Aufgaben:Exercise 1.2Z: Bit Error Measurement - Revision history

Guenter at 16:16, 29 April 2022

2022-04-29T16:16:56Z

Guenter at 13:37, 29 April 2022

2022-04-29T13:37:01Z

Guenter at 12:56, 29 April 2022

2022-04-29T12:56:00Z

Hwang at 15:51, 2 March 2022

2022-03-02T15:51:59Z

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Guenter at 16:27, 8 February 2022

2022-02-08T16:27:06Z

Guenter at 15:23, 2 February 2022

2022-02-02T15:23:58Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.2Z: Bitfehlermessung to Aufgaben:Exercise 1.2Z: Bit Error Measurement

2022-02-02T15:19:36Z

Guenter moved page Aufgabe 1.2Z: Bitfehlermessung to Exercise 1.2Z: Bit Error Measurement

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

2021-03-23T12:47:50Z

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

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

2020-07-09T12:57:34Z

Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

Guenter at 11:40, 30 January 2019

2019-01-30T11:40:27Z

@@ Line 4: / Line 4: @@
-[[File:EN_Dig_Z_1_2.png|right|frame|Simulated bit error frequencies]]
+[[File:EN_Dig_Z_1_2.png|right|frame|Simulated bit error frequencies&nbsp; $(h_{\rm B})$; &nbsp; &nbsp; in last column &nbsp;$(N \to \infty) $: &nbsp; &nbsp; $h_{\rm B} \to p_{\rm B}$ &nbsp; &rArr; &nbsp; bit error probability.]]
 The bit error probability
 :$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$
@@ Line 114: / Line 114: @@
-'''(6)'''&nbsp; The number of measured bit errors should be&nbsp; $n_{\rm B} \ge 100$.&nbsp; Therefore, approximately (rounding errors should be taken into account):
+'''(6)'''&nbsp; The number of measured bit errors should be&nbsp; $n_{\rm B} \ge 100$.&nbsp; Therefore,&nbsp; approximately (rounding errors should be taken into account):
 :$$n_{\rm B} =  {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$
-*It further follows that in the simulation for $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$ still a sufficient number of bit errors occurred $(n_{\rm B} =315)$, while for $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ on average only $n_{\rm B} =52$ errors are to be expected.
+*It further follows that in the simulation for&nbsp; $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$&nbsp; still a sufficient number of bit errors occurred&nbsp; $(n_{\rm B} =1.6 \cdot 10^{6}\cdot 0.197 \cdot 10^{-3}= 315)$,&nbsp; while for&nbsp; $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$&nbsp; on average only&nbsp; $n_{\rm B} =52$&nbsp; errors are to be expected.
-*For this dB value, about twice the number of bits would have to be simulated.
+*For this dB value,&nbsp; about twice the number of bits would have to be simulated.
 {{ML-Fuß}}

