Aufgaben:Exercise 4.5: Irrelevance Theorem - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]		[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]
	[[de:Aufgaben:4.~~5_Theorem_der_Irrelevanz~~]]		[[de:Aufgaben:Aufgabe 4.5: Theorem der Irrelevanz]]

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	*The  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"complementary Gaussian error function"]]  gives the following results:		*The  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"complementary Gaussian error function"]]  gives the following results:
	:$${\rm Q}(0) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5\hspace{0.05cm},\hspace{1.35cm}{\rm Q}(2^{0.5}) = 0.786 \cdot 10^{-1}\hspace{0.05cm},\hspace{1.1cm}{\rm Q}(2) = 0.227 \cdot 10^{-1}\hspace{0.05cm},$$		:$${\rm Q}(0) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5\hspace{0.05cm},\hspace{1.35cm}{\rm Q}(2^{0.5}) = 0.786 \cdot 10^{-1}\hspace{0.05cm},\hspace{1.1cm}{\rm Q}(2) = 0.227 \cdot 10^{-1}\hspace{0.05cm},$$
	:$${\rm Q}(2 \cdot 2^{0.5}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.234 \cdot 10^{-2}\hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4) = 0.317 \cdot 10^{-4}		:$${\rm Q}(2 \cdot 2^{0.5}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.234 \cdot 10^{-2}\hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4) = 0.317 \cdot 10^{-4}\hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4 \cdot 2^{0.5}) = 0.771 \cdot 10^{-8}\hspace{0.05cm}.$$
	\hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4 \cdot 2^{0.5}) = 0.771 \cdot 10^{-8}\hspace{0.05cm}.$$


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	*With bipolar  (antipodal)  transmitted signal and AWGN noise,  the error probability of the optimal receiver  (whether implemented as a correlation or matched filter receiver)  is equal to		*With bipolar  (antipodal)  transmitted signal and AWGN noise,  the error probability of the optimal receiver  (whether implemented as a correlation or matched filter receiver)  is equal to
	:$$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm} error} ) = {\rm Q} \left ( \sqrt{ E_s /{\sigma}}\right )		:$$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm} error} ) = {\rm Q} \left ( \sqrt{ E_s /{\sigma}}\right )= {\rm Q} \left ( \sqrt{2 E_s /N_0}\right ) \hspace{0.05cm}.$$
	= {\rm Q} \left ( \sqrt{2 E_s /N_0}\right ) \hspace{0.05cm}.$$

