Aufgaben:Exercise 3.7: Bit Error Rate (BER) - Revision history

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:39Z

Text replacement - "rms value" to "standard deviation"

Bene: Text replacement - "root mean square" to "standard deviation"

2022-02-17T11:00:51Z

Text replacement - "root mean square" to "standard deviation"

Guenter at 11:13, 2 February 2022

2022-02-02T11:13:10Z

Guenter at 10:43, 2 February 2022

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Guenter at 15:46, 1 February 2022

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Noah at 18:51, 15 January 2022

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Noah at 20:37, 3 January 2022

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Noah at 20:36, 3 January 2022

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Noah at 00:33, 2 January 2022

2022-01-02T00:33:57Z

Guenter at 10:18, 20 December 2021

2021-12-20T10:18:56Z

@@ Line 58: / Line 58: @@
-{What values can the bit error rate&nbsp; $h_{\rm B}$&nbsp; take?&nbsp; Show that the linear mean&nbsp; $m_{h{\rm B}}$&nbsp; of this random variable is equal to the bit error probability&nbsp; $p_{\rm B}$&nbsp; What is its rms value?
+{What values can the bit error rate&nbsp; $h_{\rm B}$&nbsp; take?&nbsp; Show that the linear mean&nbsp; $m_{h{\rm B}}$&nbsp; of this random variable is equal to the bit error probability&nbsp; $p_{\rm B}$&nbsp; What is its standard deviation?
 |type="{}"}
 $\sigma_{h{\rm B}} \ = \ $ { 0.0001 3% }
-{Under certain conditions,&nbsp; a binomially distributed random variable can be approximated by a Gaussian distribution <br>with equal mean&nbsp; $(m_{h{\rm B}})$&nbsp; and equal rms value&nbsp; $(\sigma_{h{\rm B}})$.&nbsp; Which statement is true?
+{Under certain conditions,&nbsp; a binomially distributed random variable can be approximated by a Gaussian distribution <br>with equal mean&nbsp; $(m_{h{\rm B}})$&nbsp; and equal standard deviation&nbsp; $(\sigma_{h{\rm B}})$.&nbsp; Which statement is true?
 |type="()"}
 + ${\rm Pr}(\hspace{0.05cm}|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}| \le \varepsilon)=1- 2\cdot \rm Q({\varepsilon}/{\sigma_{\it h}{\rm B}}).$
@@ Line 108: / Line 108: @@
 :$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$
-*The rms value results in
+*The standard deviation results in
 :$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$

@@ Line 53: / Line 53: @@
-{How large is the root mean square of the random variable&nbsp; $n_{\rm B}$&nbsp; with&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$?
+{How large is the standard deviation of the random variable&nbsp; $n_{\rm B}$&nbsp; with&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$?
 |type="{}"}
 $\sigma_{n{\rm B}} \ = \ $ { 10 3% }
@@ Line 97: / Line 97: @@
-'''(2)'''&nbsp; Für the root mean square of the binomial distribution holds with good approximation:
+'''(2)'''&nbsp; Für the standard deviation of the binomial distribution holds with good approximation:
 :$$\sigma_{n{\rm B}}=\sqrt{N\cdot p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}
 \hspace{0.15cm}\underline{\approx 10}.$$

