Aufgaben:Exercise 2.5: Ternary Signal Transmission - Revision history

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	:*from which the equivalent bit rate can be calculated according to  $R_{\rm B} = H/T$,		:*from which the equivalent bit rate can be calculated according to  $R_{\rm B} = H/T$,
	:*here applies with probabilities  $p_0 = 0.2$  and  $p_{–} = p_+ = 0.4$:		:*here applies with probabilities  $p_0 = 0.2$  and  $p_{–} = p_+ = 0.4$:
	::$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\~~rmbit~~/ternary \ symbol}}\hspace{0.05cm}.$$		::$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm bit/ternary \ symbol}}\hspace{0.05cm}.$$

	*Thus,  the equivalent bit rate here is  $\approx 4 \ \%$  smaller than the maximum possible equivalent bit rate for  $M = 3$.		*Thus,  the equivalent bit rate here is  $\approx 4 \ \%$  smaller than the maximum possible equivalent bit rate for  $M = 3$.
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	[[Category:Digital Signal Transmission: Exercises\|^2.2 Redundancy-Free Coding^]]		[[Category:Digital Signal Transmission: Exercises\|^2.2 Redundancy-Free Coding^]]
	[[de:Aufgaben:2.~~5_Ternäre_Signalübertragung~~]]		[[de:Aufgaben:Aufgabe 2.5: Ternäre Signalübertragung]]

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	*First, the occurrence probabilities of the three input symbols are assumed to be equally probable:		*First, the occurrence probabilities of the three input symbols are assumed to be equally probable:
	:$$p_{\rm -} = {\rm Pr}(-s_0) = {1}/{ 3}, \hspace{0.15cm} p_{\rm 0} = {\rm Pr}(0) = {1}/{ 3},		:$$p_{\rm -} = {\rm Pr}(-s_0) = {1}/{ 3}, \hspace{0.15cm} p_{\rm 0} = {\rm Pr}(0) = {1}/{ 3},\hspace{0.15cm} p_{\rm +} = {\rm Pr}(+s_0) ={1}/{ 3}\hspace{0.05cm}.$$
	\hspace{0.15cm} p_{\rm +} = {\rm Pr}(+s_0) ={1}/{ 3}\hspace{0.05cm}.$$

	*For the time being,  the decision thresholds are centered at  $E_{–} = \, –s_0/2$ and $E_{+} = +s_0/2$.		*For the time being,  the decision thresholds are centered at  $E_{–} = \, –s_0/2$ and $E_{+} = +s_0/2$.
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	:*with equally probable input symbols		:*with equally probable input symbols
	:*and threshold values exactly in the middle between two adjacent amplitude levels holds:		:*and threshold values exactly in the middle between two adjacent amplitude levels holds:
	:$$p_{\rm S} =		:$$p_{\rm S} =\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot \sigma_d}}\right)\hspace{0.05cm}.$$
	\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot \sigma_d}}\right)
	\hspace{0.05cm}.$$
	* You can numerically determine the error probability values according to our applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]].		* You can numerically determine the error probability values according to our applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]].

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  According to the given equation,  with  $M = 3$  and  $\sigma_d/s_0 = 0.25$:		'''(1)'''  According to the given equation,  with  $M = 3$  and  $\sigma_d/s_0 = 0.25$:
	:$$p_{\rm S} =		:$$p_{\rm S} =\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot\sigma_d}}\right)= {4}/{ 3}\cdot {\rm Q}(2) ={4}/{ 3}\cdot 0.0228\hspace{0.15cm}\underline {\approx 3 \,\%}\hspace{0.05cm}.$$
	\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot
	\sigma_d}}\right)= {4}/{ 3}\cdot {\rm Q}(2) ={4}/{ 3}\cdot 0.0228\hspace{0.15cm}\underline {\approx 3 \,\%}
	\hspace{0.05cm}.$$


