Aufgaben:Exercise 1.1Z: Sum of Two Ternary Signals: Difference between revisions

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Einige grundlegende Definitionen}}
{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Some_Basic_Definitions}}


[[File:P_ID146__Sto_Z1_1.png|right|framed|Summe $S$ zweier <br>Ternärsignale&nbsp; $X$&nbsp; und&nbsp; $Y$]]
[[File:P_ID146__Sto_Z1_1.png|right|framed|Sum $S$ of two <br>ternary signals&nbsp; $X$&nbsp; and&nbsp; $Y$]]
Gegeben seien zwei dreistufige Nachrichtenquellen&nbsp; $X$&nbsp; und&nbsp; $Y$, deren Ausgangssignale jeweils nur die Werte&nbsp; $-1$,&nbsp; $0$&nbsp; und&nbsp; $+1$&nbsp; annehmen können. Die Signalquellen sind statistisch voneinander unabhängig.  
Let two three-stage message sources&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp;  be given,&nbsp; whose output signals can only assume the values&nbsp; $-1$,&nbsp; $0$&nbsp; and&nbsp; $+1$&nbsp; respectively.&nbsp; The signal sources are statistically independent of each other.  


*Eine einfache Schaltung bildet nun das Summensignal&nbsp; $S = X + Y$.
*A simple circuit now forms the sum signal&nbsp; $S = X + Y$.
*Bei der Signalquelle&nbsp; $X$&nbsp; treten die Werte&nbsp; $-1$,&nbsp; $0$&nbsp; und&nbsp; $+1$&nbsp; mit gleicher Wahrscheinlichkeit auf.  
*At the signal source&nbsp; $X$,&nbsp; the values&nbsp; $-1$,&nbsp; $0$&nbsp; and&nbsp; $+1$&nbsp; occur with equal probability.  
*Bei der Quelle&nbsp; $Y$&nbsp; ist der Signalwert&nbsp; $0$&nbsp; doppelt so wahrscheinlich wie die beiden anderen Werte&nbsp; $-1$&nbsp; bzw.&nbsp; $+1$.
*For source&nbsp; $Y$,&nbsp; the signal value&nbsp; $0$&nbsp; is twice as likely as the other two values&nbsp; $-1$&nbsp; and&nbsp; $+1$, respectively.








 
Hints:  
 
*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Einige_grundlegende_Definitionen | Some basic definitions of probability theory]].
 
''Hinweise:''
*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Einige_grundlegende_Definitionen | Einige grundlegende Definitionen der Wahrscheinlichkeitsrechnung]].
   
   
*Lösen Sie die Teilaufgaben&nbsp; '''(3)'''&nbsp; und&nbsp; '''(4)'''&nbsp; nach der klassischen Definition.  
*Solve the subtasks&nbsp; '''(3)'''&nbsp; and&nbsp; '''(4)'''&nbsp; according to the classical definition.
*Berücksichtigen Sie trotzdem die unterschiedlichen Auftrittshäufigkeiten des Signals&nbsp; $Y$.
*Nevertheless,&nbsp; consider the different occurrence frequencies of the signal&nbsp; $Y$.
*Der Inhalt dieses Abschnitts ist im Lernvideo&nbsp; [[Klassische_Definition_der_Wahrscheinlickeit_(Lernvideo)|Klassische Definition der Wahrscheinlichkeit]]&nbsp; zusammengefasst.
*The topic of this section is illustrated with examples in the (German language) learning video &nbsp; <br>[[Klassische_Definition_der_Wahrscheinlickeit_(Lernvideo)|Klassische Definition der Wahrscheinlichkeit]]&nbsp; $\Rightarrow$ "Classical definition of probability".




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===Fragebogen===
===Questions===


<quiz display=simple>
<quiz display=simple>
{Wie groß sind die Auftrittswahrscheinlichkeiten der Signalwerte von&nbsp; $Y$? Wie groß ist die Wahrscheinlichkeit, dass&nbsp; $Y = 0$&nbsp; ist ?
{What are the probabilities of occurrence of the signal values of&nbsp; $Y$?&nbsp; What is the probability that&nbsp; $Y = 0$&nbsp;?
|type="{}"}
|type="{}"}
${\rm Pr}(Y=0) \ = \ $ { 0.5 3% }
${\rm Pr}(Y=0) \ = \ $ { 0.5 3% }




{Wieviele unterschiedliche Signalwerte&nbsp; $(I)$&nbsp; kann das Summensignal&nbsp; $S$&nbsp; annehmen? Welche sind dies?
{How many different signal values&nbsp; $(I)$&nbsp; can the sum signal&nbsp; $S$&nbsp; assume?&nbsp; Which are these?
|type="{}"}
|type="{}"}
$ I \ = \ $ { 5 3% }
$ I \ = \ $ { 5 3% }




