Difference between revisions of "Aufgaben:Exercise 3.09: Correlation Receiver for Unipolar Signaling"

From LNTwww
m (Text replacement - "”" to """)
 
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Digitalsignalübertragung/Optimale_Empfängerstrategien}}
+
{{quiz-Header|Buchseite=Digital_Signal_Transmission/Optimal_Receiver_Strategies}}
  
[[File:P_ID1464__Dig_A_3_9.png|right|frame|Beispielhafte Korrelationswerte]]
+
[[File:P_ID1464__Dig_A_3_9.png|right|frame|Example correlation values]]
Betrachtet wird die gemeinsame Entscheidung von  $N = 3$  Binärsymbolen (Bit) mittels des Korrelationsempfängers. Die  $M = 8$  möglichen Quellensymbolfolgen  $Q_i$  besitzen alle die gleiche Wahrscheinlichkeit und sie sind durch die folgenden unipolaren Amplitudenkoeffizienten festgelegt:
+
The joint decision of  $N = 3$  binary symbols  ("bits")  by means of the correlation receiver is considered. 
 +
 
 +
The  $M = 8$  possible source symbol sequences  $Q_i$  all have the same probability and they are defined by the following unipolar amplitude coefficients:
 
:$$Q_0 = 000, \hspace{0.15cm}Q_1 = 001,\hspace{0.15cm}Q_2 = 010,\hspace{0.15cm}Q_3 = 011
 
:$$Q_0 = 000, \hspace{0.15cm}Q_1 = 001,\hspace{0.15cm}Q_2 = 010,\hspace{0.15cm}Q_3 = 011
 
  \hspace{0.05cm},\hspace{0.15cm}
 
  \hspace{0.05cm},\hspace{0.15cm}
Line 9: Line 11:
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
Weiter gilt:
+
Further applies:
*Die möglichen Sendesignale  $s_i(t)$  – jeweils mit der Dauer  $3T$  – sind alle rechteckförmig mit Ausnahme von  $s_0(t) \equiv 0$.  
+
*The possible transmitted signals  $s_i(t)$  – each with duration  $3T$  – are all rectangular with the exception of  $s_0(t) \equiv 0$.
*Die Signale  $s_1(t)$,  $s_2(t)$  und  $s_4(t)$  mit nur jeweils einer "$1$" besitzen die Signalenergie  $E_{\rm B}$  (steht für "Energie pro Bit"), während zum Beispiel die Energie von  $s_7(t)$  gleich  $3E_{\rm B}$ ist.
+
 +
*The signals  $s_1(t)$,  $s_2(t)$  and  $s_4(t)$  with only one  "$1$"  each have the signal energy  $E_{\rm B}$  (stands for "energy per bit"),  while for example the energy of  $s_7(t)=3E_{\rm B}$.
  
  
Der Korrelationsempfänger bildet aus dem verrauschten Empfangssignal  $r(t) = s(t) + n(t)$  insgesamt  $2^3 = 8$  Entscheidungsgrößen (Metriken)
+
The correlation receiver forms from the noisy received signal  $r(t) = s(t) + n(t)$  a total of  $2^3 = 8$  decision variables (metrics)
:$$W_i  =  I_i  - {E_i}/{2 }\hspace{0.3cm}{\rm mit}\hspace{0.3cm}
+
:$$W_i  =  I_i  - {E_i}/{2 }\hspace{0.3cm}{\rm with}\hspace{0.3cm}
 
I_i =\int_{0}^{3T} r(t) \cdot s_i(t) \,{\rm d} t
 
I_i =\int_{0}^{3T} r(t) \cdot s_i(t) \,{\rm d} t
 
\hspace{0.3cm}( i = 0,\text{...} , 7)$$
 
\hspace{0.3cm}( i = 0,\text{...} , 7)$$
  
und setzt die Sinkensymbolfolge  $V = Q_j$, falls  $W_j$  größer ist als alle anderen  $W_{i \ne j}$. Damit trifft er eine optimale Entscheidung im Sinne von Maximum–Likelihood.
+
and sets the sink symbol sequence  $V = Q_j$,  if  $W_j$  is larger than all other  $W_{i \ne j}$.  Thus,  it makes an optimal decision in the sense of  "maximum likelihood".
  