@@ Line 90: / Line 90: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The <u>last two statements</u> are true:
+'''(1)'''&nbsp; The&nbsp; <u>last two statements</u>&nbsp; are true:
 *Relative to the random variable&nbsp; $n_{\rm B}$&nbsp; there is the classical case of a binomial distribution.
-*The sum over&nbsp; $N$&nbsp; binary random variables is formed.
+*The sum over&nbsp; $N$&nbsp; binary random variables is formed.&nbsp; The possible values of&nbsp; $n_{\rm B}$&nbsp; thus lie between&nbsp; $0$&nbsp; and&nbsp; $N$.
 *The linear mean gives &nbsp; $m_{n{\rm B}}=p_{\rm B}\cdot N=\rm 10^{-3}\cdot 10^{5}=\rm 100.$
-'''(2)'''&nbsp; Für the rms of the binomial distribution holds with good approximation:
+'''(2)'''&nbsp; Für the root mean square of the binomial distribution holds with good approximation:
 :$$\sigma_{n{\rm B}}=\sqrt{N\cdot p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}
 \hspace{0.15cm}\underline{\approx 10}.$$
@@ Line 106: / Line 105: @@
 '''(3)'''&nbsp; Possible values of&nbsp; $h_{\rm B}$&nbsp; are all integer multiples of&nbsp; $1/N$.&nbsp; These all lie between&nbsp; $0$&nbsp; and&nbsp; $1$.
-*For the mean value, one obtains:
+*For the mean value,&nbsp; one obtains:
 :$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$
-*The rms results in
+*The rms value results in
 :$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$
-'''(4)'''&nbsp; Correct is <u>the first proposition</u>. It holds:
+'''(4)'''&nbsp; Correct is&nbsp; <u>the first proposition</u>.&nbsp; It holds:
 :$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}}),$$
 :$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \varepsilon {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}}{\rm )}$$
 :$$\Rightarrow \hspace{0.5cm}\rm Pr(\it |h_{\rm B} - p_{\rm B}| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}}).$$
@@ Line 124: / Line 122: @@
 :$$p_{\varepsilon}=\rm 1-\rm 2\cdot \rm Q(\frac{\rm 10^{\rm -4}}{\rm 10^{\rm -4}} {\rm )}=\rm 1-\rm 2\cdot\rm Q(\rm 1)\hspace{0.15cm}\underline{\approx\rm 0.684}.$$
-In words:
+In words: &nbsp; If one determines the bit error rate by simulation over&nbsp; $10^5$&nbsp; symbols,&nbsp; with a confidence level of&nbsp; $\underline{68.4\%}$&nbsp; <br>one obtains a value between&nbsp; $0.9 \cdot 10^{-3}$&nbsp; and&nbsp; $1.1 \cdot 10^{-3}$,&nbsp; if&nbsp; $p_{\rm B} = 10^{-3}$.
@@ Line 136: / Line 131: @@
-'''(7)'''&nbsp; It must&nbsp; $\alpha = \varepsilon/\sigma_{h{\rm B}}$&nbsp; With the result of the subtask&nbsp; '''(2)'''&nbsp; then follows:
+'''(7)'''&nbsp; It must&nbsp; $\alpha = \varepsilon/\sigma_{h{\rm B}}$.&nbsp; With the result of the subtask&nbsp; '''(2)'''&nbsp; then follows:
 :$$\frac{\varepsilon}{\sqrt{p_{\rm B}\cdot(\rm 1-\it p_{\rm B})/N}}\ge {\rm 2} \hspace{0.5cm}\Rightarrow\hspace{0.5cm}
 N\ge \frac{\rm 4\cdot \it p_{\rm B}\cdot(\rm 1-\it p_{\rm B})}{\varepsilon^{\rm 2}}\approx \frac{\rm 4\cdot 10^{-3}}{10^{-8}}\hspace{0.15cm}\underline{=\rm 400\hspace{0.08cm}000}.$$

