Exercise 3.7Z: Spread Spectrum in UMTS

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 transmitted signal $s(t)$.
This is by the spreading factor $J$ higher frequent (=more frequent) 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 diagram shows exemplary signal characteristics of $q(t)$ und $c(t)$.

Notes:

This exercise belongs to the chapter Die Charakteristika von UMTS.
Reference is made to the chapter Nachrichtentechnische Aspekte von UMTS in the book„Beispiele von Nachrichtensystemen”.
For the calculation of the chip duration $T_{\rm C}$ , please refer to page Physikalische Kanäle .
There you will find, among other things, the information important for this task, which is transmitted on the so-called Dedicated Physical Channel (DPCH) in ten milliseconds exactly $15 \cdot 2560 \ \rm Chips$ .
In subtask (5), the system asks for transmit chips. For example, the "sending chip" $s_{3}$ denotes the constant signal value of $s(t)$ in the time interval $2T_{\rm C}$ ... $3T_{\rm C}$.

	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.

$R_{\rm C} \ = \ $

$\ \rm Mchip/s $

$T_{\rm C} \hspace{0.18cm} = \ $

$ \ \rm µ s $

$J \ = \ $

$R_{\rm B} \ = \ $

$\ \rm kbit/s $

$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 is $R_{\rm C} = 100 \cdot 38400 \ {\rm Chips/s} \ \underline{= 3.84 \ \rm Mchip/s}$.
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 $J = 4$ zu $R_{\rm B} = R_{\rm C}/J \ \underline{= 960 \ \rm kbit/s}$.

With the maximum spreading factor $J = 512$ for UMTS, the bit rate is only 5s $7.5 \ \rm kbit/s$.

(5) The following applies to 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 ($q_{1} = +1)$:

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

On the other hand, the two other transmission chips sought are to be assigned to the second data bit $(q_{2} = -1)$:

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