Union Bound":
	:$$p_{\rm UB} = \sum\limits_{i = 0 }^{M-1} {\rm Pr}(\boldsymbol{ s }_i) \cdot p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \ge {\rm Pr}({ \cal E}) \hspace{0.05cm}.$$		:$$p_{\rm UB} = \sum\limits_{i = 0 }^{M-1} {\rm Pr}(\boldsymbol{ s }_i) \cdot p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_i} \ge {\rm Pr}({ \cal E}) \hspace{0.05cm}.$$

	*The graph shows three different signal space constellations, each with  $M = 3$  signal space points  $\boldsymbol{s}_0$,  $\boldsymbol{s}_1$  and  $\boldsymbol{s}_2$  in two-dimensional space  $(N = 2)$.		*The graph shows three different signal space constellations, <br>each with  $M = 3$  signal space points  $\boldsymbol{s}_0$,  $\boldsymbol{s}_1$  and  $\boldsymbol{s}_2$  in two-dimensional space  $(N = 2)$.

	*The basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$  are suitably normalized.		*The basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$  are suitably normalized.
	*Thus, the signal space coordinates are also pure numerical values without unit:
			*Thus,  the signal space coordinates are also pure numerical values without unit:
	:$$\boldsymbol{ s }_1 = (-1, \hspace{0.1cm}+1)\hspace{0.05cm}, \hspace{0.2cm}		:$$\boldsymbol{ s }_1 = (-1, \hspace{0.1cm}+1)\hspace{0.05cm}, \hspace{0.2cm}
	\boldsymbol{ s }_2 = (+1, \hspace{0.1cm}+1)\hspace{0.05cm}.$$		\boldsymbol{ s }_2 = (+1, \hspace{0.1cm}+1)\hspace{0.05cm}.$$

	The signal space point  $\boldsymbol{s}_0$  in configuration  $\rm A$  is located such that  $\boldsymbol{s}_0$,  $\boldsymbol{s}_1$,  $\boldsymbol{s}_2$  describe an equilateral triangle~~. In contrast, in configurations  $\rm B$  and  $\rm C$,    $\boldsymbol{s}_0 = (0, 0)$  and  $\boldsymbol{s}_0 = (0, \ -1)$, respectively~~.		*The signal space point  $\boldsymbol{s}_0$  in configuration  $\rm A$  is located such that  $\boldsymbol{s}_0$,  $\boldsymbol{s}_1$,  $\boldsymbol{s}_2$  describe an equilateral triangle.

			*In contrast,  in configurations  $\rm B$  and  $\rm C$,    $\boldsymbol{s}_0 = (0,\ 0)$  resp.  $\boldsymbol{s}_0 = (0, \ –1)$.




	''Notes:''
	* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability\|"Approximation of the Error Probability"]].		Notes:
	*Use the AWGN rms value  $\sigma_n = 0.5$ for all calculations.		* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability\|"Approximation of the Error Probability"]].

			*Use the AWGN rms value  $\sigma_n = 0.5$  for all calculations.

	* Given the following values of the complementary Gaussian error function:		* Given the following values of the complementary Gaussian error function:
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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Which of the three configurations leads to the smallest error probability (at least according to the ''Union Bound'' approximation)?		{Which of the three configurations leads to the smallest error probability  $($at least according to the Union Bound approximation$)$?
	\|type="()"}		\|type="()"}
	- Configuration  $\rm A$,		- Configuration  $\rm A$,
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	+ Configuration  $\rm C$.		+ Configuration  $\rm C$.

	{Calculate the "averaged Union Bound"  $(p_{\rm UB})$  for configuration  $\rm A$.		{Calculate the  "averaged Union Bound"  $(p_{\rm UB})$  for configuration  $\rm A$.
	\|type="{}"}		\|type="{}"}
	$p_{\rm UB} \ = \ ${ 4.6 3% } $\ \%$		$p_{\rm UB} \ = \ ${ 4.6 3% } $\ \%$

	{Calculate the "averaged Union Bound" for configuration  $\rm B$.		{Calculate the  "averaged Union Bound"  for configuration  $\rm B$.
	\|type="{}"}		\|type="{}"}
	$p_{\rm UB} \ = \ ${ 12.1 3% } $\ \%$		$p_{\rm UB} \ = \ ${ 12.1 3% } $\ \%$

