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:EN_Mod_Z_5_3.png|right|frame|Two possible realizations for PN generators]]
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[[File:EN_Mod_Z_5_3neu.png|right|frame|Two PN generator realizations]]
 
The diagram shows two possible generators for generating PN sequences in unipolar representation:   $u_ν ∈ \{0, 1\}$.  
 
The diagram shows two possible generators for generating PN sequences in unipolar representation:   $u_ν ∈ \{0, 1\}$.  
 
*The upper generator with the coefficients
 
*The upper generator with the coefficients
 
:$$ g_0 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_1 = 0 \hspace{0.05cm}, \hspace{0.2cm}g_2 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_3 = 1 \hspace{0.05cm}$$
 
:$$ g_0 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_1 = 0 \hspace{0.05cm}, \hspace{0.2cm}g_2 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_3 = 1 \hspace{0.05cm}$$
:is denoted by the octal identifier   $(g_3, g_2, g_1, g_0)_{\rm oktal} = (15)$. 
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: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 equal to  $(17)$.
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*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:  
 
*One speaks of an M-sequence if for the period length of the sequence   $〈u_ν〉$   holds:  
 
:$$P = 2^G – 1.$$  
 
:$$P = 2^G – 1.$$  
:Here   $G$  denotes the degree of the shift register, which is equal to the number of memory cells.
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:Here,  $G$  denotes the degree of the shift register,  which is equal to the number of memory cells.
  
  
  
  
 
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Notes:  
 
 
 
 
 
 
 
 
''Note:''
 
 
*The exercise belongs to the chapter   [[Modulation_Methods/Spreading_Sequences_for_CDMA|Spreading Sequences for CDMA]].
 
*The exercise belongs to the chapter   [[Modulation_Methods/Spreading_Sequences_for_CDMA|Spreading Sequences for CDMA]].
*However, 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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*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".  
* We would also like to draw your attention to the learning video  [[Erläuterung_der_PN–Generatoren_an_einem_Beispiel_(Lernvideo)|Explanation of PN generators using an example]].   
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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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$G \ = \ $  { 3 }  
 
$G \ = \ $  { 3 }  
  
{Give the period length &nbsp;$P$&nbsp; of the PN generator with the octal identifier &nbsp;$(15)$&nbsp; an.
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{Give the period length &nbsp;$P$&nbsp; of the PN generator with the octal identifier &nbsp;$(15)$.
 
|type="{}"}
 
|type="{}"}
 
$P\ = \ $  { 7 }  
 
$P\ = \ $  { 7 }  
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{Which of the following statements are true for each M-sequence?
 
{Which of the following statements are true for each M-sequence?
 
|type="[]"}
 
|type="[]"}
- The number of zeros and ones is the same.
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- The number of&nbsp; "zeros"&nbsp; and&nbsp; "ones"&nbsp; is the same.
+ In each period there is one more one than zeros.
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+ In each period there is one more&nbsp; "ones"&nbsp; than&nbsp; "zeros".
+ The maximum number of consecutive ones is &nbsp;$G$.
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+ The maximum number of consecutive&nbsp; "ones"&nbsp; is &nbsp;$G$.
 
+ The sequence &nbsp;$1 0 1 0 1 0$ ... &nbsp; is not possible.
 
+ The sequence &nbsp;$1 0 1 0 1 0$ ... &nbsp; is not possible.
  
{Specify the period length &nbsp;$P$&nbsp; of the PN generator with the octal identifier&nbsp;$(17)$&nbsp; an.
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{Specify the period length &nbsp;$P$&nbsp; of the PN generator with the octal identifier&nbsp;$(17)$.
 
|type="{}"}
 
|type="{}"}
 
$P\ = \ $ { 1 }
 
$P\ = \ $ { 1 }
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'''(3)'''&nbsp; <u>Solutions 2, 3 and 4</u> are correct:
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'''(3)'''&nbsp; <u>Solutions 2, 3 and 4</u>&nbsp; are correct:
*The maximum number of consecutive ones is&nbsp; $G$&nbsp; (whenever there is a one in all&nbsp; $G$&nbsp; memory cells).  
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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).  
*On the other hand, it is not possible that all memory cells are filled with zeros&nbsp; (otherwise only zeros would be generated).  
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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).  
*Therefore, there is always one more one than zeros.
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*Therefore,&nbsp; there is always one more&nbsp; "ones"&nbsp; than zeros.
*The period length of the last sequence is&nbsp; $P = 2$.&nbsp; For an M-sequence&nbsp; $P = 2^G –1$.&nbsp; For no value of&nbsp; $G$&nbsp; is&nbsp; $P = 2$&nbsp; possible.
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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.
  
  
  
'''(4)'''&nbsp; If all memory cells are occupied with ones, the generator with the octal identifier&nbsp; $(17)$&nbsp; returns a&nbsp; $1$ again:
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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:
 
:$$u_{\nu} \big [ u_{\nu-1} + u_{\nu-2} + u_{\nu-3} \big ] \,\,{\rm mod} \,\,2 =1 \hspace{0.05cm}.$$
 
:$$u_{\nu} \big [ u_{\nu-1} + u_{\nu-2} + u_{\nu-3} \big ] \,\,{\rm mod} \,\,2 =1 \hspace{0.05cm}.$$
*Since this does not change the memory allocation, all further binary values generated will also be&nbsp; $1$&nbsp; each &nbsp; ⇒ &nbsp; $\underline{P = 1}$.
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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; $\underline{P = 1}$.
  
  
  
'''(5)'''&nbsp; <u>Answer 1</u> is correct:  
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'''(5)'''&nbsp; <u>Answer 1</u>&nbsp; is correct:  
 
*One speaks of an M-sequence only if&nbsp; $P = 2^G –1$&nbsp; holds.  
 
*One speaks of an M-sequence only if&nbsp; $P = 2^G –1$&nbsp; holds.  
*Here, "M" stands for "maximum".
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*Here,&nbsp; "M"&nbsp; stands for "maximum".
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}

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".