	'''(2)'''  When the noise rms value is doubled,  the error probability increases significantly:		'''(2)'''  When the noise rms value is doubled,  the error probability increases significantly:
	:$$p_{\rm S} = {4}/{ 3}\cdot {\rm Q}(1)= {4}/{ 3}\cdot 0.1587 \hspace{0.15cm}\underline {\approx 21.2 \,\%}		:$$p_{\rm S} = {4}/{ 3}\cdot {\rm Q}(1)= {4}/{ 3}\cdot 0.1587 \hspace{0.15cm}\underline {\approx 21.2 \,\%}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	*The falsification probability of the symbol  $0$  is twice as large  (it is limited by two thresholds).		*The falsification probability of the symbol  $0$  is twice as large  (it is limited by two thresholds).
	* Considering the individual symbol probabilities,  we obtain:		* Considering the individual symbol probabilities,  we obtain:
	:$$p_{\rm S} = {1}/{ 4}\cdot p + {1}/{ 2}\cdot 2p +{1}/{ 4}\cdot p = 1.5 \cdot p = 1.5 \cdot 0.1587		:$$p_{\rm S} = {1}/{ 4}\cdot p + {1}/{ 2}\cdot 2p +{1}/{ 4}\cdot p = 1.5 \cdot p = 1.5 \cdot 0.1587\hspace{0.15cm}\underline {\approx23.8 \,\%}\hspace{0.05cm}.$$
	\hspace{0.15cm}\underline {\~~approx~~
	23.8 \,\%}
	\hspace{0.05cm}.$$


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	[[File:P_ID1328__Dig_A_2_5e.png\|right\|frame\|Optimal thresholds for subtask  '''(4)''']]		[[File:P_ID1328__Dig_A_2_5e.png\|right\|frame\|Optimal thresholds for subtask  '''(4)''']]

	:$$\frac{ 1/2}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ E_{\rm +}^2}{2 \cdot \sigma_d^2}\right]		:$$\frac{ 1/2}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ E_{\rm +}^2}{2 \cdot \sigma_d^2}\right]= \frac{ 1/4}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ (s_0 -E_{\rm +})^2}{2 \cdot \sigma_d^2}\right]$$
	= \frac{ 1/4}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ (s_0 -E_{\rm +})^2}{2 \cdot \sigma_d^2}\right]$$		:$$\Rightarrow \hspace{0.3cm} {\rm exp} \left[ \frac{ (s_0 -E_{\rm +})^2 - E_{\rm +}^2}{2 \cdot\sigma_d^2}\right]= {1}/{ 2}\Rightarrow \hspace{0.3cm} {\rm exp} \left[ \frac{ 1 -2 \cdot E_{\rm +}/s_0}{2 \cdot\sigma_d^2/s_0^2}\right]= {1}/{ 2}$$
	:$$\Rightarrow \hspace{0.3cm} {\rm exp} \left[ \frac{ (s_0 -E_{\rm +})^2 - E_{\rm +}^2}{2 \cdot		:$$\Rightarrow \hspace{0.3cm}\frac{ E_{\rm +}}{s_0}= \frac{1}{ 2}+ \frac{\sigma_d^2} {s_0^2} \cdot {\rm ln}(2)\hspace{0.15cm}\underline {=0.673}\hspace{0.15cm}\approx{2}/ {3} \hspace{0.05cm}.$$
	\sigma_d^2}\right]= {1}/{ 2}
	\Rightarrow \hspace{0.3cm} {\rm exp} \left[ \frac{ 1 -2 \cdot E_{\rm +}/s_0}{2 \cdot
	\sigma_d^2/s_0^2}\right]= {1}/{ 2}$$
	:$$\Rightarrow \hspace{0.3cm}\frac{ E_{\rm +}}{s_0}= \frac{1}
	{ 2}+ \frac{\sigma_d^2} {s_0^2} \cdot {\rm ln}(2)\hspace{0.15cm}\underline {=0.673}\hspace{0.15cm}\approx
	{2}/ {3} \hspace{0.05cm}.$$


