Aufgaben:Exercise 3.8: OVSF Codes

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Tree diagram to construct
an OVSF–Code

The spreading codes for UMTS should

• be orthogonal, in order to avoid mutual influence of the participants,
• at the same time also allow a flexible realization of different spreading factors  $J$ .

An example are the  "Orthogonal Variable Spreading Factor Codes"  $\rm (OVSF)$, which provide the spreading codes of lengths from  $J = 4$  to  $J = 512$ .

As shown in the graphic, these can be created with the help of a code tree.  In doing so, each branching from a code   $\mathcal{C}$ results in two new codes   $(+\mathcal{C}\ +\mathcal{C})$  and  $(+\mathcal{C} \ –\mathcal{C})$.

The diagram illustrates the principle given here using the following example  $J = 4$.  If you number the spreading sequences from  $0$  to  $J -1$, the spreading sequences result

$$\langle c_\nu^{(0)}\rangle = \ {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm},$$

$$ \langle c_\nu^{(1)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$

$$\langle c_\nu^{(2)}\rangle = \ {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$

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

According to this nomenclature, there are the spreading sequences  $\langle c_\nu^{(0)}\rangle, \text{...} ,\langle c_\nu^{(7)}\rangle$  for the spreading factor  $J = 8$. 

It should be noted that no predecessor or successor of a code may be used by other participants.

• 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:

• This task belongs to the chapter  Spreizfolgen für CDMA.
• Particular reference is made to the page  Codes mit variablem Spreizfaktor (OVSF–Code).

Questionnaire

1

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

	$\langle c_\nu^{(1)}\rangle = +\hspace{-0.05cm}1 \ +\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1$,
	$\langle c_\nu^{(3)}\rangle = +\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1$,
	$\langle c_\nu^{(5)}\rangle = +\hspace{-0.05cm}1 \ -\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1$,
	$\langle c_\nu^{(7)}\rangle = +\hspace{-0.05cm}1 \ -\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ +\hspace{-0.08cm}1 \ -\hspace{-0.08cm}1$.

2

How many UMTS users can be served with  $J = 8$  at maximum?

$K_{\rm max} \ = \ $

3

How many users can be supplied with  $J = 8$  if three of them should use a spreading code with  $J = 4$ ?

$K \ = \ $

4

The tree structure applies to  $J = 32$.  Is the following assignment feasible:  
twice   $J = 4$, once  $J = 8$, once  $J = 164$  and eight times  $J = 32$?

	Yes.
	No.

Solution

Solution

OVSF tree structure for $J = 8$

(1)  The following graphic shows the OVSF tree structure for $J = $8 users.

• From this it can be seen that the solutions 1, 3 and 4 apply, but not the second.

(2)  If each user is assigned a spreading code with the spreading degree $J = 8$, $K_{\rm max} \ \underline{= 8}$ users can be supplied.

(3)  If three users are supplied with $J = 4$, only two users can be served by a spreading sequence with $J = 8$ (see example yellow background in the graphic) $\ \Rightarrow \ \ \underline{K = 5}$.

(4)  We denote

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

Then the following condition must be fulfilled:

$$K_4 \cdot \frac{32}{4} + K_8 \cdot \frac{32}{8} +K_{16} \cdot \frac{32}{16} +K_{32} \cdot \frac{32}{32} \le 32\hspace{0.3cm} \Rightarrow \hspace{0.3cm} K_4 \cdot8 + K_8 \cdot 4 +K_{16} \cdot 2 +K_{32} \cdot1 \le 32 \hspace{0.05cm}.$$

• Because $2 \cdot 8 + 1 \cdot 4 + 2 \cdot 2 + 8 = 32$ the desired assignment is just allowed   ⇒   The answer is YES.
• For example, providing the $J = 4$ twice blocks the upper half of the tree, after providing a $J = 8$ spreading code, $3$ of the $8$ branches remain to be occupied at the $J = 8$ level, and so on and so forth.