@@ Line 83: / Line 83: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Only the <u>second solution</u> is correct:
+'''(1)'''&nbsp; Only the&nbsp; <u>second solution</u>&nbsp; is correct:
-*Of course, the accuracy of the BER measurement is influenced by the parameter $N$ to a large extent. On statistical average, the BER measurement naturally becomes better when $N$ is increased.
+*Of course,&nbsp; the accuracy of the BER measurement is influenced by the parameter&nbsp; $N$&nbsp; to a large extent.&nbsp; On statistical average,&nbsp; the BER measurement naturally becomes better when&nbsp; $N$&nbsp; is increased.
-*However, there is no deterministic relationship between the number of simulated bits and the accuracy of the BER measurement, as shown, for example, by the results for $10 \cdot \lg \ E_{\rm B}/N_0 = 6 \ \rm dB$:
+*However,&nbsp; there is no deterministic relationship between the number of simulated bits and the accuracy of the BER measurement,&nbsp; as shown,&nbsp; for example,&nbsp; by the results for&nbsp; $10 \cdot \lg \ E_{\rm B}/N_0 = 6 \ \rm dB$:
-*For $N = 6.4 \cdot 10^4\  (n_{\rm B} = 0.258 \cdot 10^{-2})$, the deviation from the true value $(0.239 \cdot 10^{-2})$ is smaller than for $N = 1.28 \cdot 10^5\  (n_{\rm B} = 0.272 \cdot 10^{-2})$.
+*For&nbsp; $N = 6.4 \cdot 10^4\  (h_{\rm B} = 0.258 \cdot 10^{-2})$,&nbsp; the deviation from the true value&nbsp; $(0.239 \cdot 10^{-2})$&nbsp; is smaller than for&nbsp; $N = 1.28 \cdot 10^5\  (h_{\rm B} = 0.272 \cdot 10^{-2})$.
-'''(2)'''&nbsp; At $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$, i.e. $E_{\rm B} = N_0$, the following values are obtained:
+'''(2)'''&nbsp; At&nbsp; $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$,&nbsp; i.e.&nbsp; $E_{\rm B} = N_0$,&nbsp; the following values are obtained:
 :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{64000}}\hspace{0.1cm}\underline {\approx 1.1
    \cdot10^{-3}}\hspace{0.05cm},$$
@@ Line 97: / Line 97: @@
-'''(3)'''&nbsp; For this, $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$ yields the following values:
+'''(3)'''&nbsp; For this,&nbsp; $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$&nbsp; yields the following values:
 :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}-  p_{\rm B}}{h_{\rm B}}
 = \frac{0.0779-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.9\% }\hspace{0.05cm}$$
@@ Line 103: / Line 103: @@
-'''(4)'''&nbsp; Due to the smaller error probability, the values are now smaller than in subtask (2):
+'''(4)'''&nbsp; Due to the smaller error probability,&nbsp; the values are now smaller than in subtask&nbsp; '''(2)''':
 :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{6.4 \cdot 10^{4}}}\hspace{0.1cm}\underline {\approx
 .3 \cdot 10^{-5}}\hspace{0.05cm},$$
@@ Line 109: / Line 109: @@
-'''(5)'''&nbsp; Despite the much smaller standard deviation $\sigma_h$, the smaller error probability results in larger relative deviations for $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ than for $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$:
+'''(5)'''&nbsp; Despite the much smaller standard deviation&nbsp; $\sigma_h$,&nbsp; the smaller error probability results in larger relative deviations for&nbsp; $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$&nbsp; <br> &nbsp; &nbsp; &nbsp; &nbsp; than for&nbsp; $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$:
 :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}-  p_{\rm B}}{h_{\rm B}}= \frac{0.625 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline { \approx 86\% } \hspace{0.05cm},$$
 :$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}-  p_{\rm B}}{h_{\rm B}}= \frac{0.325 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline {\approx -3.3\%}\hspace{0.05cm}.$$
-'''(6)'''&nbsp; The number of measured bit errors should be $n_{\rm B} \ge 100$. Therefore, approximately (rounding errors should be taken into account):
+'''(6)'''&nbsp; The number of measured bit errors should be&nbsp; $n_{\rm B} \ge 100$.&nbsp; Therefore, approximately (rounding errors should be taken into account):
 :$$n_{\rm B} =  {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$

