Aufgaben:Exercise 1.4: 2S/3E Channel Model - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence/Independence^]]		[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence/Independence^]]
	[[de:Aufgaben:1.~~4_2S~~/3E-Kanalmodell]]		[[de:Aufgaben:Aufgabe 1.4: 2S/3E-Kanalmodell]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence/Independence^]]		[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence/Independence^]]
			[[de:Aufgaben:1.4_2S/3E-Kanalmodell]]

Guenter at 14:17, 30 November 2021

2021-11-30T14:17:04Z

Guenter at 14:12, 30 November 2021

2021-11-30T14:12:42Z

← Older revision		Revision as of 16:12, 30 November 2021
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	[[File:EN_Sto_A_1_4.png\|right\|frame\|$\rm 2S/3E$  channel model]]		[[File:EN_Sto_A_1_4.png\|right\|frame\|$\rm 2S/3E$  channel model]]
	A transmitter  (German:  "Sender"   ⇒   subscript  "S")  emits the binary symbols  $\rm L$  $($event  $S_{\rm L})$  and  $\rm H$  $($event  $S_{\rm H})$ .		A transmitter  (German:  "Sender"   ⇒   subscript  "S")  emits the binary symbols  $\rm L$  $($event  $S_{\rm L})$  and  $\rm H$  $($event  $S_{\rm H})$ .
	*If conditions are good, the digital receiver  (German:  "Empfänger"   ⇒   subscript  "E")  also decides only on the binary symbols  $\rm L$  $($event  $E_{\rm L})$  or  $\rm H$  $($event  $E_{\rm H})$.		*If conditions are good,  the digital receiver  (German:  "Empfänger"   ⇒   subscript  "E")  also decides only on the binary symbols  $\rm L$  $($event  $E_{\rm L})$  or  $\rm H$  $($event  $E_{\rm H})$.
	*However,  if the receiver can suspect that an error has occurred during transmission,  it makes no decision  $($event  $E_{\rm K})$;  <br>$\rm K$  here stands for  "No decision".		*However,  if the receiver can suspect that an error has occurred during transmission,  it makes no decision  $($event  $E_{\rm K})$;  <br>$\rm K$  here stands for  "No decision".

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	*It can be seen that a transmitted  $\rm L$  may well be received as a symbol  $\rm H$.		*It can be seen that a transmitted  $\rm L$  may well be received as a symbol  $\rm H$.
	*In contrast,  the transition from  $\rm H$  to  $\rm L$  is not possible.		*In contrast,  the transition from  $\rm H$  to  $\rm L$  is not possible.
	*Let the symbol ~~appearance~~ probabilities at the transmitter be  ${\rm Pr}(S_{\rm L}) = 0.3$  and  ${\rm Pr}(S_{\rm H}) = 0.7$.		*Let the symbol probabilities at the transmitter be  ${\rm Pr}(S_{\rm L}) = 0.3$  and  ${\rm Pr}(S_{\rm H}) = 0.7$.


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	*The topic of this chapter is illustrated with examples in the  (German language)  learning video		*The topic of this chapter is illustrated with examples in the  (German language)  learning video
	:[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)\|Statistische Abhängigkeit und Unabhängigkeit]]   $\Rightarrow$   "Statistical dependence and independence".		::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)\|Statistische Abhängigkeit und Unabhängigkeit]]   $\Rightarrow$   "Statistical dependence and independence".

Guenter at 14:06, 30 November 2021

2021-11-30T14:06:50Z

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	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}}		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}}

	[[File:EN_Sto_A_1_4.png\|right\|frame\|$\rm 2S/3E$ channel model]]		[[File:EN_Sto_A_1_4.png\|right\|frame\|$\rm 2S/3E$  channel model]]
	A transmitter emits the binary symbols  $\rm L$  $($event  $S_{\rm L})$  and  $\rm H$  $($event  $S_{\rm H})$ .		A transmitter  (German:  "Sender"   ⇒   subscript  "S")  emits the binary symbols  $\rm L$  $($event  $S_{\rm L})$  and  $\rm H$  $($event  $S_{\rm H})$ .
	*If conditions are good, the digital receiver also decides only on the binary symbols  $\rm L$  $($~~Ereignis~~  $E_{\rm L})$  or  $\rm H$  $($event  $E_{\rm H})$.		*If conditions are good, the digital receiver  (German:  "Empfänger"   ⇒   subscript  "E")  also decides only on the binary symbols  $\rm L$  $($event  $E_{\rm L})$  or  $\rm H$  $($event  $E_{\rm H})$.
	*However, if the receiver can suspect that an error has occurred during transmission, it makes no decision  $($event  $E_{\rm K})$;  $\rm K$  here stands for "No decision").		*However,  if the receiver can suspect that an error has occurred during transmission,  it makes no decision  $($event  $E_{\rm K})$;  <br>$\rm K$  here stands for  "No decision".


