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

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]
	[[de:Aufgaben:3.~~7_Bitfehlerquote_~~(BER)]]		[[de:Aufgaben:Aufgabe 3.7: Bitfehlerquote (BER)]]

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	'''(2)'''  Für the standard deviation of the binomial distribution holds with good approximation:		'''(2)'''  Für the standard deviation of the binomial distribution holds with good approximation:
	:$$\sigma_{n{\rm B}}=\sqrt{N\cdot p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}		:$$\sigma_{n{\rm B}}=\sqrt{N\cdot p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}\hspace{0.15cm}\underline{\approx 10}.$$
	\hspace{0.15cm}\underline{\approx 10}.$$


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	'''(7)'''  It must  $\alpha = \varepsilon/\sigma_{h{\rm B}}$.  With the result of the subtask  '''(2)'''  then follows:		'''(7)'''  It must  $\alpha = \varepsilon/\sigma_{h{\rm B}}$.  With the result of the subtask  '''(2)'''  then follows:
	:$$\frac{\varepsilon}{\sqrt{p_{\rm B}\cdot(\rm 1-\it p_{\rm B})/N}}\ge {\rm 2} \hspace{0.5cm}\Rightarrow\hspace{0.5cm}		:$$\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}.$$
	N\ge \frac{\rm 4\cdot \it p_{\rm B}\cdot(\rm 1-\it p_{\rm B})}{\varepsilon^{\rm 2}}\approx \frac{\rm 4\cdot 10^{-3}}{10^{-8}}\hspace{0.15cm}\underline{=\rm 400\hspace{0.08cm}000}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]
			[[de:Aufgaben:3.7_Bitfehlerquote_(BER)]]

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

2022-02-17T11:11:39Z

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	{What values can the bit error rate  $h_{\rm B}$  take?  Show that the linear mean  $m_{h{\rm B}}$  of this random variable is equal to the bit error probability  $p_{\rm B}$  What is its ~~rms value~~?		{What values can the bit error rate  $h_{\rm B}$  take?  Show that the linear mean  $m_{h{\rm B}}$  of this random variable is equal to the bit error probability  $p_{\rm B}$  What is its standard deviation?
	\|type="{}"}		\|type="{}"}
	$\sigma_{h{\rm B}} \ = \ $ { 0.0001 3% }		$\sigma_{h{\rm B}} \ = \ $ { 0.0001 3% }


	{Under certain conditions,  a binomially distributed random variable can be approximated by a Gaussian distribution <br>with equal mean  $(m_{h{\rm B}})$  and equal ~~rms value~~  $(\sigma_{h{\rm B}})$.  Which statement is true?		{Under certain conditions,  a binomially distributed random variable can be approximated by a Gaussian distribution <br>with equal mean  $(m_{h{\rm B}})$  and equal standard deviation  $(\sigma_{h{\rm B}})$.  Which statement is true?
	\|type="()"}		\|type="()"}
	+ ${\rm Pr}(\hspace{0.05cm}\|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}\| \le \varepsilon)=1- 2\cdot \rm Q({\varepsilon}/{\sigma_{\it h}{\rm B}}).$		+ ${\rm Pr}(\hspace{0.05cm}\|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}\| \le \varepsilon)=1- 2\cdot \rm Q({\varepsilon}/{\sigma_{\it h}{\rm B}}).$
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	:$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$		:$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$

	*The ~~rms value~~ results in		*The standard deviation results in
	:$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$		:$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$

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

2022-02-17T11:00:51Z

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

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

Guenter at 11:13, 2 February 2022

2022-02-02T11:13:10Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The <u>last two statements</u> are true:		'''(1)'''  The  <u>last two statements</u>  are true:
	*Relative to the random variable  $n_{\rm B}$  there is the classical case of a binomial distribution.		*Relative to the random variable  $n_{\rm B}$  there is the classical case of a binomial distribution.
	*The sum over  $N$  binary random variables is formed.		*The sum over  $N$  binary random variables is formed.  The possible values of  $n_{\rm B}$  thus lie between  $0$  and  $N$.
	*The possible values of  $n_{\rm B}$  thus lie between  $0$  and  $N$.
	*The linear mean gives   $m_{n{\rm B}}=p_{\rm B}\cdot N=\rm 10^{-3}\cdot 10^{5}=\rm 100.$		*The linear mean gives   $m_{n{\rm B}}=p_{\rm B}\cdot N=\rm 10^{-3}\cdot 10^{5}=\rm 100.$



