Aufgaben:Exercise 5.5: Multi-User Interference - Revision history

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	[[Category:Modulation Methods: Exercises\|^5.4 BER of the PN Modulation^]]		[[Category:Modulation Methods: Exercises\|^5.4 BER of the PN Modulation^]]
	[[de:Aufgaben:5.~~5_Mehrteilnehmer–Interferenzen~~]]		[[de:Aufgaben:Aufgabe 5.5: Mehrteilnehmer–Interferenzen]]

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	[[Category:Modulation Methods: Exercises\|^5.4 BER of the PN Modulation^]]		[[Category:Modulation Methods: Exercises\|^5.4 BER of the PN Modulation^]]
			[[de:Aufgaben:5.5_Mehrteilnehmer–Interferenzen]]

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	* The bit error probability without interfering subscribers in the same frequency band is:		* The bit error probability without interfering subscribers in the same frequency band is:
	:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
	* The value of the signal at sampling time, without noise or interfering subscribers, is equal to $±s_0$   ("Nyquist system").  As a result, the bit error probability for noise ~~rms value~~  $σ_d$  before the decision stage $($ originating from the AWGN noise$)$  can be stated as follows:		* The value of the signal at sampling time, without noise or interfering subscribers, is equal to $±s_0$   ("Nyquist system").  As a result, the bit error probability for noise standard deviation  $σ_d$  before the decision stage $($ originating from the AWGN noise$)$  can be stated as follows:
	:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

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	<quiz display=simple>		<quiz display=simple>
	{What is the  (normalized)  noise ~~rms value~~ at the decision maker?		{What is the  (normalized)  noise standard deviation at the decision maker?
	\|type="{}"}		\|type="{}"}
	$σ_d/s_0 \ = \ $ { 0.4 3% }		$σ_d/s_0 \ = \ $ { 0.4 3% }

Hwang at 14:50, 20 December 2021

2021-12-20T14:50:25Z

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	* The bit error probability without interfering subscribers in the same frequency band is:		* The bit error probability without interfering subscribers in the same frequency band is:
	:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
	* The value of the signal at sampling time, without noise or interfering subscribers, is equal to $±s_0$   ("Nyquist system").  As a result, the bit error probability for noise rms value  $σ_d$  before the decision stage $($~~ ~~ originating from the AWGN noise$)$  can be stated as follows:		* The value of the signal at sampling time, without noise or interfering subscribers, is equal to $±s_0$   ("Nyquist system").  As a result, the bit error probability for noise rms value  $σ_d$  before the decision stage $($ originating from the AWGN noise$)$  can be stated as follows:
	:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

Hwang at 14:49, 20 December 2021

2021-12-20T14:49:35Z

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	* The bit error probability without interfering subscribers in the same frequency band is:		* The bit error probability without interfering subscribers in the same frequency band is:
	:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
	* ~~Since in~~ the ~~absence of interfering subscribers all detection~~ signal ~~samples  (~~without ~~input~~ noise~~)  are~~ equal to  $±s_0$   ("Nyquist system"),  the bit error probability ~~with the~~ noise rms value  $σ_d$  $($~~before the decision maker,~~  originating from the AWGN noise$)$  ~~is given~~ as follows:		* The value of the signal at sampling time, without noise or interfering subscribers, is equal to $±s_0$   ("Nyquist system").  As a result, the bit error probability for noise rms value  $σ_d$  before the decision stage $($  originating from the AWGN noise$)$  can be stated as follows:
	:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

Guenter at 16:26, 17 December 2021

2021-12-17T16:26:52Z

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	* The bit error probability without interfering subscribers in the same frequency band is:		* The bit error probability without interfering subscribers in the same frequency band is:
	:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
	* Since in the absence of interfering subscribers all ~~payload~~ samples are equal to  $±s_0$   ("Nyquist system"),  the bit error probability with the noise ~~effective~~ value  $σ_d$  before the decision maker  ~~$($~~originating from the AWGN noise$)$  is given as follows:		* Since in the absence of interfering subscribers all detection signal samples  (without input noise)  are equal to  $±s_0$   ("Nyquist system"),  the bit error probability with the noise rms value  $σ_d$  $($before the decision maker,  originating from the AWGN noise$)$  is given as follows:
	:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

Guenter at 14:14, 17 December 2021

2021-12-17T14:14:18Z

Guenter at 13:53, 17 December 2021

2021-12-17T13:53:46Z

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	*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].		*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
	*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Two_users_with_M-sequence_spreading\|Two users with M-sequence spreading]].		*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Two_users_with_M-sequence_spreading\|Two users with M-sequence spreading]].
	*For the so-called [[Applets:Complementary_Gaussian_Error_Functions\|Q-function]],  the following approximations can be assumed:		*For the so-called  [[Applets:Complementary_Gaussian_Error_Functions\|Q-function]],  the following approximations can be assumed:
	:$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$		:$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$

