Difference between revisions of "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]]
 
[[File:P_ID2260__Bei_Z_4_5.png|right|frame|Source signal and spread signal]]
For UMTS/CDMA, the so-called "PN modulation" is applied:  
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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  $s(t)$.  
 
*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".
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*This is by the spreading factor  $J$  of higher frequency than  $q(t)$;  this is called  "band spreading".
  
  
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*This task refers to the chapter  [[Mobile_Communications/Characteristics_of_UMTS|Characteristics of UMTS]].
 
*This task refers to the chapter  [[Mobile_Communications/Characteristics_of_UMTS|Characteristics of UMTS]].
 
*Reference is made to the chapter  [[Examples_of_Communication_Systems/Telecommunications_Aspects_of_UMTS|Telecommunications Aspects of UMTS]]  in the book "Examples of Communication_Systems".  
 
*Reference is made to the chapter  [[Examples_of_Communication_Systems/Telecommunications_Aspects_of_UMTS|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  [[Examples_of_Communication_Systems/UMTS_Network_Architecture#Physical_channels|Physical channels]] .  
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*For the calculation of the chip duration   $T_{\rm C}$,  please refer to section  [[Examples_of_Communication_Systems/UMTS_Network_Architecture#Physical_channels|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  $15 \cdot 2560 \ \rm Chips$.
 
*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 "sending chips".  For example, the "sending chip"  $s_{3}$  denotes the constant signal value of  $s(t)$  in the time interval  $2T_{\rm C}$ ... $3T_{\rm C}$.
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*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; $2T_{\rm C}$ ... $3T_{\rm C}$.
  
 
   
 
   
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$T_{\rm C} \hspace{0.18cm} = \ $ {  0.26 3% } $ \ \rm &micro; s $
 
$T_{\rm C} \hspace{0.18cm} = \ $ {  0.26 3% } $ \ \rm &micro; s $
  
{Which spreading factor can be read from the graph on the data page?
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{Which spreading factor can be read from the graph in the data section?
 
|type="{}"}
 
|type="{}"}
 
$J \ = \ $ { 4 }
 
$J \ = \ $ { 4 }
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$R_{\rm B} \ = \ $ { 960 3% } $\ \rm kbit/s $
 
$R_{\rm B} \ = \ $ { 960 3% } $\ \rm kbit/s $
  
{What are the&nbsp; "chips"&nbsp; of the signal&nbsp; $s(t)$?
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{What are the&nbsp; "transmitting chips"&nbsp; &rArr; &nbsp; of the signal&nbsp; $s(t)$?
 
|type="{}"}
 
|type="{}"}
 
$s_{3} \ = \ $ { -1.03--0.97 }
 
$s_{3} \ = \ $ { -1.03--0.97 }
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'''(1)'''&nbsp; Correct is the <u>solution 2</u>:
 
'''(1)'''&nbsp; Correct is the <u>solution 2</u>:
*For UMTS, the chip duration $T_{\rm C}$, which is still to be calculated in the subtask '''(2)''', is predefined.  
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*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 $J$, the greater the bit duration.
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*The greater the degree of spreading&nbsp; $J$, the greater the bit duration.
 
   
 
   
  
  
'''(2)'''&nbsp; According to the note on the information page, in $10 \ \rm ms$ exactly $15 \cdot 2560 = 38400 \ \rm Chips$ are transferred.  
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'''(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 is &nbsp; $R_{\rm C} = 100 \cdot 38400 \ {\rm Chips/s} \ \underline{= 3.84 \ \rm Mchip/s}$.  
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*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; $T_{\rm C} \ \underline{\approx 0.26 \ \rm &micro; s}$.
 
*The chip duration is the reciprocal of this: &nbsp; $T_{\rm C} \ \underline{\approx 0.26 \ \rm &micro; s}$.
  
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'''(4)'''&nbsp; The bit rate is calculated with the spreading factor $J = 4$ zu $R_{\rm B} = R_{\rm C}/J \ \underline{= 960 \ \rm  kbit/s}$.
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'''(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 $J = 512$ for UMTS, the bit rate is only 5s $7.5 \ \rm kbit/s$.
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* 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 transmitted signal $s(t) = q(t) \cdot c(t)$.  
+
'''(5)'''&nbsp; The following applies to the transmission 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  ($q_{1} = +1)$:
+
*The chips&nbsp; $s_{3}$&nbsp; and&nbsp; $s_{4}$&nbsp; of the transmission signal belong to the first data bit&nbsp; $(q_{1} = +1)$:
 
:$$s_3 = c_3 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_4 = c_4 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$
 
:$$s_3 = c_3 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_4 = c_4 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$
 
*On the other hand, the two other transmission chips sought are to be assigned to the second data bit $(q_{2} = -1)$:
 
*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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:$$s_5 = -c_5= -c_1 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_6 = -c_6= -c_2 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Exercises for Mobile Communications|^3.4 Characteristics of UMTS^]]
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[[Category:Mobile Communications: Exercises|^3.4 Characteristics of UMTS^]]

Latest revision as of 00:42, 13 November 2022

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  $c(t)$.



Notes:

  • 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 section  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  $2T_{\rm C}$ ... $3T_{\rm C}$.



Questionnaire

1

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.

2

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 $

3

Which spreading factor can be read from the graph in the data section?

$J \ = \ $

4

What is the bit rate with this spreading factor?

$R_{\rm B} \ = \ $

$\ \rm kbit/s $

5

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

$s_{3} \ = \ $

$s_{4} \ = \ $

$s_{5} \ = \ $

$s_{6} \ = \ $


Musterlösung

(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 in the information section, 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 $(q_{2} = -1)$:
$$s_5 = -c_5= -c_1 \hspace{0.15cm}\underline {= -1},\hspace{0.4cm}s_6 = -c_6= -c_2 \hspace{0.15cm}\underline {= +1}\hspace{0.05cm}.$$