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

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:56:58Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:56, 16 March 2026
Line 120:		Line 120:

	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]
	[[de:Aufgaben:1.~~2Z_Bitfehlermessung~~]]		[[de:Aufgaben:Aufgabe 1.2Z: Bitfehlermessung]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:30:21Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:30, 16 March 2026
Line 91:		Line 91:

	'''(2)'''  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)'''  At  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$,  i.e.  $E_{\rm B} = N_0$,  the following values are obtained:
	:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{64000}}\hspace{0.1cm}\underline {\approx 1.1		:$$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},$$
	\cdot10^{-3}}\hspace{0.05cm},$$		:$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{1600000}}\hspace{0.1cm}\underline {\approx0.22 \cdot10^{-3}}\hspace{0.05cm}.$$
	:$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{1600000}}\hspace{0.1cm}\underline {\~~approx~~
	0.22 \cdot10^{-3}}\hspace{0.05cm}.$$


	'''(3)'''  For this,  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$  yields the following values:		'''(3)'''  For this,  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$  yields the following values:
	:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}		:$$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}$$
	= \frac{0.0779-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.9\% }\hspace{0.05cm}$$
	:$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.0782-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.5\% } \hspace{0.05cm}.$$		:$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.0782-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.5\% } \hspace{0.05cm}.$$


	'''(4)'''  Due to the smaller error probability,  the values are now smaller than in subtask  '''(2)''':		'''(4)'''  Due to the smaller error probability,  the values are now smaller than in subtask  '''(2)''':
	:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{6.4 \cdot 10^{4}}}\hspace{0.1cm}\underline {\~~approx~~		:$$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 {\approx2.3 \cdot 10^{-5}}\hspace{0.05cm},$$
	2.3 \cdot 10^{-5}}\hspace{0.05cm},$$
	:$$ N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{1.6 \cdot 10^{6}}}\hspace{0.1cm}\underline {\approx 0.46 \cdot10^{-5}}\hspace{0.05cm}.$$		:$$ N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{1.6 \cdot 10^{6}}}\hspace{0.1cm}\underline {\approx 0.46 \cdot10^{-5}}\hspace{0.05cm}.$$

Line 115:		Line 111:

	'''(6)'''  The number of measured bit errors should be  $n_{\rm B} \ge 100$.  Therefore,  approximately (rounding errors should be taken into account):		'''(6)'''  The number of measured bit errors should be  $n_{\rm B} \ge 100$.  Therefore,  approximately (rounding errors should be taken into account):
	:$$n_{\rm B} = {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$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}.$$
	p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$
	*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  $(n_{\rm B} =1.6 \cdot 10^{6}\cdot 0.197 \cdot 10^{-3}= 315)$,  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  $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$  still a sufficient number of bit errors occurred  $(n_{\rm B} =1.6 \cdot 10^{6}\cdot 0.197 \cdot 10^{-3}= 315)$,  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.
	*For this dB value,  about twice the number of bits would have to be simulated.		*For this dB value,  about twice the number of bits would have to be simulated.

Maintenance script: Fix spaces in LaTeX braces to prevent Quiz misparse

2026-03-16T13:21:25Z

Fix spaces in LaTeX braces to prevent Quiz misparse

← Older revision		Revision as of 15:21, 16 March 2026
Line 51:		Line 51:
	{Give the standard deviation  $\sigma_h$  for different  $N$.  Let  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$.		{Give the standard deviation  $\sigma_h$  for different  $N$.  Let  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$.
	\|type="{}"}		\|type="{}"}
	$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 1.1 3% } $\ \cdot 10^{ -3 }\ $		$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 1.1 3% } $\ \cdot 10^{-3}\ $
	$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} σ_h \ = \ $ { 0.22 3% } $\ \cdot 10^{ -3 }\ $		$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} σ_h \ = \ $ { 0.22 3% } $\ \cdot 10^{-3}\ $


Line 63:		Line 63:
	{Give the standard deviation  $\sigma_h$  for different  $N$.  Let $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$.		{Give the standard deviation  $\sigma_h$  for different  $N$.  Let $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$.
	\|type="{}"}		\|type="{}"}
	$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 2.3 3% } $\ \cdot 10^{ -5 }\ $		$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 2.3 3% } $\ \cdot 10^{-5}\ $
	$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} σ_h \ = \ $ { 0.46 3% } $\ \cdot 10^{ -5 }\ $		$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} σ_h \ = \ $ { 0.46 3% } $\ \cdot 10^{-5}\ $

