Aufgaben:Exercise 4.09: Decision Regions at Laplace - 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:Aufgaben:4.~~09_Entscheidungsregionen_bei_Laplace~~]]		[[de:Aufgaben:Aufgabe 4.09: Entscheidungsregionen bei Laplace]]

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	[[File:P_ID2044__Dig_A_4_9.png\|right\|frame\|Three Laplace decision regions]]		[[File:P_ID2044__Dig_A_4_9.png\|right\|frame\|Three Laplace decision regions]]
	We consider a transmission system based on the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  The two equally probable transmitted signals are given by the signal points		We consider a transmission system based on the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  The two equally probable transmitted signals are given by the signal points
	:$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm}		:$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm}\boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}.$$
	\boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}.$$

	In the following we normalize the energy parameter to  $E = 1$  for simplification and thus obtain		In the following we normalize the energy parameter to  $E = 1$  for simplification and thus obtain
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	Let the noise components  $n_1(t)$  and  $n_2(t)$  be independent of each other and each be Laplace distributed with parameter  $a = 1$:		Let the noise components  $n_1(t)$  and  $n_2(t)$  be independent of each other and each be Laplace distributed with parameter  $a = 1$:
	:$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- \| \eta_1\|} \hspace{0.05cm}, \hspace{0.2cm}		:$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- \| \eta_1\|} \hspace{0.05cm}, \hspace{0.2cm}p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- \| \eta_2\|} \hspace{0.05cm} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \boldsymbol{ p }_{\boldsymbol{ n }} (\eta_1, \eta_2) = {1}/{4} \cdot {\rm e}^{- \| \eta_1\|- \| \eta_2\|} \hspace{0.05cm}. $$
	p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- \| \eta_2\|} \hspace{0.05cm} \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} \boldsymbol{ p }_{\boldsymbol{ n }} (\eta_1, \eta_2) = {1}/{4} \cdot {\rm e}^{- \| \eta_1\|- \| \eta_2\|} \hspace{0.05cm}. $$

	The properties of such a Laplace noise will be discussed in detail in  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]].		The properties of such a Laplace noise will be discussed in detail in  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]].

	The received signal  $\boldsymbol{r}$  is composed additively of the transmitted signal  $\boldsymbol{s}$  and the  noise component  $\boldsymbol{n}$:		The received signal  $\boldsymbol{r}$  is composed additively of the transmitted signal  $\boldsymbol{s}$  and the  noise component  $\boldsymbol{n}$:
	:$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n }		:$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n }\hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2)\hspace{0.05cm},\hspace{0.45cm}\boldsymbol{ s } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} ( s_1, s_2)\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n } = ( n_1, n_2)\hspace{0.05cm}. $$
	\hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2)
	\hspace{0.05cm},\hspace{0.45cm}
	\boldsymbol{ s } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} ( s_1, s_2)
	\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n } = ( n_1, n_2)
	\hspace{0.05cm}. $$

	The corresponding realizations are denoted as follows:		The corresponding realizations are denoted as follows:
	:$$\boldsymbol{ s }\text{:} \hspace{0.4cm} (s_{01},s_{02}){\hspace{0.15cm}\rm and \hspace{0.15cm}} (s_{11},s_{12})		:$$\boldsymbol{ s }\text{:} \hspace{0.4cm} (s_{01},s_{02}){\hspace{0.15cm}\rm and \hspace{0.15cm}} (s_{11},s_{12})\hspace{0.05cm},\hspace{0.8cm} \boldsymbol{ r }\text{:} \hspace{0.4cm} (\rho_{1},\rho_{2})\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n }\text{:} \hspace{0.4cm} (\eta_{1},\eta_{2})\hspace{0.05cm}.$$
	\hspace{0.05cm},\hspace{0.8cm} \boldsymbol{ r }\text{:} \hspace{0.4cm} (\rho_{1},\rho_{2})
	\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n }\text{:} \hspace{0.4cm} (\eta_{1},\eta_{2})
	\hspace{0.05cm}.$$