@@ Line 7: / Line 7: @@
 The bit error probability
 :$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$
-of a binary system was simulatively determined by a measurement of the bit error rate (BER)
+of a binary system was simulatively determined by a measurement of the bit error rate&nbsp; $\rm (BER)$:
-:$$h_{\rm B} = {n_{\rm B}}/{N}$$.
+:$$h_{\rm B} = {n_{\rm B}}/{N}.$$
-Often, &nbsp;$h_{\rm B}$&nbsp; is also called bit error frequency.
+Often, &nbsp;$h_{\rm B}$&nbsp; is also called&nbsp; "bit error frequency".
@@ Line 22: / Line 22: @@
 The following properties are referred to in the exercise questionnaire:
-*The bit error frequency &nbsp;$h_{\rm B}$&nbsp; is, to a first approximation, a Gaussian distributed random variable with mean &nbsp;$m_h = p_{\rm B}$&nbsp; and variance &nbsp;$\sigma_h^2  \approx p_{\rm B}$.
+*The bit error frequency &nbsp;$h_{\rm B}$&nbsp; is&nbsp; (to a first approximation)&nbsp; a Gaussian distributed random variable with mean &nbsp;$m_h = p_{\rm B}$&nbsp; and variance &nbsp;$\sigma_h^2  \approx p_{\rm B}/N$.
 *The relative deviation of the bit error frequency from the probability is
 :$$\varepsilon_{\rm rel}= \frac {h_{\rm B}-p_{\rm B}}{p_{\rm B}}\hspace{0.05cm}.$$
-*As a rough rule of thumb on the required accuracy, the number of measured bit errors should be &nbsp;$n_{\rm B} \ge 100$.&nbsp;
+*As a rough rule of thumb on the required accuracy,&nbsp; the number of measured bit errors should be &nbsp;$n_{\rm B} \ge 100$.&nbsp;
@@ Line 33: / Line 33: @@
-''Note:''
+Note:
-*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|Error Probability for Baseband Transmission]].
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|"Error Probability for Baseband Transmission"]].
@@ Line 45: / Line 45: @@
 |type="[]"}
 - The accuracy of the BER measurement is independent of &nbsp;$N$.
-+ The larger &nbsp;$N$&nbsp; is, the more accurate the BER measurement is on average.
++ The larger &nbsp;$N$&nbsp; is,&nbsp; the more accurate the BER measurement is on average.
-- The larger &nbsp;$N$&nbsp; is, the more accurate each individual BER measurement is.
+- The larger &nbsp;$N$&nbsp; is,&nbsp; the more accurate each individual BER measurement is.
@@ Line 61: / Line 61: @@
-{Give the standard deviation &nbsp;$\sigma_h$&nbsp; for different &nbsp;$N$&nbsp; Let $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$.
+{Give the standard deviation &nbsp;$\sigma_h$&nbsp; for different &nbsp;$N$.&nbsp; Let $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$.
 |type="{}"}
 $N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}  σ_h \ = \ $  { 2.3 3% }  $\ \cdot 10^{ -5 }\ $
@@ Line 73: / Line 73: @@
-{Up to what (logarithmic) &nbsp;$E_{\rm B}/N_0$ value is &nbsp;$N = 1.6 \cdot 10^6$&nbsp; sufficient due to the condition &nbsp;$n_{\rm B} \ge 100$?&nbsp;
+{Up to what (logarithmic) &nbsp;$E_{\rm B}/N_0$&nbsp; value is &nbsp;$N = 1.6 \cdot 10^6$&nbsp; sufficient due to the condition &nbsp;$n_{\rm B} \ge 100$?&nbsp;
 |type="{}"}
 $\text{Maximum} \ \big [10 \cdot \lg \ E_{\rm B}/N_0 \big]  \ = \ $ { 8 3% }  $\ \rm dB $

@@ Line 4: / Line 4: @@
-[[File:P_ID1265__Dig_Z_1_2.png|right|frame|Simulierte Bitfehlerhäufigkeiten]]
+[[File:EN_Dig_Z_1_2.png|right|frame|Simulierte Bitfehlerhäufigkeiten]]
 Die Bitfehlerwahrscheinlichkeit
 :$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$

← Older revision		Revision as of 15:23, 2 February 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER ~~bei Basisbandsystemen~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]

← Older revision	Revision as of 15:19, 2 February 2022
(No difference)

← Older revision		Revision as of 12:47, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^1.2 BER bei Basisbandsystemen^]]	+	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER bei Basisbandsystemen^]]