Maintenance script: Add German interlanguage link

2026-03-16T12:30:12Z

Add German interlanguage link

← Older revision		Revision as of 14:30, 16 March 2026
Line 125:		Line 125:

	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]
			[[de:Aufgaben:1.2Z_Bitfehlermessung]]

Guenter at 16:16, 29 April 2022

2022-04-29T16:16:56Z

← Older revision		Revision as of 18:16, 29 April 2022
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  $(h_{\rm B})$;     in last column  $(N \to \infty) $:     $h_{\rm B} \to p_{\rm B}$   ⇒   bit error probability.]]
	The bit error probability		The bit error probability
	:$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$		:$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$
Line 114:		Line 114:


	'''(6)'''  The number of measured bit errors should be  $n_{\rm B} \ge 100$.  Therefore, approximately (rounding errors should be taken into account):		'''(6)'''  The number of measured bit errors should be  $n_{\rm B} \ge 100$.  Therefore,  approximately (rounding errors should be taken into account):
	:$$n_{\rm B} = {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$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}.$$		p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$
	*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  $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$  still a sufficient number of bit errors occurred  $(n_{\rm B} =1.6 \cdot 10^{6}\cdot 0.197 \cdot 10^{-3}= 315)$,  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.
	*For this dB value, about twice the number of bits would have to be simulated.		*For this dB value,  about twice the number of bits would have to be simulated.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 13:37, 29 April 2022

2022-04-29T13:37:01Z

← Older revision		Revision as of 15:37, 29 April 2022
Line 83:		Line 83:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Only the <u>second solution</u> is correct:		'''(1)'''  Only the  <u>second solution</u>  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,  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.
	*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,  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  $10 \cdot \lg \ E_{\rm B}/N_0 = 6 \ \rm dB$:
	*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  $N = 6.4 \cdot 10^4\ (h_{\rm B} = 0.258 \cdot 10^{-2})$,  the deviation from the true value  $(0.239 \cdot 10^{-2})$  is smaller than for  $N = 1.28 \cdot 10^5\ (h_{\rm B} = 0.272 \cdot 10^{-2})$.



	'''(2)'''  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)'''  At  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$,  i.e.  $E_{\rm B} = N_0$,  the following values are obtained:
	:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{64000}}\hspace{0.1cm}\underline {\approx 1.1		:$$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},$$		\cdot10^{-3}}\hspace{0.05cm},$$
Line 97:		Line 97:


	'''(3)'''  For this, $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$ yields the following values:		'''(3)'''  For this,  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$  yields the following values:
	:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}		:$$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}$$		= \frac{0.0779-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.9\% }\hspace{0.05cm}$$
Line 103:		Line 103:


	'''(4)'''  Due to the smaller error probability, the values are now smaller than in subtask (2):		'''(4)'''  Due to the smaller error probability,  the values are now smaller than in subtask  '''(2)''':
	:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{6.4 \cdot 10^{4}}}\hspace{0.1cm}\underline {\approx		:$$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
	2.3 \cdot 10^{-5}}\hspace{0.05cm},$$		2.3 \cdot 10^{-5}}\hspace{0.05cm},$$
Line 109:		Line 109:


	'''(5)'''  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)'''  Despite the much smaller standard deviation  $\sigma_h$,  the smaller error probability results in larger relative deviations for  $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$  <br>         than for  $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$:
	:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.625 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline { \approx 86\% } \hspace{0.05cm},$$		:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.625 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline { \approx 86\% } \hspace{0.05cm},$$
	:$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.325 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline {\approx -3.3\%}\hspace{0.05cm}.$$		:$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.325 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline {\approx -3.3\%}\hspace{0.05cm}.$$


	'''(6)'''  The number of measured bit errors should be $n_{\rm B} \ge 100$. Therefore, approximately (rounding errors should be taken into account):		'''(6)'''  The number of measured bit errors should be  $n_{\rm B} \ge 100$.  Therefore, approximately (rounding errors should be taken into account):
	:$$n_{\rm B} = {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$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}.$$		p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$