	The decision rule of the MAP and ML receivers  $($both are identical due to the same symbol probabilities$)$  are:		The decision rule of the MAP and ML receivers  $($both are identical due to the same symbol probabilities$)$  are:
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	⇒   With the further conditions for the  "decision for  $m_0$"  can also be written:		⇒   With the further conditions for the  "decision for  $m_0$"  can also be written:
	:$${1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.1cm} \right ] >		:$${1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.1cm} \right ] >{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.1cm} \right ] $$
	{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.1cm} \right ] $$		:$$\Rightarrow \hspace{0.3cm} \| \rho_1 +1\|+ \| \rho_2 +1\| <\| \rho_1 -1\|+ \| \rho_2 -1\|$$
	:$$\Rightarrow \hspace{0.3cm} \| \rho_1 +1\|+ \| \rho_2 +1\| <		:$$\Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = \| \rho_1 +1\|+ \| \rho_2 +1\| -\| \rho_1 -1\|- \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$
	\| \rho_1 -1\|+ \| \rho_2 -1\|$$
	:$$\Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = \| \rho_1 +1\|+ \| \rho_2 +1\| -
	\| \rho_1 -1\|- \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$

	⇒   This function  $L(\rho_1, \rho_2)$  is frequently referred to in the questions.		⇒   This function  $L(\rho_1, \rho_2)$  is frequently referred to in the questions.
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	'''(1)'''  <u>Solutions 1 and 2</u>  are correct:		'''(1)'''  <u>Solutions 1 and 2</u>  are correct:
	*The joint probability densities under conditions  $m_0$  and  $m_1$,  respectively,  are:		*The joint probability densities under conditions  $m_0$  and  $m_1$,  respectively,  are:
	:$$p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 )		:$$p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 )\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.05cm} \right ]\hspace{0.05cm},$$
	\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.05cm} \right ]\hspace{0.05cm},$$		:$$p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_1 )\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.05cm} \right ]\hspace{0.05cm}.$$
	:$$p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_1 )
	\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.05cm} \right ]\hspace{0.05cm}.$$
	*For equally probable symbols   ⇒   ${\rm Pr}(m_0) = {\rm Pr}(m_1) = 0.5$,  the MAP decision rule is:   Decide for symbol  $m_0$    ⇔   signal  $s_0$,  if		*For equally probable symbols   ⇒   ${\rm Pr}(m_0) = {\rm Pr}(m_1) = 0.5$,  the MAP decision rule is:   Decide for symbol  $m_0$    ⇔   signal  $s_0$,  if
	:$$p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 ) >		:$$p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 ) >p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } (\rho_{1},\rho_{2} \|m_1 ) \hspace{0.05cm}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.05cm} \right ] >{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\|\hspace{0.05cm} \right ] $$
	p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } (\rho_{1},\rho_{2} \|m_1 ) \hspace{0.05cm}\hspace{0.3cm}		:$$\Rightarrow \hspace{0.3cm} \| \rho_1 +1\|+ \| \rho_2 +1\| <\| \rho_1 -1\|+ \| \rho_2 -1\|\hspace{0.3cm}\Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = \| \rho_1 +1\|-\| \rho_1 -1\|+ \| \rho_2 +1\| - \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} {1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.05cm} \right ] >
	{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\|\hspace{0.05cm} \right ] $$
	:$$\Rightarrow \hspace{0.3cm} \| \rho_1 +1\|+ \| \rho_2 +1\| <
	\| \rho_1 -1\|+ \| \rho_2 -1\|\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = \| \rho_1 +1\|-
	\| \rho_1 -1\|+ \| \rho_2 +1\| - \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$