@@ Line 23: / Line 23: @@
 $\rm (B)$&nbsp; For the metrological determination of the transmission quality or for a system simulation,&nbsp; it is necessary to rely on the&nbsp; '''Bit Error Rate'''&nbsp; $\rm (BER)$:
 ::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$
+:*$n_{\rm B}$&nbsp; indicates the number of bit errors occurred when a total of&nbsp; $N$&nbsp; binary symbols&nbsp; ("bits")&nbsp; were transmitted.
-:*$h_{\rm B}$&nbsp; is a&nbsp; [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission#Definition_der_Bitfehlerquote|relative frequency]].&nbsp; $n_{\rm B}$&nbsp; indicates the number of bit errors occurred when a total of&nbsp; $N$&nbsp; symbols (bits) transmitted.
+:*In the limiting case&nbsp; $N \to \infty$&nbsp; the relative frequency&nbsp; $h_{\rm B}$&nbsp; coincides with the probability&nbsp; $p_{\rm B}$.&nbsp; Here now the question shall be clarified,&nbsp; which statistical uncertainty has to be expected with finite&nbsp; $N$.
@@ Line 41: / Line 34: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables|Gaussian random variables]].
+*Solve this exercise as far as possible in general.&nbsp; Use the parameter values&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$ for control input.
-*Read the exercises as far as possible in general.
+*The following are some values of the so-called&nbsp; "Q-function":
 :$$\rm Q(\rm 1.00)=\rm 0.159,\hspace{0.5cm}\rm Q(\rm 1.65)=\rm 0.050,\hspace{0.5cm}\rm Q(\rm 1.96)=\rm 0.025,\hspace{0.5cm}\rm Q(\rm 2.59)=\rm 0.005.$$
@@ Line 62: / Line 53: @@
-{How large; is the rms of the random variable&nbsp; $n_{\rm B}$&nbsp; for&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$?
+{How large is the root mean square of the random variable&nbsp; $n_{\rm B}$&nbsp; with&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$?
 |type="{}"}
 $\sigma_{n{\rm B}} \ = \ $ { 10 3% }
-{What values can the bit error rate&nbsp; $h_{\rm B}$&nbsp; take?&nbsp; <br>Show that the linear mean&nbsp; $m_{h{\rm B}}$&nbsp; of this random variable is equal to the actual bit error probability&nbsp; $p_{\rm B}$&nbsp; What is its dispersion?
+{What values can the bit error rate&nbsp; $h_{\rm B}$&nbsp; take?&nbsp; Show that the linear mean&nbsp; $m_{h{\rm B}}$&nbsp; of this random variable is equal to the bit error probability&nbsp; $p_{\rm B}$&nbsp; What is its rms value?
 |type="{}"}
 $\sigma_{h{\rm B}} \ = \ $ { 0.0001 3% }
-{Under certain conditions, a binomially distributed random variable can be approximated by a Gaussian distribution with equal mean&nbsp; $(m_{h{\rm B}})$&nbsp; and equal dispersion&nbsp; $(\sigma_{h{\rm B}})$&nbsp; Which statement is true?
+{Under certain conditions,&nbsp; a binomially distributed random variable can be approximated by a Gaussian distribution <br>with equal mean&nbsp; $(m_{h{\rm B}})$&nbsp; and equal rms value&nbsp; $(\sigma_{h{\rm B}})$.&nbsp; Which statement is true?
 |type="()"}
 + ${\rm Pr}(\hspace{0.05cm}|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}| \le \varepsilon)=1- 2\cdot \rm Q({\varepsilon}/{\sigma_{\it h}{\rm B}}).$
@@ Line 79: / Line 70: @@
-{For abbreviation, we use the confidence level&nbsp; $p_\varepsilon = {\rm Pr}(\hspace{0.05cm}|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}| \le \varepsilon)$. &nbsp; Which&nbsp; $p_\varepsilon$&nbsp; results with&nbsp; $\varepsilon = 10^{-4}$,&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$&nbsp;?
+{For abbreviation,&nbsp; we use the confidence level&nbsp; $p_\varepsilon = {\rm Pr}(\hspace{0.05cm}|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}| \le \varepsilon)$. &nbsp; Which&nbsp; $p_\varepsilon$&nbsp; results with&nbsp; $\varepsilon = 10^{-4}$, &nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$&nbsp;?
 |type="{}"}
 $p_\varepsilon \ = \ $ { 0.684 3% }
-{Let the argument of the Q-function be&nbsp; $\alpha$.&nbsp; What is the minimum &nbsp; $\alpha$&nbsp; that must be chosen for the confidence level&nbsp; $p_\varepsilon = 95\%$&nbsp; to be&nbsp;?
+{Let the argument of the Q-function be&nbsp; $\alpha$.&nbsp; What is the minimum value of&nbsp; $\alpha$&nbsp; that must be chosen for the confidence level&nbsp; $p_\varepsilon = 95\%$&nbsp;?
 |type="{}"}
 $\alpha_{\rm min} \ = \ $ { 1.96 3% }
-{It still holds&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $p_\varepsilon = 95\%$. &nbsp; Over how many symbols&nbsp; $(N_\text{min})$&nbsp; must be averaged at least, <br>so that the determined bit error rate in the range between&nbsp; $0. 9 \cdot 10^{-3}$&nbsp; and&nbsp; $1.1 \cdot 10^{-3}$&nbsp; lies &nbsp; $(\varepsilon = 10^{-4}, \ \text{10% of nominal value)}$&nbsp;?
+{It still holds&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $p_\varepsilon = 95\%$. &nbsp; Over how many symbols&nbsp; $(N_\text{min})$&nbsp; must be averaged at least, <br>so that the determined bit error rate lies in the range between&nbsp; $0. 9 \cdot 10^{-3}$&nbsp; and&nbsp; $1.1 \cdot 10^{-3}$&nbsp;  $(\varepsilon = 10^{-4}, \ \text{10% of its nominal value)}$&nbsp;?
 |type="{}"}
 $N_\text{min} \ = \ ${ 400000 3% }