{Mit welchen Wahrscheinlichkeiten treten die in der Teilaufgabe&nbsp; '''(2)'''&nbsp; ermittelten Werte auf? Wie wahrscheinlich ist der Maximalwert&nbsp; $S_{\rm max}$?  
{What are the probabilities of the values determined in subtask&nbsp; '''(2)'''?&nbsp; How probable is the maximum value&nbsp; $S_{\rm max}$?  
|type="{}"}
|type="{}"}
$ {\rm Pr}(S = S_{\rm max} ) \ = \ $ { 0.0833 3% }
$ {\rm Pr}(S = S_{\rm max} ) \ = \ $ { 0.0833 3% }




{Wie ändern sich die Wahrscheinlichkeiten, wenn nun anstelle der Summe die Differenz&nbsp; $D = X - Y$&nbsp; betrachtet wird?&nbsp; Begründen Sie Ihre Antwort.
{How do the probabilities change,&nbsp; if now instead of the sum the difference&nbsp; $D = X - Y$&nbsp; is considered?&nbsp; Give reasons for your answer.
|type="[]"}
|type="[]"}
+ Die Wahrscheinlichkeiten bleiben gleich.
+ The probabilities remain the same.
- Die Wahrscheinlichkeiten ändern sich.&nbsp; Wie ändern sie sich?
- The probabilities change.&nbsp; How do they change?




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</quiz>
</quiz>


===Musterlösung===
===Solution===
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{{ML-Kopf}}
'''(1)'''&nbsp; Da die Wahrscheinlichkeiten von&nbsp; $ \pm 1$&nbsp; gleich sind und&nbsp; ${\rm Pr}(Y = 0) = 2 \cdot {\rm Pr}(Y = 1)$&nbsp; gilt, erhält man:
'''(1)'''&nbsp; Since the probabilities of&nbsp; $ \pm 1$&nbsp; are the same and&nbsp; ${\rm Pr}(Y = 0) = 2 \cdot {\rm Pr}(Y = 1)$&nbsp; holds, we get:


:$${\rm Pr}(Y = 1) + {\rm Pr}(Y = 0) + {\rm Pr}(Y = -1) = 1/2 \cdot {\rm Pr}(Y = 0) + {\rm Pr}(Y = 0) + 1/2\cdot {\rm Pr}(Y = 0) = 1\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}(Y = 0)\;\underline { = 0.5}. $$
:$${\rm Pr}(Y = 1) + {\rm Pr}(Y = 0) + {\rm Pr}(Y = -1) = 1/2 \cdot {\rm Pr}(Y = 0) + {\rm Pr}(Y = 0) + 1/2\cdot {\rm Pr}(Y = 0) = 1\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}(Y = 0)\;\underline { = 0.5}. $$




'''(2)'''&nbsp; $S$&nbsp; kann insgesamt&nbsp; $\underline {I =5}$&nbsp; Werte annehmen, nämlich&nbsp; $0$,&nbsp; $\pm 1$&nbsp; und&nbsp; $\pm 2$.
[[File:EN_Sto_Z1_1_c_neu.png|right|frame|400px|Sum and difference of ternary random variables]]
'''(2)'''&nbsp; $S$&nbsp; can take a total of&nbsp; $\underline {I =5}$&nbsp; values, namely&nbsp; $0$,&nbsp; $\pm 1$&nbsp; and&nbsp; $\pm 2$.




[[File:P_ID192__Sto_Z1_1_c.png|right|frame|400px|Summe und Differenz ternärer Zufallsgrößen]]
'''(3)'''&nbsp; Since&nbsp; $Y$&nbsp; is not equally distributed, one cannot (actually) apply the "Classical Definition of Probability" here.
'''(3)'''&nbsp; Da&nbsp; $Y$&nbsp; nicht gleichverteilt ist, kann man hier  (eigentlich) die &bdquo;Klassische Definition der Wahrscheinlichkeit&rdquo; nicht anwenden.


*Teilt man&nbsp; $Y$&nbsp; jedoch gemäß der Grafik in vier Bereiche auf, wobei man zwei der Bereiche dem Ereignis&nbsp; $Y = 0$&nbsp; zuordnet, so kann man trotzdem gemäß der klassischen Definition vorgehen.  
*However,&nbsp; if we divide&nbsp; $Y$&nbsp; into four ranges according to the graph,&nbsp; assigning two of the ranges to the event&nbsp; $Y = 0$,&nbsp; we can still proceed according to the classical definition.
*Man erhält dann:
*One then obtains:


:$${\rm Pr}(S = 0) = {4}/{12} = {1}/{3},$$
:$${\rm Pr}(S = 0) = {4}/{12} = {1}/{3},$$
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'''(4)'''&nbsp; Aus der Grafik ist auch ersichtlich, dass das Differenzsignal&nbsp; $D$&nbsp; und das Summensignal&nbsp; $S$&nbsp; die gleichen Werte mit gleichen Wahrscheinlichkeiten annehmen.  
'''(4)'''&nbsp; It is also evident from the graph that the difference signal&nbsp; $D$&nbsp; and the sum signal&nbsp; $S$&nbsp; take the same values with equal probabilities.


