Exercise 3.7Z: Spread Spectrum in UMTS

From LNTwww

Source signal and spread signal

For UMTS/CDMA, the so-called  "PN modulation"  is applied:

  • The rectangular digital signal  $q(t)$  is multiplied by the spread signal  $c(t)$  and gives the transmission signal  $s(t)$.
  • This is by the spreading factor  $J$  of higher frequency than  $q(t)$;  this is called  "band spreading".

At the receiver the same spreading signal  $c(t)$  is added (namely in phase!).  This reverses the band spreading   ⇒   "band compression".

The graphic shows exemplary signal characteristics of  $q(t)$  und  $c(t)$.


  • This task refers to the chapter  Characteristics of UMTS.
  • Reference is made to the chapter  Telecommunications Aspects of UMTS  in the book "Examples of Communication_Systems".
  • For the calculation of the chip duration   $T_{\rm C}$,  please refer to page  Physical Channels .
  • There you will find, among other things, the information important for this task, which is transmitted on the so-called  "Dedicated Physical Channel"  $\rm (DPCH)$  in ten milliseconds exactly  $15 \cdot 2560 \ \rm Chips$.
  • In subtask  (5), the system asks for "transmitting chips". 
    For example, the "transmitting chip"  $s_{3}$  denotes the constant signal value of  $s(t)$  in the time interval  $2T_{\rm C}$ ... $3T_{\rm C}$.



Which of the following statements are true?

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


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

$R_{\rm C} \ = \ $

$\ \rm Mchip/s $
$T_{\rm C} \hspace{0.18cm} = \ $

$ \ \rm µ s $


Which spreading factor can be read from the graph on the data page?

$J \ = \ $


What is the bit rate with this spreading factor?

$R_{\rm B} \ = \ $

$\ \rm kbit/s $


What are the  "transmitting chips"  ⇒   of the signal  $s(t)$?

$s_{3} \ = \ $

$s_{4} \ = \ $

$s_{5} \ = \ $

$s_{6} \ = \ $


(1)  Correct is the solution 2:

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

(2)  According to the note on the information page, in  $10 \ \rm ms$  exactly  $15 \cdot 2560 = 38400 \ \rm chips$  are transferred.

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

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

(4)  The bit rate is calculated with the spreading factor  $J = 4$  to  $R_{\rm B} = R_{\rm C}/J \ \underline{= 960 \ \rm kbit/s}$.

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

(5)  The following applies to the transmission signal $s(t) = q(t) \cdot c(t)$.

  • The chips  $s_{3}$  and  $s_{4}$  of the transmission signal belong to the first data bit  $(q_{1} = +1)$:
$$s_3 = c_3 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_4 = c_4 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$
  • On the other hand, the two other transmission chips sought are to be assigned to the second data bit $(q_{2} = -1)$:
$$s_5 = -c_5= -c_1 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_6 = -c_6= -c_2 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$