Aufgaben:Exercise 4.6: OVSF Codes

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Tree structure for the construction
of an OVSF code

The spreading codes for UMTS should

• all be orthogonal to each other to avoid mutual interference between subscribers,
• be as flexible as possible in order to realize different spreading factors  $J$ .

An example of this are the so-called  Orthogonal Variable Spreading Factor  (OVSF) codes, which provide spreading codes of lengths from  $J = 4$  to  $J = 512$ . These can be created using a code tree, as shown in the diagram. Thereby, at each branch from one code  $\mathcal{C}$  two new codes are created.

• $(+\mathcal{C} \ +\hspace{-0.05cm}\mathcal{C})$,
• $(+\mathcal{C}\ -\hspace{-0.05cm}\mathcal{C})$.

The diagram illustrates the principle given here by the example  $J = 4$.

If we number the spreading sequences from  $0$  to  $J -1$  we get here the spreading sequences

$$ \langle c_\nu^{(0)}\rangle \ = \ {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm},\hspace{0.3cm} \langle c_\nu^{(1)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$

$$\langle c_\nu^{(2)}\rangle \ = \ {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},\hspace{0.3cm} \langle c_\nu^{(3)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm}.$$

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

Note that no predecessor and successor of a code may be used by other participants.

• 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  $J = 4$.

Hints:

• This exercise belongs to the chapter  "Telecommunications Aspects of UMTS".
• Reference is made in particular to the page  "Spreading codes and scrambling with UMTS".

Questions

1

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

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

2

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

$K_{\rm max} \ = \ $

3

How many subscribers can be served if three of them 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  $J = 16$,  eight times $J = 32$ ?

	Yes.
	No.

Solution

Solution

OVSF tree structure for  $J = 8$

(1)  The graph shows the OVSF tree structure for  $J = 8$  user.

• From this it can be seen that proposed solutions 1, 3, and 4 are true, but not the second one.

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

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

(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:

$$ K_4 \cdot \frac{32}{4} + K_8 \cdot \frac{32}{8} +K_{16} \cdot \frac{32}{16} +K_{32} \cdot \frac{32}{32} \le 32 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K_4 \cdot8 + K_8 \cdot 4 +K_{16} \cdot2 +K_{32} \cdot1 \le 32 \hspace{0.05cm}.$$

• Because  $2 \cdot 8 + 1 \cdot 4 + 2 \cdot 2 + 8 = 32$  the desired allocation is just allowed   ⇒   Answer YES.
• Providing the degree of spread twice  $J = 4$  blocks, for example, the top half of the tree, after providing one spread with  $J = 8$, three of the eight branches remain to be occupied at the  $J = 8$ level, and so on.