	'''(5)'''  Using the approximate result from subtask  '''(4)''',  we obtain:		'''(5)'''  Using the approximate result from subtask  '''(4)''',  we obtain:
	:$$p_{\rm S} \ = \		:$$p_{\rm S} \ = \{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{\sigma_d}}\right)+ 2 \cdot { 1}/{2} \cdot {\rm Q} \left( {\frac{2s_0/3}{\sigma_d}}\right) +{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{\sigma_d}}\right)$$
	{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{
	\sigma_d}}\right)+ 2 \cdot { 1}/{2} \cdot {\rm Q} \left( {\frac{2s_0/3}{
	\sigma_d}}\right) +{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{
	\sigma_d}}\right)$$
	[[File:P_ID1329__Dig_A_2_5g.png\|right\|frame\|Optimal thresholds for subtask  '''(6)''']]		[[File:P_ID1329__Dig_A_2_5g.png\|right\|frame\|Optimal thresholds for subtask  '''(6)''']]


	:$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \ = { 1}/{2} \cdot {\rm Q} \left( 2/3 \right)+ {\rm Q} \left( 4/3		:$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \ = { 1}/{2} \cdot {\rm Q} \left( 2/3 \right)+ {\rm Q} \left( 4/3\right)={ 1}/{2} \cdot 0.251 + 0.092 \hspace{0.15cm}\underline {\approx 21.7 \,\%}\hspace{0.05cm}.$$
	\right)=
	{ 1}/{2} \cdot 0.251 + 0.092 \hspace{0.15cm}\underline {\approx 21.7 \,\%}
	\hspace{0.05cm}.$$


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	'''(7)'''  Similar to the solution for subtask  '''(5)''',  we now obtain:		'''(7)'''  Similar to the solution for subtask  '''(5)''',  we now obtain:
	:$$p_{\rm S} \ = \ 0.4 \cdot {\rm Q} \left( 4/3 \right)+ 2 \cdot 0.2 \cdot{\rm Q} \left( 2/3		:$$p_{\rm S} \ = \ 0.4 \cdot {\rm Q} \left( 4/3 \right)+ 2 \cdot 0.2 \cdot{\rm Q} \left( 2/3\right)+0.4 \cdot {\rm Q} \left( 4/3 \right)$$
	\right)+0.4 \cdot {\rm Q} \left( 4/3 \right)$$		:$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \0.4 \cdot (0.092 + 0.251 + 0.092)\hspace{0.15cm}\underline {\approx 17.4 \,\%}\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \
	0.4 \cdot (0.092 + 0.251 + 0.092)
	\hspace{0.15cm}\underline {\approx 17.4 \,\%}
	\hspace{0.05cm}.$$

	$\text{Discussion of the result:}$		$\text{Discussion of the result:}$
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	:*from which the equivalent bit rate can be calculated according to  $R_{\rm B} = H/T$,		:*from which the equivalent bit rate can be calculated according to  $R_{\rm B} = H/T$,
	:*here applies with probabilities  $p_0 = 0.2$  and  $p_{–} = p_+ = 0.4$:		:*here applies with probabilities  $p_0 = 0.2$  and  $p_{–} = p_+ = 0.4$:
	::$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm		::$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rmbit/ternary \ symbol}}\hspace{0.05cm}.$$
	~~bit~~/ternary \ symbol}}
	\hspace{0.05cm}.$$

	*Thus,  the equivalent bit rate here is  $\approx 4 \ \%$  smaller than the maximum possible equivalent bit rate for  $M = 3$.		*Thus,  the equivalent bit rate here is  $\approx 4 \ \%$  smaller than the maximum possible equivalent bit rate for  $M = 3$.