 +
In the table,  the  (uncorrected)  correlation values  $I_0, \ \text{...} \ , I_7$  for three different systems differing in terms of noise  $n(t)$  and labeled  $\rm A$,  $\rm B$  or  $\rm C$. 
 +
*One of these columns stands for  "no noise",
 +
*one for  "minor noise",  and
 +
*another one for  "strong noise".
  
In der Tabelle sind die (unkorrigierten) Korrelationswerte  $I_0, \ \text{...} \ , I_7$  für drei verschiedene Systeme angegeben, die sich hinsichtlich der Störungen  $n(t)$  unterscheiden und mit  $\rm A$,  $\rm B$  oder  $\rm C$  bezeichnet werden.
 
*Eine dieser Spalten steht für "keine Störung",
 
*eine für "geringe Störungen" und
 
*eine weitere für "starke Störungen".
 
  
  
Zur Bestimmung der Metriken für die drei Systemvarianten wurde stets die gleiche Quellensymbolfolge gesendet.
 
  
  
 +
Note:
 +
*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimal_Receiver_Strategies|"Optimal Receiver Strategies"]].
  
 +
*The same source symbol sequence was always sent to determine the metrics for the three system variants.
  
 
 
''Hinweis:''
 
*Die Aufgabe gehört zum  Kapitel  [[Digitalsignal%C3%BCbertragung/Optimale_Empf%C3%A4ngerstrategien|Optimale Empfängerstrategien]].
 
 
   
 
   
  
  
  
===Fragebogen===
+
===Question===
 
<quiz display=simple>
 
<quiz display=simple>
{Bei welchem System gibt es keine Störungen &nbsp;$n(t)$? Bei
+
{For which system is there&nbsp; "no noise" &nbsp; &rArr; &nbsp; $n(t)=0$? At
 
|type="()"}
 
|type="()"}
 
- $\rm System \ A$,
 
- $\rm System \ A$,
Line 49: Line 50:
 
- $\rm System \ C$.
 
- $\rm System \ C$.
  
{Welche Quellensymbolfolge &nbsp;$Q_k &#8712; {Q_0, \ \text{...} \ , Q_7}$&nbsp; wurde tatsächtlich gesendet?
+
{Which source symbol sequence &nbsp;$Q_k &#8712; {Q_0, \ \text{...} \ , Q_7}$&nbsp; was actually sent?
 
|type="{}"}
 
|type="{}"}
 
$k \ = \ $ { 2 }
 
$k \ = \ $ { 2 }
  
{Welcher Entscheidungswert &nbsp;$W_j$&nbsp; ist bei System &nbsp;$\rm A$&nbsp; am größten?
+
{Which decision value &nbsp;$W_j$&nbsp; is largest for system &nbsp;$\rm A$?&nbsp;  
 
|type="{}"}
 
|type="{}"}
 
${\rm System \ A} \text{:} \hspace{0.2cm} j \ = \ $ { 2 }
 
${\rm System \ A} \text{:} \hspace{0.2cm} j \ = \ $ { 2 }
  
{Welcher Entscheidungswert &nbsp;$W_j$&nbsp; ist beim System &nbsp;$\rm C$&nbsp; am größten?
+
{Which decision value &nbsp;$W_j$&nbsp; is largest for system &nbsp;$\rm C$?&nbsp;  
 
|type="{}"}
 
|type="{}"}
 
${\rm System \ C} \text{:} \hspace{0.2cm} j \ = \ $ { 6 }
 
${\rm System \ C} \text{:} \hspace{0.2cm} j \ = \ $ { 6 }
  
{Bei welchem System treten die größten Störungen auf? Bei
+
{For which system do the largest noise occur? At
 
|type="()"}
 
|type="()"}
 
- $\rm System \ A$,
 
- $\rm System \ A$,
Line 67: Line 68:
 
+ $\rm System \ C$.
 
