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

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[[File:|right|]]
[[File:EN_Mod_Z_5_4.png|right|frame|Construction of an OVSF code]]
The spreading codes for &nbsp;[[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_UMTS|UMTS]]&nbsp; should
* all be orthogonal to each other in order to avoid mutual interference between subscribers,
* additionally allow a flexible realization of different spreading factors &nbsp;$J$.&nbsp;




===Fragebogen===
One example of this is the so-called &nbsp;[[Modulation_Methods/Spreading_Sequences_for_CDMA#Codes_with_variable_spreading_factor_.28OVSF_codes.29|"Orthogonal Variable Spreading Factor" code]],&nbsp; which provide spreading codes with lengths from &nbsp;$J = 4$&nbsp; to &nbsp;$J = 512$.&nbsp;
 
These can be created using a code tree,&nbsp; as shown in the diagram.&nbsp; In this process,&nbsp; two new codes &nbsp;$(+C \ +C)$&nbsp; and &nbsp;$(+C \ -C)$&nbsp; are created from a code&nbsp; $C$&nbsp; at each branching.
 
The diagram illustrates the principle given here with the example &nbsp;$J = 4$:&nbsp;
*If the spreading sequences are numbered from &nbsp;$0$&nbsp; to &nbsp;$J -1$&nbsp; 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,&nbsp; for the spreading factor &nbsp;$J = 8$&nbsp; there are the spreading sequences &nbsp;$\langle c_\nu^{(0)}\rangle $,&nbsp; ... ,&nbsp; $\langle c_\nu^{(7)}\rangle $.
*Note that no predecessor and successor of a code may be used for another participant.
*So,&nbsp; in the example,&nbsp; four spreading codes with spreading factor &nbsp;$J = 4$&nbsp; could be used or the three codes highlighted in yellow – once with &nbsp;$J = 2$&nbsp; and twice with &nbsp;$J = 4$.
 
 
 
 
 
 
Notes:
*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Spreizfolgen_für_CDMA|Spreading Sequences for CDMA]].
*Reference is made in particular to the section&nbsp; [[Modulation_Methods/Spreading_Sequences_for_CDMA#Codes_with_variable_spreading_factor_.28OVSF_codes.29|Codes with variable spreading factor (OVSF codes)]]&nbsp; in the theory part.
* We would also like to draw your attention to the&nbsp; (German language)&nbsp; interactive SWF module &nbsp;[[Applets:OVSF-Codes_(Applet)|OVSF]].&nbsp;
 
 
===Questions===


<quiz display=simple>
<quiz display=simple>
{Multiple-Choice Frage
{Construct the tree diagram for &nbsp;$J = 8$.&nbsp; What are the resulting OVSF codes?
|type="[]"}
|type="[]"}
- Falsch
+ '''Code word 1:''' &nbsp; $ \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},$
+ Richtig
- '''Code word 3:''' &nbsp; $ \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}$ ,
+ '''Code word 5:''' &nbsp; $ \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}$,
+ '''Code word 7:''' &nbsp; $ \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}$.




{Input-Box Frage
{What is the maximum number of UMTS subscribers &nbsp;$(K_{\rm max})$&nbsp; that can be served with &nbsp;$J = 8$&nbsp;?
|type="{}"}
|type="{}"}
$\alpha$ = { 0.3 }
$K_{\rm max} \ = \ $ { 8 }
 


{How many subscribers &nbsp;$(K)$&nbsp; can be served if three of these subscribers are to use a spreading code with &nbsp;$J = 4$&nbsp;?
|type="{}"}
$K \ = \ $ { 5 }


{Assume a tree structure for &nbsp;$J = 32$.&nbsp;&nbsp;  Is the following assignment feasible:<br>Twice &nbsp;$J = 4$,&nbsp; once &nbsp;$J = 8$,&nbsp; twice &nbsp;$J = 16$&nbsp; and eight times &nbsp;$J = 32$?
|type="()"}
+ Yes.
- No.
</quiz>
</quiz>


===Musterlösung===
===Solution===
{{ML-Kopf}}
{{ML-Kopf}}
'''1.'''
[[File:P_ID1892__Mod_Z_5_4a.png|right|frame|OVSF tree structure for &nbsp;$J = 8$]]
'''2.'''
'''(1)'''&nbsp; The diagram shows the OVSF tree structure for &nbsp;$J = 8$ users.&nbsp; From this it can be seen that&nbsp; <u>solutions 1, 3 and 4</u>&nbsp; apply,&nbsp; but not the second one.
'''3.'''
 
'''4.'''
 
'''5.'''
 
'''6.'''
'''(2)'''&nbsp; If each user is assigned a spreading code with&nbsp; $J = 8$,&nbsp;&nbsp; $K_{\rm max}\hspace{0.15cm}\underline{ = 8}$&nbsp; subscribers can be served.
'''7.'''
 
 
 
'''(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; $K\hspace{0.15cm}\underline{ = 5}$.
 
 
 
'''(4)'''&nbsp; We denote by
* $K_4 = 2$&nbsp; the number of spreading sequences with&nbsp; $J = 4$,
* $K_8 = 1$&nbsp; the number of spreading sequences with&nbsp; $J = 8$,
* $K_{16} = 2$&nbsp; the number of spreading sequences with&nbsp; $J = 16$,
* $K_{32} = 8$&nbsp; the number of spreading sequences with&nbsp; $J = 32$.
 
 
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 &nbsp;$2 · 8 + 1 · 4 + 2 · 2 + 8 = 32$,&nbsp; the desired occupancy is just allowed &nbsp; ⇒ &nbsp; <u>answer YES</u>.  
*For example,&nbsp; supplying the spreading factor &nbsp;$J = 4$&nbsp; twice blocks the upper half of the tree.
*After providing one spreading &nbsp;$J = 8$,&nbsp; three of the eight branches remain to be occupied on the&nbsp; $J = 8$&nbsp; level, and so on.
 
{{ML-Fuß}}
{{ML-Fuß}}






[[Category:Aufgaben zu Modulationsverfahren|^5.3 Spreizfolgen für CDMA^]]
[[Category:Modulation Methods: Exercises|^5.3 Spread Sequences for CDMA^]]
[[de:Aufgaben:Aufgabe 5.4Z: OVSF–Codes]]

Latest revision as of 17:56, 16 March 2026

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?

Code word 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},$
Code word 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}$ ,
Code word 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}$,
Code word 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.