	*With   $E_s = 8 \cdot 10^{\rm –6} \ \rm Ws$ and $N_0 = 10^{\rm –6} \ \rm W/Hz$,  we further obtain:		*With   $E_s = 8 \cdot 10^{\rm –6} \ \rm Ws$ and $N_0 = 10^{\rm –6} \ \rm W/Hz$,  we further obtain:
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	'''(4)'''  Because of  ${\rm Pr}(m = m_0) = {\rm Pr}(m = m_1)$,  the decision rule of the optimal receiver  (whether realized as MAP or as ML)  is:		'''(4)'''  Because of  ${\rm Pr}(m = m_0) = {\rm Pr}(m = m_1)$,  the decision rule of the optimal receiver  (whether realized as MAP or as ML)  is:
	:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm arg} \max_i \hspace{0.1cm} [ p_{m \hspace{0.05cm}\|\hspace{0.05cm}r_1, \hspace{0.05cm}r_2 } \hspace{0.05cm} (m_i \hspace{0.05cm}\|\hspace{0.05cm}\rho_1, \hspace{0.05cm}\rho_2 ) ] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}\|\hspace{0.05cm} m} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}\|\hspace{0.05cm} m_i ) \cdot {\rm Pr} (m = m_i)] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}\|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}\|\hspace{0.05cm} s_i ) ]		:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm arg} \max_i \hspace{0.1cm} [ p_{m \hspace{0.05cm}\|\hspace{0.05cm}r_1, \hspace{0.05cm}r_2 } \hspace{0.05cm} (m_i \hspace{0.05cm}\|\hspace{0.05cm}\rho_1, \hspace{0.05cm}\rho_2 ) ] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}\|\hspace{0.05cm} m} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}\|\hspace{0.05cm} m_i ) \cdot {\rm Pr} (m = m_i)] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}\|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}\|\hspace{0.05cm} s_i ) ]\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*This composite probability density can be rewritten as follows:		*This composite probability density can be rewritten as follows:
	:$$\hat{m} ={\rm arg} \max_i \hspace{0.1cm} [ p_{r_1 \hspace{0.05cm}\|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1 \hspace{0.05cm}\|\hspace{0.05cm} s_i ) \cdot p_{r_2 \hspace{0.05cm}\|\hspace{0.05cm} r_1, \hspace{0.05cm}s} \hspace{0.05cm} ( \rho_2 \hspace{0.05cm}\|\hspace{0.05cm} \rho_1, \hspace{0.05cm}s_i ) ]		:$$\hat{m} ={\rm arg} \max_i \hspace{0.1cm} [ p_{r_1 \hspace{0.05cm}\|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1 \hspace{0.05cm}\|\hspace{0.05cm} s_i ) \cdot p_{r_2 \hspace{0.05cm}\|\hspace{0.05cm} r_1, \hspace{0.05cm}s} \hspace{0.05cm} ( \rho_2 \hspace{0.05cm}\|\hspace{0.05cm} \rho_1, \hspace{0.05cm}s_i ) ]\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Now,  since the second multiplicand also depends on the message  ($s_i$),  $r_2$  should definitely be included in the decision process.  Thus,  the correct answer is:  <u>YES</u>.		*Now,  since the second multiplicand also depends on the message  ($s_i$),  $r_2$  should definitely be included in the decision process.  Thus,  the correct answer is:  <u>YES</u>.
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	'''(5)'''  For AWGN noise with variance  $\sigma^2$,  the two composite densities introduced in  '''(4)'''  together with their product  $P$  give:		'''(5)'''  For AWGN noise with variance  $\sigma^2$,  the two composite densities introduced in  '''(4)'''  together with their product  $P$  give:
	:$$p_{r_1\hspace{0.05cm}\|\hspace{0.05cm} s}(\rho_1\hspace{0.05cm}\|\hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_1 - s_i)^2}{2 \sigma^2}\right ]\hspace{0.05cm},\hspace{1cm}		:$$p_{r_1\hspace{0.05cm}\|\hspace{0.05cm} s}(\rho_1\hspace{0.05cm}\|\hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_1 - s_i)^2}{2 \sigma^2}\right ]\hspace{0.05cm},\hspace{1cm}p_{r_2\hspace{0.05cm}\|\hspace{0.05cm}r_1,\hspace{0.05cm} s}(\rho_2\hspace{0.05cm}\|\hspace{0.05cm}\rho_1, \hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_2 - (\rho_1 - s_i))^2}{2 \sigma^2}\right ]\hspace{0.05cm}$$
	p_{r_2\hspace{0.05cm}\|\hspace{0.05cm}r_1,\hspace{0.05cm} s}(\rho_2\hspace{0.05cm}\|\hspace{0.05cm}\rho_1, \hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_2 - (\rho_1 - s_i))^2}{2 \sigma^2}\right ]\hspace{0.05cm}$$		:$$ \Rightarrow \hspace{0.3cm}P \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{{2\pi} \cdot \sigma^2}\cdot {\rm exp} \left [ - \frac{1}{2 \sigma^2} \cdot \left \{ (\rho_1 - s_i)^2+ (\rho_2 - (\rho_1 - s_i))^2\right \}\right ]\hspace{0.05cm}.$$
	:$$ \Rightarrow \hspace{0.3cm}
	P \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{{2\pi} \cdot \sigma^2}\cdot {\rm exp} \left [ - \frac{1}{2 \sigma^2} \cdot \left \{ (\rho_1 - s_i)^2
	+ (\rho_2 - (\rho_1 - s_i))^2\right \}\right ]\hspace{0.05cm}.$$

	*We are looking for the argument that maximizes this product  $P$,  which at the same time means that the expression in the curly brackets should take the smallest possible value:		*We are looking for the argument that maximizes this product  $P$,  which at the same time means that the expression in the curly brackets should take the smallest possible value:
	:$$\mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \max_i \hspace{0.1cm} P = {\rm arg} \min _i \hspace{0.1cm} \left \{ (\rho_1 - s_i)^2		:$$\mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \max_i \hspace{0.1cm} P = {\rm arg} \min _i \hspace{0.1cm} \left \{ (\rho_1 - s_i)^2+ (\rho_2 - (\rho_1 - s_i))^2\right \} $$
	+ (\rho_2 - (\rho_1 - s_i))^2\right \} $$		:$$\Rightarrow \hspace{0.3cm} \mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \min _i \hspace{0.1cm}\left \{ \rho_1^2 - 2\rho_1 s_i + s_i^2+ \rho_2^2 - 2\rho_1 \rho_2 + 2\rho_2 s_i+ \rho_1^2 - 2\rho_1 s_i + s_i^2\right \}\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} \mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \min _i \hspace{0.1cm}\left \{ \rho_1^2 - 2\rho_1 s_i + s_i^2
	+ \rho_2^2 - 2\rho_1 \rho_2 + 2\rho_2 s_i+ \rho_1^2 - 2\rho_1 s_i + s_i^2\right \}
	\hspace{0.05cm}.$$