	The diagram shows a simple channel model in terms of transition probabilities.  It can be seen that a transmitted  $\rm L$  may well be received as a symbol  $\rm H$~~ ~~.  In contrast, the transition from  $\rm H$  to  $\rm L$  is not possible.		The diagram shows a simple channel model in terms of transition probabilities.
			*It can be seen that a transmitted  $\rm L$  may well be received as a symbol  $\rm H$.
	Let the symbol appearance probabilities at the transmitter be  ${\rm Pr}(S_{\rm L}) = 0.3$  and  ${\rm Pr}(S_{\rm H}) = 0.7$.		*In contrast,  the transition from  $\rm H$  to  $\rm L$  is not possible.
			*Let the symbol appearance probabilities at the transmitter be  ${\rm Pr}(S_{\rm L}) = 0.3$  and  ${\rm Pr}(S_{\rm H}) = 0.7$.


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	<quiz display=simple>		<quiz display=simple>
	{What is the probability that the receiver chooses the symbol  $\rm L$~~ ~~?		{What is the probability that the receiver chooses the symbol  $\rm L$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(E_{\rm L}) \ = \ $ { 0.21 3% }		${\rm Pr}(E_{\rm L}) \ = \ $ { 0.21 3% }


	{What is the probability that the receiver chooses the ~~symbo~~  $\rm H$~~ ~~?		{What is the probability that the receiver chooses the symbol  $\rm H$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(E_{\rm H}) \ = \ $ { 0.66 3% }		${\rm Pr}(E_{\rm H}) \ = \ $ { 0.66 3% }
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	{What is the probability that the receiver makes a wrong decision?		{What is the probability that the receiver makes a wrong decision?
	\|type="{}"}		\|type="{}"}
	$\text{Pr(~~falsche Entscheidung~~)} \ = \ $ { 0.03 3% }		$\text{Pr(wrong decision)} \ = \ $ { 0.03 3% }


	{What is the probability that symbol  $\rm L$  was actually sent if the receiver decided to use symbol  $\rm L$~~ ~~?		{What is the probability that symbol  $\rm L$  was actually sent if the receiver decided to use symbol  $\rm L$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(S_{\rm L}\hspace{0.05cm}\|\hspace{0.05cm}E_{\rm L} ) \ = \ $ { 1 3% }		${\rm Pr}(S_{\rm L}\hspace{0.05cm}\|\hspace{0.05cm}E_{\rm L} ) \ = \ $ { 1 3% }

Noah at 19:56, 28 November 2021

2021-11-28T19:56:06Z

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Guenter at 16:07, 12 October 2021

2021-10-12T16:07:18Z

← Older revision		Revision as of 18:07, 12 October 2021
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	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence ~~and~~ Independence^]]		[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence/Independence^]]

Guenter at 16:06, 12 October 2021

2021-10-12T16:06:20Z

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	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 ~~Statistische (Un-)Abhängigkeit~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence and Independence^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 1.4: 2S/3E-Kanalmodell to Aufgaben:Exercise 1.4: 2S/3E Channel Model

2021-10-12T15:50:26Z

Guenter moved page Aufgaben:Aufgabe 1.4: 2S/3E-Kanalmodell to Aufgaben:Exercise 1.4: 2S/3E Channel Model

← Older revision	Revision as of 17:50, 12 October 2021
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Javier: Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

2021-03-23T12:44:35Z

Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

← Older revision		Revision as of 14:44, 23 March 2021
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	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^1.3 Statistische (Un-)Abhängigkeit^]]		[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistische (Un-)Abhängigkeit^]]

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Only if the symbol  $\rm L$  was sent, the receiver can decide for the symbol  $\rm L$  at the given channel.		'''(1)'''  Only if the symbol  $\rm L$  was sent,  the receiver can decide for the symbol  $\rm L$  at the given channel.