	'''(2)'''  Für the ~~rms~~ of the binomial distribution holds with good approximation:		'''(2)'''  Für the root mean square of the binomial distribution holds with good approximation:
	:$$\sigma_{n{\rm B}}=\sqrt{N\cdot p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}		:$$\sigma_{n{\rm B}}=\sqrt{N\cdot p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}
	\hspace{0.15cm}\underline{\approx 10}.$$		\hspace{0.15cm}\underline{\approx 10}.$$
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	'''(3)'''  Possible values of  $h_{\rm B}$  are all integer multiples of  $1/N$.  These all lie between  $0$  and  $1$.		'''(3)'''  Possible values of  $h_{\rm B}$  are all integer multiples of  $1/N$.  These all lie between  $0$  and  $1$.

	*For the mean value, one obtains:		*For the mean value,  one obtains:
	:$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$		:$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$

	*The rms results in		*The rms value results in
	:$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$		:$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$


	'''(4)'''  Correct is <u>the first proposition</u>. It holds:		'''(4)'''  Correct is  <u>the first proposition</u>.  It holds:
	:$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}}),$$		:$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}}),$$
	:$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \varepsilon {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}}{\rm )}$$		:$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \varepsilon {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}}{\rm )}$$
	:$$\Rightarrow \hspace{0.5cm}\rm Pr(\it \|h_{\rm B} - p_{\rm B}\| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}}).$$		:$$\Rightarrow \hspace{0.5cm}\rm Pr(\it \|h_{\rm B} - p_{\rm B}\| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}}).$$



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	:$$p_{\varepsilon}=\rm 1-\rm 2\cdot \rm Q(\frac{\rm 10^{\rm -4}}{\rm 10^{\rm -4}} {\rm )}=\rm 1-\rm 2\cdot\rm Q(\rm 1)\hspace{0.15cm}\underline{\approx\rm 0.684}.$$		:$$p_{\varepsilon}=\rm 1-\rm 2\cdot \rm Q(\frac{\rm 10^{\rm -4}}{\rm 10^{\rm -4}} {\rm )}=\rm 1-\rm 2\cdot\rm Q(\rm 1)\hspace{0.15cm}\underline{\approx\rm 0.684}.$$

	In words:		In words:   If one determines the bit error rate by simulation over  $10^5$  symbols,  with a confidence level of  $\underline{68.4\%}$  <br>one obtains a value between  $0.9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$,  if  $p_{\rm B} = 10^{-3}$.
	*If one determines the bit error rate by simulation over  $10^5$  symbols,
	*with a confidence level of  $\underline{68.4\%}$  one obtains a value between  $0.9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$,
	*if  $p_{\rm B} = 10^{-3}$~~  is~~.


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	'''(7)'''  It must  $\alpha = \varepsilon/\sigma_{h{\rm B}}$  With the result of the subtask  '''(2)'''  then follows:		'''(7)'''  It must  $\alpha = \varepsilon/\sigma_{h{\rm B}}$.  With the result of the subtask  '''(2)'''  then follows:
	:$$\frac{\varepsilon}{\sqrt{p_{\rm B}\cdot(\rm 1-\it p_{\rm B})/N}}\ge {\rm 2} \hspace{0.5cm}\Rightarrow\hspace{0.5cm}		:$$\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}.$$		N\ge \frac{\rm 4\cdot \it p_{\rm B}\cdot(\rm 1-\it p_{\rm B})}{\varepsilon^{\rm 2}}\approx \frac{\rm 4\cdot 10^{-3}}{10^{-8}}\hspace{0.15cm}\underline{=\rm 400\hspace{0.08cm}000}.$$