@@ Line 4: / Line 4: @@
 [[File:EN_Sto_A_3_7.png|right|frame|To illustrate the bit error rate]]
-We consider a binary transmission system with.
+We consider a binary transmission system with
-*the source symbol sequence&nbsp; $\langle q_\nu \rangle $&nbsp; and
+*the source symbol sequence&nbsp; $\langle q_\nu \rangle $,&nbsp; and
 *the sink symbol sequence&nbsp; $\langle v_\nu \rangle $.
-If sink symbol&nbsp; $v_\nu$&nbsp; and source symbol&nbsp; $q_\nu$&nbsp; do not match, there is a bit error &nbsp; &rArr; &nbsp; $e_\nu = 1$.
+If the sink symbol&nbsp; $v_\nu$&nbsp; and source symbol&nbsp; $q_\nu$&nbsp; do not match,&nbsp; there is a&nbsp; "bit error" &nbsp; &rArr; &nbsp; $e_\nu = 1$.&nbsp; Otherwise&nbsp; $e_\nu = 0$&nbsp; holds.
-$\rm (A)$&nbsp; The most important evaluation criterion of such a digital system is
+$\rm (A)$&nbsp; The most important evaluation criterion of such a digital system is the&nbsp; '''Bit Error Probability''':
 :*With the expected value&nbsp; ${\rm E}\big[\text{ ...} \big]$&nbsp; this is defined as follows:
 :: $$\it p_{\rm B} = \rm E\big[\rm Pr(\it v_{\nu} \ne q_{\nu} \rm )\big]=\rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\lim_{{\it N}\to\infty}\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N}\rm Pr(\it e_{\nu}=\rm 1). $$
-:*The right part of this equation describes a time averaging;&nbsp;this must always be applied, for example, to time-varying channels.
+:*The right part of this equation describes a time averaging;&nbsp;this must always be applied to time-varying channels.
-:*If the error probability is the same for all symbols (which is assumed here), the above equation can be simplified:
+:*If the error probability is the same for all symbols&nbsp; (which is assumed here),&nbsp; the above equation can be simplified:
 ::$$\it p_{\rm B} = \rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\rm E\big[\it e_{\nu} \rm \big].$$
+$\rm (B)$&nbsp; For the metrological determination of the transmission quality or for a system simulation,&nbsp; it is necessary to rely on the&nbsp; '''Bit Error Rate'''&nbsp; $\rm (BER)$:
 ::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$
@@ Line 34: / Line 30: @@
 :*In the limiting case&nbsp; $N \to \infty$&nbsp; the relative frequency&nbsp; $h_{\rm B}$&nbsp; coincides with the probability&nbsp; $p_{\rm B}$&nbsp;.
 :*Here now the question shall be clarified, which statistical uncertainty has to be expected with finite&nbsp; $N$&nbsp;.

