Difference between revisions of "Aufgaben:Exercise 5.3Z: Realization of a PN Sequence"

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{{quiz-Header|Buchseite=Modulationsverfahren/Spreizfolgen für CDMA
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{{quiz-Header|Buchseite=Modulation_Methods/Spreading_Sequences_for_CDMA
 
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}}
  
[[File:P_ID1886__Mod_Z_5_3.png|right|]]
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[[File:EN_Mod_Z_5_3neu.png|right|frame|Two PN generator realizations]]
Die Grafik zeigt zwei mögliche Generatoren zur Erzeugung von PN–Sequenzen in unipolarer Darstellung: $u_ν$ ∈ {0, 1}. Der obere Generator mit den Koeffizienten
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The diagram shows two possible generators for generating PN sequences in unipolar representation:   $u_ν ∈ \{0, 1\}$.  
$$ g_0 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_1 = 0 \hspace{0.05cm}, \hspace{0.2cm}g_2 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_3 = 1 \hspace{0.05cm}.$$
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*The upper generator with the coefficients
wird durch die Oktalkennung $(g_3, g_2, g_1, g_0)_{oktal} = (15)$ bezeichnet. Entsprechend ist die Oktalkennung des zweiten PN–Generators gleich (17).
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:$$ g_0 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_1 = 0 \hspace{0.05cm}, \hspace{0.2cm}g_2 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_3 = 1 \hspace{0.05cm}$$
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:is denoted by the octal identifier   $(g_3,\ g_2,\ g_1,\ g_0)_{\rm octal} = (15)$. 
  
Man spricht von einer M–Sequenz, wenn für die Periodenlänge der Folge 〈$u_ν$〉 gilt: $P = 2^G – 1$. Hierbei bezeichnet G den Grad des Schieberegisters, der gleich der Anzahl der Speicherzellen ist.
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*Accordingly,  the octal identifier of the second PN generator is  $(17)$.
  
'''Hinweis:''' Die Aufgabe bezieht sich auf das [http://en.lntwww.de/Modulationsverfahren/Spreizfolgen_f%C3%BCr_CDMA Kapitel 5.3] dieses Buches sowie auf das [http://en.lntwww.de/Stochastische_Signaltheorie/Erzeugung_von_diskreten_Zufallsgr%C3%B6%C3%9Fen Kapitel 2.5] im Buch „Stochastische Signaltheorie”. Wir möchten Sie gerne auch auf das folgende Lehrvideo hinweisen:
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*One speaks of an M-sequence if for the period length of the sequence   $〈u_ν〉$   holds:  
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:$$P = 2^G – 1.$$
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:Here,  $G$  denotes the degree of the shift register,  which is equal to the number of memory cells.
  
Verdeutlichung der PN–Generatoren (Dateigröße 982 kB – Dauer 5:08)
 
  
===Fragebogen===
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Notes:
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*The exercise belongs to the chapter   [[Modulation_Methods/Spreading_Sequences_for_CDMA|Spreading Sequences for CDMA]].
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*Reference is also made to the chapter   [[Theory_of_Stochastic_Signals/Erzeugung_von_diskreten_Zufallsgr%C3%B6%C3%9Fen |Generation of Discrete Random Variables]]  in the book "Theory of Stochastic Signals".
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* We would also like to draw your attention to the&nbsp;  (German language)&nbsp;  learning video <br> &nbsp; [[Erläuterung_der_PN–Generatoren_an_einem_Beispiel_(Lernvideo)|Erläuterung der PN–Generatoren an einem Beispiel]] &nbsp; &rArr;&nbsp;  "Explanation of PN generators using an example".&nbsp;
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
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{What is the degree &nbsp;$G$&nbsp; of the two PN generators considered here?
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|type="{}"}
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$G \ = \ $  { 3 }
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{Give the period length &nbsp;$P$&nbsp; of the PN generator with the octal identifier &nbsp;$(15)$.
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|type="{}"}
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$P\ = \ $  { 7 }
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{Which of the following statements are true for each M-sequence?
 
|type="[]"}
 
|type="[]"}
- Falsch
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- The number of&nbsp; "zeros"&nbsp; and&nbsp; "ones"&nbsp; is the same.
+ Richtig
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+ In each period there is one more&nbsp; "ones"&nbsp; than&nbsp; "zeros".
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+ The maximum number of consecutive&nbsp; "ones"&nbsp; is &nbsp;$G$.
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+ The sequence &nbsp;$1 0 1 0 1 0$ ... &nbsp; is not possible.
  