+ $\rm System \ C$.
  
{Welche Aussagen gelten unter der Annahme, dass &nbsp;$Q_2$&nbsp; gesendet wurde und der Korrelationsempfänger sich normalerweise auch für &nbsp;$Q_2$&nbsp; entscheidet?
+
{Which statements are valid under the assumption that &nbsp;$Q_2$&nbsp; was sent and the correlation receiver normally chooses &nbsp;$Q_2$&nbsp; as well?
 
|type="[]"}
 
|type="[]"}
+ Die Differenz zwischen &nbsp;$W_2$&nbsp; und dem nächstgrößten Wert &nbsp;$W_{i \ne 2}$&nbsp; ist um so kleiner, je stärker die Störungen sind.
+
+ The difference between &nbsp;$W_2$&nbsp; and the next largest value &nbsp;$W_{i \ne 2}$&nbsp; is smaller the stronger the noise is.
- Wenn es zu einer Verfälschung kommt, dann entscheidet sich der Empfänger am wahrscheinlichsten für die Symbolfolge &nbsp;$Q_6$.
+
- When falsification occurs,&nbsp; the receiver is most likely to decide in favor of the symbol sequence &nbsp;$Q_6$.
+ Die Wahrscheinlichkeiten für fehlerhafte Entscheidungen zugunsten von &nbsp;$Q_0$, &nbsp;$Q_3$&nbsp; bzw. &nbsp;$Q_6$&nbsp; sind gleich.
+
+ The probabilities for erroneous decisions in favor of &nbsp;$Q_0$, &nbsp;$Q_3$&nbsp; and &nbsp;$Q_6$,&nbsp; respectively, are equal.
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 2</u>:  
+
'''(1)'''&nbsp; <u>Solution 2</u>&nbsp; is correct:  
*Beim System &nbsp;$\rm B$&nbsp; treten viermal die Metriken $0$ und viermal die Metriken $1$ auf.  
+
*For system &nbsp;$\rm B$,&nbsp; metrics&nbsp; "$0$"&nbsp; occur four times and metrics&nbsp; "$1$"&nbsp; occur four times.
*Dies weist auf $n(t) = 0$ hin, da sich sonst &ndash; wie bei den Systemen &nbsp;$\rm A$&nbsp; und &nbsp;$\rm C$&nbsp; &ndash; alle $I_i$ unterscheiden müssten.
+
*This points to&nbsp; $n(t) = 0$,&nbsp; otherwise &ndash; as in systems &nbsp;$\rm A$&nbsp; and &nbsp;$\rm C$&nbsp; &ndash; all&nbsp; $I_i$&nbsp; would have to differ.
  
  
'''(2)'''&nbsp; Beim System &nbsp;$\rm B$&nbsp; ergeben sich folgende Entscheidungswerte $W_i = I_i \ &ndash; E_i/2$, jeweils normiert auf $E_{\rm B}$:
+
'''(2)'''&nbsp; For system &nbsp;$\rm B$,&nbsp; the decision values&nbsp; $W_i = I_i \ - E_i/2$,&nbsp; each normalized to&nbsp; $E_{\rm B}$,&nbsp; are as follows:
 
:$$W_0 = 0 - 0 = 0, \hspace{0.2cm}W_1 = 0 - 0.5 = -0.5
 
:$$W_0 = 0 - 0 = 0, \hspace{0.2cm}W_1 = 0 - 0.5 = -0.5
 
  \hspace{0.05cm},$$
 
  \hspace{0.05cm},$$
Line 92: Line 93:
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
*Der maximale Wert $W_2 = 0.5$ &#8658; $i = 2$.  
+
*The maximum value&nbsp; $W_2 = 0.5$ &nbsp; &#8658; &nbsp; $i = 2$.  
*Der Korrelationsempfänger entscheidet sich also für $V = Q_2$.  
+
*Thus,&nbsp; the correlation receiver decides to use&nbsp; $V = Q_2$.  
*Da keine Störungen auftreten, wurde tatsächtlich auch $Q_2 =$ "$\rm 010$" gesendet &nbsp; &#8658; &nbsp; $\underline { k= 2}$.
+
*Since there is no noise,&nbsp; $Q_2 =$ "$\rm 010$"&nbsp; was indeed also sent &nbsp; &#8658; &nbsp; $\underline { k= 2}$.
  
