Difference between revisions of "Aufgaben:Exercise 1.2: Decimal/Binary Converter"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Mengentheoretische Grundlagen}}
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Set_Theory_Basics}}
==A1.2 Schaltlogik (D/B-Wandler)==
 
[[File:P_ID61__Sto_A_1_2.png|right|]]
 
Ein Zahlengenerator $Z$ liefert Dezimalwerte im Bereich von 1 bis 15. Diese werden in Binärzahlen umgewandelt (rot umrandeter Block). Der Ausgang besteht aus den vier Binärwerten $A$, $B$, $C$ und $D$ mit abnehmender Wertigkeit. Beispielsweise liefert $Z = 11$ die Binärwerte
 
$$ A = 1, B = 0, C = 1, D = 1. $$
 
Mengentheoretisch lässt sich dies wie folgt darstellen:
 
$$ Z = 11\qquad\widehat{=}\qquad A \cap\bar{ B} \cap C \cap D$$
 
Aus den binären Größen A, B, C und D werden drei weitere Boolsche Ausdrücke gebildet, deren Vereinigungsmenge mit X bezeichnet wird:
 
:<math> U = A \cap \bar{D} </math>
 
:<math> V = \bar{A} \cap B \cap \bar{D} </math>
 
$$W, wobei \, \bar{W} = \bar{A} \cup \bar{D} \cup (\bar{B} \cap C) \cup (B \cap \bar{C}). $$
 
Für die folgenden Fragen ist zu berücksichtigen, dass $Z = 0  ⇒  A = B = C = D = 0$ bereits durch den Zahlengenerator ausgeschlossen ist. Beachten Sie ferner, dass nicht alle Eingangsgrößen $A$, $B$, $C$ und $D$ zur Berechnung aller Zwischengrößen $U$, $V$ und $W$ herangezogen werden.
 
Hinweis: Diese Aufgabe bezieht sich auf den Lehrstoff von Kapitel 1.2. Eine Zusammenfassung der theoretischen Grundlagen mit Beispielen bringt das nachfolgende Lernvideo:
 
  
===Fragebogen zu "A1.2 Schaltlogik (D/B-Wandler)"===
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[[File:EN_Sto_A_1_2.png|right|frame|Logical circuit for D/B converting]]
 +
A number generator&nbsp; $Z$&nbsp; supplies decimal values in the range&nbsp; $1$&nbsp; to&nbsp; $15$.
 +
*These are converted into binary numbers&nbsp; (block outlined in red).
 +
*The output consists of the four binary values&nbsp; $A$,&nbsp; $B$,&nbsp; $C$&nbsp; and&nbsp; $D$&nbsp; with decreasing significance.
 +
*For example&nbsp; $Z = 11$&nbsp; delivers the binary values
 +
:$$ A = 1, \ B = 0, \ C = 1, \ D = 1. $$
 +
Set-theoretically,&nbsp; this can be represented as follows:
 +
:$$ Z = 11\qquad\widehat{=}\qquad A \cap\overline{ B} \cap C \cap D.$$
 +
 
 +
Three more Boolean expressions are formed from the binary quantities&nbsp; $A$,&nbsp; $B$,&nbsp; $C$&nbsp; and&nbsp; $D$&nbsp; and their union set is denoted&nbsp; $X$&nbsp;:
 +
::<math> U = A \cap \overline{D} </math>
 +
::<math> V = \overline{A} \cap B \cap \overline{D} </math>
 +
:$$W,\;  {\rm where} \; \, \overline{W} = \overline{A} \cup \overline{D} \cup (\overline{B} \cap C) \cup (B \cap \overline{C}). $$
 +
*Note that&nbsp; $Z = 0 \ ⇒ \ A = B = C = D = 0$&nbsp; is already excluded by the number generator.
 +
*Note also that not all input quantities&nbsp; $A$,&nbsp; $B$,&nbsp; $C$&nbsp; and&nbsp; $D$&nbsp; are used to calculate all intermediate quantities&nbsp; $U$,&nbsp; $V$&nbsp; and&nbsp; $W$,&nbsp; resp.
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Hints:
 +
*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Mengentheoretische_Grundlagen|Set theory basic]].
 +
*The topic of this chapter is illustrated with examples in the (German language) learning video:&nbsp;
 +
 