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	[[Category:Digital Signal Transmission: Exercises\|^2.2 Redundancy-Free Coding^]]		[[Category:Digital Signal Transmission: Exercises\|^2.2 Redundancy-Free Coding^]]
			[[de:Aufgaben:2.5_Ternäre_Signalübertragung]]

Guenter at 15:19, 3 June 2022

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← Older revision		Revision as of 17:19, 3 June 2022
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	'''(4)'''  Since the symbol  $0$  occurs more frequently and can also be ~~biased '''KORREKTUR:~~ falsified~~'''~~ in both directions,  the thresholds should be shifted outward.		'''(4)'''  Since the symbol  $0$  occurs more frequently and can also be falsified in both directions,  the thresholds should be shifted outward.
	*The optimal decision threshold  $E_{\rm +, \ opt}$  is obtained from the intersection of the two Gaussian functions shown in the graph.  It must hold:		*The optimal decision threshold  $E_{\rm +, \ opt}$  is obtained from the intersection of the two Gaussian functions shown in the graph.  It must hold:

Hwang at 08:25, 24 May 2022

2022-05-24T08:25:28Z

← Older revision		Revision as of 10:25, 24 May 2022
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	'''(4)'''  Since the symbol  $0$  occurs more frequently and can also be biased in both directions,  the thresholds should be shifted outward.		'''(4)'''  Since the symbol  $0$  occurs more frequently and can also be biased '''KORREKTUR: falsified''' in both directions,  the thresholds should be shifted outward.
	*The optimal decision threshold  $E_{\rm +, \ opt}$  is obtained from the intersection of the two Gaussian functions shown in the graph.  It must hold:		*The optimal decision threshold  $E_{\rm +, \ opt}$  is obtained from the intersection of the two Gaussian functions shown in the graph.  It must hold:

Hwang at 08:24, 24 May 2022

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← Older revision		Revision as of 10:24, 24 May 2022
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	*While for equally probable ternary symbols		*While for equally probable ternary symbols
	:*the entropy is  $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary symbol}$		:*the entropy is  $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary \ symbol}$
	:*from which the equivalent bit rate can be calculated according to  $R_{\rm B} = H/T$,		:*from which the equivalent bit rate can be calculated according to  $R_{\rm B} = H/T$,
	:*here applies with probabilities  $p_0 = 0.2$  and  $p_{–} = p_+ = 0.4$:		:*here applies with probabilities  $p_0 = 0.2$  and  $p_{–} = p_+ = 0.4$:

Guenter at 15:19, 17 May 2022

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← Older revision		Revision as of 17:19, 17 May 2022
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	*However,  redundancy-free coding is no longer present,  even if the amplitude coefficients are statistically independent of each other.		*However,  redundancy-free coding is no longer present,  even if the amplitude coefficients are statistically independent of each other.

	*While for equally probable ternary symbols the entropy is $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary \ symbol}$.		*While for equally probable ternary symbols
			:*the entropy is  $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary symbol}$
	*~~The~~ equivalent bit rate~~  (the information flow)~~  $R_{\rm B} = H/T$,  with probabilities $p_0 = 0.2$ and $p_{–} = p_+ = 0.4$:		:*from which the equivalent bit rate can be calculated according to  $R_{\rm B} = H/T$,
	:$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm		:*here applies with probabilities  $p_0 = 0.2$  and  $p_{–} = p_+ = 0.4$:
			::$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm
	bit/ternary \ symbol}}		bit/ternary \ symbol}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

Guenter at 14:41, 17 May 2022

2022-05-17T14:41:19Z

← Older revision		Revision as of 16:41, 17 May 2022
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	$p_0 = 1/2, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} E_{\rm +, \ opt} \ = \ ${ 0.673 3% }		$p_0 = 1/2, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} E_{\rm +, \ opt} \ = \ ${ 0.673 3% }

