Aufgaben:Exercise 1.2: Bit Error Rate - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:55:08Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:55, 16 March 2026
Line 84:		Line 84:

	'''(5)'''  This condition can be formulated with  $\varepsilon = 5 \cdot 10^{-6}$  as follows:		'''(5)'''  This condition can be formulated with  $\varepsilon = 5 \cdot 10^{-6}$  as follows:
	:$${\rm Q} \left( {\varepsilon}/{\sigma} \right) < 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\varepsilon}/{\sigma} > {\rm Q}^{-1}(0.05) \approx 1.64\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{\varepsilon^2}{\sigma^2}\approx \frac{\varepsilon^2 \~~cdotN~~}{p}> 1.64^2 = 2.69$$		:$${\rm Q} \left( {\varepsilon}/{\sigma} \right) < 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\varepsilon}/{\sigma} > {\rm Q}^{-1}(0.05) \approx 1.64\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{\varepsilon^2}{\sigma^2}\approx \frac{\varepsilon^2 \cdot N}{p}> 1.64^2 = 2.69$$
	:$$\Rightarrow \hspace{0.3cm} N >\frac{2.69 \cdot p}{\varepsilon^2}= \frac{2.69 \cdot 10^{-4}}{25\cdot10^{-12}}\hspace{0.1cm}\underline {\approx 10.8 \cdot 10^{6}}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} N >\frac{2.69 \cdot p}{\varepsilon^2}= \frac{2.69 \cdot 10^{-4}}{25\cdot10^{-12}}\hspace{0.1cm}\underline {\approx 10.8 \cdot 10^{6}}\hspace{0.05cm}.$$

Line 93:		Line 93:

	[[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.~~2_Bitfehlerquote_~~(BER)]]		[[de:Aufgaben:Aufgabe 1.2: Bitfehlerquote (BER)]]

Maintenance script: Join 1 split formula blocks

2026-03-16T14:15:56Z

Join 1 split formula blocks

← Older revision		Revision as of 16:15, 16 March 2026
Line 11:		Line 11:
	$ \langle q_\nu \rangle $  and finding the error sequence  $ \langle e_\nu \rangle $.  Thereby holds:		$ \langle q_\nu \rangle $  and finding the error sequence  $ \langle e_\nu \rangle $.  Thereby holds:

	:$$e_\nu = \left\{ \begin{array}{c} 0 \\		:$$e_\nu = \left\{ \begin{array}{c} 0 \\1 \\ \end{array} \right.\quad\begin{array}{{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array}\begin{array}{{20}c}v_\nu = q_\nu \hspace{0.05cm}, \\v_\nu \ne q_\nu . \\\end{array}$$
	1 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{for}}
	\\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c}
	v_\nu = q_\nu \hspace{0.05cm}, \\
	v_\nu \ne q_\nu . \\
	\end{array}$$
	The bit error rate is an approximation for the bit error probability  $p$:		The bit error rate is an approximation for the bit error probability  $p$:
	:$${\rm BER} = \frac{1}{N}\cdot\sum_{\nu=1}^N e_\nu.$$		:$${\rm BER} = \frac{1}{N}\cdot\sum_{\nu=1}^N e_\nu.$$

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

2026-03-16T13:47:26Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:47, 16 March 2026
Line 85:		Line 85:

	'''(4)'''  With  $p = 10^{-4}$,  the comparable probability is:		'''(4)'''  With  $p = 10^{-4}$,  the comparable probability is:
	:$${\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)		:$${\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)= 2 \cdot {\rm Q} \left( {\varepsilon}/{\sigma}\right);\hspace{0.5cm}\text{with}\hspace{0.5cm}\sigma \approx \sqrt{{ p}/{N}}=10^{-5}\hspace{0.05cm}, \hspace{0.3cm}\varepsilon = 5 \cdot10^{-6}\text{:}$$
	= 2 \cdot {\rm Q} \left( {\varepsilon}/{\sigma}
	\right);\hspace{0.5cm}
	\text{with}\hspace{0.5cm}\sigma \approx \sqrt{{ p}/{N}}=
	10^{-5}\hspace{0.05cm}, \hspace{0.3cm}\varepsilon = 5 \~~cdot~~
	10^{-6}\text{:}$$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)= 2 \cdot {\rm Q} \left( \frac{5 \cdot 10^{-6}}{10^{-5}} \right) = 2 \cdot 0.309 \hspace{0.1cm}\underline {= 0.618} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}{\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)= 2 \cdot {\rm Q} \left( \frac{5 \cdot 10^{-6}}{10^{-5}} \right) = 2 \cdot 0.309 \hspace{0.1cm}\underline {= 0.618} \hspace{0.05cm}.$$