Guenter at 10:43, 2 February 2022

2022-02-02T10:43:34Z

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	$\rm (B)$  For the metrological determination of the transmission quality or for a system simulation,  it is necessary to rely on the  '''Bit Error Rate'''  $\rm (BER)$:		$\rm (B)$  For the metrological determination of the transmission quality or for a system simulation,  it is necessary to rely on the  '''Bit Error Rate'''  $\rm (BER)$:
			:*The bit error rate is an  "a posteriori parameter"  derived from a performed statistical experiment as a  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission#Definition_der_Bitfehlerquote\|relative frequency]].
	::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$		::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$
			:*$n_{\rm B}$  indicates the number of bit errors occurred when a total of  $N$  binary symbols  ("bits")  were transmitted.
	:*~~$h_{\rm B}$  is a  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission#Definition_der_Bitfehlerquote\|relative frequency]]. ~~ $n_{\rm B}$  indicates the number of bit errors occurred when a total of  $N$  symbols (bits) transmitted.		:*In the limiting case  $N \to \infty$  the relative frequency  $h_{\rm B}$  coincides with the probability  $p_{\rm B}$.  Here now the question shall be clarified,  which statistical uncertainty has to be expected with finite  $N$.

	:*In the limiting case  $N \to \infty$  the relative frequency  $h_{\rm B}$  coincides with the probability  $p_{\rm B}$ .
	:*Here now the question shall be clarified, which statistical uncertainty has to be expected with finite  $N$~~ .~~
	~~:it is necessary to rely on the comparable ''A-posteriori parameter''  '''Bit Error Rate'''  must be ignored~~.






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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian random variables]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian random variables]].
			*Solve this exercise as far as possible in general.  Use the parameter values  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$ for control input.
	*~~Read the exercises~~ as far as possible in general.		*The following are some values of the so-called  "Q-function":
	*Use the parameter values  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$ for control input.
	*The following are some values of the so-called Q-function:
	:$$\rm Q(\rm 1.00)=\rm 0.159,\hspace{0.5cm}\rm Q(\rm 1.65)=\rm 0.050,\hspace{0.5cm}\rm Q(\rm 1.96)=\rm 0.025,\hspace{0.5cm}\rm Q(\rm 2.59)=\rm 0.005.$$		:$$\rm Q(\rm 1.00)=\rm 0.159,\hspace{0.5cm}\rm Q(\rm 1.65)=\rm 0.050,\hspace{0.5cm}\rm Q(\rm 1.96)=\rm 0.025,\hspace{0.5cm}\rm Q(\rm 2.59)=\rm 0.005.$$

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	{How large; is the ~~rms~~ of the random variable  $n_{\rm B}$  ~~for~~  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$?		{How large is the root mean square of the random variable  $n_{\rm B}$  with  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$?
	\|type="{}"}		\|type="{}"}
	$\sigma_{n{\rm B}} \ = \ $ { 10 3% }		$\sigma_{n{\rm B}} \ = \ $ { 10 3% }


	{What values can the bit error rate  $h_{\rm B}$  take?  ~~<br>~~Show that the linear mean  $m_{h{\rm B}}$  of this random variable is equal to the ~~actual~~ bit error probability  $p_{\rm B}$  What is its ~~dispersion~~?		{What values can the bit error rate  $h_{\rm B}$  take?  Show that the linear mean  $m_{h{\rm B}}$  of this random variable is equal to the bit error probability  $p_{\rm B}$  What is its rms value?
	\|type="{}"}		\|type="{}"}
	$\sigma_{h{\rm B}} \ = \ $ { 0.0001 3% }		$\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  $(m_{h{\rm B}})$  and equal ~~dispersion~~  $(\sigma_{h{\rm B}})$  Which statement is true?		{Under certain conditions,  a binomially distributed random variable can be approximated by a Gaussian distribution <br>with equal mean  $(m_{h{\rm B}})$  and equal rms value  $(\sigma_{h{\rm B}})$.  Which statement is true?
	\|type="()"}		\|type="()"}
	+ ${\rm Pr}(\hspace{0.05cm}\|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}\| \le \varepsilon)=1- 2\cdot \rm Q({\varepsilon}/{\sigma_{\it h}{\rm B}}).$		+ ${\rm Pr}(\hspace{0.05cm}\|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}\| \le \varepsilon)=1- 2\cdot \rm Q({\varepsilon}/{\sigma_{\it h}{\rm B}}).$
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	{For abbreviation, we use the confidence level  $p_\varepsilon = {\rm Pr}(\hspace{0.05cm}\|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}\| \le \varepsilon)$.   Which  $p_\varepsilon$  results with  $\varepsilon = 10^{-4}$,  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$ ?		{For abbreviation,  we use the confidence level  $p_\varepsilon = {\rm Pr}(\hspace{0.05cm}\|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}\| \le \varepsilon)$.   Which  $p_\varepsilon$  results with  $\varepsilon = 10^{-4}$,   $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$ ?
	\|type="{}"}		\|type="{}"}
	$p_\varepsilon \ = \ $ { 0.684 3% }		$p_\varepsilon \ = \ $ { 0.684 3% }


