Difference between revisions of "Aufgaben:Exercise 3.7: PN Modulation"

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'''(1)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:
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'''(1)'''&nbsp; Correct is the <u>solution 3</u>:
*Es handelt sich hier um einen optimalen Empfänger.  
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*This is an optimal receiver.  
*Ohne Rauschen ist das Signal $b(t)$ innerhalb eines jeden Bits konstant gleich $+1$ oder $-1$.  
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*Without noise, the signal $b(t)$ within each bit is constantly equal to $+1$ or $-1$.  
*Aus der angegebenen Gleichung für den Integrator
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*From the equation given for the integrator
 
:$$d (\nu T) = \frac{1}{T} \cdot \hspace{-0.03cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t$$
 
:$$d (\nu T) = \frac{1}{T} \cdot \hspace{-0.03cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t$$
:folgt, dass $d(\nu T)$ nur die Werte $±1$ annehmen kann.  
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:follows that $d(\nu T)$ can only take the values $±1$.  
  
  
'''(2)'''&nbsp; Richtig ist wieder der <u>Lösungsvorschlag 3</u>:
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'''(2)'''&nbsp; Again, the  <u>solution 3</u> is correct:
 
* Im rauschfreien Fall &nbsp; &rArr; &nbsp; $n(t) = 0$ kann auf die zweifache Multiplikation mit $c(t) ∈ \{+1, -1\}$ &nbsp; &rArr; &nbsp; $c(t)^{2} = 1$ verzichtet werden, so dass das obere Modell mit dem unteren Modell identisch ist.
 
* Im rauschfreien Fall &nbsp; &rArr; &nbsp; $n(t) = 0$ kann auf die zweifache Multiplikation mit $c(t) ∈ \{+1, -1\}$ &nbsp; &rArr; &nbsp; $c(t)^{2} = 1$ verzichtet werden, so dass das obere Modell mit dem unteren Modell identisch ist.
  
  
'''(3)'''&nbsp; Zutreffend ist der  <u>Lösungsvorschlag 1</u>:
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'''(3)'''&nbsp; The <u>solution 1</u> is applicable:
*Da beide Modelle im rauschfreien Fall identisch sind, muss nur das Rauschsignal angepasst werden: $n'(t) = n(t) \cdot c(t)$.  
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*Since both models are identical in the noiseless case, only the noise signal has to be adjusted: $n'(t) = n(t) \cdot c(t)$.  
*Die Lösungsvorschläge 2 und 3 sind dagegen nicht zutreffend: Die Integration muss auch weiterhin über $T = J \cdot T_{c}$ erfolgen (nicht über $J \cdot T$) und die PN–Modulation verringert das AWGN–Rauschen nicht.  
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*The solutions 2 and 3, on the other hand, are not applicable: Integration must still be done via $T = J \cdot T_{c}$ (not via $J \cdot T$) and PN modulation does not reduce the AWGN noise.  
  
  
'''(4)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:
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'''(4)'''&nbsp; The <u>solution 3</u> is correct:
*Die für BPSK und AWGN–Kanal gültige Gleichung
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*The equation valid for BPSK and AWGN channel
 
:$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { \frac{2 \cdot E_{\rm B}}{N_{\rm 0}} } \hspace{0.05cm} \right )$$
 
:$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { \frac{2 \cdot E_{\rm B}}{N_{\rm 0}} } \hspace{0.05cm} \right )$$
:ist somit auch bei der PN–Modulation anwendbar und zwar unabhängig vom Spreizfaktor $J$ und von der spezifischen Spreizfolge.  
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:is therefore also applicable to PN modulation, independent of the spreading factor $J$ and the specific spreading sequence.  
*Bei AWGN–Rauschen wird die Fehlerwahrscheinlichkeit durch Bandspreizung weder vergrößert noch verkleinert.  
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*For AWGN noise, the error probability is neither increased nor decreased by band spreading.  
  
