Aufgaben:Exercise 5.3: AWGN and BSC Model - Revision history

Guenter at 13:30, 5 September 2022

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Guenter at 13:18, 5 September 2022

2022-09-05T13:18:36Z

Guenter at 15:28, 1 September 2022

2022-09-01T15:28:47Z

Hwang at 13:15, 26 August 2022

2022-08-26T13:15:23Z

Hwang at 12:55, 26 August 2022

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Hwang at 11:56, 26 August 2022

2022-08-26T11:56:27Z

Guenter at 16:37, 23 August 2022

2022-08-23T16:37:17Z

Guenter: Guenter moved page Aufgaben:Aufgabe 5.3: AWGN- und BSC-Modell to Aufgaben:Exercise 5.3: AWGN and BSC Model

2022-08-23T16:32:57Z

Guenter moved page Aufgabe 5.3: AWGN- und BSC-Modell to Exercise 5.3: AWGN and BSC Model

Javier: Text replacement - "”" to """

2021-05-28T14:43:19Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:39:36Z

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@@ Line 64: / Line 64: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The average error probability is $p_{\rm M} = {\rm Q}(s_0/\sigma) = 0.01$.
+'''(1)'''&nbsp; The mean error probability is&nbsp; $p_{\rm M} = {\rm Q}(s_0/\sigma) = 0.01$.
 *From this it follows for the quotient of the detection useful sample value and the detection noise rms value:
 :$${s_0}/{\sigma}= {\rm Q}^{-1} \left ( 0.01 \right ) \hspace{0.15cm}\underline {\approx 2.32}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; With $E = 0$, the probabilities of the given digital channel model are given by:
+'''(2)'''&nbsp; With&nbsp; $E = 0$,&nbsp; the probabilities of the digital channel model are given by:
 :$$p_2 = p_3 = p = 0.01 \hspace{0.05cm}, \hspace{0.2cm}p_1 = p_4 = 1-p = 0.99\hspace{0.05cm}.$$
-*A comparison with the theory part shows that this channel model corresponds to the BSC model, independent of the statistics of the source symbols.
+*A comparison with the theory part shows that this channel model corresponds to the BSC model,&nbsp; independent of the statistics of the source symbols.
-*Thus, <u>both solutions</u> are correct.
+*Thus,&nbsp; <u>both solutions</u>&nbsp; are correct.
-'''(3)'''&nbsp; The transition probability $p_2$ now describes the case where the decision threshold $E = 0.25 \cdot s_0$ was mistakenly undershot.
+'''(3)'''&nbsp; The transition probability&nbsp; $p_2$&nbsp; now describes the case where the decision threshold&nbsp; $E = 0.25 \cdot s_0$&nbsp; was mistakenly undershot.
-*Then $v_{\nu} = \mathbf{L}$, although $q_{\nu} = \mathbf{H}$ was sent. Thus, the distance from the threshold is only $0.75 \cdot s_0$ and it holds:
+*Then&nbsp; $v_{\nu} = \mathbf{L}$,&nbsp; although $q_{\nu} = \mathbf{H}$&nbsp; was sent.&nbsp; Thus,&nbsp; the distance from the threshold is only&nbsp; $0.75 \cdot s_0$&nbsp; and it holds:
 :$$p_{\rm 2}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Q} \left ( \frac{0.75 \cdot s_0}{\sigma} \right ) = {\rm Q} \left ( 0.75 \cdot 2.32 \right )
   = {\rm Q} \left ( 1.74 \right )\hspace{0.15cm}\underline {\approx 0.041}\hspace{0.05cm}, \hspace{0.5cm}
@@ Line 84: / Line 84: @@
 .959}\hspace{0.05cm}.$$
-*Similarly, the transition probabilities $p_3$ and $p_4$ can be calculated, now assuming the threshold distance $1.25 \cdot s_0$:
+*Similarly, the transition probabilities&nbsp; $p_3$&nbsp; and&nbsp; $p_4$&nbsp; can be calculated,&nbsp; now assuming the threshold distance&nbsp; $1.25 \cdot s_0$:
 :$$p_{\rm 3}  = {\rm Q} \left ( 1.25 \cdot 2.32 \right )
   = {\rm Q} \left ( 2.90 \right )\hspace{0.15cm}\underline {\approx  0.002}\hspace{0.05cm}, \hspace{0.2cm}
@@ Line 91: / Line 91: @@
-'''(4)'''&nbsp; <u>Neither</u> of the two solutions applies:
+'''(4)'''&nbsp; <u>Neither</u>&nbsp; of the two solutions applies:
-*With the decision threshold $E &ne; 0$, the BSC model is not applicable regardless of the symbol statistic,
+*With the decision threshold&nbsp; $E &ne; 0$,&nbsp; the BSC model is not applicable regardless of the symbol statistic,
-'''(5)'''&nbsp; <u>Statements 1 and 3</u> are true, but statement 2 is not:
-*In the BSC model, $p_{\rm M} = 1\%$ is independent of the symbol probabilities $p_{\rm L}$ and $p_{\rm H}$.
+'''(5)'''&nbsp; <u>Statements 1 and 3</u>&nbsp; are true,&nbsp; but not statement 2:
-*In contrast, for $p_{\rm L} = 0.9$, $p_{\rm H} = 0.1$ and $E = +s_0/4$:
+*In the BSC model,&nbsp; $p_{\rm M} = 1\%$&nbsp; is independent of the symbol probabilities&nbsp; $p_{\rm L}$&nbsp; and&nbsp; $p_{\rm H}$.
 :$$p_{\rm M}  = 0.9 \cdot p_{\rm 3} + 0.1 \cdot p_{\rm 2}= 0.9 \cdot 0.2\% + 0.1 \cdot 4.1\%
   \approx 0.59\% \hspace{0.05cm}.$$
-*The minimum results for $p_{\rm L} = 0.93$ and $p_{\rm H} = 0.07$ to $p_{\rm M} \approx 0.45\%$.
+*The minimum results for&nbsp; $p_{\rm L} = 0.93$&nbsp; and&nbsp; $p_{\rm H} = 0.07$&nbsp; to
 {{ML-Fuß}}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}}
-[[File:EN_Dig_A_5_3.png|right|frame|AWGN channel and BSC model]]
+[[File:EN_Dig_A_5_3.png|right|frame|AWGN and BSC model]]
-The graphic above shows the analog channel model of a digital transmission system, where the additive noise signal&nbsp; $n(t)$&nbsp; with the (two-sided) noise power density&nbsp; $N_0/2$&nbsp; is effective. This is AWGN noise. The variance of the noise component before the decision (after the matched filter) is then
+The upper graphic shows the analog channel model of a digital transmission system,&nbsp; where the additive noise signal&nbsp; $n(t)$&nbsp; with the&nbsp; $($two-sided$)$&nbsp; noise power density&nbsp; $N_0/2$&nbsp; is effective.&nbsp; This is AWGN noise.&nbsp; The variance of the noise component before the decision&nbsp; $($after the matched filter$)$&nbsp; is then
 :$$\sigma^2 = \frac{N_0}{2T} \hspace{0.05cm}.$$
 Further, let hold:
-* No intersymbol interference occurs. If the symbol&nbsp; $q_{\nu} = \mathbf{H}$&nbsp; was sent, the useful component of the detection signal is equal to&nbsp; $+s_0$, while for&nbsp; $q_{\nu} = \mathbf{L}$,&nbsp; it is equal to&nbsp; $-s_0$.
+* No intersymbol interference occurs.&nbsp; If the symbol&nbsp; $q_{\nu} = \mathbf{H}$&nbsp; was sent,&nbsp; the useful component of the detection signal is equal to&nbsp; $+s_0$,&nbsp; while for&nbsp; $q_{\nu} = \mathbf{L}$,&nbsp; it is equal to&nbsp; $-s_0$.
-* The threshold decision takes into account a threshold drift, that is, the threshold&nbsp; $E$&nbsp; may well deviate from the optimal value&nbsp; $E = 0$.&nbsp; The <i>decision rule</i> is:
 :$$\upsilon_\nu =
   \left\{ \begin{array}{c} \mathbf{H} \\
@@ Line 17: / Line 18: @@
 :$$p_{\rm M} = {\rm Q} \left ( {s_0}/{\sigma} \right ) = 0.01\hspace{0.05cm}.$$
-The bottom graph shows a digital channel model characterized by the four transition probabilities&nbsp; $p_1,&nbsp; p_2,&nbsp; p_3$&nbsp; and&nbsp; $p_4$.&nbsp; This is to be fitted to the analog channel model.
+&rArr; &nbsp; The bottom graph shows a digital channel model characterized by the four transition probabilities&nbsp; $p_1, &nbsp; p_2, &nbsp; p_3$ &nbsp; and &nbsp; $p_4$.&nbsp; This is to be fitted to the analog channel model.
@@ Line 23: / Line 24: @@
+<u>Notes:</u>
 * The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)| "Binary Symmetric Channel"]].
-''Notes:''
 * Numerical values of the Q&ndash;function can be determined with the interactive applet&nbsp; [[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian Error Functions"]].&nbsp;
@@ Line 38: / Line 38: @@
 $s_0/\sigma\ = \ ${ 2.32 3% }
-{For the threshold, let&nbsp; $E = 0$. Is the digital transmission system at hand describable by the BSC model, assuming that
+{For the threshold, let&nbsp; $E = 0$.&nbsp; Is the digital transmission system at hand describable by the BSC model,&nbsp; assuming that
 |type="[]"}
 + the source symbols&nbsp; $\mathbf{L}$&nbsp; and&nbsp; $\mathbf{H}$&nbsp; are equally probable,
-+ the source symbol&nbsp; $\mathbf{L}$&nbsp; occurs significantly more frequently than $\mathbf{H}$?
++ the source symbol&nbsp; $\mathbf{L}$&nbsp; occurs significantly more frequently than&nbsp; $\mathbf{H}$?
 {Calculate the transition probabilities for&nbsp; $E = +s_0/4$.
@@ Line 50: / Line 50: @@
 $p_4 \ = \ $ { 0.998 3% }
-{Now let&nbsp; $E = +s_0/4$. Is the present digital transmission system describable by the BSC model under the condition that
+{Now let&nbsp; $E = +s_0/4$.&nbsp; Is the present digital transmission system describable by the BSC model under the condition that
 |type="[]"}
 - the source symbols&nbsp; $\mathbf{L}$&nbsp; and&nbsp; $\mathbf{H}$&nbsp; are equally probable,
 - the source symbol&nbsp; $\mathbf{L}$&nbsp; occurs significantly more frequently than&nbsp; $\mathbf{H}$?
-{Let&nbsp; $p_{\rm L} = {\rm Pr}(q_{\nu} = \mathbf{L})$&nbsp; and&nbsp; $p_{\rm H} = {\rm Pr}(q_{\nu} = \mathbf{H})$. Which of the following statements is then true for the mean error probability&nbsp; $p_{\rm M}$?&nbsp;
+{Let&nbsp; $p_{\rm L} = {\rm Pr}(q_{\nu} = \mathbf{L})$&nbsp; and&nbsp; $p_{\rm H} = {\rm Pr}(q_{\nu} = \mathbf{H})$.&nbsp; Which of the following statements are then true for the mean error probability&nbsp; $p_{\rm M}$?&nbsp;
 |type="[]"}
 + $p_{\rm M}$&nbsp; in the BSC model &nbsp;$($valid for &nbsp;$E = 0)$&nbsp; is independent of&nbsp; $p_{\rm L}$&nbsp; and &nbsp;$p_{\rm H}$.
 - $p_{\rm M}$&nbsp; in the BSC model &nbsp;$($valid for &nbsp;$E = 0)$&nbsp; is smallest for&nbsp; $p_{\rm L} = p_{\rm H}$.&nbsp;
-+ For&nbsp; $p_{\rm L} = 0.9$,&nbsp; $p_{\rm H} = 0.1$&nbsp; and&nbsp; $E = +s_0/4$&nbsp; is&nbsp; $p_{\rm M} < 1\%$.
++ For&nbsp; $p_{\rm L} = 0.9$,&nbsp; $p_{\rm H} = 0.1$&nbsp; and&nbsp; $E = +s_0/4$&nbsp; &nbsp; &rArr; &nbsp; $p_{\rm M} < 1\%$.
 </quiz>

