Aufgaben:Exercise 1.3Z: Threshold Optimization - Revision history

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	'''(1)'''  From the Q-function,  substitution  $t^{2} = u^{2}/2$  yields:		'''(1)'''  From the Q-function,  substitution  $t^{2} = u^{2}/2$  yields:
	:$$\rm Q (\it x)= \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it		:$$\rm Q (\it x)= \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\itx}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u =\frac{\rm 1}{\sqrt{\rm \pi}}\int_{\itx/\sqrt{\rm 2}}^{+\infty}\rm e^{\it -t^{\hspace{0.05cm}\rm 2}}\,{d \it t} ={1}/{2} \cdot {\rm erfc} \left( {x}/{\sqrt{2}}\right)\hspace{0.05cm}.$$
	x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u =
	\frac{\rm 1}{\sqrt{\rm \pi}}\int_{\it
	x/\sqrt{\rm 2}}^{+\infty}\rm e^{\it -t^{\hspace{0.05cm}\rm 2}}\,{d \it t} =
	{1}/{2} \cdot {\rm erfc} \left( {x}/{\sqrt{2}}
	\right)\hspace{0.05cm}.$$
	From this it follows that the  <u>first solution</u>  is correct:		From this it follows that the  <u>first solution</u>  is correct:
	:$${\rm erfc} (x)= 2 \cdot {\rm Q} (\sqrt{2} \cdot x) \hspace{0.05cm}.$$		:$${\rm erfc} (x)= 2 \cdot {\rm Q} (\sqrt{2} \cdot x) \hspace{0.05cm}.$$

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	:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$
	Here,  the following functions are used:		Here,  the following functions are used:
	:$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it		:$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\itx}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u}\hspace{0.05cm},\hspace{1cm}{\rm erfc} ({\it x}) = \frac{\rm 2}{\sqrt{\rm\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u\hspace{0.05cm}.$$
	x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u}
	\hspace{0.05cm},\hspace{1cm}{\rm erfc} ({\it x}) = \frac{\rm 2}{\sqrt{\rm
	\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u
	\hspace{0.05cm}.$$
	*The above equation holds for the threshold  $E = 0$,  regardless of the symbol probabilities  $p_{\rm L}$  and  $p_{\rm H}$.		*The above equation holds for the threshold  $E = 0$,  regardless of the symbol probabilities  $p_{\rm L}$  and  $p_{\rm H}$.
	*However,  a smaller error probability can be obtained with a different threshold  $E$  if the two occurrence probabilities are different   $(p_{\rm L} ≠ p_{\rm H})$.		*However,  a smaller error probability can be obtained with a different threshold  $E$  if the two occurrence probabilities are different   $(p_{\rm L} ≠ p_{\rm H})$.
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	*The following applies to the derivative of the Q-function:		*The following applies to the derivative of the Q-function:
	:$$\frac{{\rm d\hspace{0.05cm}Q} ({\it x})} {{\rm d}\hspace{0.05cm}x} = \frac{\rm 1}{\sqrt{\rm 2\pi}}		:$$\frac{{\rm d\hspace{0.05cm}Q} ({\it x})} {{\rm d}\hspace{0.05cm}x} = \frac{\rm 1}{\sqrt{\rm 2\pi}}\cdot \rm e^{\it -x^{\rm 2}/\rm 2} \hspace{0.05cm}.$$
	\cdot \rm e^{\it -x^{\rm 2}/\rm 2} \hspace{0.05cm}.$$


