Aufgaben:Exercise 2.5: Ternary Signal Transmission - Revision history

Guenter at 15:19, 3 June 2022

2022-06-03T15:19:27Z

Hwang at 08:25, 24 May 2022

2022-05-24T08:25:28Z

Hwang at 08:24, 24 May 2022

2022-05-24T08:24:08Z

Guenter at 15:19, 17 May 2022

2022-05-17T15:19:59Z

Guenter at 14:41, 17 May 2022

2022-05-17T14:41:19Z

Guenter at 14:19, 17 May 2022

2022-05-17T14:19:49Z

Guenter at 14:14, 17 May 2022

2022-05-17T14:14:11Z

Hwang at 15:49, 16 May 2022

2022-05-16T15:49:03Z

Hwang at 14:00, 19 April 2022

2022-04-19T14:00:38Z

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Guenter at 12:57, 4 April 2022

2022-04-04T12:57:41Z

← Older revision		Revision as of 15:19, 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:

← Older revision		Revision as of 08:25, 24 May 2022
Line 91:		Line 91:


−	'''(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:

@@ Line 142: / Line 142: @@
 *While for equally probable ternary symbols
-:*the entropy is&nbsp; $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary symbol}$
+:*the entropy is&nbsp; $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary \ symbol}$
 :*from which the equivalent bit rate can be calculated according to&nbsp; $R_{\rm B} = H/T$,&nbsp;
 :*here applies with probabilities&nbsp; $p_0 = 0.2$&nbsp; and&nbsp; $p_{&ndash;} = p_+ = 0.4$:

@@ Line 141: / Line 141: @@
 *However,&nbsp; redundancy-free coding is no longer present,&nbsp; 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&nbsp; $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary symbol}$
-*The equivalent bit rate&nbsp; (the information flow)&nbsp; $R_{\rm B} = H/T$,&nbsp; with probabilities $p_0 = 0.2$ and $p_{&ndash;} = p_+ = 0.4$:
+:*from which the equivalent bit rate can be calculated according to&nbsp; $R_{\rm B} = H/T$,&nbsp;
-:$$H  \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)=  0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm
+:*here applies with probabilities&nbsp; $p_0 = 0.2$&nbsp; and&nbsp; $p_{&ndash;} = p_+ = 0.4$:
 bit/ternary \ symbol}}
   \hspace{0.05cm}.$$