@@ Line 11: / Line 11: @@
 If sink symbol&nbsp; $v_\nu$&nbsp; and source symbol&nbsp; $q_\nu$&nbsp; do not match, there is a bit error &nbsp; &rArr; &nbsp; $e_\nu = 1$.
- <br>Otherwise&nbsp; $e_\nu = 0$ holds.
+<br>Otherwise&nbsp; $e_\nu = 0$ holds.
@@ Line 121: / Line 121: @@
 :$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$
-*The scatter results in
+*The rms results in
-:$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$
+:$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$
 '''(4)'''&nbsp; Correct is <u>the first proposition</u>. It holds:
-:$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}),$$
+:$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}}),$$
-:$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \itvarepsilon {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}{\rm )}$$
+:$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \varepsilon {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}}{\rm )}$$
-:$$\Rightarrow \hspace{0.5cm}\rm Pr(\it |h_{\rm B} - p_{\rm B}| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}).$$
+:$$\Rightarrow \hspace{0.5cm}\rm Pr(\it |h_{\rm B} - p_{\rm B}| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}}).$$

@@ Line 27: / Line 27: @@
 $\rm (B)$&nbsp; On the other hand, for the metrological determination of the transmission quality or for the system simulation, it is necessary to rely on
-:the comparable ''A-posteriori parameter''&nbsp; '''Bit error rate'''&nbsp; must be ignored.
+:the comparable ''A-posteriori parameter''&nbsp; '''Bit Error Rate'''&nbsp; must be ignored.
 ::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$
@@ Line 65: / Line 65: @@
-{How large; is the dispersion of the random variable&nbsp; $n_{\rm B}$&nbsp; for&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$?
+{How large; is the rms of the random variable&nbsp; $n_{\rm B}$&nbsp; for&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $N = 10^{5}$?
 |type="{}"}
 $\sigma_{n{\rm B}} \ = \ $ { 10 3% }
@@ Line 92: / Line 92: @@
-{It still holds&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $p_\varepsilon = 95\%$. &nbsp; &Uuml;over how many symbols&nbsp; $(N_\text{min})$&nbsp; must be averaged at least, <br>so that the determined bit error rate in the range between&nbsp; $0. 9 \cdot 10^{-3}$&nbsp; and&nbsp; $1.1 \cdot 10^{-3}$&nbsp; lies &nbsp; $(\varepsilon = 10^{-4}, \ \text{10% of nominal value)}$&nbsp;?
+{It still holds&nbsp; $p_{\rm B} = 10^{-3}$&nbsp; and&nbsp; $p_\varepsilon = 95\%$. &nbsp; Over how many symbols&nbsp; $(N_\text{min})$&nbsp; must be averaged at least, <br>so that the determined bit error rate in the range between&nbsp; $0. 9 \cdot 10^{-3}$&nbsp; and&nbsp; $1.1 \cdot 10^{-3}$&nbsp; lies &nbsp; $(\varepsilon = 10^{-4}, \ \text{10% of nominal value)}$&nbsp;?
 |type="{}"}
 $N_\text{min} \ = \ ${ 400000 3% }