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	'''(3)'''  Now the general equation for the bit error probability,  where  $d_{\rm N}$  denotes the noise component of  $d(t)$,  is:		'''(3)'''  Now the general equation for the bit error probability,  where  $d_{\rm N}$  denotes the noise component of  $d(t)$,  is:
	:$$p_{\rm B} \ = \ p_{\rm L} \cdot {\rm Pr}( d_{\rm N}> E + s_0)+		:$$p_{\rm B} \ = \ p_{\rm L} \cdot {\rm Pr}( d_{\rm N}> E + s_0)+p_{\rm H} \cdot {\rm Pr}( d_{\rm N}< E - s_0) \ = \ p_{\rm L} \cdot {\rm Q} \left( \frac{s_0 + E}{\sigma_d}\right)+p_{\rm H} \cdot {\rm Q} \left( \frac{s_0 - E}{\sigma_d}\right)\hspace{0.05cm}.$$
	p_{\rm H} \cdot {\rm Pr}( d_{\rm N}< E - s_0) \ = \ p_{\rm L} \cdot {\rm Q} \left( \frac{s_0 + E}{\sigma_d}\right)+
	p_{\rm H} \cdot {\rm Q} \left( \frac{s_0 - E}{\sigma_d}\right)\hspace{0.05cm}.$$
	*Here,  the PDF symmetry is taken into account.  With  $p_{\rm L} = 0.88$   ⇒   $p_{\rm H} = 0.12$  and  $E = 0.1 \hspace{0.05cm} \rm V $  one obtains:		*Here,  the PDF symmetry is taken into account.  With  $p_{\rm L} = 0.88$   ⇒   $p_{\rm H} = 0.12$  and  $E = 0.1 \hspace{0.05cm} \rm V $  one obtains:
	:$$p_{\rm B} \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} + 0.1\, {\rm V}}{0.5\, {\rm V}}\right)\hspace{-0.05cm}+\hspace{-0.05cm}		:$$p_{\rm B} \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} + 0.1\, {\rm V}}{0.5\, {\rm V}}\right)\hspace{-0.05cm}+\hspace{-0.05cm}0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} - 0.1\, {\rm V}}{0.5\, {\rm V}}\right) \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} {\rm Q}(2.2) \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \cdot {\rm Q}(1.8) \hspace{-0.05cm}= \hspace{-0.05cm}0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 1.39\,\% \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 3.59\,\%\ \hspace{0.1cm}\underline { = 1.65\,\% } \hspace{0.05cm}.$$
	0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} - 0.1\, {\rm V}}{0.5\, {\rm V}}\right) \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} {\rm Q}(2.2) \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \cdot {\rm Q}(1.8) \hspace{-0.05cm}= \hspace{-0.05cm}0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 1.39\,\% \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 3.59\,\%\ \hspace{0.1cm}\underline { = 1.65\,\% } \hspace{0.05cm}.$$
	*Threshold shifting to the right by  $E = s_{0}/10$  results in an improvement from  $p_{\rm B} = 2.27 \ \%$  to  $p_{\rm B} = 1.65 \ \%$.		*Threshold shifting to the right by  $E = s_{0}/10$  results in an improvement from  $p_{\rm B} = 2.27 \ \%$  to  $p_{\rm B} = 1.65 \ \%$.


