Difference between revisions of "Aufgaben:Exercise 5.3Z: Analysis of the BSC Model"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Binary Symmetric Channel (BSC)}}
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}}
  
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[[File:P_ID1832__Dig_Z_5_3.png|right|frame|Gegebene Fehlerfolge]]
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[[File:P_ID1832__Dig_Z_5_3.png|right|frame|The given error sequence]]
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Wir betrachten zwei verschiedene BSC–Modelle mit den folgenden Parametern:
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We consider two different BSC models with the following parameters:
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* Modell $M_1 \text{:} \hspace{0.4cm} p = 0.01$,
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* Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$,
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* Modell $M_2 \text{:} \hspace{0.4cm} p = 0.02$.
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* Model $M_2 \text{:} \hspace{0.4cm} p = 0.02$.
  
  
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Die Grafik zeigt eine Fehlerfolge der Länge $N = 1000$, wobei allerdings nicht bekannt ist, von welchem der beiden Modelle diese Folge stammt.
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The graph shows an error sequence of length  $N = 1000$,  but it is not known from which of the two models this sequence originates.
  
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Die beiden Modelle sollen analysiert werden anhand
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The two models are to be analyzed on the basis of
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* der Fehlerabstandswahrscheinlichkeiten
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* the  '''error distance probabilities'''
 
:$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$
 
:$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$
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* der Fehlerabstandsverteilung
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* the  '''error distance distribution'''  $\rm (EDD)$
 
:$$V_a(k) =  {\rm Pr}(a \ge k) =  (1-p)^{k-1}\hspace{0.05cm},$$
 
:$$V_a(k) =  {\rm Pr}(a \ge k) =  (1-p)^{k-1}\hspace{0.05cm},$$
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* der Fehlerkorrelationsfunktion
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* the  '''error correlation function'''  $\rm (ECF)$
 
:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm
 
:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {\rm
−
E}[e_{\nu} \cdot e_{\nu + k}] \ \ = \ \
+
E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \
 
  \left\{ \begin{array}{c} p \\
 
  \left\{ \begin{array}{c} p \\
 
  p^2 \end{array} \right.\quad
 
  p^2 \end{array} \right.\quad
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\begin{array}{*{1}c} f{\rm \ddot{u}r }\hspace{0.15cm}k = 0  \hspace{0.05cm},
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\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0  \hspace{0.05cm},
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\\  f{\rm \ddot{u}r }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$
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\\  f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$
  
  
  
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''Hinweise:''
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Notes:
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* Die Aufgabe gehört zum Kapitel [[Digitalsignal%C3%BCbertragung/Binary_Symmetric_Channel_(BSC)| Binary Symmetric Channel (BSC)]].
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* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)| "Binary Symmetric Channel"]].
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*Durch Abzählen erkennt man, dass die Fehlerfolge der Länge $N = 1000$ genau $22$ Einsen enthält.
 
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* Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
 
  
 +
*By counting,  we would see that the error sequence of length  $N = 1000$  contains exactly  $22$  "ones".
 +
  
  
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===Fragebogen===
+
 
 +
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
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{Aus welchen Kenngrößen kann auf die mittlere Fehlerwahrscheinlichkeit $p_{\rm M}$ des BSC&ndash;Modells zurückgeschlossen werden?
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{Which parameters can be used to infer the mean error probability&nbsp; $p_{\rm M}$&nbsp; of the BSC model?
 
|type="[]"}
 
|type="[]"}
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+ FKF&ndash;Wert $\varphi_e(k = 0)$,
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+ ECF value&nbsp; $\varphi_e(k = 0)$,
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+ FKF&ndash;Wert $\varphi_e(k = 10)$,
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+ ECF value&nbsp; $\varphi_e(k = 10)$,
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- FAV&ndash;Wert $V_a(k = 1)$,
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- EDD value&nbsp; $V_a(k = 1)$,
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+ FAV&ndash;Wert $V_a(k = 2)$,
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+ EDD value&nbsp; $V_a(k = 2)$,
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+ FAV&ndash;Wert $V_a(k = 10)$.
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+ EDD value&nbsp; $V_a(k = 10)$.
  