	*However, the probability for a received  $\rm L$  is smaller by a factor of  $0.7$  than for a sent one. From this follows:		*However,  the probability for a received  $\rm L$  is smaller by a factor of  $0.7$  than for a sent one.  From this follows:
	:$${\rm Pr} (E_{\rm L}) = {\rm Pr}(S_{\rm L}) \cdot {\rm Pr} (E_{\rm L}\hspace{0.05cm}\|\hspace{0.05cm}S_{\rm L}) = 0.3 \cdot 0.7 \hspace{0.15cm}\underline {= \rm 0.21}.$$		:$${\rm Pr} (E_{\rm L}) = {\rm Pr}(S_{\rm L}) \cdot {\rm Pr} (E_{\rm L}\hspace{0.05cm}\|\hspace{0.05cm}S_{\rm L}) = 0.3 \cdot 0.7 \hspace{0.15cm}\underline {= \rm 0.21}.$$



	'''(2)'''  To the event  $E_{\rm H}$  one comes from  $S_{\rm H}$  as well as from  $S_{\rm L}$ . Therefore holds:		'''(2)'''  To the event  $E_{\rm H}$  one comes from  $S_{\rm H}$  as well as from  $S_{\rm L}$.  Therefore holds:
	:$${\rm Pr} (E_{\rm H}) = {\rm Pr} (S_{\rm H}) \cdot {\rm Pr} (E_{\rm H}\hspace{0.05cm}\|\hspace{0.05cm}S_{\rm H}) + {\rm Pr} (S_{\rm L}) \cdot {\rm Pr} (E_{\rm H}\hspace{0.05cm}\|\hspace{0.05cm} S_{\rm L})= \rm 0.7 \cdot 0.9 + 0.3 \cdot 0.1\hspace{0.15cm}\underline { = \rm 0.66}.$$		:$${\rm Pr} (E_{\rm H}) = {\rm Pr} (S_{\rm H}) \cdot {\rm Pr} (E_{\rm H}\hspace{0.05cm}\|\hspace{0.05cm}S_{\rm H}) + {\rm Pr} (S_{\rm L}) \cdot {\rm Pr} (E_{\rm H}\hspace{0.05cm}\|\hspace{0.05cm} S_{\rm L})= \rm 0.7 \cdot 0.9 + 0.3 \cdot 0.1\hspace{0.15cm}\underline { = \rm 0.66}.$$



	'''(3)'''  The events  $E_{\rm H}$,  $E_{\rm L}$  and  $E_{\rm K}$  together form a complete system. It follows that:		'''(3)'''  The events  $E_{\rm H}$,  $E_{\rm L}$  and  $E_{\rm K}$  together form a complete system.  It follows that:
	:$${\rm Pr} (E_{\rm K}) = 1 - {\rm Pr} (E_{\rm L}) - {\rm Pr} (E_{\rm H}) \hspace{0.15cm}\underline {= \rm 0.13}.$$		:$${\rm Pr} (E_{\rm K}) = 1 - {\rm Pr} (E_{\rm L}) - {\rm Pr} (E_{\rm H}) \hspace{0.15cm}\underline {= \rm 0.13}.$$

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	'''(5)'''  If the symbol  $\rm L$  was received, only  $\rm L$  could have been sent. It follows that:		'''(5)'''  If the symbol  $\rm L$  was received,  only  $\rm L$  could have been sent. It follows that:
	:$${\rm Pr} (S_{\rm L} \hspace{0.05cm}\|\hspace{0.05cm} E_{\rm L}) \hspace{0.15cm}\underline {= \rm 1}.$$		:$${\rm Pr} (S_{\rm L} \hspace{0.05cm}\|\hspace{0.05cm} E_{\rm L}) \hspace{0.15cm}\underline {= \rm 1}.$$



	'''(6)'''  For example, Bayes' theorem is suitable for solving this problem:		'''(6)'''  For example,  Bayes' theorem is suitable for solving this problem:
	:$${\rm Pr} (S_{\rm L}\hspace{0.05cm}\|\hspace{0.05cm} E_{\rm K}) =\frac{ {\rm Pr} ( E_{\rm K} \hspace{0.05cm}\|\hspace{0.05cm} S_{\rm L}) \cdot {\rm Pr} (S_{\rm L})}{{\rm Pr} (E_{\rm K})} =\frac{ \rm 0.2 \cdot 0.3}{\rm 0.13} = \frac{\rm 6}{\rm 13}\hspace{0.15cm}\underline { \approx \rm 0.462}.$$		:$${\rm Pr} (S_{\rm L}\hspace{0.05cm}\|\hspace{0.05cm} E_{\rm K}) =\frac{ {\rm Pr} ( E_{\rm K} \hspace{0.05cm}\|\hspace{0.05cm} S_{\rm L}) \cdot {\rm Pr} (S_{\rm L})}{{\rm Pr} (E_{\rm K})} =\frac{ \rm 0.2 \cdot 0.3}{\rm 0.13} = \frac{\rm 6}{\rm 13}\hspace{0.15cm}\underline { \approx \rm 0.462}.$$
	{{ML-Fuß}}		{{ML-Fuß}}