	'''(4)'''  This optimization task is solved by setting the derivative to zero,  taking into account the note on the information section:		'''(4)'''  This optimization task is solved by setting the derivative to zero,  taking into account the note on the information section:
	:$$\frac{{\rm d\hspace{0.05cm}} p_{\rm B}(E)} {{\rm d}\hspace{0.05cm}E} = - \frac{\rm p_{\rm L}}{\sqrt{\rm 2\pi}\cdot {\sigma_d}}		:$$\frac{{\rm d\hspace{0.05cm}} p_{\rm B}(E)} {{\rm d}\hspace{0.05cm}E} = - \frac{\rm p_{\rm L}}{\sqrt{\rm 2\pi}\cdot {\sigma_d}}\cdot {\rm exp}\left(- \frac{(s_0 + E)^2}{{\rm 2}\cdot{\sigma_d}^2}\right)+\frac{\rm p_{\rm H}}{\sqrt{\rm 2\pi}\cdot{\sigma_d}} \cdot {\rm exp}\left(- \frac{(s_0 - E)^2}{{\rm 2}\cdot{\sigma_d}^2}\right) = 0$$
	\cdot {\rm exp}\left(- \frac{(s_0 + E)^2}{{\rm 2}\cdot		:$$\Rightarrow \hspace{0.3cm} \frac{p_{\rm L}} {p_{\rm H}} = - \frac{{\rm exp} \left(-\frac{(s_0 - E)^2}{{\rm 2}\cdot{\sigma_d}^2}\right)}{{\rm exp} \left(-\frac{(s_0 + E)^2}{{\rm2}\cdot {\sigma_d}^2}\right)} = {\rm exp} \left( \frac{2 \cdot E\cdot s_0 }{ {\sigma_d}^2}\right)\hspace{0.05cm}.$$
	{\sigma_d}^2}\right)+\frac{\rm p_{\rm H}}{\sqrt{\rm 2\pi}\cdot
	{\sigma_d}} \cdot {\rm exp}\left(- \frac{(s_0 - E)^2}{{\rm 2}\cdot
	{\sigma_d}^2}\right) = 0$$
	:$$\Rightarrow \hspace{0.3cm} \frac{p_{\rm L}} {p_{\rm H}} = - \frac{
	{\rm exp} \left(-\frac{(s_0 - E)^2}{{\rm 2}\cdot
	{\sigma_d}^2}\right)}{{\rm exp} \left(-\frac{(s_0 + E)^2}{{\rm
	2}\cdot {\sigma_d}^2}\right)} = {\rm exp} \left( \frac{2 \cdot E
	\cdot s_0 }{ {\sigma_d}^2}\right)\hspace{0.05cm}.$$
	*Thus,  we obtain for the optimal threshold value in general:		*Thus,  we obtain for the optimal threshold value in general:
	:$$E_{\rm opt} = \frac{\sigma_d^2} {2 \cdot s_0} \cdot {\rm ln}\hspace{0.05cm}\frac{p_{\rm L}} {p_{\rm H}} \hspace{0.05cm}.$$		:$$E_{\rm opt} = \frac{\sigma_d^2} {2 \cdot s_0} \cdot {\rm ln}\hspace{0.05cm}\frac{p_{\rm L}} {p_{\rm H}} \hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^1.2 BER for Baseband Systems^]]
			[[de:Aufgaben:1.3Z_Schwellenwertoptimierung]]

Guenter at 13:53, 30 April 2022

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Guenter at 13:36, 30 April 2022

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Hwang at 17:36, 2 March 2022

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	*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]].

	*You can use the interactive applet  [[Applets:~~Komplementäre_Gaußsche_Fehlerfunktionen~~\|~~Komplementäre Gaußsche Fehlerfunktionen~~]]  to determine error probabilities.		*You can use the interactive applet  [[Applets:Complementary_Gaussian_Error_Functions\|Complementary Gaussian Error Functions]]  to determine error probabilities.
	*The following applies to the derivative of the Q-function:		*The following applies to the derivative of the Q-function:
	:$$\frac{{\rm d\hspace{0.05cm}Q} ({\it x})} {{\rm d}\hspace{0.05cm}x} = \frac{\rm 1}{\sqrt{\rm 2\pi}}		:$$\frac{{\rm d\hspace{0.05cm}Q} ({\it x})} {{\rm d}\hspace{0.05cm}x} = \frac{\rm 1}{\sqrt{\rm 2\pi}}

Hwang at 17:27, 2 March 2022

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Guenter at 16:29, 8 February 2022

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	[[File:~~P_ID1268__Dig_Z_1_3~~.png\|right\|frame\|Zur Optimierung des Schwellenwertes]]		[[File:EN_Dig_Z_1_3.png\|right\|frame\|Zur Optimierung des Schwellenwertes]]
	In dieser Aufgabe wird ein bipolares Binärsystem mit AWGN–Rauschen (''Additive White Gaussian Noise'' ) betrachtet, so dass für die Bitfehlerwahrscheinlichkeit gilt:		In dieser Aufgabe wird ein bipolares Binärsystem mit AWGN–Rauschen (''Additive White Gaussian Noise'' ) betrachtet, so dass für die Bitfehlerwahrscheinlichkeit gilt:
	:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$