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{Von welchem Model stammt die angegebene Fehlerfolge?
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{From which model does the given error sequence originate?
−
|type="[]"}
+
|type="()"}
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- Modell $M_1$,
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- Model $M_1$,
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+ Modell $M_2$.
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+ model $M_2$.
  
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{Wie groß ist der mittlere Fehlerabstand von Modell $M_1$?
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{What is the mean error distance of model&nbsp; $M_1$?
 
|type="{}"}
 
|type="{}"}
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$E[a] \ = \ ${ 10 3% }  
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${\rm E}\big[a\big] \ = \ ${ 10 3% }  
  
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{Wie groß sind für das Modell $M_1$ die folgenden Wahrscheinlichkeiten?
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{What are the following probabilities for model&nbsp; $M_1$?&nbsp;
 
|type="{}"}
 
|type="{}"}
 
${\rm Pr}(a = 1) \ = \ ${ 0.1 3% }  
 
${\rm Pr}(a = 1) \ = \ ${ 0.1 3% }  
 
${\rm Pr}(a = 2) \ = \ ${ 0.09 3% }  
 
${\rm Pr}(a = 2) \ = \ ${ 0.09 3% }  
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${\rm Pr}(a = E[a]) \ = \ ${ 0.0387 3% }  
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${\rm Pr}(a = {\rm E}\big[a\big]) \ = \ ${ 0.0387 3% }  
  
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{Berechnen Sie für das Modell $M_1$ folgende Werte der Fehlerabstandsverteilung:
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{Calculate the following values of the error distance distribution for model&nbsp; $M_1$:&nbsp;
 
|type="{}"}
 
|type="{}"}
 
$V_a(k = 2) \ = \ ${ 0.9 3% }
 
$V_a(k = 2) \ = \ ${ 0.9 3% }
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</quiz>
 
</quiz>
  
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===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
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'''(1)'''&nbsp; Beim BSC&ndash;Modell ist die mittlere Fehlerwahrscheinlichkeit $p_{\rm M}$ stets gleich der charakteristischen Wahrscheinlichkeit $p$. Für die Fehlerkorrelationsfunktion und die Fehlerabstandsverteilung gelten
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'''(1)'''&nbsp; In the BSC model,&nbsp; the mean error probability $p_{\rm M}$&nbsp; is always equal to the characteristic probability&nbsp; $p$.
 +
 +
*For the error correlation function and the error distance distribution are valid
 
:$$\varphi_{e}(k) =
 
:$$\varphi_{e}(k) =
 
  \left\{ \begin{array}{c} p \\
 
  \left\{ \begin{array}{c} p \\
 
  p^2 \end{array} \right.\quad
 
  p^2 \end{array} \right.\quad
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\begin{array}{*{1}c} f{\rm \ddot{u}r }\hspace{0.15cm}k = 0  \hspace{0.05cm},
+
\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0  \hspace{0.05cm},
−
\\  f{\rm \ddot{u}r }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}
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\\  f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}$$
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\hspace{0.4cm}V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$
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:$$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$
 +
 
 +
*$p$&nbsp; can be determined from all the given characteristics,&nbsp; except $V_a(k = 1)$.&nbsp; This EDD value is independent of&nbsp; $p$ &nbsp; equal to &nbsp; $(1&ndash;p)^0 = 1$.
 +
 +
*Therefore, the&nbsp; <u>solutions 1, 2, 4 and 5</u>&nbsp; are correct.
 +
 
  
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$p$ lässt sich aus allen angegebenen Kenngrößen ermitteln, nur nicht aus $V_a(k = 1)$. Dieser FAV&ndash;Wert ist unabhängig von $p$ gleich $(1&ndash;p)^0 = 1$. Zutreffend sind somit die <u>Lösungsvorschläge 1, 2, 4 und 5</u>.
 