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

2026-03-16T13:30:20Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:30, 16 March 2026
Line 80:		Line 80:
	'''(3)'''  The probability that the  $\rm BER$  takes a value outside the range		'''(3)'''  The probability that the  $\rm BER$  takes a value outside the range
	$0.95 \cdot p$ ... $1.05 \cdot p$  is obtained with  $\varepsilon = 5 \cdot 10^{-4}$  $(p = 10^{-2})$  as follows.		$0.95 \cdot p$ ... $1.05 \cdot p$  is obtained with  $\varepsilon = 5 \cdot 10^{-4}$  $(p = 10^{-2})$  as follows.
	:$${\rm Pr} \left( {\rm BER} < 0.95 \cdot 10^{-2} \right)		:$${\rm Pr} \left( {\rm BER} < 0.95 \cdot 10^{-2} \right)= {\rm Pr} \left( {\rm BER} > 1.05 \cdot 10^{-2} \right)= {\rm Q} \left({\varepsilon}/{\sigma} \right)$$
	= {\rm Pr} \left( {\rm BER} > 1.05 \cdot 10^{-2} \right)		:$$\Rightarrow \hspace{0.3cm}{\rm Pr} \left( \|{\rm BER} - p\| > \varepsilon \right)= 2 \cdot {\rm Q} \left( \frac{5 \cdot 10^{-4}}{10^{-4}} \right) = 2 \cdot 0.287 \cdot 10^{-6}\hspace{0.1cm}\underline {= 0.00574 \cdot 10^{-4}}\hspace{0.05cm}.$$
	= {\rm Q} \left({\varepsilon}/{\sigma} \right)$$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr} \left( \|{\rm BER} - p\| > \varepsilon \right)
	= 2 \cdot {\rm Q} \left( \frac{5 \cdot 10^{-4}}{10^{-4}} \right) = 2 \cdot 0.287 \cdot 10^{-6}\hspace{0.1cm}\underline {= 0.00574 \cdot 10^{-4}}\hspace{0.05cm}.$$


Line 94:		Line 91:
	10^{-5}\hspace{0.05cm}, \hspace{0.3cm}\varepsilon = 5 \cdot		10^{-5}\hspace{0.05cm}, \hspace{0.3cm}\varepsilon = 5 \cdot
	10^{-6}\text{:}$$		10^{-6}\text{:}$$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)		:$$\Rightarrow \hspace{0.3cm}{\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)= 2 \cdot {\rm Q} \left( \frac{5 \cdot 10^{-6}}{10^{-5}} \right) = 2 \cdot 0.309 \hspace{0.1cm}\underline {= 0.618} \hspace{0.05cm}.$$
	= 2 \cdot {\rm Q} \left( \frac{5 \cdot 10^{-6}}{10^{-5}} \right) = 2 \cdot 0.309 \hspace{0.1cm}\underline {= 0.618} \hspace{0.05cm}.$$