Guenter at 15:25, 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^]]

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	:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$
	Here,  the following functions are used:		Here,  the following functions are used:
	:$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\~~itx~~}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u}\hspace{0.05cm},\hspace{1cm}{\rm erfc} ({\it x}) = \frac{\rm 2}{\sqrt{\rm\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u\hspace{0.05cm}.$$		:$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u}\hspace{0.05cm},\hspace{1cm}{\rm erfc} ({\it x}) = \frac{\rm 2}{\sqrt{\rm\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u\hspace{0.05cm}.$$
	*The above equation holds for the threshold  $E = 0$,  regardless of the symbol probabilities  $p_{\rm L}$  and  $p_{\rm H}$.		*The above equation holds for the threshold  $E = 0$,  regardless of the symbol probabilities  $p_{\rm L}$  and  $p_{\rm H}$.
	*However,  a smaller error probability can be obtained with a different threshold  $E$  if the two occurrence probabilities are different   $(p_{\rm L} ≠ p_{\rm H})$.		*However,  a smaller error probability can be obtained with a different threshold  $E$  if the two occurrence probabilities are different   $(p_{\rm L} ≠ p_{\rm H})$.
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	'''(1)'''  From the Q-function,  substitution  $t^{2} = u^{2}/2$  yields:		'''(1)'''  From the Q-function,  substitution  $t^{2} = u^{2}/2$  yields:
	:$$\rm Q (\it x)= \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\~~itx~~}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u =\frac{\rm 1}{\sqrt{\rm \pi}}\int_{\~~itx~~/\sqrt{\rm 2}}^{+\infty}\rm e^{\it -t^{\hspace{0.05cm}\rm 2}}\,{d \it t} ={1}/{2} \cdot {\rm erfc} \left( {x}/{\sqrt{2}}\right)\hspace{0.05cm}.$$		:$$\rm Q (\it x)= \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u =\frac{\rm 1}{\sqrt{\rm \pi}}\int_{\it x/\sqrt{\rm 2}}^{+\infty}\rm e^{\it -t^{\hspace{0.05cm}\rm 2}}\,{d \it t} ={1}/{2} \cdot {\rm erfc} \left( {x}/{\sqrt{2}}\right)\hspace{0.05cm}.$$
	From this it follows that the  <u>first solution</u>  is correct:		From this it follows that the  <u>first solution</u>  is correct:
	:$${\rm erfc} (x)= 2 \cdot {\rm Q} (\sqrt{2} \cdot x) \hspace{0.05cm}.$$		:$${\rm erfc} (x)= 2 \cdot {\rm Q} (\sqrt{2} \cdot x) \hspace{0.05cm}.$$
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	[[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.~~3Z_Schwellenwertoptimierung~~]]		[[de:Aufgaben:Aufgabe 1.3Z: Schwellenwertoptimierung]]

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From the Q-function, substitution $t^{2} = u^{2}/2$ yields:		'''(1)'''  From the Q-function,  substitution  $t^{2} = u^{2}/2$  yields:
	:$$\rm Q (\it x)= \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it		:$$\rm Q (\it x)= \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it
	x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u =		x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u =
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	{1}/{2} \cdot {\rm erfc} \left( {x}/{\sqrt{2}}		{1}/{2} \cdot {\rm erfc} \left( {x}/{\sqrt{2}}
	\right)\hspace{0.05cm}.$$		\right)\hspace{0.05cm}.$$
	From this it follows that the <u>first solution</u> is correct:		From this it follows that the  <u>first solution</u>  is correct:
	:$${\rm erfc} (x)= 2 \cdot {\rm Q} (\sqrt{2} \cdot x) \hspace{0.05cm}.$$		:$${\rm erfc} (x)= 2 \cdot {\rm Q} (\sqrt{2} \cdot x) \hspace{0.05cm}.$$