  
 +
'''(2)'''&nbsp; The relative error frequency of the given sequence is equal to&nbsp; $h_{\rm F} = 22/1000 \approx 0.022$.
 +
*It is quite obvious that the error sequence was generated by the model&nbsp; $M_2$&nbsp; &nbsp; &#8658; &nbsp; $p_{\rm M} = 0.02$.
 +
 +
*Because of the short sequence,&nbsp; $h_{\rm F}$&nbsp; does not match&nbsp; $p_{\rm M}$&nbsp; exactly,&nbsp; but at least approximates &nbsp;&#8658;&nbsp; <u>solution 2</u>.
  
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'''(2)'''&nbsp; Die relative Fehlerhäufigkeit der angegebenen Folge ist gleich $h_{\rm F} = 22/1000 \approx 0.022$. Es ist ganz offensichtlich, dass die Fehlerfolge vom Modell $M_2$ &nbsp;&#8658;&nbsp; $p_{\rm M} = 0.02$ generiert wurde. Aufgrund der kurzen Folge stimmt $h_{\rm F}$ mit $p_{\rm M}$ zwar nicht exakt überein, aber zumindest näherungsweise &nbsp;&#8658;&nbsp; <u>Vorschlag 2</u>.
 
  
  
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'''(3)'''&nbsp; Der mittlere Fehlerabstand &ndash; also der Erwartungswert der Zufallsgröße $a$ &ndash; ist gleich dem Kehrwert der mittleren Fehlerwahrscheinlichkeit &#8658; $E[a] = 1/0.1 \ \underline {= 10}$.
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'''(3)'''&nbsp; The mean error distance &ndash; that is,&nbsp; the expected value of the random variable&nbsp; $a$&nbsp; &ndash; is equal to the inverse of the mean error probability &#8658;  
 +
:$${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}.$$
  
  
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'''(4)'''&nbsp; Entsprechend der Gleichung ${\rm Pr}(a = k) = (1&ndash;p)^{k&ndash;1} \cdot p$ erhält man:
+
 
 +
'''(4)'''&nbsp; According to the equation &nbsp; ${\rm Pr}(a = k) = (1&ndash;p)^{k&ndash;1} \cdot p$ &nbsp; we obtain:
 
:$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {=
 
:$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {=
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0.1}\hspace{0.05cm}, \hspace{0.2cm}{\rm Pr}(a = 2) = 0.9 \cdot 0.1
+
0.1}\hspace{0.05cm},$$
 +
:$${\rm Pr}(a = 2) = 0.9 \cdot 0.1
 
\hspace{0.15cm}\underline {= 0.09}\hspace{0.05cm},$$
 
\hspace{0.15cm}\underline {= 0.09}\hspace{0.05cm},$$
 
:$${\rm Pr}(a = {\rm E}[a]) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}
 
:$${\rm Pr}(a = {\rm E}[a]) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}
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'''(5)'''&nbsp; Aus der Beziehung $V_a(k) = (1&ndash;p)^{k&ndash;1}$ erhält man
+
 
 +
'''(5)'''&nbsp; From the relation&nbsp; $V_a(k) = (1&ndash;p)^{k&ndash;1}$&nbsp; we obtain
 
:$$V_a(k = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^1 \hspace{0.15cm}\underline {= 0.9 } \hspace{0.3cm}
 
:$$V_a(k = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^1 \hspace{0.15cm}\underline {= 0.9 } \hspace{0.3cm}
 
\Rightarrow \hspace{0.3cm}{\rm Pr}(a = 1) = V_a(k =
 
\Rightarrow \hspace{0.3cm}{\rm Pr}(a = 1) = V_a(k =
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\hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10}   
 
\hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10}   
 
\hspace{0.15cm}\underline {=0.3487}.$$
 
\hspace{0.15cm}\underline {=0.3487}.$$
−
Zur Kontrolle im Vergleich zur Teilaufgabe (4):
+
*To check in comparison with subtask&nbsp; '''(4)''':
 
:$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k =
 
:$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k =
 
11) = 0.3874 - 0.3487  {= 0.0387}\hspace{0.05cm}.$$
 
11) = 0.3874 - 0.3487  {= 0.0387}\hspace{0.05cm}.$$
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[[Category:Aufgaben zu Digitalsignalübertragung|^5.2 Binary Symmetric Channel (BSC)^]]
+
[[Category:Digital Signal Transmission: Exercises|^5.2 Binary Symmetric Channel^]]

Latest revision as of 14:52, 5 September 2022

The given error sequence

We consider two different BSC models with the following parameters:

  • Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$,
  • Model $M_2 \text{:} \hspace{0.4cm} p = 0.02$.


