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 | + | '''(3)''' When three subscribers are served by J=4 <br> ⇒ only two subscribers can still be served by a spreading sequence with J=8 (see exemplary yellow background in the diagram) ⇒ K=5_. |
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*Because of 2·8+1·4+2·2+8=32, the desired occupancy is just allowed ⇒ <u>answer YES</u>. | *Because of 2·8+1·4+2·2+8=32, the desired occupancy is just allowed ⇒ <u>answer YES</u>. | ||
| − | *For example, supplying the spreading factor J=4 twice blocks the upper half of the tree. | + | *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. | + | *After providing one spreading J=8, three of the eight branches remain to be occupied on the J=8 level, and so on. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 17:23, 13 December 2021
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
- ⟨c(0)ν⟩=+1+1+1+1,⟨c(1)ν⟩=+1+1−1−1,
- ⟨c(2)ν⟩=+1−1+1−1,⟨c(3)ν⟩=+1−1−1+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:
- The exercise belongs to the chapter Spreading Sequences for CDMA.
- Reference is made in particular to the section Codes with variable spreading factor (OVSF codes) in the theory part.
- We would also like to draw your attention to the (German language) interactive SWF module OVSF.
Questions
Solution
(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:
- K4⋅324+K8⋅328+K16⋅3216+K32⋅3232≤32⇒K4⋅8+K8⋅4+K16⋅2+K32⋅1≤32.
- 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.