	'''(2)'''  Independently of the symbol probabilities, with the decision threshold $E = 0$, we obtain:		'''(2)'''  Independently of the symbol probabilities,  with the decision threshold  $E = 0$,  we obtain:
	:$$p_{\rm B} = {\rm Q} \left( {s_0}/{\sigma_d}\right)= {\rm Q} (2) \hspace{0.1cm}\underline {= 2.27 \, \% }\hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( {s_0}/{\sigma_d}\right)= {\rm Q} (2) \hspace{0.1cm}\underline {= 2.27 \, \% }\hspace{0.05cm}.$$


	'''(3)'''  Now the general equation for the bit error probability, where $d_{\rm N}$ denotes the noise component of $d(t)$, is:		'''(3)'''  Now the general equation for the bit error probability,  where  $d_{\rm N}$  denotes the noise component of  $d(t)$,  is:
	:$$p_{\rm B} \ = \ p_{\rm L} \cdot {\rm Pr}( d_{\rm N}> E + s_0)+		:$$p_{\rm B} \ = \ p_{\rm L} \cdot {\rm Pr}( d_{\rm N}> E + s_0)+
	p_{\rm H} \cdot {\rm Pr}( d_{\rm N}< E - s_0) \ = \ p_{\rm L} \cdot {\rm Q} \left( \frac{s_0 + E}{\sigma_d}\right)+		p_{\rm H} \cdot {\rm Pr}( d_{\rm N}< E - s_0) \ = \ p_{\rm L} \cdot {\rm Q} \left( \frac{s_0 + E}{\sigma_d}\right)+
	p_{\rm H} \cdot {\rm Q} \left( \frac{s_0 - E}{\sigma_d}\right)\hspace{0.05cm}.$$		p_{\rm H} \cdot {\rm Q} \left( \frac{s_0 - E}{\sigma_d}\right)\hspace{0.05cm}.$$
	*Here, PDF symmetry is taken into account. With $p_{\rm L} = 0.88$   ⇒   $p_{\rm H} = 0.12$ and $E = 0.1 \hspace{0.05cm} \rm V $ one obtains:		*Here,  the PDF symmetry is taken into account.  With  $p_{\rm L} = 0.88$   ⇒   $p_{\rm H} = 0.12$  and  $E = 0.1 \hspace{0.05cm} \rm V $  one obtains:
	:$$p_{\rm B} \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} + 0.1\, {\rm V}}{0.5\, {\rm V}}\right)\hspace{-0.05cm}+\hspace{-0.05cm}		:$$p_{\rm B} \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} + 0.1\, {\rm V}}{0.5\, {\rm V}}\right)\hspace{-0.05cm}+\hspace{-0.05cm}
	0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} - 0.1\, {\rm V}}{0.5\, {\rm V}}\right) \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} {\rm Q}(2.2) \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \cdot {\rm Q}(1.8) \hspace{-0.05cm}= \hspace{-0.05cm}0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 1.39\,\% \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 3.59\,\%\ \hspace{0.1cm}\underline { = 1.65\,\% } \hspace{0.05cm}.$$		0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} - 0.1\, {\rm V}}{0.5\, {\rm V}}\right) \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} {\rm Q}(2.2) \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \cdot {\rm Q}(1.8) \hspace{-0.05cm}= \hspace{-0.05cm}0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 1.39\,\% \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 3.59\,\%\ \hspace{0.1cm}\underline { = 1.65\,\% } \hspace{0.05cm}.$$
	*Threshold shifting to the right by $E = s_{0}/10$ results in an improvement from $p_{\rm B} = 2.27 \ \%$ to $p_{\rm B} = 1.65 \ \%$.		*Threshold shifting to the right by  $E = s_{0}/10$  results in an improvement from  $p_{\rm B} = 2.27 \ \%$  to  $p_{\rm B} = 1.65 \ \%$.