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	'''(3)'''  <u>Solution 1</u>  is correct:		'''(3)'''  <u>Solution 1</u>  is correct:
	*The result of subtask  '''(1)'''  was:  Decide for the symbol  $m_0$,  if		*The result of subtask  '''(1)'''  was:  Decide for the symbol  $m_0$,  if
	:$$L (\rho_1, \rho_2) = \| \rho_1 +1\| -		:$$L (\rho_1, \rho_2) = \| \rho_1 +1\| -\| \rho_1 -1\|+ \| \rho_2 +1\| - \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$
	\| \rho_1 -1\|+ \| \rho_2 +1\| - \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$
	*In the considered  (inner)  range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$  holds with the result of subtask  '''(2)''':		*In the considered  (inner)  range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$  holds with the result of subtask  '''(2)''':
	:$$\| \rho_1+1\| - \| \rho_1 -1\| = 2\rho_1 \hspace{0.05cm}, \hspace{0.2cm} \| \rho_2+1\| - \| \rho_2 -1\| = 2\rho_2 \hspace{0.05cm}.$$		:$$\| \rho_1+1\| - \| \rho_1 -1\| = 2\rho_1 \hspace{0.05cm}, \hspace{0.2cm} \| \rho_2+1\| - \| \rho_2 -1\| = 2\rho_2 \hspace{0.05cm}.$$

	*If we insert this result above,  we have to decide for  $m_0$  exactly if		*If we insert this result above,  we have to decide for  $m_0$  exactly if
	:$$L (\rho_1, \rho_2) = 2 \cdot ( \rho_1+\rho_2) < 0		:$$L (\rho_1, \rho_2) = 2 \cdot ( \rho_1+\rho_2) < 0\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \rho_1+\rho_2 < 0\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \rho_1+\rho_2 < 0\hspace{0.05cm}.$$


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	'''(5)'''  <u>Solution 1</u>  is correct:		'''(5)'''  <u>Solution 1</u>  is correct:
	*After similar calculation as in subtask  '''(3)''',  one arrives at the result:		*After similar calculation as in subtask  '''(3)''',  one arrives at the result:
	:$$L (\rho_1, \rho_2) = -2 + D_2 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$L (\rho_1, \rho_2) = -2 + D_2 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$
	{\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$


	'''(6)'''  <u>Solution 2</u>  is correct:  Decision on  $m_1$.		'''(6)'''  <u>Solution 2</u>  is correct:  Decision on  $m_1$.
	*Similar to subtask  '''(4)''',  the following holds here:		*Similar to subtask  '''(4)''',  the following holds here:
	:$$D_1 = \| \rho_1 +1\| - \| \rho_1 -1\| \in \{-2, ... \hspace{0.05cm} , +2 \}		:$$D_1 = \| \rho_1 +1\| - \| \rho_1 -1\| \in \{-2, ... \hspace{0.05cm} , +2 \}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}L (\rho_1, \rho_2) = 2 + D_1 \ge 0\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}L (\rho_1, \rho_2) = 2 + D_1 \ge 0
	\hspace{0.05cm}.$$


	'''(7)'''  <u>Solution 1</u>  is correct:  Decision on  $m_0$.		'''(7)'''  <u>Solution 1</u>  is correct:  Decision on  $m_0$.
	*After similar reasoning as in the last subtask,  we arrive at the result:		*After similar reasoning as in the last subtask,  we arrive at the result:
	:$$L (\rho_1, \rho_2) = -2 + D_1 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$L (\rho_1, \rho_2) = -2 + D_1 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$
	{\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]
			[[de:Aufgaben:4.09_Entscheidungsregionen_bei_Laplace]]

Guenter at 15:49, 29 July 2022

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Guenter at 14:12, 29 July 2022

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← Older revision		Revision as of 16:12, 29 July 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_ID2044__Dig_A_4_9.png\|right\|frame\|Three decision regions ~~<br>for Laplace~~]]		[[File:P_ID2044__Dig_A_4_9.png\|right\|frame\|Three Laplace decision regions]]
	We consider a transmission system based on the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$. The two equally probable transmitted signals are given by the signal points		We consider a transmission system based on the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  The two equally probable transmitted signals are given by the signal points
	:$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm}		:$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm}
	\boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}$$.		\boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}$$.
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	The messages  $m_0$  and  $m_1$  are uniquely assigned to the signals  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$  defined in this way.		The messages  $m_0$  and  $m_1$  are uniquely assigned to the signals  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$  defined in this way.