  
'''(3)'''&nbsp; Für die Entscheidungswerte von System &nbsp;$\rm A$&nbsp; gilt:
+
'''(3)'''&nbsp; For the decision values of system &nbsp;$\rm A$&nbsp; holds:
 
:$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -0.07 - 0.50 = -0.57, $$
 
:$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -0.07 - 0.50 = -0.57, $$
 
:$$W_2 = 1.13 - 0.50 = 0.63, \hspace{0.2cm}W_3 = 1.06 - 1.00 = 0.06 \hspace{0.05cm},$$
 
:$$W_2 = 1.13 - 0.50 = 0.63, \hspace{0.2cm}W_3 = 1.06 - 1.00 = 0.06 \hspace{0.05cm},$$
Line 103: Line 104:
 
:$$W_6 = 1.18 - 1.00 = 0.18, \hspace{0.2cm}W_7 = 1.11 - 1.50  = -0.39 \hspace{0.05cm}.$$
 
:$$W_6 = 1.18 - 1.00 = 0.18, \hspace{0.2cm}W_7 = 1.11 - 1.50  = -0.39 \hspace{0.05cm}.$$
  
*Das Maximum ist $W_j = W_2$ &nbsp; &#8658; &nbsp; $\underline { j= 2}$.  
+
*The maximum is&nbsp; = $W_j = W_2$ &nbsp; &#8658; &nbsp; $\underline { j= 2}$.  
*Das heißt, dass der Korrelationsempfänger auch bei System &nbsp;$\rm A$&nbsp; die richtige Entscheidung $V = Q_2$ trifft.
+
*This means that the correlation receiver also makes the correct decision&nbsp; $V = Q_2$&nbsp; for system &nbsp;$\rm A$.&nbsp;  
*Ohne den Korrekturterm $(&ndash; E_i/2)$ hätte der Empfänger allerdings die falsche Entscheidung $V = Q_6$ getroffen.
+
*However,&nbsp; without the correction term&nbsp; $(&ndash; E_i/2)$,&nbsp; the receiver would have made the wrong decision&nbsp; $V = Q_6$.
  
  
'''(4)'''&nbsp; Der Korrelationsempfänger &nbsp;$\rm C$&nbsp; hat folgende Werte zu vergleichen:
+
'''(4)'''&nbsp; The correlation receiver &nbsp;$\rm C$&nbsp; has to compare the following values:
 
:$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -1.31 - 0.50 =
 
:$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -1.31 - 0.50 =
 
  -1.81
 
  -1.81
Line 122: Line 123:
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
Die Maximierung ergibt hier $\underline {j = 6}$ &nbsp; &#8658; &nbsp; $V = Q_6$.  
+
The maximization here gives&nbsp; $\underline {j = 6}$ &nbsp; &#8658; &nbsp; $V = Q_6$.  
*Da aber $Q_2$ gesendet wurde, entscheidet hier der Korrelationsempfänger falsch.  
+
*But since&nbsp; $Q_2$&nbsp; was sent,&nbsp; the correlation receiver decides wrong here.
*Die Störungen sind zu stark.
+
*The noise is too strong.
  
  
  
'''(5)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:  
+
'''(5)'''&nbsp; <u>Solution 3</u>&nbsp; is correct:  
*Die Störungen sind bei System &nbsp;$\rm C$&nbsp; am größten und für die aktuellen Empfangswerte sogar so groß, dass der Korrelationsempfänger eine Fehlentscheidung trifft.
+
*The noise is greatest for system &nbsp;$\rm C$&nbsp; and is even so great for the current received values that the correlation receiver makes an incorrect decision.
 +
 