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Gaußverteilte Zufallsgröße
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables
 }}
-[[File:EN_Sto_A_3_7.png|right|frame|Zur Verdeutlichung der Bitfehlerquote]]
+[[File:EN_Sto_A_3_7.png|right|frame|To illustrate the bit error rate]]
-Wir betrachten ein bin&auml;res &Uuml;bertragungssystem mit
+We consider a binary transmission system with.
-*der Quellensymbolfolge&nbsp; $\langle q_\nu \rangle $&nbsp; und
+*the source symbol sequence&nbsp; $\langle q_\nu \rangle $&nbsp; and
-*der Sinkensymbolfolge&nbsp; $\langle v_\nu \rangle $.
+*the sink symbol sequence&nbsp; $\langle v_\nu \rangle $.
-Stimmen Sinkensymbol&nbsp; $v_\nu$&nbsp; und Quellensymbol&nbsp; $q_\nu$&nbsp; nicht &uuml;berein, so liegt ein Bitfehler vor &nbsp; &rArr; &nbsp;  $e_\nu = 1$. <br>Ansonsten gilt&nbsp; $e_\nu = 0$.
+If sink symbol&nbsp; $v_\nu$&nbsp; and source symbol&nbsp; $q_\nu$&nbsp; do not match, there is a bit error &nbsp; &rArr; &nbsp; $e_\nu = 1$.
-$\rm (A)$&nbsp; Das wichtigste Beurteilungskriterium eines solchen Digitalsystems ist
+$\rm (A)$&nbsp; The most important evaluation criterion of such a digital system is
+:the&nbsp; '''Bit Error Probability'''&nbsp;.
 :*Mit dem Erwartungswert&nbsp; ${\rm E}\big[\text{ ...} \big]$&nbsp; ist diese ist wie folgt definiert:
-::$$\it p_{\rm B} = \rm E\big[\rm Pr(\it v_{\nu} \ne q_{\nu} \rm )\big]=\rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\lim_{{\it N}\to\infty}\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N}\rm Pr(\it e_{\nu}=\rm 1).$$
+:: $$\it p_{\rm B} = \rm E\big[\rm Pr(\it v_{\nu} \ne q_{\nu} \rm )\big]=\rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\lim_{{\it N}\to\infty}\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N}\rm Pr(\it e_{\nu}=\rm 1). $$
 :*Der rechte Teil dieser Gleichung beschreibt eine Zeitmittelung;&nbsp;diese muss zum Beispiel bei zeitvarianten Kan&auml;len stets angewandt werden.
@@ Line 23: / Line 23: @@
 ::$$\it p_{\rm B} = \rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\rm E\big[\it e_{\nu} \rm \big].$$
-:Die Bitfehlerwahrscheinlichkeit ist eine ''A-priori-Kenngr&ouml;&szlig;e'', erlaubt also eine Vorhersage f&uuml;r das zu erwartende Resultat.
+:Die Bitfehlerwahrscheinlichkeit ist eine ''A-priori-Kenngröße'', erlaubt also eine Vorhersage für das zu erwartende Resultat.
 $\rm (B)$&nbsp; Dagegen muss zur messtechnischen Ermittlung der &Uuml;bertragungsqualit&auml;t oder bei der Systemsimulation auf
-:die vergleichbare ''A-posteriori-Kenngr&ouml;&szlig;e''&nbsp; '''Bitfehlerquote'''&nbsp; (englisch: &nbsp; <i>Bit Error Rate</i>)&nbsp; &uuml;bergegangen werden:
+:die vergleichbare ''A-posteriori-Kenngröße''&nbsp; '''Bitfehlerquote'''&nbsp; (englisch: &nbsp; <i>Bit Error Rate</i>)&nbsp; übergegangen werden:
 ::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$
-:*$h_{\rm B}$&nbsp; ist eine&nbsp; [[Digital_Signal_Transmission/Fehlerwahrscheinlichkeit_bei_Basisbandübertragung#Definition_der_Bitfehlerquote|relative H&auml;ufigkeit]].&nbsp; $n_{\rm B}$&nbsp; gibt die Anzahl der aufgetretenen Bitfehler an, wenn insgesamt&nbsp; $N$&nbsp; Symbole (Bit) &uuml;bertragen wurden.
+:*$h_{\rm B}$&nbsp; is a&nbsp; [[Digital_Signal_Transmission/Error_Probability_in_Baseband_Transmission#Definition_of_Bit_Error_Ratio|relative H&auml;ufigkeit]].&nbsp; $n_{\rm B}$&nbsp; indicates the number of bit errors occurred when a total of&nbsp; $N$&nbsp; symbols (bits) &uuml;btransmitted.
-:*Im Grenzfall&nbsp; $N \to \infty$&nbsp; stimmt die relative H&auml;ufigkeit&nbsp; $h_{\rm B}$&nbsp; mit der Wahrscheinlichkeit&nbsp; $p_{\rm B}$&nbsp; &uuml;berein.
+:*In the limiting case&nbsp; $N \to \infty$&nbsp; the relative frequency&nbsp; $h_{\rm B}$&nbsp; coincides with the probability&nbsp; $p_{\rm B}$&nbsp;.
-:*Hier soll nun die Frage gekl&auml;rt werden, mit welcher statistischen Unsicherheit bei endlichem&nbsp; $N$&nbsp; gerechnet werden muss.
+:*Here now the question shall be clarified, which statistical uncertainty has to be expected with finite&nbsp; $N$&nbsp;.

← Older revision		Revision as of 10:18, 20 December 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^3.5 ~~Gaußverteilte Zufallsgröße~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]