	*Here  $\mu$  denotes the estimated value of the message.  In this minimization,  all terms that do not depend on the message  $s_i$  can now be omitted.  Likewise,  the terms  $s_i^2$  are disregarded,  since  $s_0^2 = s_1^2$  holds.  Thus,  the much simpler decision rule is obtained:		*Here  $\mu$  denotes the estimated value of the message.  In this minimization,  all terms that do not depend on the message  $s_i$  can now be omitted.  Likewise,  the terms  $s_i^2$  are disregarded,  since  $s_0^2 = s_1^2$  holds.  Thus,  the much simpler decision rule is obtained:
	:$$\mu = {\rm arg} \min _i \hspace{0.1cm}\left \{ - 4\rho_1 s_i + 2\rho_2 s_i \right \}={\rm arg} \min _i \hspace{0.1cm}\left \{ (\rho_2 - 2\rho_1) \cdot s_i \right \}		:$$\mu = {\rm arg} \min _i \hspace{0.1cm}\left \{ - 4\rho_1 s_i + 2\rho_2 s_i \right \}={\rm arg} \min _i \hspace{0.1cm}\left \{ (\rho_2 - 2\rho_1) \cdot s_i \right \}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*So,  correct is already the proposed solution 2.  But after multiplication by  $–1/2$,  we also get the last mentioned decision rule:		*So,  correct is already the proposed solution 2.  But after multiplication by  $–1/2$,  we also get the last mentioned decision rule:
	:$$\mu = {\rm arg} \max_i \hspace{0.1cm}\left \{ (\rho_1 - \rho_2/2) \cdot s_i \right \}		:$$\mu = {\rm arg} \max_i \hspace{0.1cm}\left \{ (\rho_1 - \rho_2/2) \cdot s_i \right \}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Thus,  the <u>solutions 2 and 3</u>  are correct.		*Thus,  the <u>solutions 2 and 3</u>  are correct.
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	'''(6)'''  Setting $K_1 = 1$  and  $\underline {K_2 = \, -0.5}$,  the optimal decision rule with realization  $\rho = \rho_1 \, – \rho_2/2$ is:		'''(6)'''  Setting $K_1 = 1$  and  $\underline {K_2 = \, -0.5}$,  the optimal decision rule with realization  $\rho = \rho_1 \, – \rho_2/2$ is:
	:$$\mu =		:$$\mu =\left\{ \begin{array}{c} m_0 \\m_1 \end{array} \right.\quad\begin{array}{*{1}c} {\rm f{or}} \hspace{0.15cm} \rho > 0 \hspace{0.05cm},\\ {\rm f{or}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$
	\left\{ \begin{array}{c} m_0 \\
	m_1 \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm f{or}} \hspace{0.15cm} \rho > 0 \hspace{0.05cm},
	\\ {\rm f{or}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$

	*Since  $\rho = 0$  only occurs with probability  $0$,  it does not matter in the sense of probability theory whether one assigns the message  $\mu = m_0$  or  $\mu = m_1$  to this event  "$\rho = 0$".		*Since  $\rho = 0$  only occurs with probability  $0$,  it does not matter in the sense of probability theory whether one assigns the message  $\mu = m_0$  or  $\mu = m_1$  to this event  "$\rho = 0$".
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	'''(7)'''  With  $K_2 = \, -0.5$  one obtains for the input value of the decision:		'''(7)'''  With  $K_2 = \, -0.5$  one obtains for the input value of the decision:
	:$$r = r_1 - r_2/2 = s + n_1 - (n_1 + n_2)/2 = s + n \hspace{0.3cm}		:$$r = r_1 - r_2/2 = s + n_1 - (n_1 + n_2)/2 = s + n \hspace{0.3cm}\Rightarrow \hspace{0.3cm} n = \frac{n_1 - n_2}{2}\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} n = \frac{n_1 - n_2}{2}\hspace{0.05cm}.$$

	*The variance of this random variable is		*The variance of this random variable is
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	*From this,  the error probability is analogous to subtask  '''(2)''':		*From this,  the error probability is analogous to subtask  '''(2)''':
	:$${\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} \left ( \sqrt{\frac{E_s}{N_0/4}}\right ) =		:$${\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} \left ( \sqrt{\frac{E_s}{N_0/4}}\right ) ={\rm Q} \left ( 4 \cdot \sqrt{2}\right ) = {1}/{2} \cdot {\rm erfc}(4) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$
	{\rm Q} \left ( 4 \cdot \sqrt{2}\right ) = {1}/{2} \cdot {\rm erfc}(4) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}
	\hspace{0.05cm}.$$