	'''(5)'''  This condition can be formulated with  $\varepsilon = 5 \cdot 10^{-6}$  as follows:		'''(5)'''  This condition can be formulated with  $\varepsilon = 5 \cdot 10^{-6}$  as follows:
	:$${\rm Q} \left( {\varepsilon}/{\sigma} \right) < 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$${\rm Q} \left( {\varepsilon}/{\sigma} \right) < 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\varepsilon}/{\sigma} > {\rm Q}^{-1}(0.05) \approx 1.64\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{\varepsilon^2}{\sigma^2}\approx \frac{\varepsilon^2 \cdotN}{p}> 1.64^2 = 2.69$$
	{\varepsilon}/{\sigma} > {\rm Q}^{-1}(0.05) \approx 1.64		:$$\Rightarrow \hspace{0.3cm} N >\frac{2.69 \cdot p}{\varepsilon^2}= \frac{2.69 \cdot 10^{-4}}{25\cdot10^{-12}}\hspace{0.1cm}\underline {\approx 10.8 \cdot 10^{6}}\hspace{0.05cm}.$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	\frac{\varepsilon^2}{\sigma^2}\approx \frac{\varepsilon^2 \~~cdot~~
	N}{p}> 1.64^2 = 2.69$$
	:$$\Rightarrow \hspace{0.3cm} N >
	\frac{2.69 \cdot p}{\varepsilon^2}= \frac{2.69 \cdot 10^{-4}}{25
	\cdot10^{-12}}\hspace{0.1cm}\underline {\approx 10.8 \cdot 10^{6}}\hspace{0.05cm}.$$

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

2026-03-16T13:21:24Z

Fix spaces in LaTeX braces to prevent Quiz misparse

← Older revision		Revision as of 15:21, 16 March 2026
Line 48:		Line 48:
	{Calculate the standard deviation  $\sigma$  for  $N = 10^6$  and  $p = 10^{-2}$.		{Calculate the standard deviation  $\sigma$  for  $N = 10^6$  and  $p = 10^{-2}$.
	\|type="{}"}		\|type="{}"}
	$\sigma \ =\ $ { 1 3% } $\ \cdot 10^{ -4 }\ $		$\sigma \ =\ $ { 1 3% } $\ \cdot 10^{-4}\ $

	{What is the probability that the bit error rate differs in magnitude by more than  $5\%$  from the probability  $\underline{p = 10^{-2}}$?		{What is the probability that the bit error rate differs in magnitude by more than  $5\%$  from the probability  $\underline{p = 10^{-2}}$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(\|{\rm BER} – p\| > 0.05 · p) \ =\ ${ 0.00574 10% } $\ \cdot 10^{ -4 }\ $		${\rm Pr}(\|{\rm BER} – p\| > 0.05 · p) \ =\ ${ 0.00574 10% } $\ \cdot 10^{-4}\ $

	{What is the same probability with  $\underline{p = 10^{-4}}$?		{What is the same probability with  $\underline{p = 10^{-4}}$?
Line 60:		Line 60:
	{What would be the minimum size of  $N$  for  $\underline{p = 10^{-4}}$  to be no more than  $10\%$  outside the interval of  $0.95 \cdot 10^{-4}$ ... $1.05 \cdot 10^{-4}$?		{What would be the minimum size of  $N$  for  $\underline{p = 10^{-4}}$  to be no more than  $10\%$  outside the interval of  $0.95 \cdot 10^{-4}$ ... $1.05 \cdot 10^{-4}$?
	\|type="{}"}		\|type="{}"}
	$N_{\rm min} \ =\ $ { 10.8 10% } $\ \cdot 10^{ 6 }\ $		$N_{\rm min} \ =\ $ { 10.8 10% } $\ \cdot 10^{6}\ $

Maintenance script: Add German interlanguage link

2026-03-16T12:30:15Z

Add German interlanguage link

← Older revision		Revision as of 14:30, 16 March 2026
Line 114:		Line 114:

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

Hwang at 14:51, 9 May 2022

2022-05-09T14:51:01Z

← Older revision		Revision as of 16:51, 9 May 2022
Line 22:		Line 22:
	The larger the simulation parameter  $N$  is chosen,  the more accurate is this approximation.		The larger the simulation parameter  $N$  is chosen,  the more accurate is this approximation.

	From  [[Aufgaben:Exercise_3.7:_Bit_Error_Rate_(BER)\|Exercise 3.7]]  in the book  "Stochastic Signal Theory"  it is known that the random variable  "BER"  is actually binomially distributed,  but can be approximated with good approximation by a  (discrete)  Gaussian distribution with mean  $p$  and standard deviation  $\sigma$:		From  [[Aufgaben:Exercise_3.7:_Bit_Error_Rate_(BER)\|"Exercise 3.7"]]  in the book  "Stochastic Signal Theory"  it is known that the random variable  "BER"  is actually binomially distributed,  but can be approximated with good approximation by a  (discrete)  Gaussian distribution with mean  $p$  and standard deviation  $\sigma$:
	:$$\sigma = \sqrt{\frac{ p\cdot (\rm 1- \it p)}{N}}.$$		:$$\sigma = \sqrt{\frac{ p\cdot (\rm 1- \it p)}{N}}.$$