	{What is the error probability for optimal thresholds?		{What is the symbol error probability for optimal thresholds?
	\|type="{}"}		\|type="{}"}
	${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 21.7 3% } $\ \%$		${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 21.7 3% } $\ \%$
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  According to the given equation, with $M = 3$ and $\sigma_d/s_0 = 0.25$:		'''(1)'''  According to the given equation,  with  $M = 3$  and  $\sigma_d/s_0 = 0.25$:
	:$$p_{\rm S} =		:$$p_{\rm S} =
	\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot		\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot
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	'''(2)'''  When the noise rms value is doubled, the error probability ~~also~~ increases significantly:		'''(2)'''  When the noise rms value is doubled,  the error probability increases significantly:
	:$$p_{\rm S} = {4}/{ 3}\cdot {\rm Q}(1)= {4}/{ 3}\cdot 0.1587 \hspace{0.15cm}\underline {\approx 21.2 \,\%}		:$$p_{\rm S} = {4}/{ 3}\cdot {\rm Q}(1)= {4}/{ 3}\cdot 0.1587 \hspace{0.15cm}\underline {\approx 21.2 \,\%}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(3)'''  The two outer symbols are each ~~distorted~~ with probability $p = {\rm Q}(s_0/(2 \cdot \sigma_d)) = 0.1587$.		'''(3)'''  The two outer symbols are each falsified with probability  $p = {\rm Q}(s_0/(2 \cdot \sigma_d)) = 0.1587$.
	*The ~~distortion~~ probability of the symbol  $0$  is twice as large (it is limited by two thresholds).		*The falsification probability of the symbol  $0$  is twice as large  (it is limited by two thresholds).
	* Considering the individual symbol probabilities, we obtain:		* Considering the individual symbol probabilities,  we obtain:
	:$$p_{\rm S} = {1}/{ 4}\cdot p + {1}/{ 2}\cdot 2p +{1}/{ 4}\cdot p = 1.5 \cdot p = 1.5 \cdot 0.1587		:$$p_{\rm S} = {1}/{ 4}\cdot p + {1}/{ 2}\cdot 2p +{1}/{ 4}\cdot p = 1.5 \cdot p = 1.5 \cdot 0.1587
	\hspace{0.15cm}\underline {\approx		\hspace{0.15cm}\underline {\approx
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	'''(4)'''  Since the symbol  $0$  occurs more frequently and can also be biased in both directions, the thresholds should be shifted outward.		'''(4)'''  Since the symbol  $0$  occurs more frequently and can also be biased in both directions,  the thresholds should be shifted outward.
	*The optimal decision threshold $E_{\rm +, \ opt}$ is obtained from the intersection of the two Gaussian functions shown in the graph. It must hold:		*The optimal decision threshold  $E_{\rm +, \ opt}$  is obtained from the intersection of the two Gaussian functions shown in the graph.  It must hold:

	[[File:P_ID1328__Dig_A_2_5e.png\|right\|frame\|Optimal thresholds for '''(4)''']]		[[File:P_ID1328__Dig_A_2_5e.png\|right\|frame\|Optimal thresholds for subtask  '''(4)''']]

	:$$\frac{ 1/2}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ E_{\rm +}^2}{2 \cdot \sigma_d^2}\right]		:$$\frac{ 1/2}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ E_{\rm +}^2}{2 \cdot \sigma_d^2}\right]
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	'''(5)'''  Using the approximate result from '''(4)''', we obtain:		'''(5)'''  Using the approximate result from subtask  '''(4)''',  we obtain:
	:$$p_{\rm S} \ = \		:$$p_{\rm S} \ = \
	{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{		{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{
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	\sigma_d}}\right) +{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{		\sigma_d}}\right) +{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{
	\sigma_d}}\right)$$		\sigma_d}}\right)$$
			[[File:P_ID1329__Dig_A_2_5g.png\|right\|frame\|Optimal thresholds for subtask  '''(6)''']]