	Let the ~~two~~ noise components  $n_1(t)$  and  $n_2(t)$  be independent of each other and each be Laplace distributed with parameter  $a = 1$:		Let the noise components  $n_1(t)$  and  $n_2(t)$  be independent of each other and each be Laplace distributed with parameter  $a = 1$:
	:$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- \| \eta_1\|} \hspace{0.05cm}, \hspace{0.2cm}		:$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- \| \eta_1\|} \hspace{0.05cm}, \hspace{0.2cm}
	p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- \| \eta_2\|} \hspace{0.05cm} \hspace{0.3cm}		p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- \| \eta_2\|} \hspace{0.05cm} \hspace{0.3cm}
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	The properties of such a Laplace noise will be discussed in detail in  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]].		The properties of such a Laplace noise will be discussed in detail in  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]].

	The received signal  $\boldsymbol{r}$  is composed additively of the transmitted signal  $\boldsymbol{s}$  and the  noise ~~signal~~  $\boldsymbol{n}$:		The received signal  $\boldsymbol{r}$  is composed additively of the transmitted signal  $\boldsymbol{s}$  and the  noise component  $\boldsymbol{n}$:
	:$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n }		:$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n }
	\hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2)		\hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2)
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The decision rule of the MAP and ML receivers (both are identical due to the same symbol probabilities) are:		The decision rule of the MAP and ML receivers  $($both are identical due to the same symbol probabilities$)$  are:

	:Decide for the symbol  $m_0$, if   $p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 ) >		⇒   Decide for the symbol  $m_0$, if   $p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 ) >
	p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } (\rho_{1},\rho_{2} \|m_1 ) \hspace{0.05cm}.$		p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } (\rho_{1},\rho_{2} \|m_1 ) \hspace{0.05cm}.$

	With the further conditions for ~~this $($~~decision for  $m_0)$  can also be written:		⇒   With the further conditions for the  "decision for  $m_0$"  can also be written:
	:$${1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.1cm} \right ] >		:$${1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.1cm} \right ] >
	{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.1cm} \right ] $$		{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.1cm} \right ] $$
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	\| \rho_1 -1\|- \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$		\| \rho_1 -1\|- \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$

	This function  $L(\rho_1, \rho_2)$  is frequently referred to in the questions.		⇒   This function  $L(\rho_1, \rho_2)$  is frequently referred to in the questions.


	The graph shows three different decision regions  $(I_0, \ I_1)$.		The graph shows three different decision regions  $(I_0, \ I_1)$.
	*For AWGN noise, only the upper variant  $\rm A$  would be optimal.		:*For AWGN noise,  only the upper variant  $\rm A$  would be optimal.
	*Also for the considered Laplace noise, variant  $\rm A$  leads to the smallest possible error probability, see  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]]:
	~~:$$p_{\rm min} = {\rm Pr}({\cal{E}} \hspace{0.05cm}\|\hspace{0.05cm} {\rm optimal\hspace{0.15cm} receiver}) = {\rm e}^{-2} \approx 13.5\,\%\hspace{0.05cm}.$$~~

	*It is to be examined whether variant  $\rm B$  or variant  $\rm C$  is also optimal, i.e. whether their error probabilities are also as small as possible equal to  $p_{\rm min}$.


			:*Also for the considered Laplace noise, variant  $\rm A$  leads to the smallest possible error probability, see  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]]:
			::$$p_{\rm min} = {\rm Pr}({\cal{E}} \hspace{0.05cm}\|\hspace{0.05cm} {\rm optimal\hspace{0.15cm} receiver}) = {\rm e}^{-2} \approx 13.5\,\%\hspace{0.05cm}.$$
			:*It is to be examined whether variant  $\rm B$  or variant  $\rm C$  is also optimal, i.e. whether their error probabilities are also as small as possible equal to  $p_{\rm min}$.



	''Note:''		Note:   The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability\|"Approximation of the Error Probability"]].
	* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability\|"Approximation of the Error Probability"]].