 +
 
 +
 
 +
'''(6)'''&nbsp; <u>Statements 1 and 3</u> are correct:
 +
*In the error-free case&nbsp; $($system &nbsp;$\rm B)$,&nbsp; the difference between&nbsp; $W_2 = 0.5$&nbsp; and the next largest values&nbsp; $W_0 = W_3 = W_6 = 0$&nbsp; is equal to&nbsp; $D_{\hspace{0.02cm}\rm min} =0.5$ in each case.
 +
 
 +
*In system &nbsp;$\rm A$&nbsp; (light noise),&nbsp; the difference between&nbsp; $W_2 = 0.63$&nbsp; and the next largest value&nbsp; $W_6 = 0.18$&nbsp; is still&nbsp; $D_{\hspace{0.02cm}\rm min} = 0.45$.
 +
 +
*If the noise power is increased by factor&nbsp; $50$,&nbsp; the correlation receiver still decides correctly,&nbsp; but then the minimum difference&nbsp; $D_{\hspace{0.02cm}\rm min} = 0.16$&nbsp; is significantly smaller.
  
 +
*For system &nbsp;$\rm C$,&nbsp; where the correlation receiver is overcharged &nbsp; &rArr; &nbsp; subtask&nbsp; '''(4)''',&nbsp; a noise power larger by a factor of&nbsp; $400$&nbsp; compared to system &nbsp;$\rm A$&nbsp; was used as a basis.
  
 +
*If the correlation receiver decides the transmitted sequence&nbsp; $Q_2$&nbsp; incorrectly,&nbsp; a falsification to the sequences&nbsp; $Q_0$,&nbsp; $Q_3$&nbsp; resp.&nbsp; $Q_6$&nbsp; is most likely,&nbsp; <br>since all these three sequences differ from&nbsp; $Q_2$&nbsp; only in one bit each.
  
'''(6)'''&nbsp; Richtig sind die <u>Aussagen 1 und 3</u>:
+
*The fact that&nbsp; $W_6$&nbsp; is always larger than&nbsp; $W_0$&nbsp; or&nbsp; $W_3$&nbsp; in the described simulation is "coincidence" and should not be overinterpreted.
*Im fehlerfreien Fall (System &nbsp;$\rm B$) ist die Differenz zwischen $W_2 = 0.5$ und den nächstgrößten Werten $W_0 = W_3 = W_6 = 0$ jeweils gleich $D_{\hspace{0.02cm}\rm min} =0.5$.
 
*Bei System &nbsp;$\rm A$&nbsp; (leichte Störungen) ist die Differenz zwischen $W_2 = 0.63$ und dem nächstgrößeren Wert $W_6 = 0.18$ immerhin noch $D_{\hspace{0.02cm}\rm min} = 0.45$.
 
*Erhöht man die Rauschleistung um den Faktor $50$, so entscheidet der Korrelationsempfänger immer noch richtig, doch ist dann die minimale Differenz $D_{\hspace{0.02cm}\rm min} = 0.16$ deutlich kleiner.
 
*Für das System &nbsp;$\rm C$, bei dem der Korrelationsempfänger überfordert ist &nbsp; &rArr; &nbsp; Teilaufgabe '''(4)''', wurde eine gegenüber dem System &nbsp;$\rm A$&nbsp; um den Faktor 400 größere Rauschleistung zugrundegelegt.
 
*Entscheidet der Korrelationsempfänger die gesendete Folge $Q_2$ falsch, so ist eine Verfälschung zu den Folgen $Q_0$, $Q_3$ bzw. $Q_6$ am wahrscheinlichsten, da sich alle diese drei Folgen von $Q_2$ nur jeweils in einem Bit unterscheiden.
 
*Dass bei der beschriebenen Simulation $W_6$ stets größer ist als $W_0$ bzw. $W_3$, ist "Zufall" und sollte nicht überinterpretiert werden.  
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Digital Signal Transmission: Exercises|^3.7 Optimale Empfängerstrategien^]]
+
[[Category:Digital Signal Transmission: Exercises|^3.7 Optimal Receiver Strategies^]]

Latest revision as of 17:55, 30 June 2022

Example correlation values

The joint decision of  $N = 3$  binary symbols  ("bits")  by means of the correlation receiver is considered. 