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}}
-[[File:P_ID1831__Dig_A_5_3.png|right|frame|AWGN channel and BSC model]]
+[[File:EN_Dig_A_5_3.png|right|frame|AWGN channel and BSC model]]
 The graphic above shows the analog channel model of a digital transmission system, where the additive noise signal&nbsp; $n(t)$&nbsp; with the (two-sided) noise power density&nbsp; $N_0/2$&nbsp; is effective. This is AWGN noise. The variance of the noise component before the decision (after the matched filter) is then
 :$$\sigma^2 = \frac{N_0}{2T} \hspace{0.05cm}.$$

@@ Line 72: / Line 72: @@
 :$$p_2 = p_3 = p = 0.01 \hspace{0.05cm}, \hspace{0.2cm}p_1 = p_4 = 1-p = 0.99\hspace{0.05cm}.$$
-*Ein Vergleich mit dem Theorieteil zeigt, dass dieses Kanalmodell dem BSC&ndash;Modell entspricht, und zwar unabhängig von der Statistik der Quellensymbole.
+*A comparison with the theory part shows that this channel model corresponds to the BSC model, independent of the statistics of the source symbols.
-*Richtig sind also <u>beide Lösungsvorschläge</u>.
+*Thus, <u>both solutions</u> are correct.
-'''(3)'''&nbsp; Die Übergangswahrscheinlichkeit $p_2$ beschreibt nun den Fall, dass die Enscheiderschwelle $E = 0.25 \cdot s_0$ fälschlicherweise unterschritten wurde.
+'''(3)'''&nbsp; The transition probability $p_2$ now describes the case where the decision threshold $E = 0.25 \cdot s_0$ was mistakenly undershot.
-*Dann ist $v_{\nu} = \mathbf{L}$, obwohl $q_{\nu} = \mathbf{H}$ gesendet wurde. Der Abstand von der Schwelle beträgt somit nur $0.75 \cdot s_0$ und es gilt:
+*Then $v_{\nu} = \mathbf{L}$, although $q_{\nu} = \mathbf{H}$ was sent. Thus, the distance from the threshold is only $0.75 \cdot s_0$ and it holds:
 :$$p_{\rm 2}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Q} \left ( \frac{0.75 \cdot s_0}{\sigma} \right ) = {\rm Q} \left ( 0.75 \cdot 2.32 \right )
   = {\rm Q} \left ( 1.74 \right )\hspace{0.15cm}\underline {\approx 0.041}\hspace{0.05cm}, \hspace{0.5cm}
@@ Line 84: / Line 84: @@
 .959}\hspace{0.05cm}.$$
-*In ähnlicher Weise können die Übergangswahrscheinlichkeiten $p_3$ und $p_4$ berechnet werden, wobei nun vom Schwellenabstand $1.25 \cdot s_0$ auszugehen ist:
+*Similarly, the transition probabilities $p_3$ and $p_4$ can be calculated, now assuming the threshold distance $1.25 \cdot s_0$:
 :$$p_{\rm 3}  = {\rm Q} \left ( 1.25 \cdot 2.32 \right )
   = {\rm Q} \left ( 2.90 \right )\hspace{0.15cm}\underline {\approx  0.002}\hspace{0.05cm}, \hspace{0.2cm}
@@ Line 91: / Line 91: @@
-'''(4)'''&nbsp; <u>Keiner</u> der beiden Lösungsvorschläge trifft zu:
+'''(4)'''&nbsp; <u>Neither</u> of the two solutions applies:
-*Mit der Entscheiderschwelle $E &ne; 0$ ist das BSC&ndash;Modell unabhängig von der Symbolstatistik nicht anwendbar,
+*With the decision threshold $E &ne; 0$, the BSC model is not applicable regardless of the symbol statistic,
-*da die Symmetrieeigenschaft des Kanals (das Kennzeichen "S" in "BSC") nicht gegeben ist.
+*since the symmetry property of the channel (the "S" flag in "BSC") does not hold.
-'''(5)'''&nbsp; Die <u>Aussagen 1 und 3</u> treffen zu, nicht aber die Aussage 2:
+'''(5)'''&nbsp; <u>Statements 1 and 3</u> are true, but statement 2 is not:
-*Beim BSC&ndash;Modell ist $p_{\rm M} = 1\%$ unabhängig von den Symbolwahrscheinlichkeiten $p_{\rm L}$ und $p_{\rm H}$.
+*In the BSC model, $p_{\rm M} = 1\%$ is independent of the symbol probabilities $p_{\rm L}$ and $p_{\rm H}$.
-*Dagegen gilt für $p_{\rm L} = 0.9$, $p_{\rm H} = 0.1$ sowie $E = +s_0/4$:
+*In contrast, for $p_{\rm L} = 0.9$, $p_{\rm H} = 0.1$ and $E = +s_0/4$:
 :$$p_{\rm M}  = 0.9 \cdot p_{\rm 3} + 0.1 \cdot p_{\rm 2}= 0.9 \cdot 0.2\% + 0.1 \cdot 4.1\%
   \approx 0.59\% \hspace{0.05cm}.$$
-*Das Minimum ergibt sich für $p_{\rm L} = 0.93$ und $p_{\rm H} = 0.07$ zu $p_{\rm M} \approx 0.45\%$.
+*The minimum results for $p_{\rm L} = 0.93$ and $p_{\rm H} = 0.07$ to $p_{\rm M} \approx 0.45\%$.
 {{ML-Fuß}}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}}
-[[File:P_ID1831__Dig_A_5_3.png|right|frame|AWGN–Kanal und BSC–Modell]]
+[[File:P_ID1831__Dig_A_5_3.png|right|frame|AWGN channel and BSC model]]
-Die Grafik zeigt oben das analoge Kanalmodell eines digitalen Übertragungssystems, wobei das additive Rauschsignal&nbsp; $n(t)$&nbsp; mit der (zweiseitigen) Rauschleistungsdichte&nbsp; $N_0/2$&nbsp; wirksam ist. Es handelt sich um AWGN&ndash;Rauschen. Die Varianz des Rauschanteils vor dem Entscheider (nach dem Matched&ndash;Filter) ist dann
+The graphic above shows the analog channel model of a digital transmission system, where the additive noise signal&nbsp; $n(t)$&nbsp; with the (two-sided) noise power density&nbsp; $N_0/2$&nbsp; is effective. This is AWGN noise. The variance of the noise component before the decision (after the matched filter) is then
 :$$\sigma^2 = \frac{N_0}{2T} \hspace{0.05cm}.$$
-Weiter soll gelten:
+Further, let hold:
-* Es treten keine Impulsinterferenzen auf. Wurde das Symbol&nbsp; $q_{\nu} = \mathbf{H}$&nbsp; gesendet, so ist der Nutzanteil des Detektionssignal gleich&nbsp; $+s_0$, bei&nbsp; $q_{\nu} = \mathbf{L}$&nbsp; dagegen&nbsp; $-s_0$.
+* No intersymbol interference occurs. If the symbol&nbsp; $q_{\nu} = \mathbf{H}$&nbsp; was sent, the useful component of the detection signal is equal to&nbsp; $+s_0$, while for&nbsp; $q_{\nu} = \mathbf{L}$,&nbsp; it is equal to&nbsp; $-s_0$.
-* Der Schwellenwertentscheider berücksichtigt eine Schwellendrift, das heißt, die Schwelle&nbsp; $E$&nbsp; kann durchaus vom Optimalwert&nbsp; $E = 0$&nbsp; abweichen. Die <i>Entscheidungsregel</i> lautet:
+* The threshold decision takes into account a threshold drift, that is, the threshold&nbsp; $E$&nbsp; may well deviate from the optimal value&nbsp; $E = 0$.&nbsp; The <i>decision rule</i> is:
 :$$\upsilon_\nu =
   \left\{ \begin{array}{c} \mathbf{H} \\
   \mathbf{L} \end{array} \right.\quad
-\begin{array}{*{1}c} {\rm falls}\hspace{0.15cm}d (\nu \cdot T) > E  \hspace{0.05cm},
+\begin{array}{*{1}c} {\rm if}\hspace{0.15cm}d (\nu \cdot T) > E  \hspace{0.05cm},
-\\  {\rm falls} \hspace{0.15cm} d (\nu \cdot T) \le E\hspace{0.05cm}.\\ \end{array}$$
+\\  {\rm if} \hspace{0.15cm} d (\nu \cdot T) \le E\hspace{0.05cm}.\\ \end{array}$$
-* Mit dem Schwellenwert&nbsp; $E = 0$&nbsp; ergibt sich die mittlere Fehlerwahrscheinlichkeit zu
