Aufgaben:Exercise 4.5: Irrelevance Theorem - Revision history

Guenter at 11:38, 18 July 2022

2022-07-18T11:38:12Z

Show changes

Guenter at 16:01, 15 July 2022

2022-07-15T16:01:42Z

Guenter at 15:11, 8 July 2022

2022-07-08T15:11:45Z

Hwang at 13:19, 28 June 2022

2022-06-28T13:19:09Z

Hwang at 15:08, 20 June 2022

2022-06-20T15:08:49Z

Show changes

Guenter at 12:00, 8 June 2022

2022-06-08T12:00:21Z

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 Aufgabe 4.5: Theorem der Irrelevanz to Exercise 4.5: Irrelevance Theorem

Javier: Text replacement - "”" to """

2021-05-28T14:40:08Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:33:46Z

Text replacement - "„" to """

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

2021-03-23T12:51:57Z

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

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver}}
-[[File:EN_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&nbsp; "detector"&nbsp; and&nbsp; "decision"]]
-The communication system given by the graph is to be investigated. The binary message&nbsp; $m &#8712; \{m_0, m_1\}$&nbsp; with equal occurrence probabilities
+The communication system given by the graph is to be investigated.&nbsp; The binary message&nbsp; $m &#8712; \{m_0, m_1\}$&nbsp; with equal occurrence probabilities
 :$${\rm Pr} (m_0 ) = {\rm Pr} (m_1 ) = 0.5$$
@@ Line 11: / Line 11: @@
 where the assignments&nbsp; $m_0 &#8660; s_0$&nbsp; and&nbsp; $m_1 &#8660; s_1$&nbsp; are one-to-one.
-The detector (highlighted in green in the figure) provides two decision values
+The detector&nbsp; $($highlighted in green in the figure$)$&nbsp; provides two decision values
 :$$r_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} s + n_1\hspace{0.05cm},$$
 :$$r_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} n_1 + n_2\hspace{0.05cm},$$
-from which the decision forms the estimated values&nbsp; $\mu &#8712; \{m_0, m_1\}$&nbsp; for the transmitted message&nbsp; $m$.&nbsp; The decision includes
+from which the decision forms the estimated values&nbsp; $\mu &#8712; \{m_0,\ m_1\}$&nbsp; for the transmitted message&nbsp; $m$.&nbsp; The decision includes
 *two weighting factors&nbsp; $K_1$&nbsp; and&nbsp; $K_2$,
 *a summation point, and
-*a threshold decision with the threshold at $0$.
+*a summation point,&nbsp; and
 Three evaluations are considered in this exercises:
-* Decision based on&nbsp; $r_1$&nbsp;   ($K_1 &ne; 0, K_2 = 0$),
+# Decision based on&nbsp; $r_1$&nbsp;   $(K_1 &ne; 0,\ K_2 = 0)$,
-* decision based on&nbsp; $r_2$&nbsp;   ($K_1 = 0, K_2 &ne; 0$),
+# decision based on&nbsp; $r_2$&nbsp;   $(K_1 = 0,\ K_2 &ne; 0)$,
-* joint evaluation of&nbsp; $r_1$&nbsp; und&nbsp; $r_2$&nbsp;   $(K_1 &ne; 0, K_2 &ne; 0)$.
+# joint evaluation of&nbsp; $r_1$&nbsp; und&nbsp; $r_2$&nbsp;   $(K_1 &ne; 0,\ K_2 &ne; 0)$.
-Let the two noise sources&nbsp; $n_1$&nbsp; and&nbsp; $n_2$&nbsp; be independent of each other and also independent of the transmitted signal&nbsp; $s &#8712; \{s_0, s_1\}$.
+:where a binary message signal&nbsp; $s &#8712; \{s_0,\ s_1\}$&nbsp; with&nbsp; $s_0 = \sqrt{E_s} \hspace{0.05cm},\hspace{0.2cm}s_1 = -\sqrt{E_s}$&nbsp; is assumed.
-$n_1$&nbsp; and&nbsp; $n_2$&nbsp; can each be modeled by AWGN noise sources $($white, Gaussian distributed, mean-free, variance&nbsp; $\sigma^2 = N_0/2)$.&nbsp; For numerical calculations, use the values
+*$n_1$&nbsp; and&nbsp; $n_2$&nbsp; can each be modeled by AWGN noise sources&nbsp; $($white,&nbsp; Gaussian distributed,&nbsp; mean-free,&nbsp; variance&nbsp; $\sigma^2 = N_0/2)$. &nbsp;
 :$$E_s = 8 \cdot 10^{-6}\,{\rm Ws}\hspace{0.05cm},\hspace{0.2cm}N_0 = 10^{-6}\,{\rm W/Hz} \hspace{0.05cm}.$$
-The&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"complementary Gaussian error function"]]&nbsp; gives the following results:
+*The&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"complementary Gaussian error function"]]&nbsp; 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}(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}
@@ Line 40: / Line 59: @@
@@ Line 63: / Line 66: @@
 {What statements apply here regarding the receiver?
 |type="()"}
-- The ML receiver is better than the MAP receiver here.
+- The ML receiver is better than the MAP receiver.
-- The MAP receiver is better than the ML receiver here.
+- The MAP receiver is better than the ML receiver.
-+ Both receivers deliver the same result here.
++ Both receivers deliver the same result.
 {What is the error probability with&nbsp; $K_2 = 0$?
 |type="{}"}
-${\rm Pr(symbol error)}\ = \ $  { 0.00317 3% } $\ \%$
+${\rm Pr(symbol\hspace{0.15cm} error)}\ = \ $  { 0.00317 3% } $\ \%$
 {What is the error probability with&nbsp; $K_1 = 0$?
 |type="{}"}
-${\rm Pr(symbol error)}\ = \ $ { 50 3% } $\ \%$
+${\rm Pr(symbol\hspace{0.15cm}error)}\ = \ $ { 50 3% } $\ \%$
 {Can an improvement be achieved by using&nbsp; $r_1$&nbsp; <b>and</b> &nbsp;$r_2$?&nbsp;
@@ Line 86: / Line 89: @@
 + $\mu = {\rm arg \ max} \, \big[(\rho_1 - \rho_2/2) \cdot s_i \big]$.
-{How can this rule be implemented exactly with the given decision (threshold at zero)? Let &nbsp;$K_1 = 1$.
+{How can this rule be implemented exactly with the given decision&nbsp; (threshold at zero)?&nbsp; Let &nbsp;$K_1 = 1$.
 |type="{}"}
 $K_2 \ = \ $ { -0.515--0.485 }
-{What is the (minimum) error probability with the realization according to subtask '''(6)'''?
+{What is the&nbsp; (minimum)&nbsp; error probability with the realization according to subtask&nbsp; '''(6)'''?
 |type="{}"}
-${\rm Minimum \ \big[Pr(symbol error)\big]} \ = \ $ { 0.771 3% } $\ \cdot 10^{\rm -8}$
+${\rm Minimum \ \big[Pr(symbol\hspace{0.15cm}error)\big]} \ = \ $ { 0.771 3% } $\ \cdot 10^{\rm -8}$
 </quiz>

