View source for Aufgaben:Exercise 3.7Z: Spread Spectrum in UMTS


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[[File:P_ID2260__Bei_Z_4_5.png|right|frame|Source signal and spread signal]]
For UMTS/CDMA, the so-called&nbsp; "PN modulation"&nbsp; is applied: 
*The rectangular digital signal&nbsp; $q(t)$&nbsp; is multiplied by the spread signal&nbsp; $c(t)$&nbsp; and gives the transmission signal&nbsp; $s(t)$. 
*This is by the spreading factor&nbsp; $J$&nbsp; of higher frequency than&nbsp; $q(t)$;&nbsp; this is called&nbsp; "band spreading".


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

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





''Notes:''

*This task refers to the chapter&nbsp; [[Mobile_Communications/Characteristics_of_UMTS|Characteristics of UMTS]].
*Reference is made to the chapter&nbsp; [[Examples_of_Communication_Systems/Telecommunications_Aspects_of_UMTS|Telecommunications Aspects of UMTS]]&nbsp; in the book "Examples of Communication_Systems". 
*For the calculation of the chip duration &nbsp; $T_{\rm C}$,&nbsp; please refer to section&nbsp; [[Examples_of_Communication_Systems/UMTS_Network_Architecture#Physical_channels|Physical Channels]]&nbsp;. 
*There you will find, among other things, the information important for this task, which is transmitted on the so-called&nbsp; "Dedicated Physical Channel"&nbsp; $\rm (DPCH)$&nbsp; in ten milliseconds exactly&nbsp; $15 \cdot 2560 \ \rm Chips$.
*In subtask&nbsp; '''(5)''', the system asks for "transmitting chips".&nbsp; <br>For example, the "transmitting chip"&nbsp; $s_{3}$&nbsp; denotes the constant signal value of&nbsp; $s(t)$&nbsp; in the time interval&nbsp; $2T_{\rm C}$ ... $3T_{\rm C}$.

 



===Questionnaire===

<quiz display=simple>

{Which of the following statements are true?
|type="[]"}
- For UMTS the bit duration&nbsp; $T_{\rm B}$&nbsp; is fixed.
+ For UMTS, the chip duration&nbsp; $T_{\rm C}$&nbsp; is fixed.
- Both values depend on the channel conditions.

{Specify the chip duration&nbsp; $T_{\rm C}$&nbsp; and the chip rate&nbsp; $R_{\rm C}$&nbsp; in the downlink.
|type="{}"}
$R_{\rm C} \ = \ $ { 3.84 3% } $\ \rm Mchip/s $
$T_{\rm C} \hspace{0.18cm} = \ $ {  0.26 3% } $ \ \rm &micro; s $

{Which spreading factor can be read from the graph in the data section?
|type="{}"}
$J \ = \ $ { 4 }

{What is the bit rate with this spreading factor?
|type="{}"}
$R_{\rm B} \ = \ $ { 960 3% } $\ \rm kbit/s $

{What are the&nbsp; "transmitting chips"&nbsp; &rArr; &nbsp; of the signal&nbsp; $s(t)$?
|type="{}"}
$s_{3} \ = \ $ { -1.03--0.97 }
$s_{4} \ = \ $ { 1 3% }
$s_{5} \ = \ $ { -1.03--0.97 }
$s_{6} \ = \ $ { 1 3% }

</quiz>

===Musterlösung===
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'''(1)'''&nbsp; Correct is the <u>solution 2</u>:
*For UMTS, the chip duration&nbsp; $T_{\rm C}$, which is still to be calculated in the subtask&nbsp; '''(2)''', is predefined. 
*The greater the degree of spreading&nbsp; $J$, the greater the bit duration.
 


'''(2)'''&nbsp; According to the note in the information section, in&nbsp; $10 \ \rm ms$&nbsp; exactly&nbsp; $15 \cdot 2560 = 38400 \ \rm chips$&nbsp; are transferred. 
*With this the chip rate:&nbsp; $R_{\rm C} = 100 \cdot 38400 \ {\rm Chips/s} \ \underline{= 3.84 \ \rm Mchip/s}$. 
*The chip duration is the reciprocal of this: &nbsp; $T_{\rm C} \ \underline{\approx 0.26 \ \rm &micro; s}$.



'''(3)'''&nbsp; Each data bit consists of four spreading chips  &nbsp; &rArr; &nbsp;   $\underline{J = 4}$.



'''(4)'''&nbsp; The bit rate is calculated with the spreading factor&nbsp; $J = 4$&nbsp; to&nbsp; $R_{\rm B} = R_{\rm C}/J \ \underline{= 960 \ \rm  kbit/s}$.
* With the maximum spreading factor&nbsp; $J = 512$&nbsp; for UMTS, the bit rate is only&nbsp; $7.5 \ \rm kbit/s$.



'''(5)'''&nbsp; The following applies to the transmission signal $s(t) = q(t) \cdot c(t)$. 
*The chips&nbsp; $s_{3}$&nbsp; and&nbsp; $s_{4}$&nbsp; of the transmission signal belong to the first data bit&nbsp;  $(q_{1} = +1)$:
:$$s_3 = c_3 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_4 = c_4 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$
*On the other hand, the two other transmission chips sought are to be assigned to the second data bit $(q_{2} = -1)$:
:$$s_5 = -c_5= -c_1 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_6 = -c_6= -c_2 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$

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[[Category:Mobile Communications: Exercises|^3.4 Characteristics of UMTS^]]