Aufgaben:Exercise 4.5Z: About Spread Spectrum with UMTS

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Source signal and spread signal

With UMTS/CDMA,  the so-called  "Pseudo-noise modulation"  is applied.  Or:  "Direct Sequence Spread Spectrum":

• The rectangular digital signal  $q(t)$  is thereby multiplied by the spreading signal  $c(t)$  to give the transmitted signal  $s(t)$.

• This is higher in frequency than  $q(t)$  by the spreading factor  $J$,  and is referred to as  "spread spectrum".

• At the receiver,  the same spreading signal  $c(t)$  is multiplied in phase synchronism,  reversing the spreading process  ⇒   "despreading".

The graph shows example signal waveforms for  $q(t)$  and  $c(t)$.

In the subtask  (5)  is asked about transmit chips. For example, the "transmit chip"  $s_{3}$  denotes the constant signal value of  $s(t)$  in the time interval  $2 T_{\rm C} ... 3 T_{\rm C}$.

Hints:

• This exercise mostly refers to the page  "Telecommunications Aspects of UMTS".
• For calculation of chip duration  $T_{\rm C}$  reference is made to the theory page  "Physical channels"  in the chapter "UMTS network architecture".
• There you will find, among other things, the information that on the so-called  Dedicated Physical Channel  ('DPCH )  in ten milliseconds exactly  $15 \cdot 2560$ chips are transmitted.

Questions

1

Which statements are correct?

	With UMTS, the bit duration  $T_{\rm B}$  is fixed.
	For UMTS, the chip duration  $T_{\rm C}$  is fixed.
	Both quantities depend on the channel conditions.

2

Specify the chip duration  $T_{\rm C}$  and chip rate  $R_{\rm C}$  in the downlink.

$T_{\rm C} \hspace{0.28cm} = \ $ $ \ \rm µ s$

$R_{\rm C} \hspace{0.2cm} = \ $ $ \ \rm Mchip/s$

3

What spreading factor can be read from the graph on the information page?

$J \ = \ $

4

What is the bit rate at this spreading factor?

$ R_{\rm B} \ = \ $ $ \ \rm kbit/s$

5

What are the values  $(\pm 1)$  of the "chips" of the transmitted signal  $s(t)$?

$s_{3} \ = \ $

$s_{4} \ = \ $

$s_{5} \ = \ $

$s_{6} \ = \ $

Solution

Solution

(1)  Correct is the answer 2:

• Fixed for UMTS is the chip duration  $T_{\rm C}$, which is still to be calculated in the subtask (2).
• The larger the spreading degree  $J$  is, the larger the bit duration is.

(2)  According to the note on the information page, exactly  $15 \cdot 2560 = 38400$ chips are transferred in ten milliseconds.

• Thus the chip rate  $R_{\rm C} = 100 \cdot 38400 \ {\rm chips/s} \hspace{0.15cm}\underline{= 3.84 \ \rm Mchip/s}$.
• The chip duration is the reciprocal of this:   $T_{\rm C} \hspace{0.15cm}\underline{\approx 0.26 \ \rm µ s}$.

(3)  Each data bit consists of four spreading chips   ⇒   $\underline{J = 4}$.

(4)  The bit rate is given by  $J = 4$  to  $R_{\rm B} \hspace{0.15cm}\underline{= 960 \ \rm kbit/s}$.

• With the maximum spreading factor for UMTS  $J = 512 $, on the other hand, the bit rate is only more  $7.5 \ \rm kbit/s$.

(5)  For the transmitted signal  $s(t) = q(t) \cdot c(t)$.

• The chips  $s_{3}$  and  $s_{4}$  of the transmitted signal belong to the first data bit  $(q_{1} = +1)$:

$$s_3 = c_3 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_4 = c_4 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$

• In contrast, the two other transmitting chips we are looking for are associated with the second data bit  $(q_{2} = -1)$  :

$$s_5 = -c_5= -c_1 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_6 = -c_6= -c_2 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$