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Which of the decision rules are correct? Decide for  $m_0$, if		{Which of the decision rules are correct?  Decide for  $m_0$,  if
	\|type="[]"}		\|type="[]"}
	+ $p_{\it r\hspace{0.03cm}\|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}\|\hspace{0.03cm}m_0) > p_{\it r\hspace{0.03cm}\|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}\|\hspace{0.03cm}m_1)$,		+ $p_{\it r\hspace{0.03cm}\|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}\|\hspace{0.03cm}m_0) > p_{\it r\hspace{0.03cm}\|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}\|\hspace{0.03cm}m_1)$,
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	{How can the expression  $\|x+1\| \ -\|x \ -1\|$  be transformed?		{How can the expression  $\|x+1\| \ -\|x \ -1\|$  be transformed?
	\|type="[]"}		\|type="[]"}
	+ For  $x ≥ +1$,    $\|x + 1\| \, -\|x -1\| = 2$.		+ For  $x ≥ +1$:    $\|x + 1\| \, -\|x -1\| = 2$.
	+ For  $x ≤ \, -1$,    $\|x+1\| \,-\|x \, -1\| = \, -2$.		+ For  $x ≤ \, -1$:    $\|x+1\| \,-\|x \, -1\| = \, -2$.
	+ For  $-1 ≤ x ≤ +1$,    $\|x+1\| \, -\|x \, -1\| = 2x$.		+ For  $-1 ≤ x ≤ +1$:    $\|x+1\| \, -\|x \, -1\| = 2x$.

	{What is the decision rule in the range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$?		{What is the decision rule in the range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$?
	\|type="()"}		\|type="()"}
	+ Decision for  $m_0$, if  $\rho_1 + \rho_2 < 0$.		+ Decision for  $m_0$,  if  $\rho_1 + \rho_2 < 0$.
	- Decision for  $m_1$, if  $\rho_1 + \rho_2 < 0$.		- Decision for  $m_1$,  if  $\rho_1 + \rho_2 < 0$.

	{What is the decision rule in the range  $\rho_1 > +1$?		{What is the decision rule in the range  $\rho_1 > +1$?

Hwang at 06:54, 18 July 2022

2022-07-18T06:54:45Z

Show changes

Hwang at 16:00, 11 July 2022

2022-07-11T16:00:02Z

← Older revision		Revision as of 18:00, 11 July 2022
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	{{quiz-Header\|Buchseite=~~Digitalsignalübertragung~~/~~Approximation der Fehlerwahrscheinlichkeit~~}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}

	[[File:P_ID2044__Dig_A_4_9.png\|right\|frame\|~~Drei Entscheidungsregionen~~ <br>~~für~~ Laplace]]		[[File:P_ID2044__Dig_A_4_9.png\|right\|frame\|Three decision regions <br>for Laplace]]
	~~Wir betrachten ein Übertragungssystem, basierend auf den Basisfunktionen~~  $\varphi_1(t)$  ~~und~~  $\varphi_2(t)$. ~~Die beiden gleichwahrscheinlichen Sendesignale sind durch die Signalpunkte~~		We consider a transmission system based on the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$. The two equally probable transmitted signals are given by the signal points
	:$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm}		:$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm}
	\boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}$$		\boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}$$.

	~~gegeben. Im Folgenden normieren wir zur Vereinfachung den Energieparameter zu~~  $E = 1$  ~~und erhalten somit~~		In the following we normalize the energy parameter to  $E = 1$  for simplification and thus obtain
	:$$\boldsymbol{ s }_0 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (-1, \hspace{0.1cm}-1) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_0\hspace{0.05cm}, $$		:$$\boldsymbol{ s }_0 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (-1, \hspace{0.1cm}-1) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_0\hspace{0.05cm}, $$
	:$$ \boldsymbol{ s }_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (+1, \hspace{0.1cm}+1)\hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_1\hspace{0.05cm}.$$		:$$ \boldsymbol{ s }_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (+1, \hspace{0.1cm}+1)\hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_1\hspace{0.05cm}.$$

	~~Die Nachrichten~~  $m_0$  ~~und~~  $m_1$  ~~sind den so festgelegten Signalen~~  $\boldsymbol{s}_0$  ~~und~~  $\boldsymbol{s}_1$  ~~eindeutig zugeordnet~~.		The messages  $m_0$  and  $m_1$  are uniquely assigned to the signals  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$  defined in this way.