 +
:[[Mengentheoretische_Begriffe_und_Gesetzmäßigkeiten_(Lernvideo)|"Mengentheoretische Begriffe und Gesetzmäßigkeiten"]] &nbsp; &rArr; &nbsp; "Set-theoretical terms and laws".
 +
 
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche Aussagen sind bezüglich der Zufallsgröße $U$ zutreffend?
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{Which statements are true regarding the random variable&nbsp; $U$?
 
|type="[]"}
 
|type="[]"}
- $U$ beinhaltet 2 Elemente.
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- $U$&nbsp; contains two elements.
+ $U$ beinhaltet 4 Elemente.
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+ $U$&nbsp; contains four elements.
- Das kleinste Element von $U$ ist 4.
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- The smallest element of&nbsp; $U$&nbsp; is&nbsp; $4$.
+ Das größte Element von $U$ ist 14.
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+ The largest element of&nbsp; $U$&nbsp; is&nbsp; $14$.
  
{Welche Aussagen sind bezüglich der Zufallsgröße $V$ zutreffend?
+
{Which statements are true regarding the random variable&nbsp; $V$?
 
|type="[]"}
 
|type="[]"}
+ $V$ beinhaltet 2 Elemente.
+
+ $V$&nbsp; contains two elements.
- $V$ beinhaltet 4 Elemente.
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- $V$&nbsp; contains four elements.
+ Das kleinste Element von $V$ ist 4.
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+ The smallest element of&nbsp; $V$&nbsp; is&nbsp; $4$.
- Das größte Element von $V$ ist 14.
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- The largest element of&nbsp; $V$&nbsp; is&nbsp; $14$.
 
 
{Welche Aussagen sind bezüglich der Zufallsgröße $W$ zutreffend?  
+
{Which statements are true regarding the random variable&nbsp; $W$?
 
|type="[]"}
 
|type="[]"}
+ $W$ beinhaltet 2 Elemente.
+
+ $W$&nbsp; contains two elements.
- $W$ beinhaltet 4 Elemente.
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- $W$&nbsp; contains four elements.
- Das kleinste Element von $W$ ist 4.
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- The smallest element of&nbsp; $W$&nbsp; is&nbsp; $4$.
- Das größte Element von $W$ ist 14.
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- The largest element of&nbsp; $W$&nbsp; is&nbsp; $14$.
  
{Welche Aussagen sind bezüglich der Zufallsgröße $P$ zutreffend?
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{Which statements are true regarding the random quantity&nbsp; $P$&nbsp;?
|type="[]"}
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|type="()"}
- $P$ beinhlatet alle Zweierpotenzen.
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- $P$&nbsp; contains all powers of two.
+ $P$ beinhaltet alle Primzahlen.
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+ $P$&nbsp; contains all prime numbers.
- $P$ beschreibt die leere Menge <math>\phi</math> .
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- $P$&nbsp; describes the empty set&nbsp; $\phi$.
- $P$ ist identisch mit der Grundmenge $G = {1,2, ... , 15}$.
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- $P$&nbsp; is identical with the universal set&nbsp; $G = {1,2, \ \text{...} \ , 15}$.
  
 
</quiz>
 
</quiz>
  
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
:'''1.''' Das Ereignis $U$ beinhaltet alle diejenigen Zahlen ≥ 8 (A = 1), die gerade sind (D = 0): 8, 10, 12, 14 ⇒  Richtig sind die <u>Lösungsalternativen 2 und 4</u>.
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'''(1)'''&nbsp; The event&nbsp; $U$&nbsp; contains
:'''2.''' Das Ereignis $V$ besteht aus den beiden Zahlen 4 (binär 0100) und 6 (binär 0110)  Richtig sind hier die <u>Lösungsalternativen 1 und 3</u>.
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*those numbers greater/equal to eight&nbsp; $(A = 1)$,&nbsp;
:'''3.''' [[File:P_ID2848__Sto_A_1_2c.png|frame|]]
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*which are even&nbsp; $(D = 0)$:&nbsp; $8, 10, 12, 14$ &nbsp;
Für das Ereignis W gilt mit dem Theorem von de Morgan:
+
 