	:$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \ = { 1}/{2} \cdot {\rm Q} \left( 2/3 \right)+ {\rm Q} \left( 4/3		:$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \ = { 1}/{2} \cdot {\rm Q} \left( 2/3 \right)+ {\rm Q} \left( 4/3
	\right)=		\right)=
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	~~[[File:P_ID1329__Dig_A_2_5g.png\|right\|frame\|Optimal thresholds for '''(6)''']]~~		'''(6)'''  After a similar calculation as in subtask  '''(4)'''  we get
	'''(6)'''  After a similar calculation as in ~~point~~ '''(4)''' we get
	*$E_+ = 1 \, –0.0673 \ \underline{= 0.327} \approx 1/3$.		*$E_+ = 1 \, –0.0673 \ \underline{= 0.327} \approx 1/3$.
	*$E_{–} = \, –E_+$ is still valid.		*$E_{–} = \, –E_+$ is still valid.
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	'''(7)'''  Similar to the ~~sample~~ solution for subtask '''(5)''', ~~one~~ now ~~obtains~~:		'''(7)'''  Similar to the solution for subtask  '''(5)''',  we now obtain:
	:$$p_{\rm S} \ = \ 0.4 \cdot {\rm Q} \left( 4/3 \right)+ 2 \cdot 0.2 \cdot{\rm Q} \left( 2/3		:$$p_{\rm S} \ = \ 0.4 \cdot {\rm Q} \left( 4/3 \right)+ 2 \cdot 0.2 \cdot{\rm Q} \left( 2/3
	\right)+0.4 \cdot {\rm Q} \left( 4/3 \right)$$		\right)+0.4 \cdot {\rm Q} \left( 4/3 \right)$$
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Discussion of the result:		$\text{Discussion of the result:}$
	*Accordingly, there is a smaller symbol error probability ($17.4 \ \%$ versus $21.2 \ \%$) than with equal probability amplitude coefficients.		*Accordingly,  there is a smaller symbol error probability  $(17.4 \ \%$ versus $21.2 \ \%)$  than with equal probability amplitude coefficients.
	*However, redundancy-free coding is no longer present, even if the amplitude coefficients are statistically independent of each other.
	*While for equally probable ternary symbols the entropy is $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary \ symbol}$ ~~beträgt~~   ~~⇒   equivalent bit rate~~ (the information flow) $R_{\rm B} = H/T$, with probabilities $p_0 = 0.2$ and $p_{–} = p_+ = 0.4$:		*However,  redundancy-free coding is no longer present,  even if the amplitude coefficients are statistically independent of each other.

			*While for equally probable ternary symbols the entropy is $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary \ symbol}$.

			*The equivalent bit rate  (the information flow)  $R_{\rm B} = H/T$,  with probabilities $p_0 = 0.2$ and $p_{–} = p_+ = 0.4$:
	:$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm		:$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm
	bit/ternary \ symbol}}		bit/ternary \ symbol}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*Thus, the equivalent bit rate is $4 \ \%$ smaller than the maximum possible for $M = 3$.		*Thus,  the equivalent bit rate here is  $\approx 4 \ \%$  smaller than the maximum possible equivalent bit rate for  $M = 3$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:19, 17 May 2022

2022-05-17T14:19:49Z

← Older revision		Revision as of 16:19, 17 May 2022
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	* You can numerically determine the error probability values according to our applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]].		* You can numerically determine the error probability values according to our applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]].

	* To check ~~the~~ results,&~~ndash~~; use ~~the calculation module~~  [[Applets:Fehlerwahrscheinlichkeit\|"Symbol error probability of digital communications systems"]].		* To check your results,  use our  (German language)  SWF applet  [[Applets:Fehlerwahrscheinlichkeit\|"Symbol error probability of digital communications systems"]].


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{What symbol error probability results with the (normalized) noise rms value  $\sigma_d/s_0 = 0.25$  for equally probable symbols?		{What symbol error probability results with the  (normalized)  noise rms value  $\sigma_d/s_0 = 0.25$  for equally probable symbols?
	\|type="{}"}		\|type="{}"}
	$p_0 = 1/3, \ \sigma_d = 0.25 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 3 3% } $\ \%$		$p_0 = 1/3, \ \sigma_d = 0.25 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 3 3% } $\ \%$
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	$p_0 = 1/2, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} E_{\rm +, \ opt} \ = \ ${ 0.673 3% }		$p_0 = 1/2, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} E_{\rm +, \ opt} \ = \ ${ 0.673 3% }