	~~Die zwei Rauschkomponenten~~  $n_1(t)$  ~~und~~  $n_2(t)$  ~~seien unabhängig voneinander und jeweils laplace–verteilt mit Parameter~~  $a = 1$:		Let the two noise components  $n_1(t)$  and  $n_2(t)$  be independent of each other and each be Laplace distributed with parameter  $a = 1$:
	:$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- \| \eta_1\|} \hspace{0.05cm}, \hspace{0.2cm}		:$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- \| \eta_1\|} \hspace{0.05cm}, \hspace{0.2cm}
	p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- \| \eta_2\|} \hspace{0.05cm} \hspace{0.3cm}		p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- \| \eta_2\|} \hspace{0.05cm} \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} \boldsymbol{ p }_{\boldsymbol{ n }} (\eta_1, \eta_2) = {1}/{4} \cdot {\rm e}^{- \| \eta_1\|- \| \eta_2\|} \hspace{0.05cm}. $$		\Rightarrow \hspace{0.3cm} \boldsymbol{ p }_{\boldsymbol{ n }} (\eta_1, \eta_2) = {1}/{4} \cdot {\rm e}^{- \| \eta_1\|- \| \eta_2\|} \hspace{0.05cm}. $$

	~~Die Eigenschaften eines solchen~~ Laplace~~–Rauschens werden~~ in ~~der~~  [[Aufgaben:4.~~09Z_Laplace-verteiltes_Rauschen~~\| ~~Aufgabe~~ 4.9Z]]  ~~noch eingehend behandelt.~~		The properties of such a Laplace noise will be discussed in detail in  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]].

	~~Das Empfangssignal~~  $\boldsymbol{r}$  ~~setzt sich additiv aus dem Sendesignal~~  $\boldsymbol{s}$  ~~und dem~~  ~~Rauschsignal~~  $\boldsymbol{n}$ ~~zusammen~~:		The received signal  $\boldsymbol{r}$  is composed additively of the transmitted signal  $\boldsymbol{s}$  and the  noise signal  $\boldsymbol{n}$:
	:$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n }		:$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n }
	\hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2)		\hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2)
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	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

	~~Die entsprechenden Realisierungen sind wie folgt bezeichnet~~:		The corresponding realizations are denoted as follows:
	:$$\boldsymbol{ s }\text{:} \hspace{0.4cm} (s_{01},s_{02}){\hspace{0.15cm}\rm ~~bzw.~~ \hspace{0.15cm}} (s_{11},s_{12})		:$$\boldsymbol{ s }\text{:} \hspace{0.4cm} (s_{01},s_{02}){\hspace{0.15cm}\rm and \hspace{0.15cm}} (s_{11},s_{12})
	\hspace{0.05cm},\hspace{0.8cm} \boldsymbol{ r }\text{:} \hspace{0.4cm} (\rho_{1},\rho_{2})		\hspace{0.05cm},\hspace{0.8cm} \boldsymbol{ r }\text{:} \hspace{0.4cm} (\rho_{1},\rho_{2})
	\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n }\text{:} \hspace{0.4cm} (\eta_{1},\eta_{2})		\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n }\text{:} \hspace{0.4cm} (\eta_{1},\eta_{2})
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~Die Entscheidungsregel des~~ MAP~~– und des~~ ML~~–Empfängers~~ (~~beide sind aufgrund der gleichen Symbolwahrscheinlichkeiten identisch~~) ~~lauten~~:		The decision rule of the MAP and ML receivers (both are identical due to the same symbol probabilities) are:

	:~~Entscheide für das Symbol~~  $m_0$, ~~falls~~   $p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 ) >		:Decide for the symbol  $m_0$, if   $p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} \|m_0 ) >
	p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } (\rho_{1},\rho_{2} \|m_1 ) \hspace{0.05cm}.$		p_{\boldsymbol{ r} \hspace{0.05cm}\|\hspace{0.05cm}m } (\rho_{1},\rho_{2} \|m_1 ) \hspace{0.05cm}.$