 +
 
 +
⇒  &nbsp;<u>Proposed solutions 2 and 4</u>&nbsp; are correct.
 +
 
 +
 
 +
'''(2)'''&nbsp; The event&nbsp; $V$&nbsp; consists of the two numbers&nbsp; $4$&nbsp; (binary 0100) and&nbsp; $6$&nbsp; (binary 0110)   &nbsp; ⇒ &nbsp; The correct <u>solutions are 1 and 3</u>.
 +
 
 +
 
 +
[[File:P_ID2848__Sto_A_1_2c.png|right|frame|Auxiliary Venn diagram]]
 +
'''(3)'''&nbsp; For the event&nbsp; $W$,&nbsp; de Morgan's theorem holds:
 +
 
 +
:$$\overline W = \overline A \cup \overline D \cup (\overline B \cap C) \cup (B \cap \overline C)
 +
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} W = \overline{\overline W} = A \cap D \cap (\overline{\overline B \cap C}) \cap (\overline{B \cap \overline C}).$$
  
$\bar W = \bar A \cup \bar D \cup (\bar B \cap C) \cup (B \cap \bar C)$.
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*Using de Morgan's theorems,&nbsp; it further follows:
  
$ \Rightarrow W = \bar{\bar W} = A \cap D \cap (\overline{\bar B \cap C}) \cap (\overline{B \cap \bar C})$.
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:$$ W = A \cap D \cap (B \cup \overline C) \cap (\overline B \cup C).$$
  
Mit den Sätzen von de Morgan folgt daraus weiter:
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*Finally,&nbsp; using the Boolean relation&nbsp; $(B \cup \overline C) \cap (\overline B \cup C) = (B \cap C) \cup (\overline B \cap \overline C)$&nbsp; we obtain&nbsp; (see sketch):
  
$ W = A \cap D \cap (B \cup \bar C) \cap (\bar B \cup C)$.
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:$$W = (A \cap B \cap C \cap D) \cup (A \cap \overline B \cap \overline C \cap D).$$
  
Mit der Boolschen Beziehung (siehe Skizze)
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*Thus,&nbsp; $W$&nbsp; contains the numbers&nbsp; $15$&nbsp; and&nbsp; $9$&nbsp;  &nbsp; ⇒  &nbsp;  only the&nbsp; <u>proposed solution 1</u>&nbsp; is correct.
  
$(B \cup \bar C) \cap (\bar B \cup C) = (B \cap C) \cup (\bar B \cap \bar C)$
 
  
erhält man schließlich
 
  
$W = (A \cap B \cap C \cap D) \cup (A \cap \bar B \cap \bar C \cap D)$.
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'''(4)'''&nbsp; The union of&nbsp; $U$,&nbsp; $V$&nbsp; and&nbsp; $W$&nbsp; contains the following numbers: &nbsp; $4, 6, 8, 9, 10, 12, 14, 15$.
  
Somit beinhaltet W die Zahlen 15 und 9  ⇒  <u>Lösungsvorschlag 1</u>.
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*Accordingly, the set&nbsp; $P$&nbsp; as the complement of this union is: &nbsp;
:'''4.''' Die Vereinigungsmenge von $U$, $V$ und $W$ beinhaltet folgende Zahlen: 4, 6, 8, 9, 10, 12, 14, 15. Dementsprechend gilt für die Menge $P$ als das Komplement dieser Vereinigungsmenge:
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:$$P = {\{1, 2, 3, 5, 7, 11, 13\}}.$$
$P \in {\{1, 2, 3, 5, 7, 11, 13\}}$.
 
  
Dies sind genau die mit 4 Bit darstellbaren Primzahlen ⇒  <u>Lösungsvorschlag 2</u>.
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*These are exactly the prime numbers which can be represented with four bits &nbsp; &nbsp; <u>Proposed solution 2</u>.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Stochastische Signaltheorie|^1.2 Mengentheoretische Grundlagen^]]
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[[Category:Theory of Stochastic Signals: Exercises|^1.2 Set Theory Basics^]]

Latest revision as of 20:33, 28 November 2021

Logical circuit for D/B converting

A number generator  $Z$  supplies decimal values in the range  $1$  to  $15$.