+* With the threshold value&nbsp; $E = 0$,&nbsp; the mean error probability is given by
 :$$p_{\rm M} = {\rm Q} \left ( {s_0}/{\sigma} \right ) = 0.01\hspace{0.05cm}.$$
-Die untere Grafik zeigt ein digitales Kanalmodell, das durch die vier Übergangswahrscheinlichkeiten&nbsp; $p_1,&nbsp; p_2,&nbsp; p_3$&nbsp; und&nbsp; $p_4$&nbsp; charakterisiert ist. Dieses soll an das analoge Kanalmodell angepasst werden.
+The bottom graph shows a digital channel model characterized by the four transition probabilities&nbsp; $p_1,&nbsp; p_2,&nbsp; p_3$&nbsp; and&nbsp; $p_4$.&nbsp; This is to be fitted to the analog channel model.
@@ Line 25: / Line 25: @@
-''Hinweise:''
+''Notes:''
-* Die Aufgabe gehört zum Kapitel&nbsp; [[Digitalsignal%C3%BCbertragung/Binary_Symmetric_Channel_(BSC)| Binary Symmetric Channel (BSC)]].
+* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)| "Binary Symmetric Channel (BSC)"]].
-* Zahlenwerte der Q&ndash;Funktion können mit dem interaktiven Applet&nbsp; [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Komplementäre Gaußsche Fehlerfunktionen]]&nbsp; ermittelt werden.
+* Numerical values of the Q&ndash;function can be determined with the interactive applet&nbsp; [[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian Error Functions"]].&nbsp;
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welcher Quotient&nbsp; $s_0/\sigma$&nbsp; liegt dieser Aufgabe zugrunde?
+{Which quotient&nbsp; $s_0/\sigma$&nbsp; is the basis of this exercise?
 |type="{}"}
 $s_0/\sigma\ = \ ${ 2.32 3% }
-{Für die Schwelle gelte&nbsp; $E = 0$. Ist das vorliegende digitale Übertragungssystem durch das BSC&ndash;Modell beschreibbar, unter der Voraussetzung, dass
+{For the threshold, let&nbsp; $E = 0$. Is the digital transmission system at hand describable by the BSC model, assuming that
 |type="[]"}
-+ die Quellensymbole&nbsp; $\mathbf{L}$&nbsp; und&nbsp; $\mathbf{H}$&nbsp; gleichwahrscheinlich sind,
++ the source symbols&nbsp; $\mathbf{L}$&nbsp; and&nbsp; $\mathbf{H}$&nbsp; are equally probable,
-+ das Quellensymbol&nbsp; $\mathbf{L}$&nbsp; deutlich häufiger auftritt als $\mathbf{H}$?
++ the source symbol&nbsp; $\mathbf{L}$&nbsp; occurs significantly more frequently than $\mathbf{H}$?
-{Berechnen Sie die Übergangswahrscheinlichkeiten für&nbsp; $E = +s_0/4$.
+{Calculate the transition probabilities for&nbsp; $E = +s_0/4$.
 |type="{}"}
 $p_1 \ = \ $ { 0.959 3% }
@@ Line 50: / Line 50: @@
 $p_4 \ = \ $ { 0.998 3% }
-{Nun gelte&nbsp; $E = +s_0/4$. Ist das vorliegende digitale Übertragungssystem durch das BSC&ndash;Modell beschreibbar, unter der Voraussetzung, dass
+{Now let&nbsp; $E = +s_0/4$. Is the present digital transmission system describable by the BSC model under the condition that
 |type="[]"}
-- die Quellensymbole&nbsp; $\mathbf{L}$&nbsp; und&nbsp; $\mathbf{H}$&nbsp; gleichwahrscheinlich sind,
+- the source symbols&nbsp; $\mathbf{L}$&nbsp; and&nbsp; $\mathbf{H}$&nbsp; are equally probable,
-- das Quellensymbol&nbsp; $\mathbf{L}$&nbsp; deutlich häufiger auftritt als&nbsp; $\mathbf{H}$?
+- the source symbol&nbsp; $\mathbf{L}$&nbsp; occurs significantly more frequently than&nbsp; $\mathbf{H}$?