	{Calculate the "averaged Union Bound" for configuration  $\rm C$.		{Calculate the  "averaged Union Bound"  for configuration  $\rm C$.
	\|type="{}"}		\|type="{}"}
	$p_{\rm UB} \ = \ ${ 3.2 3% } $\ \%$		$p_{\rm UB} \ = \ ${ 3.2 3% } $\ \%$

	{How would the noise rms value  $\sigma_n$  have to be changed for configuration  $\rm A$  to yield the same ~~<i>~~Union Bound~~</i>~~  as in '''(4)'''?		{How would the noise rms value  $\sigma_n$  have to be changed for configuration  $\rm A$  to yield the same  "Union Bound"  as in subtask  '''(4)'''?
	\|type="{}"}		\|type="{}"}
	$\sigma_n \ = \ ${ 0.467 3% }		$\sigma_n \ = \ ${ 0.467 3% }

Hwang at 08:40, 18 July 2022

2022-07-18T08:40:00Z

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Guenter at 10:12, 23 June 2022

2022-06-23T10:12:40Z

← Older revision		Revision as of 12:12, 23 June 2022
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	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]

Guenter: Guenter moved page Aufgaben:Exercis 4.10: Union Bound to Aufgaben:Exercise 4.10: Union Bound

2022-06-23T10:06:14Z

Guenter moved page Aufgaben:Exercis 4.10: Union Bound to Aufgaben:Exercise 4.10: Union Bound

← Older revision	Revision as of 12:06, 23 June 2022
(No difference)

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Points $\boldsymbol{s}_1$ and $\boldsymbol{s}_2$ are the same for all configurations.		'''(1)'''  Points  $\boldsymbol{s}_1$  and  $\boldsymbol{s}_2$  are the same for all configurations.
	*The smallest error probability occurs when $\boldsymbol{s}_0$ is farthest from $\boldsymbol{s}_1$ and $\boldsymbol{s}_2$.		*The smallest error probability occurs when  $\boldsymbol{s}_0$  is farthest from  $\boldsymbol{s}_1$  and  $\boldsymbol{s}_2$.
	*This is the case for configuration ~~'''~~C~~'''~~   ⇒   <u>solution 3</u>.
			*This is the case for configuration  $\rm C$   ⇒   <u>solution 3</u>.



	'''(2)'''  For configuration ~~'''~~A~~'''~~, the distance between all points is the same: $d_{01} = d_{02} = d_{12} = 2$.		'''(2)'''  For configuration  $\rm A$,  the distance between all points is the same:  $d_{01} = d_{02} = d_{12} = 2$.
	*Therefore, to calculate the ~~<i>~~Union Bound~~</i>~~, it is not necessary to average over all symbols~~, and~~ it is valid since~~, for example,~~ $\boldsymbol{s}_0$ is distorted into symbol $\boldsymbol{s}_1$ or $\boldsymbol{s}_2$ with equal probability:		*Therefore,  to calculate the  "Union Bound",  it is not necessary to average over all symbols.

			*And it is valid since e.g.  $\boldsymbol{s}_0$  is distorted into symbol  $\boldsymbol{s}_1$  or  $\boldsymbol{s}_2$  with equal probability:
	:$${\rm Pr}({ \cal E}) \le p_{\rm UB} = 2 \cdot {\rm Q} \left ( \frac{d_{ik}/2}{\sigma_n} \right ) = 2 \cdot {\rm Q}(2) \approx 2 \cdot 0.023 \hspace{0.1cm}\hspace{0.15cm}\underline {= 4.6\%} \hspace{0.05cm}. $$		:$${\rm Pr}({ \cal E}) \le p_{\rm UB} = 2 \cdot {\rm Q} \left ( \frac{d_{ik}/2}{\sigma_n} \right ) = 2 \cdot {\rm Q}(2) \approx 2 \cdot 0.023 \hspace{0.1cm}\hspace{0.15cm}\underline {= 4.6\%} \hspace{0.05cm}. $$