  • These are converted into binary numbers  (block outlined in red).
  • The output consists of the four binary values  $A$,  $B$,  $C$  and  $D$  with decreasing significance.
  • For example  $Z = 11$  delivers the binary values
$$ A = 1, \ B = 0, \ C = 1, \ D = 1. $$

Set-theoretically,  this can be represented as follows:

$$ Z = 11\qquad\widehat{=}\qquad A \cap\overline{ B} \cap C \cap D.$$

Three more Boolean expressions are formed from the binary quantities  $A$,  $B$,  $C$  and  $D$  and their union set is denoted  $X$ :

\[ U = A \cap \overline{D} \]
\[ V = \overline{A} \cap B \cap \overline{D} \]
$$W,\; {\rm where} \; \, \overline{W} = \overline{A} \cup \overline{D} \cup (\overline{B} \cap C) \cup (B \cap \overline{C}). $$
  • Note that  $Z = 0 \ ⇒ \ A = B = C = D = 0$  is already excluded by the number generator.
  • Note also that not all input quantities  $A$,  $B$,  $C$  and  $D$  are used to calculate all intermediate quantities  $U$,  $V$  and  $W$,  resp.



Hints:

  • The exercise belongs to the chapter  Set theory basic.
  • The topic of this chapter is illustrated with examples in the (German language) learning video: 
"Mengentheoretische Begriffe und Gesetzmäßigkeiten"   ⇒   "Set-theoretical terms and laws".



Questions

1

Which statements are true regarding the random variable  $U$?

$U$  contains two elements.
$U$  contains four elements.
The smallest element of  $U$  is  $4$.
The largest element of  $U$  is  $14$.

2

Which statements are true regarding the random variable  $V$?

$V$  contains two elements.
$V$  contains four elements.
The smallest element of  $V$  is  $4$.
The largest element of  $V$  is  $14$.

3

Which statements are true regarding the random variable  $W$?

$W$  contains two elements.
$W$  contains four elements.
The smallest element of  $W$  is  $4$.
The largest element of  $W$  is  $14$.

4

Which statements are true regarding the random quantity  $P$ ?

$P$  contains all powers of two.
$P$  contains all prime numbers.
$P$  describes the empty set  $\phi$.
$P$  is identical with the universal set  $G = {1,2, \ \text{...} \ , 15}$.


Solution

(1)  The event  $U$  contains

  • those numbers greater/equal to eight  $(A = 1)$, 
  • which are even  $(D = 0)$:  $8, 10, 12, 14$  


⇒  Proposed solutions 2 and 4  are correct.


(2)  The event  $V$  consists of the two numbers  $4$  (binary 0100) and  $6$  (binary 0110)   ⇒   The correct solutions are 1 and 3.


Auxiliary Venn diagram

(3)  For the event  $W$,  de Morgan's theorem holds:

$$\overline W = \overline A \cup \overline D \cup (\overline B \cap C) \cup (B \cap \overline C) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} W = \overline{\overline W} = A \cap D \cap (\overline{\overline B \cap C}) \cap (\overline{B \cap \overline C}).$$
  • Using de Morgan's theorems,  it further follows:
$$ W = A \cap D \cap (B \cup \overline C) \cap (\overline B \cup C).$$
  • Finally,  using the Boolean relation  $(B \cup \overline C) \cap (\overline B \cup C) = (B \cap C) \cup (\overline B \cap \overline C)$  we obtain  (see sketch):
$$W = (A \cap B \cap C \cap D) \cup (A \cap \overline B \cap \overline C \cap D).$$
  • Thus,  $W$  contains the numbers  $15$  and  $9$    ⇒   only the  proposed solution 1  is correct.


(4)  The union of  $U$,  $V$  and  $W$  contains the following numbers:   $4, 6, 8, 9, 10, 12, 14, 15$.

  • Accordingly, the set  $P$  as the complement of this union is:  
$$P = {\{1, 2, 3, 5, 7, 11, 13\}}.$$
  • These are exactly the prime numbers which can be represented with four bits   ⇒   Proposed solution 2.