Guenter at 13:53, 17 December 2021

2021-12-17T13:53:18Z

← Older revision		Revision as of 15:53, 17 December 2021
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	*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].		*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
	*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Two_users_with_M-sequence_spreading\|Two users with M-sequence spreading]].		*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Two_users_with_M-sequence_spreading\|Two users with M-sequence spreading]].
	*For the so-called Q-function,  the following approximations can be assumed:		*For the so-called [[Applets:Complementary_Gaussian_Error_Functions\|Q-function]],  the following approximations can be assumed:
	:$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$		:$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$

Guenter at 13:51, 17 December 2021

2021-12-17T13:51:46Z

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	Notes:		Notes:
	*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].		*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
	*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#~~Two_participants_with_M~~-sequence_spreading\|Two ~~participants~~ with M-sequence spreading]].		*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Two_users_with_M-sequence_spreading\|Two users with M-sequence spreading]].
	*For the so-called Q-function,  the following approximations can be assumed:		*For the so-called Q-function,  the following approximations can be assumed:
	:$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$		:$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From the two equations given above, it follows directly:		'''(1)'''  From the two equations given above,  it follows directly:
	:$$p_{\rm B} = {\rm Q}(2.515) = {\rm Q}({s_0}/{\sigma_d}) \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \frac{\sigma_d}{s_0} = \frac{1}{2.515} = 0.398 \hspace{0.15cm}\underline {\approx 0.4} \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q}(2.515) = {\rm Q}({s_0}/{\sigma_d}) \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \frac{\sigma_d}{s_0} = \frac{1}{2.515} = 0.398 \hspace{0.15cm}\underline {\approx 0.4} \hspace{0.05cm}.$$
	*However, one could also calculate this quantity using the more general equation		*However,  one could also calculate this quantity using the more general equation
	:$$ \sigma_d^2 = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty} \|H_{\rm I}(f) \|^2 \,\,{\rm d} {\it f}\hspace{0.05cm} = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty}{\rm si}^2(~~\pi~~ f T)\,\,{\rm d} {\it f} = \frac{N_0}{2T } \hspace{0.05cm}.$$		:$$ \sigma_d^2 = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty} \|H_{\rm I}(f) \|^2 \,\,{\rm d} {\it f}\hspace{0.05cm} = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty}{\rm sinc}^2(f T)\,\,{\rm d} {\it f} = \frac{N_0}{2T } \hspace{0.05cm}.$$
	:  Here $H_{\rm I}(f)$ describes the integrator in the frequency domain.		:  Here  $H_{\rm I}(f)$  describes the integrator in the frequency domain.
	*With $E_{\rm B}= s_0^2 · T$ the same result is obtained:		*With  $E_{\rm B}= s_0^2 · T$  the same result is obtained:
	:$$\frac{\sigma_d^2}{s_0^2} = \frac{N_0}{2 \cdot s_0^2 \cdot T } = \frac{N_0}{2 E_{\rm B} } = \frac{0.316}{2 } = 0.158\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\sigma_d}/{s_0} = 0.398 \approx 0.4 \hspace{0.05cm}.$$		:$$\frac{\sigma_d^2}{s_0^2} = \frac{N_0}{2 \cdot s_0^2 \cdot T } = \frac{N_0}{2 E_{\rm B} } = \frac{0.316}{2 } = 0.158\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\sigma_d}/{s_0} = 0.398 \approx 0.4 \hspace{0.05cm}.$$