	'''(4)'''  This optimization task is solved by setting the derivative to zero, taking into account the note on the information section:		'''(4)'''  This optimization task is solved by setting the derivative to zero,  taking into account the note on the information section:
	:$$\frac{{\rm d\hspace{0.05cm}} p_{\rm B}(E)} {{\rm d}\hspace{0.05cm}E} = - \frac{\rm p_{\rm L}}{\sqrt{\rm 2\pi}\cdot {\sigma_d}}		:$$\frac{{\rm d\hspace{0.05cm}} p_{\rm B}(E)} {{\rm d}\hspace{0.05cm}E} = - \frac{\rm p_{\rm L}}{\sqrt{\rm 2\pi}\cdot {\sigma_d}}
	\cdot {\rm exp}\left(- \frac{(s_0 + E)^2}{{\rm 2}\cdot		\cdot {\rm exp}\left(- \frac{(s_0 + E)^2}{{\rm 2}\cdot
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	2}\cdot {\sigma_d}^2}\right)} = {\rm exp} \left( \frac{2 \cdot E		2}\cdot {\sigma_d}^2}\right)} = {\rm exp} \left( \frac{2 \cdot E
	\cdot s_0 }{ {\sigma_d}^2}\right)\hspace{0.05cm}.$$		\cdot s_0 }{ {\sigma_d}^2}\right)\hspace{0.05cm}.$$
	*Thus, we obtain for the optimal threshold value in general:		*Thus,  we obtain for the optimal threshold value in general:
	:$$E_{\rm opt} = \frac{\sigma_d^2} {2 \cdot s_0} \cdot {\rm ln}\hspace{0.05cm}\frac{p_{\rm L}} {p_{\rm H}} \hspace{0.05cm}.$$		:$$E_{\rm opt} = \frac{\sigma_d^2} {2 \cdot s_0} \cdot {\rm ln}\hspace{0.05cm}\frac{p_{\rm L}} {p_{\rm H}} \hspace{0.05cm}.$$
	*With $σ_{d} = 0.5 \ \rm V$, $s_{\rm 0} = 1 \ \rm V$, $p_{\rm L} = 0.88$ and $p_{\rm H} = 0.12$, the following optimum is obtained:		*With $σ_{d} = 0.5 \ \rm V$,  $s_{\rm 0} = 1 \ \rm V$, $p_{\rm L} = 0.88$  and  $p_{\rm H} = 0.12$,  the following optimum is obtained:
	:$$E_{\rm opt} = \frac{(0.5\, {\rm V})^2} {2 \cdot 1\, {\rm V}} \cdot {\rm ln}\hspace{0.05cm}\frac{0.88} {0.12} \hspace{0.1cm}\underline { \approx 0.25\, {\rm V}}\hspace{0.05cm}.$$		:$$E_{\rm opt} = \frac{(0.5\, {\rm V})^2} {2 \cdot 1\, {\rm V}} \cdot {\rm ln}\hspace{0.05cm}\frac{0.88} {0.12} \hspace{0.1cm}\underline { \approx 0.25\, {\rm V}}\hspace{0.05cm}.$$


	'''(5)'''  The minimum error probability for the optimal threshold $E_{\rm opt} = 0.25 \hspace{0.05cm} \rm V$ is thus:		'''(5)'''  The minimum error probability for the optimal threshold  $E_{\rm opt} = 0.25 \hspace{0.05cm} \rm V$  is thus:
	:$$p_{\rm B, \hspace{0.05cm}min} = 0.88 \cdot {\rm Q}(2.5) + 0.12 \cdot {\rm Q}(1.5) = 0.88 \cdot 0.62\,\% + 0.12 \cdot 6.68\,\%\ \hspace{0.1cm}\underline {= 1.35\,\% }\hspace{0.05cm}.$$		:$$p_{\rm B, \hspace{0.05cm}min} = 0.88 \cdot {\rm Q}(2.5) + 0.12 \cdot {\rm Q}(1.5) = 0.88 \cdot 0.62\,\% + 0.12 \cdot 6.68\,\%\ \hspace{0.1cm}\underline {= 1.35\,\% }\hspace{0.05cm}.$$
	*Compared to $E = 0$, the error probability is now about $40 \%$ smaller.		*Compared to  $E = 0$,  the error probability is now about  $40 \%$  smaller.