@@ Line 2: / Line 2: @@
 {{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&nbsp; $m &#8712; \{m_0, m_1\}$&nbsp; with equal occurrence probabilities
 :$${\rm Pr} (m_0 ) = {\rm Pr} (m_1 ) = 0.5$$

@@ Line 180: / Line 180: @@
 \hspace{0.05cm}.$$
-*Durch Berücksichtigung von $r_2$ lässt sich also die Fehlerwahrscheinlichkeit von $0.317 \cdot 10^{\rm &ndash;4}$ 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 $0.317 \cdot 10^{\rm &ndash;4}$ 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 $8 \cdot 10^{\rm &ndash;6} \ \rm Ws$ auf $4 \cdot 10^{\rm &ndash;6} \ \rm Ws$, so ergibt sich hier immer noch die Fehlerwahrscheinlichkeit $0.317 \cdot 10^{\rm &ndash;4}$, wie in der Teilaufgabe (2) berechnet. Bei alleiniger Auswertung von $r_1$ würde die Fehlerwahrscheinlichkeit dagegen $0.234 \cdot 10^{\rm &ndash;2}$ betragen.
+*Halving the transmit energy from $8 \cdot 10^{\rm &ndash;6} \ \rm Ws$ to $4 \cdot 10^{\rm &ndash;6} \ \rm Ws$, we still get the error probability $0.317 \cdot 10^{\rm &ndash;4}$ here, as calculated in subtask (2). When evaluating $r_1$ alone, on the other hand, the error probability would be $0.234 \cdot 10^{\rm &ndash;2}$.
 {{ML-Fuß}}

← Older revision		Revision as of 12:00, 8 June 2022
Line 187:		Line 187:


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

← Older revision		Revision as of 14:40, 28 May 2021
Line 165:		Line 165:
	\\ {\rm f{\rm \ddot{u}r}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$		\\ {\rm f{\rm \ddot{u}r}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$

−	Da $\rho = 0$ nur mit der Wahrscheinlichkeit $0$ auftritt, ist es im Sinne der Wahrscheinlichkeitsrechnung egal, ob man diesem Ereignis "$\rho = 0$~~”~~ die Nachricht $\mu = m_0$ oder $\mu = m_1$ zuordnet.	+	Da $\rho = 0$ nur mit der Wahrscheinlichkeit $0$ auftritt, ist es im Sinne der Wahrscheinlichkeitsrechnung egal, ob man diesem Ereignis "$\rho = 0$" die Nachricht $\mu = m_0$ oder $\mu = m_1$ zuordnet.

← Older revision	Revision as of 11:58, 8 June 2022
(No difference)

← Older revision		Revision as of 14:33, 28 May 2021
Line 165:		Line 165:
	\\ {\rm f{\rm \ddot{u}r}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$		\\ {\rm f{\rm \ddot{u}r}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$

−	Da $\rho = 0$ nur mit der Wahrscheinlichkeit $0$ auftritt, ist es im Sinne der Wahrscheinlichkeitsrechnung egal, ob man diesem Ereignis ~~&bdquo;~~$\rho = 0$” die Nachricht $\mu = m_0$ oder $\mu = m_1$ zuordnet.	+	Da $\rho = 0$ nur mit der Wahrscheinlichkeit $0$ auftritt, ist es im Sinne der Wahrscheinlichkeitsrechnung egal, ob man diesem Ereignis "$\rho = 0$” die Nachricht $\mu = m_0$ oder $\mu = m_1$ zuordnet.

← Older revision		Revision as of 12:51, 23 March 2021
Line 187:		Line 187:


−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.2 Struktur des optimalen Empfängers^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.2 Struktur des optimalen Empfängers^]]