Difference between revisions of "Aufgaben:Exercise 5.4Z: OVSF Codes"

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'''(3)'''  When three subscribers are served by  $J = 4$  only two subscribers can still be served by a spreading sequence with  $J = 8$  (see exemplary yellow background in the diagram above)   ⇒   $K\hspace{0.15cm}\underline{ = 5}$.
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'''(3)'''&nbsp; When three subscribers are served by&nbsp; $J = 4$ <br>&nbsp; &rArr; &nbsp; only two subscribers can still be served by a spreading sequence with&nbsp; $J = 8$&nbsp; (see exemplary yellow background in the diagram) &nbsp; ⇒ &nbsp; $K\hspace{0.15cm}\underline{ = 5}$.
  
  
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*Because of &nbsp;$2 · 8 + 1 · 4 + 2 · 2 + 8 = 32$,&nbsp; the desired occupancy is just allowed &nbsp; ⇒ &nbsp; <u>answer YES</u>.  
 
*Because of &nbsp;$2 · 8 + 1 · 4 + 2 · 2 + 8 = 32$,&nbsp; the desired occupancy is just allowed &nbsp; ⇒ &nbsp; <u>answer YES</u>.  
*For example, supplying the spreading factor &nbsp;$J = 4$&nbsp; twice blocks the upper half of the tree.
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*For example,&nbsp; supplying the spreading factor &nbsp;$J = 4$&nbsp; twice blocks the upper half of the tree.
*After providing one spreading &nbsp;$J = 8$, three of the eight branches remain to be occupied on the &nbsp;$J = 8$ level, and so on.
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*After providing one spreading &nbsp;$J = 8$,&nbsp; three of the eight branches remain to be occupied on the&nbsp; $J = 8$&nbsp; level, and so on.
  
 
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Revision as of 16:23, 13 December 2021

Construction of an OVSF code

The spreading codes for  UMTS  should

  • all be orthogonal to each other in order to avoid mutual interference between subscribers,
  • additionally allow a flexible realization of different spreading factors  $J$. 


One example of this is the so-called  "Orthogonal Variable Spreading Factor" code,  which provide spreading codes with lengths from  $J = 4$  to  $J = 512$. 

These can be created using a code tree,  as shown in the diagram.  In this process,  two new codes  $(+C \ +C)$  and  $(+C \ -C)$  are created from a code  $C$  at each branching.

The diagram illustrates the principle given here with the example  $J = 4$: 

  • If the spreading sequences are numbered from  $0$  to  $J -1$  the spreading sequences are as follows
$$\langle c_\nu^{(0)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm},\hspace{0.3cm} \langle c_\nu^{(1)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$
$$\langle c_\nu^{(2)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},\hspace{0.3cm} \langle c_\nu^{(3)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm}.$$
  • According to this nomenclature,  for the spreading factor  $J = 8$  there are the spreading sequences  $\langle c_\nu^{(0)}\rangle $,  ... ,  $\langle c_\nu^{(7)}\rangle $.
  • Note that no predecessor and successor of a code may be used for another participant.
  • So,  in the example,  four spreading codes with spreading factor  $J = 4$  could be used or the three codes highlighted in yellow – once with  $J = 2$  and twice with  $J = 4$.




Notes:


Questions

1

Construct the tree diagram for  $J = 8$.  What are the resulting OVSF codes?

Codeword 1:   $ \langle c_\nu^{(1)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm}{-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm}{-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$
Codeword 3:   $ \langle c_\nu^{(3)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm}{-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1}$ ,
Codeword 5:   $ \langle c_\nu^{(5)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm}$,
Codeword 7:   $ \langle c_\nu^{(7)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm}{-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm}$.

2

What is the maximum number of UMTS subscribers  $(K_{\rm max})$  that can be served with  $J = 8$ ?

$K_{\rm max} \ = \ $

3

How many subscribers  $(K)$  can be served if three of these subscribers are to use a spreading code with  $J = 4$ ?

$K \ = \ $

4

Assume a tree structure for  $J = 32$.   Is the following assignment feasible:
Twice  $J = 4$,  once  $J = 8$,  twice  $J = 16$  and eight times  $J = 32$?

Yes.
No.


Solution

OVSF tree structure for  $J = 8$

(1)  The diagram shows the OVSF tree structure for  $J = 8$ users.  From this it can be seen that  solutions 1, 3 and 4  apply,  but not the second one.


(2)  If each user is assigned a spreading code with  $J = 8$,   $K_{\rm max}\hspace{0.15cm}\underline{ = 8}$  subscribers can be served.


(3)  When three subscribers are served by  $J = 4$
  ⇒   only two subscribers can still be served by a spreading sequence with  $J = 8$  (see exemplary yellow background in the diagram)   ⇒   $K\hspace{0.15cm}\underline{ = 5}$.


(4)  We denote by

  • $K_4 = 2$  the number of spreading sequences with  $J = 4$,
  • $K_8 = 1$  the number of spreading sequences with  $J = 8$,
  • $K_{16} = 2$  the number of spreading sequences with  $J = 16$,
  • $K_{32} = 8$  the number of spreading sequences with  $J = 32$.


Then the following condition must be satisfied:

$$K_4 \cdot \frac{32}{4} + K_8 \cdot \frac{32}{8} +K_{16} \cdot \frac{32}{16} +K_{32} \cdot \frac{32}{32} \le 32 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K_4 \cdot8 + K_8 \cdot 4 +K_{16} \cdot 2 +K_{32} \cdot1 \le 32 \hspace{0.05cm}.$$
  • Because of  $2 · 8 + 1 · 4 + 2 · 2 + 8 = 32$,  the desired occupancy is just allowed   ⇒   answer YES.
  • For example,  supplying the spreading factor  $J = 4$  twice blocks the upper half of the tree.
  • After providing one spreading  $J = 8$,  three of the eight branches remain to be occupied on the  $J = 8$  level, and so on.