-{Es gelte&nbsp; $p_{\rm L} = {\rm Pr}(q_{\nu} = \mathbf{L})$&nbsp; und&nbsp; $p_{\rm H} = {\rm Pr}(q_{\nu} = \mathbf{H})$. Welche der folgenden Aussagen sind dann für die mittlere Fehlerwahrscheinlichkeit&nbsp; $p_{\rm M}$&nbsp; zutreffend?
+{Let&nbsp; $p_{\rm L} = {\rm Pr}(q_{\nu} = \mathbf{L})$&nbsp; and&nbsp; $p_{\rm H} = {\rm Pr}(q_{\nu} = \mathbf{H})$. Which of the following statements is then true for the mean error probability&nbsp; $p_{\rm M}$?&nbsp;
 |type="[]"}
-+ $p_{\rm M}$&nbsp; ist beim BSC&ndash;Modell &nbsp;$($gültig für &nbsp;$E = 0)$&nbsp; unabhängig von&nbsp; $p_{\rm L}$&nbsp; und &nbsp;$p_{\rm H}$.
++ $p_{\rm M}$&nbsp; in the BSC model &nbsp;$($valid for &nbsp;$E = 0)$&nbsp; is independent of&nbsp; $p_{\rm L}$&nbsp; and &nbsp;$p_{\rm H}$.
-- $p_{\rm M}$&nbsp; ist beim BSC&ndash;Modell &nbsp;$($gültig für &nbsp;$E = 0)$&nbsp; für&nbsp; $p_{\rm L} = p_{\rm H}$&nbsp; am kleinsten.
+- $p_{\rm M}$&nbsp; in the BSC model &nbsp;$($valid for &nbsp;$E = 0)$&nbsp; is smallest for&nbsp; $p_{\rm L} = p_{\rm H}$.&nbsp;
-+ Für&nbsp; $p_{\rm L} = 0.9$,&nbsp; $p_{\rm H} = 0.1$&nbsp; und&nbsp; $E = +s_0/4$&nbsp; ist&nbsp; $p_{\rm M} < 1\%$.
++ For&nbsp; $p_{\rm L} = 0.9$,&nbsp; $p_{\rm H} = 0.1$&nbsp; and&nbsp; $E = +s_0/4$&nbsp; is&nbsp; $p_{\rm M} < 1\%$.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die mittlere Fehlerwahrscheinlichkeit beträgt $p_{\rm M} = {\rm Q}(s_0/\sigma) = 0.01$.
+'''(1)'''&nbsp; The average error probability is $p_{\rm M} = {\rm Q}(s_0/\sigma) = 0.01$.
-*Daraus folgt für den Quotienten aus Detektionsnutzabtastwert und Detektionsstöreffektivwert:
+*From this it follows for the quotient of the detection useful sample value and the detection noise rms value:
 :$${s_0}/{\sigma}= {\rm Q}^{-1} \left ( 0.01 \right ) \hspace{0.15cm}\underline {\approx 2.32}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; Mit $E = 0$ ergibt sich für die Wahrscheinlichkeiten des vorgegebenen digitalen Kanalmodells:
+'''(2)'''&nbsp; With $E = 0$, the probabilities of the given digital channel model are given by:
 :$$p_2 = p_3 = p = 0.01 \hspace{0.05cm}, \hspace{0.2cm}p_1 = p_4 = 1-p = 0.99\hspace{0.05cm}.$$

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−	{{quiz-Header\|Buchseite=~~Digitalsignalübertragung~~/~~Binary Symmetric Channel~~ (BSC)}}	+	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}}

	[[File:P_ID1831__Dig_A_5_3.png\|right\|frame\|AWGN–Kanal und BSC–Modell]]		[[File:P_ID1831__Dig_A_5_3.png\|right\|frame\|AWGN–Kanal und BSC–Modell]]

← Older revision		Revision as of 16:37, 23 August 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel ~~(BSC)~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel^]]

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 '''(4)'''&nbsp; <u>Keiner</u> der beiden Lösungsvorschläge trifft zu:
 *Mit der Entscheiderschwelle $E &ne; 0$ ist das BSC&ndash;Modell unabhängig von der Symbolstatistik nicht anwendbar,
-*da die Symmetrieeigenschaft des Kanals (das Kennzeichen &bdquo;S&rdquo; in &bdquo;BSC&rdquo;) nicht gegeben ist.
+*da die Symmetrieeigenschaft des Kanals (das Kennzeichen "S&rdquo; in "BSC&rdquo;) nicht gegeben ist.