The graph shows an error sequence of length  $N = 1000$,  but it is not known from which of the two models this sequence originates.

The two models are to be analyzed on the basis of

  • the  error distance probabilities
$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$
  • the  error distance distribution  $\rm (EDD)$
$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$
  • the  error correlation function  $\rm (ECF)$
$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \ \left\{ \begin{array}{c} p \\ p^2 \end{array} \right.\quad \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$


Notes:

  • By counting,  we would see that the error sequence of length  $N = 1000$  contains exactly  $22$  "ones".



Questions

1

Which parameters can be used to infer the mean error probability  $p_{\rm M}$  of the BSC model?

ECF value  $\varphi_e(k = 0)$,
ECF value  $\varphi_e(k = 10)$,
EDD value  $V_a(k = 1)$,
EDD value  $V_a(k = 2)$,
EDD value  $V_a(k = 10)$.

2

From which model does the given error sequence originate?

Model $M_1$,
model $M_2$.

3

What is the mean error distance of model  $M_1$?

${\rm E}\big[a\big] \ = \ $

4

What are the following probabilities for model  $M_1$? 

${\rm Pr}(a = 1) \ = \ $
${\rm Pr}(a = 2) \ = \ $
${\rm Pr}(a = {\rm E}\big[a\big]) \ = \ $

5

Calculate the following values of the error distance distribution for model  $M_1$: 

$V_a(k = 2) \ = \ $
$V_a(k = 10) \ = \ $
$V_a(k = 11) \ = \ $


Solution

Solution

(1)  In the BSC model,  the mean error probability $p_{\rm M}$  is always equal to the characteristic probability  $p$.

  • For the error correlation function and the error distance distribution are valid
$$\varphi_{e}(k) = \left\{ \begin{array}{c} p \\ p^2 \end{array} \right.\quad \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}$$
$$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$
  • $p$  can be determined from all the given characteristics,  except $V_a(k = 1)$.  This EDD value is independent of  $p$   equal to   $(1–p)^0 = 1$.
  • Therefore, the  solutions 1, 2, 4 and 5  are correct.


(2)  The relative error frequency of the given sequence is equal to  $h_{\rm F} = 22/1000 \approx 0.022$.

  • It is quite obvious that the error sequence was generated by the model  $M_2$    ⇒   $p_{\rm M} = 0.02$.
  • Because of the short sequence,  $h_{\rm F}$  does not match  $p_{\rm M}$  exactly,  but at least approximates  ⇒  solution 2.


(3)  The mean error distance – that is,  the expected value of the random variable  $a$  – is equal to the inverse of the mean error probability ⇒

$${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}.$$


(4)  According to the equation   ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$   we obtain:

$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {= 0.1}\hspace{0.05cm},$$
$${\rm Pr}(a = 2) = 0.9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.09}\hspace{0.05cm},$$
$${\rm Pr}(a = {\rm E}[a]) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(a = 10)= 0.9^9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.0387}\hspace{0.05cm}.$$


(5)  From the relation  $V_a(k) = (1–p)^{k–1}$  we obtain

$$V_a(k = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^1 \hspace{0.15cm}\underline {= 0.9 } \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}(a = 1) = V_a(k = 1) - V_a(k = 2) = 0.1\hspace{0.05cm},$$
$$V_a(k = 10)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^9 \hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10} \hspace{0.15cm}\underline {=0.3487}.$$
  • To check in comparison with subtask  (4):
$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k = 11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$

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