The  $M = 8$  possible source symbol sequences  $Q_i$  all have the same probability and they are defined by the following unipolar amplitude coefficients:

$$Q_0 = 000, \hspace{0.15cm}Q_1 = 001,\hspace{0.15cm}Q_2 = 010,\hspace{0.15cm}Q_3 = 011 \hspace{0.05cm},\hspace{0.15cm} Q_4 = 100, \hspace{0.15cm}Q_5 = 101,\hspace{0.15cm}Q_6 = 110,\hspace{0.15cm}Q_7 = 111 \hspace{0.05cm}.$$

Further applies:

  • The possible transmitted signals  $s_i(t)$  – each with duration  $3T$  – are all rectangular with the exception of  $s_0(t) \equiv 0$.
  • The signals  $s_1(t)$,  $s_2(t)$  and  $s_4(t)$  with only one  "$1$"  each have the signal energy  $E_{\rm B}$  (stands for "energy per bit"),  while for example the energy of  $s_7(t)=3E_{\rm B}$.


The correlation receiver forms from the noisy received signal  $r(t) = s(t) + n(t)$  a total of  $2^3 = 8$  decision variables (metrics)

$$W_i = I_i - {E_i}/{2 }\hspace{0.3cm}{\rm with}\hspace{0.3cm} I_i =\int_{0}^{3T} r(t) \cdot s_i(t) \,{\rm d} t \hspace{0.3cm}( i = 0,\text{...} , 7)$$

and sets the sink symbol sequence  $V = Q_j$,  if  $W_j$  is larger than all other  $W_{i \ne j}$.  Thus,  it makes an optimal decision in the sense of  "maximum likelihood".

In the table,  the  (uncorrected)  correlation values  $I_0, \ \text{...} \ , I_7$  for three different systems differing in terms of noise  $n(t)$  and labeled  $\rm A$,  $\rm B$  or  $\rm C$. 

  • One of these columns stands for  "no noise",
  • one for  "minor noise",  and
  • another one for  "strong noise".



Note:

  • The same source symbol sequence was always sent to determine the metrics for the three system variants.



Question

1

For which system is there  "no noise"   ⇒   $n(t)=0$? At

$\rm System \ A$,
$\rm System \ B$,
$\rm System \ C$.

2

Which source symbol sequence  $Q_k ∈ {Q_0, \ \text{...} \ , Q_7}$  was actually sent?

$k \ = \ $

3

Which decision value  $W_j$  is largest for system  $\rm A$? 

${\rm System \ A} \text{:} \hspace{0.2cm} j \ = \ $

4

Which decision value  $W_j$  is largest for system  $\rm C$? 

${\rm System \ C} \text{:} \hspace{0.2cm} j \ = \ $

5

For which system do the largest noise occur? At

$\rm System \ A$,
$\rm System \ B$,
$\rm System \ C$.

6

Which statements are valid under the assumption that  $Q_2$  was sent and the correlation receiver normally chooses  $Q_2$  as well?

The difference between  $W_2$  and the next largest value  $W_{i \ne 2}$  is smaller the stronger the noise is.
When falsification occurs,  the receiver is most likely to decide in favor of the symbol sequence  $Q_6$.
The probabilities for erroneous decisions in favor of  $Q_0$,  $Q_3$  and  $Q_6$,  respectively, are equal.


Solution

(1)  Solution 2  is correct:

  • For system  $\rm B$,  metrics  "$0$"  occur four times and metrics  "$1$"  occur four times.
  • This points to  $n(t) = 0$,  otherwise – as in systems  $\rm A$  and  $\rm C$  – all  $I_i$  would have to differ.