@@ Line 55: / Line 55: @@
 $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="{}"}
 ${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 21.7 3% } $\ \%$
@@ Line 70: / Line 70: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; According to the given equation, with $M = 3$ and $\sigma_d/s_0 = 0.25$:
+'''(1)'''&nbsp; According to the given equation,&nbsp; with&nbsp; $M = 3$&nbsp; and&nbsp; $\sigma_d/s_0 = 0.25$:
 :$$p_{\rm S} =
   \frac{ 2  \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot
@@ Line 77: / Line 77: @@
-'''(2)'''&nbsp; When the noise rms value is doubled, the error probability also increases significantly:
+'''(2)'''&nbsp; When the noise rms value is doubled,&nbsp; the error probability increases significantly:
 :$$p_{\rm S} = {4}/{ 3}\cdot {\rm Q}(1)= {4}/{ 3}\cdot 0.1587 \hspace{0.15cm}\underline {\approx 21.2 \,\%}
   \hspace{0.05cm}.$$
-'''(3)'''&nbsp; The two outer symbols are each distorted with probability $p = {\rm Q}(s_0/(2 \cdot \sigma_d)) = 0.1587$.
+'''(3)'''&nbsp; The two outer symbols are each falsified with probability&nbsp; $p = {\rm Q}(s_0/(2 \cdot \sigma_d)) = 0.1587$.
-*The distortion probability of the symbol &nbsp;$0$&nbsp; is twice as large (it is limited by two thresholds).
+*The falsification probability of the symbol &nbsp;$0$&nbsp; is twice as large&nbsp; (it is limited by two thresholds).
-* Considering the individual symbol probabilities, we obtain:
+* Considering the individual symbol probabilities,&nbsp; we obtain:
 :$$p_{\rm S} = {1}/{ 4}\cdot p + {1}/{ 2}\cdot 2p +{1}/{ 4}\cdot p = 1.5 \cdot p  = 1.5 \cdot 0.1587
 \hspace{0.15cm}\underline {\approx
@@ Line 91: / Line 91: @@
-'''(4)'''&nbsp; Since the symbol &nbsp;$0$&nbsp; occurs more frequently and can also be biased in both directions, the thresholds should be shifted outward.
+'''(4)'''&nbsp; Since the symbol &nbsp;$0$&nbsp; occurs more frequently and can also be biased in both directions,&nbsp; 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&nbsp; $E_{\rm +, \ opt}$&nbsp; is obtained from the intersection of the two Gaussian functions shown in the graph.&nbsp; 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&nbsp; '''(4)''']]
 :$$\frac{ 1/2}{ \sqrt{2\pi} \cdot \sigma_d} \cdot  {\rm exp} \left[ - \frac{ E_{\rm +}^2}{2 \cdot \sigma_d^2}\right]
@@ Line 107: / Line 107: @@
-'''(5)'''&nbsp; Using the approximate result from '''(4)''', we obtain:
+'''(5)'''&nbsp; Using the approximate result from subtask&nbsp; '''(4)''',&nbsp; we obtain:
 :$$p_{\rm S} \ = \
   { 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{
@@ Line 113: / Line 113: @@
   \sigma_d}}\right) +{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{
   \sigma_d}}\right)$$
 :$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \ =  { 1}/{2}  \cdot {\rm Q} \left( 2/3 \right)+ {\rm Q} \left( 4/3
   \right)=
@@ Line 119: / Line 122: @@
-[[File:P_ID1329__Dig_A_2_5g.png|right|frame|Optimal thresholds for '''(6)''']]
+'''(6)'''&nbsp;  After a similar calculation as in subtask&nbsp; '''(4)'''&nbsp; we get
 *$E_+ = 1 \, &ndash;0.0673 \ \underline{= 0.327} \approx 1/3$.
 *$E_{&ndash;} = \, &ndash;E_+$ is still valid.
@@ Line 126: / Line 128: @@
-'''(7)'''&nbsp; Similar to the sample solution for subtask '''(5)''', one now obtains:
+'''(7)'''&nbsp; Similar to the solution for subtask&nbsp; '''(5)''',&nbsp; we now obtain:
 :$$p_{\rm S} \ = \ 0.4 \cdot {\rm Q} \left( 4/3 \right)+ 2 \cdot 0.2 \cdot{\rm Q} \left( 2/3
   \right)+0.4 \cdot {\rm Q} \left( 4/3 \right)$$
@@ Line 134: / Line 136: @@
   \hspace{0.05cm}.$$
-Discussion of the result:
+$\text{Discussion of the result:}$
-*Accordingly, there is a smaller symbol error probability ($17.4 \ \%$ versus $21.2 \ \%$) than with equal probability amplitude coefficients.
+*Accordingly,&nbsp; there is a smaller symbol error probability&nbsp; $(17.4 \ \%$ versus $21.2 \ \%)$&nbsp; than with equal probability amplitude coefficients.
 *However, redundancy-free coding is no longer present, even if the amplitude coefficients are statistically independent of each other.
-*While for equally probable ternary symbols the entropy is $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary \ symbol}$ beträgt &nbsp; &rArr; &nbsp; equivalent bit rate (the information flow) $R_{\rm B} = H/T$, with probabilities $p_0 = 0.2$ and $p_{&ndash;} = p_+ = 0.4$:
+*However,&nbsp; redundancy-free coding is no longer present,&nbsp; even if the amplitude coefficients are statistically independent of each other.
 :$$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 is $4 \ \%$ smaller than the maximum possible for $M = 3$.
+*Thus,&nbsp; the equivalent bit rate here is&nbsp; $\approx 4 \ \%$&nbsp; smaller than the maximum possible equivalent bit rate for&nbsp; $M = 3$.
 {{ML-Fuß}}