 
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{Specify the period length &nbsp;$P$&nbsp; of the PN generator with the octal identifier&nbsp;$(17)$.
{Input-Box Frage
 
 
|type="{}"}
 
|type="{}"}
$\alpha$ = { 0.3 }
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$P\ = \ $ { 1 }
 
 
  
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{Which PN generator produces an M-sequence?
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|type="[]"}
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+ The generator with the octal identifier &nbsp;$(15)$.
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- The generator with the octal identifier &nbsp;$(17)$.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''
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'''(1)'''&nbsp; The degree&nbsp; $\underline{G = 3}$&nbsp; is equal to the number of memory cells of the shift register.
'''2.'''
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'''3.'''
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'''4.'''
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'''(2)'''&nbsp; From the given sequence the period length&nbsp; $\underline{P = 7}$&nbsp; can be read.&nbsp; Because of&nbsp; $P = 2^G –1$&nbsp; it is an M-sequence.
'''5.'''
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'''6.'''
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'''7.'''
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'''(3)'''&nbsp; <u>Solutions 2, 3 and 4</u>&nbsp; are correct:
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*The maximum number of consecutive&nbsp; "ones"&nbsp; is&nbsp; $G$&nbsp; (whenever there is a&nbsp; "one"&nbsp; in all&nbsp; $G$&nbsp; memory cells).
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*On the other hand,&nbsp; it is not possible that all memory cells are filled with zeros&nbsp; (otherwise only zeros would be generated).
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*Therefore,&nbsp; there is always one more&nbsp; "ones"&nbsp; than zeros.
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*The period length of the sequence&nbsp; "$1 0 1 0 1 0$ ..." &nbsp; is&nbsp; $P = 2$.&nbsp; For an M-sequence&nbsp; $P = 2^G –1$.&nbsp; For no value of&nbsp; $G$:&nbsp; &nbsp; $P = 2$&nbsp; is possible.
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'''(4)'''&nbsp; If all memory cells are occupied with ones,&nbsp; the generator with the octal identifier&nbsp; $(17)$&nbsp; returns a&nbsp; $1$&nbsp; again:
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:$$u_{\nu} \big [ u_{\nu-1} + u_{\nu-2} + u_{\nu-3} \big ] \,\,{\rm mod} \,\,2 =1 \hspace{0.05cm}.$$
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*Since this does not change the memory allocation,&nbsp; all further binary values generated will also be&nbsp; $1$&nbsp; each &nbsp; ⇒ &nbsp; $\underline{P = 1}$.
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'''(5)'''&nbsp; <u>Answer 1</u>&nbsp; is correct:
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*One speaks of an M-sequence only if&nbsp; $P = 2^G –1$&nbsp; holds.  
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*Here,&nbsp; "M"&nbsp; stands for "maximum".
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{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Modulationsverfahren|^5.3 Spreizfolgen für CDMA^]]
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[[Category:Modulation Methods: Exercises|^5.3 Spread Sequences for CDMA^]]

Latest revision as of 17:32, 20 December 2021

Two PN generator realizations

The diagram shows two possible generators for generating PN sequences in unipolar representation:   $u_ν ∈ \{0, 1\}$.

  • The upper generator with the coefficients
$$ g_0 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_1 = 0 \hspace{0.05cm}, \hspace{0.2cm}g_2 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_3 = 1 \hspace{0.05cm}$$
is denoted by the octal identifier   $(g_3,\ g_2,\ g_1,\ g_0)_{\rm octal} = (15)$. 
  • Accordingly,  the octal identifier of the second PN generator is  $(17)$.
  • One speaks of an M-sequence if for the period length of the sequence   $〈u_ν〉$  holds:
$$P = 2^G – 1.$$
Here,  $G$  denotes the degree of the shift register,  which is equal to the number of memory cells.



Notes:


Questions

1

What is the degree  $G$  of the two PN generators considered here?

$G \ = \ $

2

Give the period length  $P$  of the PN generator with the octal identifier  $(15)$.

$P\ = \ $

3

Which of the following statements are true for each M-sequence?

The number of  "zeros"  and  "ones"  is the same.
In each period there is one more  "ones"  than  "zeros".
The maximum number of consecutive  "ones"  is  $G$.
The sequence  $1 0 1 0 1 0$ ...   is not possible.

4

Specify the period length  $P$  of the PN generator with the octal identifier $(17)$.

$P\ = \ $

5

Which PN generator produces an M-sequence?

The generator with the octal identifier  $(15)$.
The generator with the octal identifier  $(17)$.


Solution

(1)  The degree  $\underline{G = 3}$  is equal to the number of memory cells of the shift register.


(2)  From the given sequence the period length  $\underline{P = 7}$  can be read.  Because of  $P = 2^G –1$  it is an M-sequence.


(3)  Solutions 2, 3 and 4  are correct:

  • The maximum number of consecutive  "ones"  is  $G$  (whenever there is a  "one"  in all  $G$  memory cells).
  • On the other hand,  it is not possible that all memory cells are filled with zeros  (otherwise only zeros would be generated).
  • Therefore,  there is always one more  "ones"  than zeros.
  • The period length of the sequence  "$1 0 1 0 1 0$ ..."   is  $P = 2$.  For an M-sequence  $P = 2^G –1$.  For no value of  $G$:    $P = 2$  is possible.


(4)  If all memory cells are occupied with ones,  the generator with the octal identifier  $(17)$  returns a  $1$  again:

$$u_{\nu} \big [ u_{\nu-1} + u_{\nu-2} + u_{\nu-3} \big ] \,\,{\rm mod} \,\,2 =1 \hspace{0.05cm}.$$
  • Since this does not change the memory allocation,  all further binary values generated will also be  $1$  each   ⇒   $\underline{P = 1}$.


(5)  Answer 1  is correct:

  • One speaks of an M-sequence only if  $P = 2^G –1$  holds.
  • Here,  "M"  stands for "maximum".