Aufgaben:Exercise 1.4: 2S/3E Channel Model

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$\rm 2S/3E$  channel model

A transmitter  (German:  "Sender"   ⇒   subscript  "S")  emits the binary symbols  $\rm L$  $($event  $S_{\rm L})$  and  $\rm H$  $($event  $S_{\rm H})$ .

• If conditions are good, the digital receiver  (German:  "Empfänger"   ⇒   subscript  "E")  also decides only on the binary symbols  $\rm L$  $($event  $E_{\rm L})$  or  $\rm H$  $($event  $E_{\rm H})$.
• However,  if the receiver can suspect that an error has occurred during transmission,  it makes no decision  $($event  $E_{\rm K})$; 
  $\rm K$  here stands for  "No decision".

The diagram shows a simple channel model in terms of transition probabilities. 

• It can be seen that a transmitted  $\rm L$  may well be received as a symbol  $\rm H$. 
• In contrast,  the transition from  $\rm H$  to  $\rm L$  is not possible.
• Let the symbol appearance probabilities at the transmitter be  ${\rm Pr}(S_{\rm L}) = 0.3$  and  ${\rm Pr}(S_{\rm H}) = 0.7$.

Hints:

• The exercise belongs to the chapter  Statistical dependence and independence.

• The topic of this chapter is illustrated with examples in the  (German language)  learning video

Statistische Abhängigkeit und Unabhängigkeit   $\Rightarrow$   "Statistical dependence and independence".

Questions

1

What is the probability that the receiver chooses the symbol  $\rm L$?

${\rm Pr}(E_{\rm L}) \ = \ $

2

What is the probability that the receiver chooses the symbol  $\rm H$?

${\rm Pr}(E_{\rm H}) \ = \ $

3

What is the probability that the receiver does not make a decision?

${\rm Pr}(E_{\rm K}) \ = \ $

4

What is the probability that the receiver makes a wrong decision?

$\text{Pr(wrong decision)} \ = \ $

5

What is the probability that symbol  $\rm L$  was actually sent if the receiver decided to use symbol  $\rm L$?

${\rm Pr}(S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}E_{\rm L} ) \ = \ $

6

What is the probability that symbol  $\rm L$  was sent if the receiver does not make a decision?

${\rm Pr}(S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}E_{\rm K} ) \ =\ $

Solution

Solution

(1)  Only if the symbol  $\rm L$  was sent, the receiver can decide for the symbol  $\rm L$  at the given channel.

• However, the probability for a received  $\rm L$  is smaller by a factor of  $0.7$  than for a sent one. From this follows:

$${\rm Pr} (E_{\rm L}) = {\rm Pr}(S_{\rm L}) \cdot {\rm Pr} (E_{\rm L}\hspace{0.05cm}|\hspace{0.05cm}S_{\rm L}) = 0.3 \cdot 0.7 \hspace{0.15cm}\underline {= \rm 0.21}.$$

(2)  To the event  $E_{\rm H}$  one comes from  $S_{\rm H}$  as well as from  $S_{\rm L}$ . Therefore holds:

$${\rm Pr} (E_{\rm H}) = {\rm Pr} (S_{\rm H}) \cdot {\rm Pr} (E_{\rm H}\hspace{0.05cm}|\hspace{0.05cm}S_{\rm H}) + {\rm Pr} (S_{\rm L}) \cdot {\rm Pr} (E_{\rm H}\hspace{0.05cm}|\hspace{0.05cm} S_{\rm L})= \rm 0.7 \cdot 0.9 + 0.3 \cdot 0.1\hspace{0.15cm}\underline { = \rm 0.66}.$$

(3)  The events  $E_{\rm H}$,  $E_{\rm L}$  and  $E_{\rm K}$  together form a complete system. It follows that:

$${\rm Pr} (E_{\rm K}) = 1 - {\rm Pr} (E_{\rm L}) - {\rm Pr} (E_{\rm H}) \hspace{0.15cm}\underline {= \rm 0.13}.$$

(4)  A wrong decision can be characterized in set-theoretic terms as follows:

$${\rm Pr} \text{(wrong decision)} = {\rm Pr} \big [(S_{\rm L} \cap E_{\rm H}) \cup (S_{\rm H} \cap E_{\rm L})\big ] = \rm 0.3 \cdot 0.1 + 0.7\cdot 0 \hspace{0.15cm}\underline {= \rm 0.03}.$$

(5)  If the symbol  $\rm L$  was received, only  $\rm L$  could have been sent. It follows that:

$${\rm Pr} (S_{\rm L} \hspace{0.05cm}|\hspace{0.05cm} E_{\rm L}) \hspace{0.15cm}\underline {= \rm 1}.$$

(6)  For example, Bayes' theorem is suitable for solving this problem:

$${\rm Pr} (S_{\rm L}\hspace{0.05cm}|\hspace{0.05cm} E_{\rm K}) =\frac{ {\rm Pr} ( E_{\rm K} \hspace{0.05cm}|\hspace{0.05cm} S_{\rm L}) \cdot {\rm Pr} (S_{\rm L})}{{\rm Pr} (E_{\rm K})} =\frac{ \rm 0.2 \cdot 0.3}{\rm 0.13} = \frac{\rm 6}{\rm 13}\hspace{0.15cm}\underline { \approx \rm 0.462}.$$