 
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Revision as of 19:49, 2 July 2020

equivalent circuit diagrams of „PN modulation” and „BPSK”

The upper figure shows the equivalent circuit diagram of PN modulation (  Direct Sequence Spread Spectrum, abbreviated DS-SS) in the equivalent low-pass range, where„ $n(t)$  stands for AWGN noise. Below, the low-pass model of binary phase modulation (BPSK) is sketched

  • The low-pass transmit signal  $s(t)$  is equal to the rectangular source signal  $q(t) ∈ \{+1, -1\}$  with rectangular duration  $T$  only for reasons of uniformity.
  • The function of the integrator can be written as follows:
$$d (\nu T) = \frac{1}{T} \cdot \int_{(\nu -1 )T }^{\nu T} \hspace{-0.03cm} b (t )\hspace{0.1cm} {\rm d}t \hspace{0.05cm}.$$
  • The two models differ in the multiplication by the  $±1$-spreading signal  $c(t)$  at the transmitter and receiver, whereas of this signal  $c(t)$  only the spread degree   $J$  is known.
  • The specification of the specific spreading sequence (M sequence or Walsh function) is not important for the solution of this task.


It has to be examined whether the lower BPSK model can also be applied with PN modulation and whether the BPSK error probability

$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { {2 \cdot E_{\rm B}}/{N_{\rm 0}} } \hspace{0.05cm} \right )$$

is also valid for PN modulation, or how the specified equation should be modified.




Notes:

  • The CDMA method used for UMTS is also known as "PN modulation".
  • The nomenclature used in this task is also partly based on the  PN–Modulation  im Buch „Modulationsverfahren”.


Questionnaire

1

Which detection values are possible with BPSK (without noise)?

$d(\nu T)$  is gauss distributed.
$d(\nu T)$  can take the values   $+1$,  $0$  und  $-1$ .
Only the values  $d(\nu T) = +1$  und  $d(\nu T) = -1$  are possible.

2

Which values are possible with PN modulation in a noise-free case??

$d(\nu T)$  is gauss distributed.
$d(\nu T)$  can take the values  $+1$,  $0$  und  $-1$ .
Only the values  $d(\nu T) = +1$  und  $d(\nu T) = -1$  are possible.

3

What modification must be made to the BPSK model so that it can also be used for PN modulation?

The noise   $n(t)$  must be replaced by  $n\hspace{0.05cm}'(t) = n(t) \cdot c(t)$ .
The integration must now be done via  $J \cdot T$ .
The noise power must be reduced by the factor  $J$ .

4

The following applies   $10 \cdot {\rm lg}\ (E_{\rm B}/N_0) = 6 \ \rm dB$.  what error probability nbsp; $p_{\rm B}$  results with PN modulation?
Hint:   For BPSK, the result is  $p_{\rm B} \approx 2.3 \cdot 10^{-3}$.

The larger  $J$  is selected, the smaller  $p_{\rm B}$ is.
The larger  $J$  is selected, the larger  $p_{\rm B}$.
It results independently from  $J$ , always the value  $p_{\rm B} = 2.3 \cdot 10^{-3}$ results.


Sample solution

(1)  Correct is the solution 3:

  • This is an optimal receiver.
  • Without noise, the signal $b(t)$ within each bit is constantly equal to $+1$ or $-1$.
  • From the equation given for the integrator
$$d (\nu T) = \frac{1}{T} \cdot \hspace{-0.03cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t$$
follows that $d(\nu T)$ can only take the values $±1$.


(2)  Again, the solution 3 is correct:

  • Im rauschfreien Fall   ⇒   $n(t) = 0$ kann auf die zweifache Multiplikation mit $c(t) ∈ \{+1, -1\}$   ⇒   $c(t)^{2} = 1$ verzichtet werden, so dass das obere Modell mit dem unteren Modell identisch ist.


(3)  The solution 1 is applicable:

  • Since both models are identical in the noiseless case, only the noise signal has to be adjusted: $n'(t) = n(t) \cdot c(t)$.
  • The solutions 2 and 3, on the other hand, are not applicable: Integration must still be done via $T = J \cdot T_{c}$ (not via $J \cdot T$) and PN modulation does not reduce the AWGN noise.


(4)  The solution 3 is correct:

  • The equation valid for BPSK and AWGN channel
$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { \frac{2 \cdot E_{\rm B}}{N_{\rm 0}} } \hspace{0.05cm} \right )$$
is therefore also applicable to PN modulation, independent of the spreading factor $J$ and the specific spreading sequence.
  • For AWGN noise, the error probability is neither increased nor decreased by band spreading.