Guenter at 12:24, 29 April 2022

2022-04-29T12:24:32Z

← Older revision		Revision as of 14:24, 29 April 2022
Line 67:		Line 67:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Only the second statement</u> is correct:		'''(1)'''  <u>Only the second statement</u>  is correct:
	*$\rm BER$ is the quotient of the number $n_{\rm B}$ of detected symbol errors and the number $N$ of all simulated symbols and thus actually a relative frequency.		*$\rm BER$  is the quotient of the number  $n_{\rm B}$  of detected symbol errors and the number  $N$  of all simulated symbols and thus actually a relative frequency.
	*The probability that ${\rm BER} = p$ is always exactly zero, since $\rm BER$ is a continuous random variable.		*The probability that  ${\rm BER} = p$  is always exactly zero,  since  $\rm BER$  is a continuous random variable.
	*However, the probability that $\rm BER$ lies in a narrow interval around $p$ increases as $N$ increases.		*However,  the probability that  $\rm BER$  lies in a narrow interval around  $p$  increases as  $N$  increases.



	'''(2)'''  The standard deviation of the Gaussian random variable $\rm BER$ with $N = 10^6$ and $p = 10^{-2}$ is given by		'''(2)'''  The standard deviation of the Gaussian random variable  $\rm BER$  with  $N = 10^6$  and  $p = 10^{-2}$  is given by
	:$$\sigma = \sqrt{{ p\cdot (\rm 1- \it p)}/{N}}\approx \sqrt{{ p}/{N}}\hspace{0.1cm}\underline {= 10^{-4}}\hspace{0.05cm}.$$		:$$\sigma = \sqrt{{ p\cdot (\rm 1- \it p)}/{N}}\approx \sqrt{{ p}/{N}}\hspace{0.1cm}\underline {= 10^{-4}}\hspace{0.05cm}.$$


	'''(3)'''  The probability that the $\rm BER$ takes a value outside the range		'''(3)'''  The probability that the  $\rm BER$  takes a value outside the range
	$0.95 \cdot p$ ... $1.05 \cdot p$ is obtained with $\varepsilon = 5 \cdot 0^{-4}$ ($p = 10^{-2}$) as follows.		$0.95 \cdot p$ ... $1.05 \cdot p$  is obtained with  $\varepsilon = 5 \cdot 10^{-4}$  $(p = 10^{-2})$  as follows.
	:$${\rm Pr} \left( {\rm BER} < 0.95 \cdot 10^{-2} \right)		:$${\rm Pr} \left( {\rm BER} < 0.95 \cdot 10^{-2} \right)
	= {\rm Pr} \left( {\rm BER} > 1.05 \cdot 10^{-2} \right)		= {\rm Pr} \left( {\rm BER} > 1.05 \cdot 10^{-2} \right)
Line 87:		Line 87:


	'''(4)'''  With $p = 10^{-4}$, the comparable probability is:		'''(4)'''  With  $p = 10^{-4}$,  the comparable probability is:
	:$${\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)		:$${\rm Pr} \left( \|{\rm BER} - 10^{-4}\| > 0.05 \cdot 10^{-4} \right)
	= 2 \cdot {\rm Q} \left( {\varepsilon}/{\sigma}		= 2 \cdot {\rm Q} \left( {\varepsilon}/{\sigma}
Line 98:		Line 98:


	'''(5)'''  This condition can be formulated with $\varepsilon = 5 \cdot 10^{-6}$ as follows:		'''(5)'''  This condition can be formulated with  $\varepsilon = 5 \cdot 10^{-6}$  as follows:
	:$${\rm Q} \left( {\varepsilon}/{\sigma} \right) < 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$${\rm Q} \left( {\varepsilon}/{\sigma} \right) < 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	{\varepsilon}/{\sigma} > {\rm Q}^{-1}(0.05) \approx 1.64		{\varepsilon}/{\sigma} > {\rm Q}^{-1}(0.05) \approx 1.64