	'''(3)'''  Here, the falsification probabilities differ for each symbol.		'''(3)'''  Here,  the falsification probabilities differ for each symbol.
	*If $\boldsymbol{s}_0$ was sent, with $d_{01} = d_{02} = 2^{0.5}$ and $\sigma = 0.5$:		*If  $\boldsymbol{s}_0$  was sent,  with $d_{01} = d_{02} = 2^{0.5}$  and  $\sigma = 0.5$:
	:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} = 2 \cdot {\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )		:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} = 2 \cdot {\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )
	= 2 \cdot {\rm Q}(\sqrt{2}) = 2 \cdot 0.079 = 0.158 \hspace{0.05cm}. $$		= 2 \cdot {\rm Q}(\sqrt{2}) = 2 \cdot 0.079 = 0.158 \hspace{0.05cm}. $$

	*In contrast, the other two conditional probabilities are smaller:		*In contrast,  the other two conditional probabilities are smaller:
	:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} = p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}		:$$p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} = p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}
	{\rm Q} \left ( \frac{{2}/2}{0.5} \right )+{\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )= {\rm Q}(2) +{\rm Q}(\sqrt{2}) = 0.023 + 0.079 = 0.102 \hspace{0.05cm}.$$		{\rm Q} \left ( \frac{{2}/2}{0.5} \right )+{\rm Q} \left ( \frac{\sqrt{2}/2}{0.5} \right )= {\rm Q}(2) +{\rm Q}(\sqrt{2}) = 0.023 + 0.079 = 0.102 \hspace{0.05cm}.$$

	By averaging, we obtain for the Union Bound considering the different distances:		*By averaging,  we obtain for the Union Bound considering the different distances:
	:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{3} \cdot \left [ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} + p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} +p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2}\right ]=		:$$p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{3} \cdot \left [ p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_0} + p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_1} +p_{{\rm UB}\hspace{0.05cm}\| \hspace{0.05cm}\boldsymbol{ s }_2}\right ]=
	{1}/{3} \cdot \left [ 2 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot ({\rm Q}({2}) + {\rm Q}(\sqrt{2})) \right ] =		{1}/{3} \cdot \left [ 2 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot ({\rm Q}({2}) + {\rm Q}(\sqrt{2})) \right ] =
	{1}/{3} \cdot \left [ 4 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot {\rm Q}({2}) \right ] $$		{1}/{3} \cdot \left [ 4 \cdot {\rm Q}(\sqrt{2})+ 2 \cdot {\rm Q}({2}) \right ] $$
	:$$ \Rightarrow \hspace{0.3cm} p_{\rm UB} =		:$$ \Rightarrow \hspace{0.3cm} p_{\rm UB} =
	{1}/{3} \cdot \~~left~~ [ 4 \cdot 0.079+ 2 \cdot 0.023 \~~right~~ ] \hspace{0.1cm}\hspace{0.12cm}\underline {\approx 12.1\% } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm UB}\ge {\rm Pr}({ \cal E})\hspace{0.05cm}.$$		{1}/{3} \cdot \big [ 4 \cdot 0.079+ 2 \cdot 0.023 \big ] \hspace{0.1cm}\hspace{0.12cm}\underline {\approx 12.1\% } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm UB}\ge {\rm Pr}({ \cal E})\hspace{0.05cm}.$$


Line 113:		Line 116:
	:$$d_{01} = d_{02} = \sqrt{2^2 + 1^2}= \sqrt{5} \approx 2.24\hspace{0.2cm}, d_{12} = 2$$		:$$d_{01} = d_{02} = \sqrt{2^2 + 1^2}= \sqrt{5} \approx 2.24\hspace{0.2cm}, d_{12} = 2$$
	:$$\Rightarrow \hspace{0.3cm} p_{\rm UB} =		:$$\Rightarrow \hspace{0.3cm} p_{\rm UB} =
	{1}/{3} \cdot \~~left~~ [ 4 \cdot {\rm Q}(\sqrt{5})+ 2 \cdot {\rm Q}({2}) \~~right~~ ] = {1}/{3} \cdot \~~left~~ [ 4 \cdot 0.013+ 2 \cdot 0.023 \~~right~~ ]\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 3.2\%} \hspace{0.05cm}. $$		{1}/{3} \cdot \big [ 4 \cdot {\rm Q}(\sqrt{5})+ 2 \cdot {\rm Q}({2}) \big ] = {1}/{3} \cdot \big [ 4 \cdot 0.013+ 2 \cdot 0.023 \big ]\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 3.2\%} \hspace{0.05cm}. $$