	{Let the argument of the Q-function be  $\alpha$.  What is the minimum   $\alpha$  that must be chosen for the confidence level  $p_\varepsilon = 95\%$~~  to be~~ ?		{Let the argument of the Q-function be  $\alpha$.  What is the minimum value of  $\alpha$  that must be chosen for the confidence level  $p_\varepsilon = 95\%$ ?
	\|type="{}"}		\|type="{}"}
	$\alpha_{\rm min} \ = \ $ { 1.96 3% }		$\alpha_{\rm min} \ = \ $ { 1.96 3% }


	{It still holds  $p_{\rm B} = 10^{-3}$  and  $p_\varepsilon = 95\%$.   Over how many symbols  $(N_\text{min})$  must be averaged at least, <br>so that the determined bit error rate in the range between  $0. 9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$  ~~lies  ~~ $(\varepsilon = 10^{-4}, \ \text{10% of nominal value)}$ ?		{It still holds  $p_{\rm B} = 10^{-3}$  and  $p_\varepsilon = 95\%$.   Over how many symbols  $(N_\text{min})$  must be averaged at least, <br>so that the determined bit error rate lies in the range between  $0. 9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$  $(\varepsilon = 10^{-4}, \ \text{10% of its nominal value)}$ ?
	\|type="{}"}		\|type="{}"}
	$N_\text{min} \ = \ ${ 400000 3% }		$N_\text{min} \ = \ ${ 400000 3% }

Guenter at 15:46, 1 February 2022

2022-02-01T15:46:17Z

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	[[File:EN_Sto_A_3_7.png\|right\|frame\|To illustrate the bit error rate]]		[[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  $\langle q_\nu \rangle $  and		*the source symbol sequence  $\langle q_\nu \rangle $,  and
	*the sink symbol sequence  $\langle v_\nu \rangle $.		*the sink symbol sequence  $\langle v_\nu \rangle $.


	If sink symbol  $v_\nu$  and source symbol  $q_\nu$  do not match, there is a bit error   ⇒   $e_\nu = 1$.		If the sink symbol  $v_\nu$  and source symbol  $q_\nu$  do not match,  there is a  "bit error"   ⇒   $e_\nu = 1$.  Otherwise  $e_\nu = 0$  holds.
	~~<br>~~Otherwise  $e_\nu = 0$ holds.


	$\rm (A)$  The most important evaluation criterion of such a digital system is		$\rm (A)$  The most important evaluation criterion of such a digital system is the  '''Bit Error Probability''':
	:the  '''Bit Error Probability'''~~ .~~
	:*With the expected value  ${\rm E}\big[\text{ ...} \big]$  this is defined as follows:		:*With the expected value  ${\rm E}\big[\text{ ...} \big]$  this is defined as follows:
	:: $$\it p_{\rm B} = \rm E\big[\rm Pr(\it v_{\nu} \ne q_{\nu} \rm )\big]=\rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\lim_{{\it N}\to\infty}\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N}\rm Pr(\it e_{\nu}=\rm 1). $$		:: $$\it p_{\rm B} = \rm E\big[\rm Pr(\it v_{\nu} \ne q_{\nu} \rm )\big]=\rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\lim_{{\it N}\to\infty}\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N}\rm Pr(\it e_{\nu}=\rm 1). $$

	:*The right part of this equation describes a time averaging; this must always be applied~~, for example,~~ to time-varying channels.		:*The right part of this equation describes a time averaging; 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  (which is assumed here),  the above equation can be simplified:
	::$$\it p_{\rm B} = \rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\rm E\big[\it e_{\nu} \rm \big].$$		::$$\it p_{\rm B} = \rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\rm E\big[\it e_{\nu} \rm \big].$$
			:*The bit error probability is an  "a priori parameter",  so it allows a prediction for the expected result.