	'''(2)'''  If the interfering participant uses the same M-sequence  $(45)$  as the considered user, <br>        then the (normalized) detection ~~utility~~ samples are equal to  $+2$  $($at $25\%)$,  $-2$  $($at $25\%)$  and  $0$  $($at $50\%)$.		'''(2)'''  If the interfering participant uses the same M-sequence  $(45)$  as the considered user, <br>        then the  (normalized)  detection samples  (without noise)  are equal to  $+2$  $($at  $25\%)$,  $-2$  $($at  $25\%)$  and  $0$  $($at  $50\%)$.
	*When  $d(νT) = ±2$,  the error probability for the considered user is significantly reduced.  In this case, both users transmit the same bit $($"$+1$" or "$-1$"$)$ and the distance from the threshold is doubled:		*When  $d(νT) = ±2$,  the error probability for the considered user is significantly reduced.
			*In this case,  both users transmit the same bit  $($"$+1$"  or  "$-1$"$)$  and the distance from the threshold is doubled:
	:$$ p_{\rm B}\,\,\big [{\rm if}\,\, d (\nu T) = \pm 2s_0 \big ] = {\rm Q} \left ( 2 \cdot 2.515 \right ) = {\rm Q} \left ( 5.03 \right ) \approx 2.45 \cdot 10^{-7} \approx 0 \hspace{0.05cm}.$$		:$$ p_{\rm B}\,\,\big [{\rm if}\,\, d (\nu T) = \pm 2s_0 \big ] = {\rm Q} \left ( 2 \cdot 2.515 \right ) = {\rm Q} \left ( 5.03 \right ) \approx 2.45 \cdot 10^{-7} \approx 0 \hspace{0.05cm}.$$
	*On the other hand, if  $d(νT) = 0$  (for example, if  $a_\text{1(s)} = +1$  and  $a_\text{1(i)} = -1$  holds or vice versa), the signals cancel completely and we obtain		*On the other hand,  if  $d(νT) = 0$  (for example, if  $a_\text{1(s)} = +1$   and  $a_\text{1(i)} = -1$   holds or vice versa),  the signals cancel completely and we obtain
	:$$p_{\rm B}~~\,\,~~\big[{\rm if}\,\, d (\nu T) = 0 \big] = {\rm Q} \left ( 0 \right ) = 0.5 \hspace{0.05cm}.$$		:$$p_{\rm B}\big[{\rm if}\,\, d (\nu T) = 0 \big] = {\rm Q} \left ( 0 \right ) = 0.5 \hspace{0.05cm}.$$
	*Averaging over these two equally probable possibilities, we thus obtain for the mean bit error probability:		*Averaging over these two equally probable possibilities,  we thus obtain for the mean bit error probability:
	:$$p_{\rm B}= 0.5 \cdot 2.45 \cdot 10^{-7}+ 0.5 \cdot 0.5 \hspace{0.15cm}\underline {\approx 25\%} \hspace{0.05cm}.$$		:$$p_{\rm B}= 0.5 \cdot 2.45 \cdot 10^{-7}+ 0.5 \cdot 0.5 \hspace{0.15cm}\underline {\approx 25\%} \hspace{0.05cm}.$$



	'''(3)'''  We first consider only the useful part   ⇒   $n(t) = 0$,  restricting ourselves to the first data symbol and assuming the coefficient $a_\text{1(s)} = +1$.		'''(3)'''  We first consider only the useful part   ⇒   $n(t) = 0$,  restricting ourselves to the first data symbol and assuming the coefficient  $a_\text{1(s)} = +1$.
	*Then within this data bit  $s(t) = c_{45}(t)$ holds.		*Then within this data bit  $s(t) = c_{45}(t)$ holds.
	*If the coefficient  $a_\text{1(i)} $  of the interfering participant is also  $+1$, then for the signals specified in front in the time interval from  $0$  to  $T$ we obtain:		*If the coefficient  $a_\text{1(i)} $  of the interfering participant is also  $+1$,  then for the signals specified in front in the time interval from  $0$  to  $T$ we obtain:
	:$$ r(t) = c_{45}(t) + c_{75}(t)\hspace{0.05cm},$$		:$$ r(t) = c_{45}(t) + c_{75}(t)\hspace{0.05cm},$$
	:$$b(t) = r(t) \cdot c_{45}(t) = \left [c_{45}(t) + c_{75}(t) \right ] \cdot c_{45}(t) = 1+ c_{45}(t) \cdot c_{75}(t) \hspace{0.05cm},$$		:$$b(t) = r(t) \cdot c_{45}(t) = \left [c_{45}(t) + c_{75}(t) \right ] \cdot c_{45}(t) = 1+ c_{45}(t) \cdot c_{75}(t) \hspace{0.05cm},$$
	:$$ d (T) = \frac{1}{T} \cdot \int_{0 }^{ T} b (t )\hspace{0.1cm} {\rm d}t = 1 + {\it \varphi}_{45,\hspace{0.05cm}75}(\lambda = 0) \hspace{0.05cm}.$$		:$$ d (T) = \frac{1}{T} \cdot \int_{0 }^{ T} b (t )\hspace{0.1cm} {\rm d}t = 1 + {\it \varphi}_{45,\hspace{0.05cm}75}(\lambda = 0) \hspace{0.05cm}.$$
	*Here,   $φ_\text{45, 75}(τ)$  denotes the PCCF between the spreading sequences with octal identifiers  $(45)$  and  $(75)$, which can be found in the table on the data page.		*Here,   $φ_\text{45, 75}(τ)$  denotes the PCCF between the spreading sequences with octal identifiers  $(45)$  and  $(75)$,  which can be found in the table on the data page.
			*Correspondingly,  for the detection sample without noise,  given  $a_\text{1(s)} = +1$  and  $a_\text{1(i)} =-1$:
	*Correspondingly, for the detection ~~utility~~ sample, given  $a_\text{1(s)} = +1$  and  $a_\text{1(i)} =-1$:
	:$$d (T) = 1 - {\it \varphi}_{45,\hspace{0.05cm}75}(\lambda = 0) \hspace{0.05cm}.$$		:$$d (T) = 1 - {\it \varphi}_{45,\hspace{0.05cm}75}(\lambda = 0) \hspace{0.05cm}.$$
	*For symmetry reasons, the coefficients  $a_\text{1(s)}  = -1$,  $a_\text{1(i)} = -1$  as well as  $a_\text{1(s)} = -1$,  $a_\text{1(i)} = +1$  provide exactly the same contributions for the bit error probability as  $a_\text{1(s)} = +1$,  $a_\text{1(i)} = +1$  and  $a_{1(s)} = +1$,  $a_{1(i)} = –1$ respectively, if we also consider the AWGN noise.		*For symmetry reasons,  the coefficients  $a_\text{1(s)}  = -1$,  $a_\text{1(i)} = -1$  as well as  $a_\text{1(s)} = -1$,  $a_\text{1(i)} = +1$  provide exactly the same contributions for the bit error probability as  $a_\text{1(s)} = +1$,  $a_\text{1(i)} = +1$  and  $a_{1(s)} = +1$,  $a_{1(i)} = -1$ respectively,  if we also consider the AWGN noise.