	*Thus,  by taking  $r_2$  into account,  the error probability can be lowered from  $0.317 \cdot 10^{\rm –4}$  to the much smaller value of  $0.771 \cdot 10^{-8}$,  although the decision component  $r_2$  contains only noise.  However,  this noise  $r_2$  allows an estimate of the noise component  $n_1$  of  $r_1$.		*Thus,  by taking  $r_2$  into account,  the error probability can be lowered from  $0.317 \cdot 10^{\rm –4}$  to the much smaller value of  $0.771 \cdot 10^{-8}$,  although the decision component  $r_2$  contains only noise.  However,  this noise  $r_2$  allows an estimate of the noise component  $n_1$  of  $r_1$.

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	[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]		[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]
			[[de:Aufgaben:4.5_Theorem_der_Irrelevanz]]

Guenter at 11:38, 18 July 2022

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Guenter at 16:01, 15 July 2022

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Guenter at 15:11, 8 July 2022

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

	[[File:~~P_ID2014__Dig_A_4_5~~.png\|right\|frame\|Considered optimal system with detector and decision]]		[[File:EN_Dig_A_4_5.png\|right\|frame\|Considered optimal system with detector and decision]]
	The communication system given by the graph is to be investigated. The binary message  $m ∈ \{m_0, m_1\}$  with equal occurrence probabilities		The communication system given by the graph is to be investigated. The binary message  $m ∈ \{m_0, m_1\}$  with equal occurrence probabilities
	:$${\rm Pr} (m_0 ) = {\rm Pr} (m_1 ) = 0.5$$		:$${\rm Pr} (m_0 ) = {\rm Pr} (m_1 ) = 0.5$$

Hwang at 13:19, 28 June 2022

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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*~~Durch Berücksichtigung von~~ $r_2$ ~~lässt sich also die Fehlerwahrscheinlichkeit von~~ $0.317 \cdot 10^{\rm –4}$ ~~auf den deutlich kleineren Wert~~ $0.771 \cdot 10^{-8}$ ~~absenken~~, ~~obwohl die Entscheidungskomponente~~ $r_2$ ~~nur Rauschen beinhaltet~~. ~~Dieses Rauschen~~ $r_2$ ~~erlaubt aber eine Schätzung der Rauschkomponente~~ $n_1$ ~~von~~ $r_1$.		*Thus, by taking $r_2$ into account, the error probability can be lowered from $0.317 \cdot 10^{\rm –4}$ to the much smaller value of $0.771 \cdot 10^{-8}$, although the decision component $r_2$ contains only noise. However, this noise $r_2$ allows an estimate of the noise component $n_1$ of $r_1$.

	*~~Halbiert man die Sendeenergie von~~ $8 \cdot 10^{\rm –6} \ \rm Ws$ ~~auf~~ $4 \cdot 10^{\rm –6} \ \rm Ws$, ~~so ergibt sich hier immer noch die Fehlerwahrscheinlichkeit~~ $0.317 \cdot 10^{\rm –4}$, ~~wie~~ in ~~der Teilaufgabe~~ (2) ~~berechnet~~. ~~Bei alleiniger Auswertung von~~ $r_1$ ~~würde die Fehlerwahrscheinlichkeit dagegen~~ $0.234 \cdot 10^{\rm –2}$ ~~betragen~~.		*Halving the transmit energy from $8 \cdot 10^{\rm –6} \ \rm Ws$ to $4 \cdot 10^{\rm –6} \ \rm Ws$, we still get the error probability $0.317 \cdot 10^{\rm –4}$ here, as calculated in subtask (2). When evaluating $r_1$ alone, on the other hand, the error probability would be $0.234 \cdot 10^{\rm –2}$.
	{{ML-Fuß}}		{{ML-Fuß}}

Hwang at 15:08, 20 June 2022

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Guenter at 12:00, 8 June 2022

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	[[Category:Digital Signal Transmission: Exercises\|^4.2 ~~Struktur des optimalen Empfängers~~^]]		[[Category:Digital Signal Transmission: Exercises\|^4.2 Structure of the Optimal Receiver^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.5: Theorem der Irrelevanz to Aufgaben:Exercise 4.5: Irrelevance Theorem

2022-06-08T11:58:53Z

Guenter moved page Aufgaben:Aufgabe 4.5: Theorem der Irrelevanz to Aufgaben:Exercise 4.5: Irrelevance Theorem

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