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

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{Construct the tree diagram for  J=8.  What are the resulting OVSF codes?
 
{Construct the tree diagram for  J=8.  What are the resulting OVSF codes?
 
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+ '''Codeword 1:'''   c(1)ν=+1+1+1+11111,
+
+ '''Code word 1:'''   c(1)ν=+1+1+1+11111,
- '''Codeword 3:'''   c(3)ν=+1+111+1+111 ,
+
- '''Code word 3:'''   c(3)ν=+1+111+1+111 ,
+ '''Codeword 5:'''   c(5)ν=+11+11+11+11,
+
+ '''Code word 5:'''   c(5)ν=+11+11+11+11,
+ '''Codeword 7:'''   c(7)ν=+111+11+1+11.
+
+ '''Code word 7:'''   c(7)ν=+111+11+1+11.
  
  

Latest revision as of 18:13, 1 November 2022

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  J1  the spreading sequences are as follows
c(0)ν=+1+1+1+1,c(1)ν=+1+111,
c(2)ν=+11+11,c(3)ν=+111+1.
  • According to this nomenclature,  for the spreading factor  J=8  there are the spreading sequences  c(0)ν,  ... ,  c(7)ν.
  • 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?

Code word 1:   c(1)ν=+1+1+1+11111,
Code word 3:   c(3)ν=+1+111+1+111 ,
Code word 5:   c(5)ν=+11+11+11+11,
Code word 7:   c(7)ν=+111+11+1+11.

2

What is the maximum number of UMTS subscribers  (Kmax)  that can be served with  J=8 ?

Kmax = 

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,   Kmax=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=5_.


(4)  We denote by

  • K4=2  the number of spreading sequences with  J=4,
  • K8=1  the number of spreading sequences with  J=8,
  • K16=2  the number of spreading sequences with  J=16,
  • K32=8  the number of spreading sequences with  J=32.


Then the following condition must be satisfied:

K4324+K8328+K163216+K32323232K48+K84+K162+K32132.
  • 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.