	~~:The bit error probability is an ''a priori parameter'', so it allows a prediction for the expected result.~~

			$\rm (B)$  For the metrological determination of the transmission quality or for a system simulation,  it is necessary to rely on the  '''Bit Error Rate'''  $\rm (BER)$:
	$\rm (B)$  ~~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''~~  '''Bit Error Rate'''  ~~must be ignored.~~
	::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$		::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$

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	:*In the limiting case  $N \to \infty$  the relative frequency  $h_{\rm B}$  coincides with the probability  $p_{\rm B}$ .		:*In the limiting case  $N \to \infty$  the relative frequency  $h_{\rm B}$  coincides with the probability  $p_{\rm B}$ .
	:*Here now the question shall be clarified, which statistical uncertainty has to be expected with finite  $N$ .		:*Here now the question shall be clarified, which statistical uncertainty has to be expected with finite  $N$ .
			:it is necessary to rely on the comparable ''A-posteriori parameter''  '''Bit Error Rate'''  must be ignored.

Noah at 18:51, 15 January 2022

2022-01-15T18:51:37Z

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	If sink symbol  $v_\nu$  and source symbol  $q_\nu$  do not match, there is a bit error   ⇒   $e_\nu = 1$.		If sink symbol  $v_\nu$  and source symbol  $q_\nu$  do not match, there is a bit error   ⇒   $e_\nu = 1$.
	<br>Otherwise  $e_\nu = 0$ holds.		<br>Otherwise  $e_\nu = 0$ holds.


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	:$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$		:$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$

	*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)'''  Correct is <u>the first proposition</u>. It holds:		'''(4)'''  Correct is <u>the first proposition</u>. It holds:
	:$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}),$$		:$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}}),$$
	:$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \~~itvarepsilon~~ {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}{\rm )}$$		:$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \varepsilon {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}}{\rm )}$$
	:$$\Rightarrow \hspace{0.5cm}\rm Pr(\it \|h_{\rm B} - p_{\rm B}\| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}).$$		:$$\Rightarrow \hspace{0.5cm}\rm Pr(\it \|h_{\rm B} - p_{\rm B}\| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}}).$$

Noah at 20:37, 3 January 2022

2022-01-03T20:37:58Z

← Older revision		Revision as of 22:37, 3 January 2022
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	$\rm (B)$  On the other hand, for the metrological determination of the transmission quality or for the system simulation, it is necessary to rely on		$\rm (B)$  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''  '''Bit ~~error rate~~'''  must be ignored.		:the comparable ''A-posteriori parameter''  '''Bit Error Rate'''  must be ignored.
	::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$		::$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$

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	{How large; is the ~~dispersion~~ of the random variable  $n_{\rm B}$  for  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$?		{How large; is the rms of the random variable  $n_{\rm B}$  for  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$?
	\|type="{}"}		\|type="{}"}
	$\sigma_{n{\rm B}} \ = \ $ { 10 3% }		$\sigma_{n{\rm B}} \ = \ $ { 10 3% }
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	{It still holds  $p_{\rm B} = 10^{-3}$  and  $p_\varepsilon = 95\%$.   ~~Üover~~ how many symbols  $(N_\text{min})$  must be averaged at least, <br>so that the determined bit error rate in the range between  $0. 9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$  lies   $(\varepsilon = 10^{-4}, \ \text{10% of nominal value)}$ ?		{It still holds  $p_{\rm B} = 10^{-3}$  and  $p_\varepsilon = 95\%$.   Over how many symbols  $(N_\text{min})$  must be averaged at least, <br>so that the determined bit error rate in the range between  $0. 9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$  lies   $(\varepsilon = 10^{-4}, \ \text{10% of nominal value)}$ ?
	\|type="{}"}		\|type="{}"}
	$N_\text{min} \ = \ ${ 400000 3% }		$N_\text{min} \ = \ ${ 400000 3% }