@@ Line 32: / Line 32: @@
 * You can numerically determine the error probability values according to  our applet &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]].
-* To check the results,&ndash; use the calculation module &nbsp;[[Applets:Fehlerwahrscheinlichkeit|"Symbol error probability of digital communications systems"]].
+* To check your results,&nbsp; use our&nbsp; (German language)&nbsp; SWF applet &nbsp;[[Applets:Fehlerwahrscheinlichkeit|"Symbol error probability of digital communications systems"]].
@@ Line 39: / Line 39: @@
 ===Questions===
 <quiz display=simple>
-{What symbol error probability results with the (normalized) noise rms value &nbsp;$\sigma_d/s_0 = 0.25$&nbsp; for equally probable symbols?
+{What symbol error probability results with the&nbsp; (normalized)&nbsp; noise rms value &nbsp;$\sigma_d/s_0 = 0.25$&nbsp; for equally probable symbols?
 |type="{}"}
 $p_0 = 1/3, \ \sigma_d = 0.25 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 3 3% } $\ \%$
@@ Line 55: / Line 55: @@
 $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="{}"}
 ${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ ${ 21.7 3% } $\ \%$

@@ Line 2: / Line 2: @@
 {{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&nbsp; $\rm (PDF)$&nbsp; of a noisy ternary signal]]
-A ternary transmission system &nbsp;$(M = 3)$&nbsp; with the possible amplitude values &nbsp;$-s_0$, &nbsp;$0$&nbsp; and &nbsp;$+s_0$ is considered.
+A ternary transmission system &nbsp;$(M = 3)$&nbsp; with the possible amplitude values &nbsp;$-s_0$, &nbsp; $0$ &nbsp; and &nbsp;$+s_0$&nbsp; is considered.
-*During transmission, additive Gaussian noise with rms value &nbsp;$\sigma_d$ is added to the signal.
+*During transmission,&nbsp; additive Gaussian noise with rms value &nbsp;$\sigma_d$&nbsp; 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 &nbsp;$E_{&ndash;}$&nbsp; and &nbsp;$E_{+}$.
@@ Line 11: / Line 12: @@
 \hspace{0.15cm} p_{\rm +} = {\rm Pr}(+s_0) ={1}/{ 3}\hspace{0.05cm}.$$
-*For the time being, the decision thresholds are centered at &nbsp;$E_{&ndash;} = \, &ndash;s_0/2$ and $E_{+} = +s_0/2$.
+*For the time being,&nbsp; the decision thresholds are centered at &nbsp;$E_{&ndash;} = \, &ndash;s_0/2$ and $E_{+} = +s_0/2$.
-From subtask '''(3)''' on, the symbol probabilities are &nbsp;$p_{&ndash;} = p_+ = 1/4$&nbsp; and &nbsp;$p_0 = 1/2$, as shown in the diagram. For this constellation, the symbol error probability &nbsp;$p_{\rm S}$&nbsp; is to be minimized by varying the decision thresholds &nbsp;$E_{&ndash;}$&nbsp; and &nbsp;$E_+$.&nbsp;
+*For this constellation,&nbsp; the symbol error probability &nbsp;$p_{\rm S}$&nbsp; is to be minimized by varying the decision thresholds &nbsp;$E_{&ndash;}$&nbsp; and &nbsp;$E_+$.&nbsp;
 * For the symbol error probability &nbsp;$p_{\rm S}$&nbsp; of a &nbsp;$M$&ndash;level transmission system
+:*with equally probable input symbols
-''Notes:''
+:*and threshold values exactly in the middle between two adjacent amplitude levels holds:
 :$$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}.$$
-* You can numerically determine the error probability values according to the &nbsp;${\rm Q}$ or &nbsp;${\rm erfc}$ function using the &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]]&nbsp; interaction module.
+* You can numerically determine the error probability values according to  our applet &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]].
 * To check the results, use the calculation module &nbsp;[[Applets:Fehlerwahrscheinlichkeit|"Symbol error probability of digital communications systems"]].

← Older revision		Revision as of 12:57, 4 April 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^2.2 ~~Redundanzfreie Codierung~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^2.2 Redundancy-Free Coding^]]