Guenter at 12:08, 29 April 2022

2022-04-29T12:08:16Z

← Older revision		Revision as of 14:08, 29 April 2022
Line 4:		Line 4:


	[[File:P_ID1264__Dig_A_1_2.png\|right\|frame\|Table of Gaussian error functions]]		[[File:P_ID1264__Dig_A_1_2.png\|right\|frame\|Table of two Gaussian error functions]]
	It is known from a digital transmission system that it can be approximated by a		It is known from a digital transmission system that it can be approximated by a
	BSC model (~~<i>~~Binary Symmetrical Channel~~</i> ~~) with error probability  $p$.		BSC model  ("Binary Symmetrical Channel") with error probability  $p$.

	For verification, the bit error rate is to be determined by comparing the sink symbol sequence  $ \langle v_\nu \rangle $  with the source symbol sequence		For verification,  the bit error rate is to be determined by comparing the sink symbol sequence  $ \langle v_\nu \rangle $  with the source symbol sequence
	$ \langle q_\nu \rangle $  and finding the error sequence  $ \langle e_\nu \rangle $.  Thereby holds:		$ \langle q_\nu \rangle $  and finding the error sequence  $ \langle e_\nu \rangle $.  Thereby holds:

	$$e_\nu = \left\{ \begin{array}{c} 0 \\		:$$e_\nu = \left\{ \begin{array}{c} 0 \\
	1 \\ \end{array} \right.\quad		1 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{for}}		\begin{array}{*{1}c} {\rm{for}}
Line 18:		Line 18:
	v_\nu \ne q_\nu . \\		v_\nu \ne q_\nu . \\
	\end{array}$$		\end{array}$$
	The bit error rate is an approximation for the bit error probability  $p$~~ ~~:		The bit error rate is an approximation for the bit error probability  $p$:
	:$${\rm BER} = \frac{1}{N}\cdot\sum_{\nu=1}^N e_\nu.$$		:$${\rm BER} = \frac{1}{N}\cdot\sum_{\nu=1}^N e_\nu.$$
	The larger the simulation parameter  $N$  is chosen, the more accurate this approximation is.		The larger the simulation parameter  $N$  is chosen,  the more accurate is this approximation.

	From  [[Aufgaben:Exercise_3.7:_Bit_Error_Rate_(BER)\|Exercise 3.7]]  in the book "Stochastic Signal Theory" it is known that the random variable "BER" is actually binomially distributed, but can be approximated with good approximation by a (discrete) Gaussian distribution with mean  $p$  and standard deviation  $\sigma$:		From  [[Aufgaben:Exercise_3.7:_Bit_Error_Rate_(BER)\|Exercise 3.7]]  in the book  "Stochastic Signal Theory"  it is known that the random variable  "BER"  is actually binomially distributed,  but can be approximated with good approximation by a  (discrete)  Gaussian distribution with mean  $p$  and standard deviation  $\sigma$:
	:$$\sigma = \sqrt{\frac{ p\cdot (\rm 1- \it p)}{N}}.$$		:$$\sigma = \sqrt{\frac{ p\cdot (\rm 1- \it p)}{N}}.$$

Line 29:		Line 29:


	''Notes:''		Notes:
	*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"]].
	*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian Distributed Random Variables]]  in the book "Stochastic Signal Theory".		*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|"Gaussian Distributed Random Variables"]]  in the book  "Stochastic Signal Theory".
	*In the table some values of the Gaussian error functions  $\rm \phi(x)$  and  $\rm Q(x)$  are given.		*In the table some values of the Gaussian error functions  ${\rm \phi}(x)$  and  ${\rm Q}(x)$  are given.


Line 43:		Line 43:
	- $\rm BER$  is a probability.		- $\rm BER$  is a probability.
	+ $\rm BER$  is a relative frequency.		+ $\rm BER$  is a relative frequency.
	- If  $N$  is sufficiently large,  $\rm BER$  coincides <b>exactly</b> with  $p$~~ ~~		- If  $N$  is sufficiently large,  $\rm BER$  coincides <b>exactly</b> with  $p$.

Hwang at 14:45, 2 March 2022

2022-03-02T14:45:56Z

Show changes