← Older revision		Revision as of 12:57, 9 July 2020
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	''Hinweis:''		''Hinweis:''
−	*Die Aufgabe gehört zum Kapitel  [[~~Digitalsignalübertragung~~/Fehlerwahrscheinlichkeit_bei_Basisbandübertragung\| Fehlerwahrscheinlichkeit bei Basisbandübertragung]].	+	*Die Aufgabe gehört zum Kapitel  [[Digital_Signal_Transmission/Fehlerwahrscheinlichkeit_bei_Basisbandübertragung\| Fehlerwahrscheinlichkeit bei Basisbandübertragung]].

@@ Line 87: / Line 87: @@
 *Es besteht jedoch kein deterministischer Zusammenhang zwischen der Anzahl der simulierten Bit und der Genauigkeit der BER&ndash;Messung, wie z. B. die Ergebnisse für $10 \cdot \lg \ E_{\rm B}/N_0 = 6 \ \rm dB$ zeigen:
 *Bei $N = 6.4 \cdot 10^4\  (n_{\rm B} = 0.258 \cdot 10^{-2})$ ist die Abweichung vom tatsächlichen Wert $(0.239 \cdot 10^{-2})$ geringer als bei $N = 1.28 \cdot 10^5\  (n_{\rm B} = 0.272 \cdot 10^{-2})$.
@@ Line 94: / Line 95: @@
 :$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{1600000}}\hspace{0.1cm}\underline {\approx
 .22 \cdot10^{-3}}\hspace{0.05cm}.$$
 '''(3)'''&nbsp; Hierfür ergeben sich mit $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$ folgende Werte:
@@ Line 99: / Line 101: @@
 = \frac{0.0779-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.9\% }\hspace{0.05cm}$$
 :$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \varepsilon_{\rm rel}= \frac{h_{\rm B}-  p_{\rm B}}{h_{\rm B}}= \frac{0.0782-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.5\% } \hspace{0.05cm}.$$
 '''(4)'''&nbsp; Aufgrund der kleineren Fehlerwahrscheinlichkeit ergeben sich nun kleinere Werte als in der Teilaufgabe (2):
@@ Line 104: / Line 107: @@
 .3 \cdot 10^{-5}}\hspace{0.05cm},$$
 :$$ N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{1.6 \cdot 10^{6}}}\hspace{0.1cm}\underline {\approx 0.46 \cdot10^{-5}}\hspace{0.05cm}.$$
 '''(5)'''&nbsp; Trotz der deutlich kleineren Streuung $\sigma_h$ ergeben sich für $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ aufgrund der kleineren Fehlerwahrscheinlichkeit größere relative Abweichungen als für $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$:
 :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}-  p_{\rm B}}{h_{\rm B}}= \frac{0.625 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline { \approx 86\% } \hspace{0.05cm},$$
 :$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}-  p_{\rm B}}{h_{\rm B}}= \frac{0.325 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline {\approx -3.3\%}\hspace{0.05cm}.$$
 '''(6)'''&nbsp; Die Anzahl  der gemessenen Bitfehler sollte $n_{\rm B} \ge 100$ sein. Deshalb gilt näherungsweise (Rundungsfehler sind zu berücksichtigen):
 :$$n_{\rm B} =  {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$
-Daraus folgt weiter, dass bei der Simulation für $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$ noch ausreichend viele Bitfehler aufgetreten sind $(n_{\rm B} =315)$, während für $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ im Mittel nur mehr $n_{\rm B} =52$ Fehler zu erwarten sind. Für diesen dB&ndash;Wert müsste etwa die doppelte Anzahl an Bits simuliert werden.
+*Daraus folgt weiter, dass bei der Simulation für $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$ noch ausreichend viele Bitfehler aufgetreten sind $(n_{\rm B} =315)$, während für $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ im Mittel nur mehr $n_{\rm B} =52$ Fehler zu erwarten sind.
 {{ML-Fuß}}