	'''(6)'''  Using the result from '''(4)''', the optimal threshold value is now:		'''(6)'''  Using the result from '''(4)''',  the optimal threshold value is now:
	:$$E_{\rm opt} = \frac{(0.5\, {\rm V})^2} {2 \cdot 1\, {\rm V}} \cdot {\rm ln}\hspace{0.05cm}\frac{0.31} {0.69}\hspace{0.1cm}\underline { \approx -0.1\, {\rm V}}\hspace{0.05cm}.$$		:$$E_{\rm opt} = \frac{(0.5\, {\rm V})^2} {2 \cdot 1\, {\rm V}} \cdot {\rm ln}\hspace{0.05cm}\frac{0.31} {0.69}\hspace{0.1cm}\underline { \approx -0.1\, {\rm V}}\hspace{0.05cm}.$$
	Please note:		Please note:
	*~~Now that~~ symbol '''L''' is less probable here, the threshold must be shifted to the left – away from the more probable symbol.		*Because the symbol  '''L'''  is less probable here,  the threshold must be shifted to the left – away from the more probable symbol.
	*The derivation of the result for '''(4)''' and the diagram on the information section shows that the optimal threshold value is to be set exactly at the point where the two Gaussian functions intersect.		*The derivation of the result for  '''(4)'''  and the diagram on the information section shows that the optimal threshold value is to be set exactly at the point where the two Gaussian functions intersect.


	'''(7)'''  Finally, using the result from '''(6)''', the following holds:		'''(7)'''  Finally,  using the result from  '''(6)''',  the following holds:
	:$$p_{\rm B, \hspace{0.05cm}min} = 0.31 \cdot {\rm Q}(1.8) + 0.69 \cdot {\rm Q}(2.2) = 0.31 \cdot 3.59\,\% + 0.69 \cdot 1.39\,\%\ \hspace{0.1cm}\underline { = 2.07\,\% } \hspace{0.05cm}.$$		:$$p_{\rm B, \hspace{0.05cm}min} = 0.31 \cdot {\rm Q}(1.8) + 0.69 \cdot {\rm Q}(2.2) = 0.31 \cdot 3.59\,\% + 0.69 \cdot 1.39\,\%\ \hspace{0.1cm}\underline { = 2.07\,\% } \hspace{0.05cm}.$$
	*Due to the less severe asymmetry, the achievable improvement of $9 \%$ is lower than calculated in ~~point~~ '''(5)'''.		*Due to the less severe asymmetry,  the achievable improvement of  $9 \%$  is lower than calculated in subtask  '''(5)'''.