	{What is the probability ~~of error~~ for optimal thresholds?		{What is the error probability for optimal thresholds?
	\|type="{}"}		\|type="{}"}
	${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 21.7 3% } $\ \%$		${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 21.7 3% } $\ \%$

Guenter at 14:14, 17 May 2022

2022-05-17T14:14:11Z

← Older revision		Revision as of 16:14, 17 May 2022
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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Redundancy-Free_Coding		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Redundancy-Free_Coding
	}}		}}
	[[File:P_ID1327__Dig_A_2_5.png\|right\|frame\|Probability density function of a noisy ternary signal]]		[[File:P_ID1327__Dig_A_2_5.png\|right\|frame\|Probability density function  $\rm (PDF)$  of a noisy ternary signal]]
	A ternary transmission system  $(M = 3)$  with the possible amplitude values  $-s_0$,  $0$  and  $+s_0$ is considered.		A ternary transmission system  $(M = 3)$  with the possible amplitude values  $-s_0$,   $0$   and  $+s_0$  is considered.
	*During transmission, additive Gaussian noise with rms value  $\sigma_d$ is added to the signal.		*During transmission,  additive Gaussian noise with rms value  $\sigma_d$  is added to the signal.

	*The recovery of the three-level digital signal at the receiver is done with the help of two decision thresholds at  $E_{–}$  and  $E_{+}$.		*The recovery of the three-level digital signal at the receiver is done with the help of two decision thresholds at  $E_{–}$  and  $E_{+}$.

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	\hspace{0.15cm} p_{\rm +} = {\rm Pr}(+s_0) ={1}/{ 3}\hspace{0.05cm}.$$		\hspace{0.15cm} p_{\rm +} = {\rm Pr}(+s_0) ={1}/{ 3}\hspace{0.05cm}.$$

	*For the time being, the decision thresholds are centered at  $E_{–} = \, –s_0/2$ and $E_{+} = +s_0/2$.		*For the time being,  the decision thresholds are centered at  $E_{–} = \, –s_0/2$ and $E_{+} = +s_0/2$.

			*From subtask  '''(3)'''  on,  the symbol probabilities are  $p_{–} = p_+ = 1/4$  and  $p_0 = 1/2$,  as shown in the diagram.

	~~From subtask '''(3)''' on~~, ~~the symbol probabilities are  $p_{–} = p_+ = 1/4$  and~~  ~~$p_0 = 1/2$, as shown in the diagram. For this constellation,~~ the symbol error probability  $p_{\rm S}$  is to be minimized by varying the decision thresholds  $E_{–}$  and  $E_+$.		*For this constellation,  the symbol error probability  $p_{\rm S}$  is to be minimized by varying the decision thresholds  $E_{–}$  and  $E_+$.




			Notes:
			* The exercise refers to the chapter  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|"Redundancy-Free Coding"]].

			* For the symbol error probability  $p_{\rm S}$  of a  $M$–level transmission system
			:*with equally probable input symbols
	~~''Notes:''~~		:*and threshold values exactly in the middle between two adjacent amplitude levels holds:
	* The exercise refers to the chapter  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|"Redundancy-Free Coding"]].
	* For the symbol error probability  $p_{\rm S}$  of a  $M$–level ~~message~~ transmission system with equally probable input symbols and threshold values exactly in the middle between two adjacent amplitude levels holds:
	:$$p_{\rm S} =		:$$p_{\rm S} =
	\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot \sigma_d}}\right)		\frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot \sigma_d}}\right)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	* You can numerically determine the error probability values according to ~~the  ${\rm Q}$ or  ${\rm erfc}$ function using the~~  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]]~~  interaction module~~.		* You can numerically determine the error probability values according to our applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]].
	* To check the results, use the calculation module  [[Applets:Fehlerwahrscheinlichkeit\|"Symbol error probability of digital communications systems"]].
			* To check the results,– use the calculation module  [[Applets:Fehlerwahrscheinlichkeit\|"Symbol error probability of digital communications systems"]].