*Dies war zu erwarten, da&nbsp; ${\rm Pr}(Y = +1) ={\rm Pr}(Y = -1)$&nbsp; vorgegeben ist &nbsp;  ⇒ &nbsp; <u>Lösungsvorschlag 1</u>.
*This was to be expected,&nbsp; since&nbsp; ${\rm Pr}(Y = +1) ={\rm Pr}(Y = -1)$&nbsp; is given &nbsp;  ⇒ &nbsp; <u>Proposed solution 1</u>.
{{ML-Fuß}}
{{ML-Fuß}}






[[Category:Aufgaben zu Stochastische Signaltheorie|^1.1 Einige grundlegende Definitionen
[[Category:Theory of Stochastic Signals: Exercises|^1.1 Some Basic Definitions
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[[de:Aufgaben:Aufgabe 1.1Z: Summe zweier Ternärsignale]]

Latest revision as of 17:53, 16 March 2026

Sum $S$ of two
ternary signals  $X$  and  $Y$

Let two three-stage message sources  $X$  and  $Y$  be given,  whose output signals can only assume the values  $-1$,  $0$  and  $+1$  respectively.  The signal sources are statistically independent of each other.

  • A simple circuit now forms the sum signal  $S = X + Y$.
  • At the signal source  $X$,  the values  $-1$,  $0$  and  $+1$  occur with equal probability.
  • For source  $Y$,  the signal value  $0$  is twice as likely as the other two values  $-1$  and  $+1$, respectively.



Hints:

  • Solve the subtasks  (3)  and  (4)  according to the classical definition.
  • Nevertheless,  consider the different occurrence frequencies of the signal  $Y$.
  • The topic of this section is illustrated with examples in the (German language) learning video  
    Klassische Definition der Wahrscheinlichkeit  $\Rightarrow$ "Classical definition of probability".



Questions

1 What are the probabilities of occurrence of the signal values of  $Y$?  What is the probability that  $Y = 0$ ?

${\rm Pr}(Y=0) \ = \ $

2 How many different signal values  $(I)$  can the sum signal  $S$  assume?  Which are these?

$ I \ = \ $

3 What are the probabilities of the values determined in subtask  (2)?  How probable is the maximum value  $S_{\rm max}$?

$ {\rm Pr}(S = S_{\rm max} ) \ = \ $

4 How do the probabilities change,  if now instead of the sum the difference  $D = X - Y$  is considered?  Give reasons for your answer.

The probabilities remain the same.
The probabilities change.  How do they change?


Solution

(1)  Since the probabilities of  $ \pm 1$  are the same and  ${\rm Pr}(Y = 0) = 2 \cdot {\rm Pr}(Y = 1)$  holds, we get:

$${\rm Pr}(Y = 1) + {\rm Pr}(Y = 0) + {\rm Pr}(Y = -1) = 1/2 \cdot {\rm Pr}(Y = 0) + {\rm Pr}(Y = 0) + 1/2\cdot {\rm Pr}(Y = 0) = 1\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}(Y = 0)\;\underline { = 0.5}. $$


Sum and difference of ternary random variables

(2)  $S$  can take a total of  $\underline {I =5}$  values, namely  $0$,  $\pm 1$  and  $\pm 2$.


(3)  Since  $Y$  is not equally distributed, one cannot (actually) apply the "Classical Definition of Probability" here.

  • However,  if we divide  $Y$  into four ranges according to the graph,  assigning two of the ranges to the event  $Y = 0$,  we can still proceed according to the classical definition.
  • One then obtains:
$${\rm Pr}(S = 0) = {4}/{12} = {1}/{3},$$
$${\rm Pr}(S = +1) = {\rm Pr}(S = -1) ={3}/{12} = {1}/{4},$$
$${\rm Pr}(S = +2) = {\rm Pr}(S = -2) ={1}/{12}$$
$$\Rightarrow \hspace{0.3cm}{\rm Pr}(S = S_{\rm max}) = {\rm Pr}(S = +2) =1/12 \;\underline {= 0.0833}.$$


(4)  It is also evident from the graph that the difference signal  $D$  and the sum signal  $S$  take the same values with equal probabilities.

  • This was to be expected,  since  ${\rm Pr}(Y = +1) ={\rm Pr}(Y = -1)$  is given   ⇒   Proposed solution 1.