(2)  For system  $\rm B$,  the decision values  $W_i = I_i \ - E_i/2$,  each normalized to  $E_{\rm B}$,  are as follows:

$$W_0 = 0 - 0 = 0, \hspace{0.2cm}W_1 = 0 - 0.5 = -0.5 \hspace{0.05cm},$$
$$W_2 = 1 - 0.5 = 0.5, \hspace{0.2cm}W_3 = 1 - 1 = 0 \hspace{0.05cm},$$
$$W_4 = 0 - 0.5 = -0.5, \hspace{0.2cm}W_5 = 0 - 1 = -1 \hspace{0.05cm}.$$
$$W_6 = 1 - 1 = 0, \hspace{0.2cm}W_7 = 1 - 1.5 = -0.5 \hspace{0.05cm}.$$
  • The maximum value  $W_2 = 0.5$   ⇒   $i = 2$.
  • Thus,  the correlation receiver decides to use  $V = Q_2$.
  • Since there is no noise,  $Q_2 =$ "$\rm 010$"  was indeed also sent   ⇒   $\underline { k= 2}$.


(3)  For the decision values of system  $\rm A$  holds:

$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -0.07 - 0.50 = -0.57, $$
$$W_2 = 1.13 - 0.50 = 0.63, \hspace{0.2cm}W_3 = 1.06 - 1.00 = 0.06 \hspace{0.05cm},$$
$$W_4 = 0.05 - 0.50 = -0.45, \hspace{0.2cm}W_5 = -0.02 - 1.00 = -1.02\hspace{0.05cm},$$
$$W_6 = 1.18 - 1.00 = 0.18, \hspace{0.2cm}W_7 = 1.11 - 1.50 = -0.39 \hspace{0.05cm}.$$
  • The maximum is  = $W_j = W_2$   ⇒   $\underline { j= 2}$.
  • This means that the correlation receiver also makes the correct decision  $V = Q_2$  for system  $\rm A$. 
  • However,  without the correction term  $(– E_i/2)$,  the receiver would have made the wrong decision  $V = Q_6$.


(4)  The correlation receiver  $\rm C$  has to compare the following values:

$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -1.31 - 0.50 = -1.81 \hspace{0.05cm},$$
$$W_2 = 3.59 - 0.50 = 3.09, \hspace{0.2cm}W_3 = 2.28 - 1.00 = 1.28 \hspace{0.05cm},$$
$$W_4 = 0.97 - 0.50 = 0.47, \hspace{0.2cm}W_5 = -0.34 - 1.00 = -1.34 \hspace{0.05cm},$$
$$W_6 = 4.56 - 1.00 = 3.56, \hspace{0.2cm}W_7 = 3.25 - 1.50 = 1.75 \hspace{0.05cm}.$$

The maximization here gives  $\underline {j = 6}$   ⇒   $V = Q_6$.

  • But since  $Q_2$  was sent,  the correlation receiver decides wrong here.
  • The noise is too strong.


(5)  Solution 3  is correct:

  • The noise is greatest for system  $\rm C$  and is even so great for the current received values that the correlation receiver makes an incorrect decision.


(6)  Statements 1 and 3 are correct:

  • In the error-free case  $($system  $\rm B)$,  the difference between  $W_2 = 0.5$  and the next largest values  $W_0 = W_3 = W_6 = 0$  is equal to  $D_{\hspace{0.02cm}\rm min} =0.5$ in each case.
  • In system  $\rm A$  (light noise),  the difference between  $W_2 = 0.63$  and the next largest value  $W_6 = 0.18$  is still  $D_{\hspace{0.02cm}\rm min} = 0.45$.
  • If the noise power is increased by factor  $50$,  the correlation receiver still decides correctly,  but then the minimum difference  $D_{\hspace{0.02cm}\rm min} = 0.16$  is significantly smaller.
  • For system  $\rm C$,  where the correlation receiver is overcharged   ⇒   subtask  (4),  a noise power larger by a factor of  $400$  compared to system  $\rm A$  was used as a basis.
  • If the correlation receiver decides the transmitted sequence  $Q_2$  incorrectly,  a falsification to the sequences  $Q_0$,  $Q_3$  resp.  $Q_6$  is most likely, 
    since all these three sequences differ from  $Q_2$  only in one bit each.
  • The fact that  $W_6$  is always larger than  $W_0$  or  $W_3$  in the described simulation is "coincidence" and should not be overinterpreted.