	~~Mit den weiteren Voraussetzungen kann hierfür~~ $($~~Entscheidung für~~  $m_0)$  ~~auch geschrieben werden~~:		With the further conditions for this $($decision for  $m_0)$  can also be written:
	:$${1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.1cm} \right ] >		:$${1}/{4} \cdot {\rm exp}\left [- \| \rho_1 +1\|- \| \rho_2 +1\| \hspace{0.1cm} \right ] >
	{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.1cm} \right ] $$		{1}/{4} \cdot {\rm exp}\left [- \| \rho_1 -1\|- \| \rho_2 -1\| \hspace{0.1cm} \right ] $$
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	\| \rho_1 -1\|- \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$		\| \rho_1 -1\|- \| \rho_2 -1\| < 0 \hspace{0.05cm}.$$

	~~Auf diese Funktion~~  $L(\rho_1, \rho_2)$  ~~wird im Fragebogen häufig Bezug genommen~~.		This function  $L(\rho_1, \rho_2)$  is frequently referred to in the questions.

	~~Die Grafik zeigt drei verschiedene Entscheidungsregionen~~  $(I_0, \ I_1)$.		The graph shows three different decision regions  $(I_0, \ I_1)$.
	*~~Bei~~ AWGN~~–Rauschen wäre nur die obere Variante~~  $\rm A$  optimal.		*For AWGN noise, only the upper variant  $\rm A$  would be optimal.
	*~~Auch beim betrachteten~~ Laplace~~–Rauschen führt die Variante~~  $\rm A$  ~~zur kleinstmöglichen Fehlerwahrscheinlichkeit~~, ~~siehe~~  [[Aufgaben:4.~~09Z_Laplace-verteiltes_Rauschen~~\| ~~Aufgabe~~ 4.9Z]]:		*Also for the considered Laplace noise, variant  $\rm A$  leads to the smallest possible error probability, see  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise\|"Exercise 4.9Z"]]:
	:$$p_{\rm min} = {\rm Pr}({\cal{E}} \hspace{0.05cm}\|\hspace{0.05cm} {\rm ~~optimaler~~\hspace{0.15cm} ~~Empf\ddot{a}nger~~}) = {\rm e}^{-2} \approx 13.5\,\%\hspace{0.05cm}.$$		:$$p_{\rm min} = {\rm Pr}({\cal{E}} \hspace{0.05cm}\|\hspace{0.05cm} {\rm optimal\hspace{0.15cm} receiver}) = {\rm e}^{-2} \approx 13.5\,\%\hspace{0.05cm}.$$

	*~~Zu untersuchen ist, ob die Variante~~  $\rm B$  ~~bzw. die Variante~~  $\rm C$  ~~ebenfalls~~ optimal ~~ist~~, ~~das heißt, ob auch deren Fehlerwahrscheinlichkeiten kleinstmöglich gleich~~  $p_{\rm min}$  ~~sind.~~		*It is to be examined whether variant  $\rm B$  or variant  $\rm C$  is also optimal, i.e. whether their error probabilities are also as small as possible equal to  $p_{\rm min}$.


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	''~~Hinweis~~:''		''Note:''
	* ~~Die Aufgabe gehört zum Kapitel~~  [[~~Digitalsignal%C3%BCbertragung~~/~~Approximation_der_Fehlerwahrscheinlichkeit~~\| Approximation ~~der Fehlerwahrscheinlichkeit~~]].		* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability\|"Approximation of the Error Probability"]].


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	===~~Fragebogen~~===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Welche der Entscheidungsregeln sind richtig? Entscheide für  $m_0$, falls		{Welche der Entscheidungsregeln sind richtig? Entscheide für  $m_0$, falls

Guenter at 10:10, 23 June 2022

2022-06-23T10:10:57Z

← Older revision		Revision as of 12:10, 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:Aufgabe 4.09: Entscheidungsregionen bei Laplace to Aufgaben:Exercise 4.09: Decision Regions at Laplace

2022-06-23T10:03:03Z

Guenter moved page Aufgaben:Aufgabe 4.09: Entscheidungsregionen bei Laplace to Aufgaben:Exercise 4.09: Decision Regions at Laplace

← Older revision	Revision as of 12:03, 23 June 2022
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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:51:36Z

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

← Older revision		Revision as of 14:51, 23 March 2021
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	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.3 BER-Approximation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]