	*Thus, using the result of subtask  '''(1)'''  and with  $φ_\text{45, 75}(λ = 0) = 7/31$,  we obtain approximately:		*Thus,  using the result of subtask  '''(1)'''  and with  $φ_\text{45, 75}(λ = 0) = 7/31$,  we obtain approximately:
	:$$p_{\rm B} = \frac{1}{2} \cdot {\rm Q} \left ( \frac{1+ 7/31}{0.4} \right ) + \frac{1}{2} \cdot {\rm Q} \left ( \frac{1- 7/31}{0.4} \right ) = \frac{1}{2} \cdot {\rm Q} \left ( \frac{1.225}{0.4} \right ) + \frac{1}{2} \cdot {\rm Q} \left ( \frac{0.775}{0.4} \right ) = \frac{1}{2} \cdot {\rm Q} \left ( 3.06 \right ) + \frac{1}{2} \cdot {\rm Q} \left ( 1.94 \right )$$		:$$p_{\rm B} = \frac{1}{2} \cdot {\rm Q} \left ( \frac{1+ 7/31}{0.4} \right ) + \frac{1}{2} \cdot {\rm Q} \left ( \frac{1- 7/31}{0.4} \right ) = \frac{1}{2} \cdot {\rm Q} \left ( \frac{1.225}{0.4} \right ) + \frac{1}{2} \cdot {\rm Q} \left ( \frac{0.775}{0.4} \right ) = \frac{1}{2} \cdot {\rm Q} \left ( 3.06 \right ) + \frac{1}{2} \cdot {\rm Q} \left ( 1.94 \right )$$
	:$$ \Rightarrow \hspace{0.3cm} p_{\rm B}\approx \frac{1}{2} \cdot \left [{\rm Q} \left ( 3 \right ) + {\rm Q} \left ( 2 \right ) \right ] = \frac{1}{2} \cdot \left [0.00135 + 0.02275 \right ] \hspace{0.15cm}\underline {= 1.2\%}\hspace{0.05cm}.$$		:$$ \Rightarrow \hspace{0.3cm} p_{\rm B}\approx \frac{1}{2} \cdot \left [{\rm Q} \left ( 3 \right ) + {\rm Q} \left ( 2 \right ) \right ] = \frac{1}{2} \cdot \left [0.00135 + 0.02275 \right ] \hspace{0.15cm}\underline {= 1.2\%}\hspace{0.05cm}.$$


	'''(4)'''  Possible solutions are <u>2 and 3</u>:
			'''(4)'''  Possible solutions are  <u>2 and 3</u>:
	* The PCCF value  $φ_\text{45, 57}(λ = 0)$  is only  $1/31$  in magnitude and thus the error probability is only slightly larger than  $0.6\%$.		* The PCCF value  $φ_\text{45, 57}(λ = 0)$  is only  $1/31$  in magnitude and thus the error probability is only slightly larger than  $0.6\%$.
	*In contrast, the sequence with octal identifiers  $(67)$  leads to the same PCCF as sequence  $(75)$.		*In contrast,  the sequence with octal identifiers  $(67)$  leads to the same PCCF as sequence  $(75)$.
	*Without interfering participants, the following applies according to the data sheet:   $p_{\rm B} = 0.6\%$.		*Without interfering participants,  the following applies according to the data sheet:   $p_{\rm B} = 0.6\%$.
	*With interference, this value cannot be undercut   ⇒   solution 1 is not possible.		*With interference,  this value cannot be undercut   ⇒   solution 1 is not possible.

	{{ML-Fuß}}		{{ML-Fuß}}