	{{ML-Fuß}}		{{ML-Fuß}}

← Older revision		Revision as of 15:36, 30 April 2022
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	[[File:EN_Dig_Z_1_3.png\|right\|frame\|Optimizing the threshold]]		[[File:EN_Dig_Z_1_3.png\|right\|frame\|Optimizing the threshold  $E$]]
	In this exercise, a bipolar binary system with AWGN noise (''Additive White Gaussian Noise'' ) is considered, so that for the bit error probability:		In this exercise,  a bipolar binary system with AWGN noise  ("Additive White Gaussian Noise")  is considered,  so that for the bit error probability:
	:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$		:$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$
	Here the following functions are used:		Here,  the following functions are used:
	:$$\rm Q (~~\it~~ x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it		:$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it
	x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u}		x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u}
	\hspace{0.05cm},\hspace{1cm}{\rm erfc} ({\it x}) = \frac{\rm 2}{\sqrt{\rm		\hspace{0.05cm},\hspace{1cm}{\rm erfc} ({\it x}) = \frac{\rm 2}{\sqrt{\rm
	\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u		\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*The above equation holds for the threshold  $E = 0$  regardless of the symbol probabilities  $p_{\rm L}$  and  $p_{\rm H}$.		*The above equation holds for the threshold  $E = 0$,  regardless of the symbol probabilities  $p_{\rm L}$  and  $p_{\rm H}$.
	*However, a smaller error probability can be obtained with a different threshold  $E$  if the two occurrence probabilities are different   $(p_{\rm L} ≠ p_{\rm H})$.		*However,  a smaller error probability can be obtained with a different threshold  $E$  if the two occurrence probabilities are different   $(p_{\rm L} ≠ p_{\rm H})$.


	The standard deviation of the noise component is always  $σ_{d} = 0.5 \ \rm V$. The two amplitudes of the detection signal component are fixed at  $±1 \ \rm V$.		The standard deviation of the noise component is always  $σ_{d} = 0.5 \ \rm V$.  The two amplitudes of the detection signal component are fixed at  $±1 \ \rm V$.

	The following symbol probabilities are to be investigated:		The following symbol probabilities are to be investigated:
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	The diagram includes this last set of parameters and the threshold  $E = 0.1 \cdot s_{\rm 0}$.  The probability density function of the detection samples  $d$ is shown.		The diagram includes this last set of parameters and the threshold  $E = 0.1 \cdot s_{\rm 0}$.  The probability density function of the detection samples  $d$  is shown.


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	''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"]].

	*You can use the interactive applet  [[Applets:Complementary_Gaussian_Error_Functions\|Complementary Gaussian Error Functions]]  to determine error probabilities.		*You can use the interactive applet  [[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian Error Functions"]]  to determine error probabilities.

	*The following applies to the derivative of the Q-function:		*The following applies to the derivative of the Q-function:
	:$$\frac{{\rm d\hspace{0.05cm}Q} ({\it x})} {{\rm d}\hspace{0.05cm}x} = \frac{\rm 1}{\sqrt{\rm 2\pi}}		:$$\frac{{\rm d\hspace{0.05cm}Q} ({\it x})} {{\rm d}\hspace{0.05cm}x} = \frac{\rm 1}{\sqrt{\rm 2\pi}}
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	<quiz display=simple>		<quiz display=simple>
	{What is the relationship between the functions  ${\rm Q}(x)$  and  ${\rm erfc}(x)$?		{What is the relationship between the functions  ${\rm Q}(x)$  and  ${\rm erfc}(x)$?
	\|type="[]"}		\|type="()"}
	+${\rm erfc}(x) =2 \cdot {\rm Q}(\sqrt{2} \cdot x)$,		+${\rm erfc}(x) =2 \cdot {\rm Q}(\sqrt{2} \cdot x)$,
	-${\rm erfc}(x) =\sqrt{2} \cdot {\rm Q}(x/\sqrt{2})$,		-${\rm erfc}(x) =\sqrt{2} \cdot {\rm Q}(x/\sqrt{2})$,
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	$E_{\rm opt} \ =\ $ { 0.25 3% } $\ \rm V$		$E_{\rm opt} \ =\ $ { 0.25 3% } $\ \rm V$

	{Let  $E = E_{\rm opt}$. What is the minimum bit error probability with  $p_{\rm L} = 0.88$?		{Let  $E = E_{\rm opt}$.  What is the minimum bit error probability with  $p_{\rm L} = 0.88$?
	\|type="{}"}		\|type="{}"}
	$p_{\rm B,\ min} \ =\ $ { 1.35 3% } $\ \%$		$p_{\rm B,\ min} \ =\ $ { 1.35 3% } $\ \%$