Guenter at 12:56, 29 April 2022

2022-04-29T12:56:00Z

← Older revision		Revision as of 14:56, 29 April 2022
Line 7:		Line 7:
	The bit error probability		The bit error probability
	:$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$		:$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$
	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  $\rm (BER)$:
	:$$h_{\rm B} = {n_{\rm B}}/{N}$$.		:$$h_{\rm B} = {n_{\rm B}}/{N}.$$
	Often,  $h_{\rm B}$  is also called bit error frequency.		Often,  $h_{\rm B}$  is also called  "bit error frequency".


Line 22:		Line 22:

	The following properties are referred to in the exercise questionnaire:		The following properties are referred to in the exercise questionnaire:
	*The bit error frequency  $h_{\rm B}$  is, to a first approximation, a Gaussian distributed random variable with mean  $m_h = p_{\rm B}$  and variance  $\sigma_h^2 \approx p_{\rm B}$.		*The bit error frequency  $h_{\rm B}$  is  (to a first approximation)  a Gaussian distributed random variable with mean  $m_h = p_{\rm B}$  and variance  $\sigma_h^2 \approx p_{\rm B}/N$.
	*The relative deviation of the bit error frequency from the probability is		*The relative deviation of the bit error frequency from the probability is
	:$$\varepsilon_{\rm rel}= \frac {h_{\rm B}-p_{\rm B}}{p_{\rm B}}\hspace{0.05cm}.$$		:$$\varepsilon_{\rm rel}= \frac {h_{\rm B}-p_{\rm B}}{p_{\rm B}}\hspace{0.05cm}.$$

	*As a rough rule of thumb on the required accuracy, the number of measured bit errors should be  $n_{\rm B} \ge 100$.		*As a rough rule of thumb on the required accuracy,  the number of measured bit errors should be  $n_{\rm B} \ge 100$.


Line 33:		Line 33:


	''Note:''		Note:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|Error Probability for Baseband Transmission]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|"Error Probability for Baseband Transmission"]].


Line 45:		Line 45:
	\|type="[]"}		\|type="[]"}
	- The accuracy of the BER measurement is independent of  $N$.		- The accuracy of the BER measurement is independent of  $N$.
	+ The larger  $N$  is, the more accurate the BER measurement is on average.		+ The larger  $N$  is,  the more accurate the BER measurement is on average.
	- The larger  $N$  is, the more accurate each individual BER measurement is.		- The larger  $N$  is,  the more accurate each individual BER measurement is.


Line 61:		Line 61:


	{Give the standard deviation  $\sigma_h$  for different  $N$  Let $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$.		{Give the standard deviation  $\sigma_h$  for different  $N$.  Let $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$.
	\|type="{}"}		\|type="{}"}
	$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 2.3 3% } $\ \cdot 10^{ -5 }\ $		$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)  $E_{\rm B}/N_0$ value is  $N = 1.6 \cdot 10^6$  sufficient due to the condition  $n_{\rm B} \ge 100$?		{Up to what (logarithmic)  $E_{\rm B}/N_0$  value is  $N = 1.6 \cdot 10^6$  sufficient due to the condition  $n_{\rm B} \ge 100$?
	\|type="{}"}		\|type="{}"}
	$\text{Maximum} \ \big [10 \cdot \lg \ E_{\rm B}/N_0 \big] \ = \ $ { 8 3% } $\ \rm dB $		$\text{Maximum} \ \big [10 \cdot \lg \ E_{\rm B}/N_0 \big] \ = \ $ { 8 3% } $\ \rm dB $

Hwang at 15:51, 2 March 2022

2022-03-02T15:51:59Z

Show changes

Guenter at 16:27, 8 February 2022

2022-02-08T16:27:06Z

← Older revision		Revision as of 18:27, 8 February 2022
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		Die Bitfehlerwahrscheinlichkeit
	:$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$		:$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$

Guenter at 15:23, 2 February 2022

2022-02-02T15:23:58Z

← Older revision		Revision as of 17:23, 2 February 2022
Line 124:		Line 124:


	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER ~~bei Basisbandsystemen~~^]]		[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]