Difference between revisions of "Aufgaben:Exercise 4.5: Pseudo Noise Modulation"

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===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Welche Detektionssignalwerte sind bei BPSK im rauschfreien Fall möglich?
+
{What detection signal values are possible with BPSK in the noise-free case?
 
|type="[]"}
 
|type="[]"}
- $d(\nu T)$&nbsp; ist gaußverteilt.
+
- $d(\nu T)$&nbsp; is Gaussian distributed.
- $d(\nu T)$&nbsp; kann die Werte&nbsp; $+1, \ 0$&nbsp; und&nbsp; $–1$&nbsp; annehmen.
+
- $d(\nu T)$&nbsp; can take the values&nbsp; $+1, \ 0$&nbsp; and&nbsp; $-1$&nbsp;.
+ Es sind nur die Werte&nbsp; $d(\nu T) = +1$&nbsp; und&nbsp; $d(\nu T) = -1$&nbsp; möglich.
+
+ Only the values&nbsp; $d(\nu T) = +1$&nbsp; and&nbsp; $d(\nu T) = -1$&nbsp; are possible.
  
{Welche Werte sind bei PN–Modulation im rauschfreien Fall möglich?
+
{What values are possible with PN modulation in the noise-free case?
 
|type="[]"}
 
|type="[]"}
- $d(\nu T)$&nbsp; ist gaußverteilt.
+
- $d(\nu T)$&nbsp; is Gaussian distributed.
- $d(\nu T)$&nbsp; kann die Werte&nbsp; $+1, \ 0$&nbsp; und&nbsp; $–1$&nbsp; annehmen.
+
- $d(\nu T)$&nbsp; can take the values&nbsp; $+1, \ 0$&nbsp; and&nbsp; $-1$&nbsp;.
+ Es sind nur die Werte&nbsp; $d(\nu T) = +1$&nbsp; und&nbsp; $d(\nu T) = -1$&nbsp; möglich.
+
+ Only the values&nbsp; $d(\nu T) = +1$&nbsp; and&nbsp; $d(\nu T) = -1$&nbsp; are possible.
  
{Welche Modifikation muss am BPSK–Modell vorgenommen werden, damit es auch für die PN–Modulation anwendbar ist?
+
{What modification must be made to the BPSK model to make it applicable to PN modulation?
 
|type="[]"}
 
|type="[]"}
+ Das Rauschen&nbsp; $n(t)$&nbsp; muss durch&nbsp; $n\hspace{0.05cm}'(t) = n(t) \cdot c(t)$&nbsp; ersetzt werden.
+
+ The noise&nbsp; $n(t)$&nbsp; must be replaced by&nbsp; $n\hspace{0.05cm}'(t) = n(t) \cdot c(t)$&nbsp;.
- Die Integration muss nun über&nbsp; $J \cdot T$&nbsp; erfolgen.
+
- The integration must now be done over&nbsp; $J \cdot T$&nbsp; .
- Die Rauschleistung muss um den Faktor&nbsp; $J$&nbsp; vermindert werden.
+
- The noise power must be reduced by a factor of&nbsp; $J$&nbsp;.
  
{Welche Bitfehlerwahrscheinlichkeit&nbsp; $p_{\rm B}$&nbsp; ergibt sich für&nbsp; $10 \cdot {\rm lg} \ (E_{\rm B}/N_{0}) = 6 \ \rm dB$&nbsp; bei PN–Modulation? <br>''Hinweis:'' &nbsp; Bei BPSK gilt in diesem Fall: &nbsp; $p_{\rm B} \approx 2.3 \cdot 10^{–3}$.
+
{What is the bit error probability&nbsp; $p_{\rm B}$&nbsp; for&nbsp; $10 \cdot {\rm lg} \ (E_{\rm B}/N_{0}) = 6 \ \rm dB$&nbsp; for PN modulation? <br>''Note:'' &nbsp; For BPSK, in this case: &nbsp; $p_{\rm B} \approx 2.3 \cdot 10^{-3}$.
 
|type="()"}
 
|type="()"}
- Je größer&nbsp; $J$&nbsp; gewählt wird, desto kleiner ist&nbsp; $p_{\rm B}$.
+
- The larger&nbsp; $J$&nbsp; is chosen, the smaller&nbsp; $p_{\rm B}$ is.
- Je größer&nbsp; $J$&nbsp; gewählt wird, desto größer ist&nbsp; $p_{\rm B}$.
+
- The larger&nbsp; $J$&nbsp; is chosen, the larger&nbsp; $p_{\rm B}$.
+ Es ergibt sich unabhängig von&nbsp; $J$&nbsp; stets der Wert&nbsp; $2.3 \cdot 10^{–3}$.
+
+ It always results in the value&nbsp; $2.3 \cdot 10^{-3}$ regardless of&nbsp; $J$&nbsp;.
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
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{{ML-Kopf}}
  
'''(1)'''&nbsp; Richtig ist der <u>letzte Lösungsvorschlag</u>:
+
'''(1)'''&nbsp; Correct is the <u>last proposed solution</u>:
*Es handelt sich hier um einen optimalen Empfänger.  
+
*This is an optimal receiver.  
*Ohne Rauschen ist Signal&nbsp; $b(t)$&nbsp; innerhalb eines jeden Bits konstant gleich $+1$ oder $-1$.  
+
*Without noise, signal&nbsp; $b(t)$&nbsp; within each bit is constant equal to $+1$ or $-1$.  
*Aus der angegebenen Gleichung für den Integrator folgt, dass&nbsp; $d(\nu T)$&nbsp; nur die Werte $±1$ annehmen kann:
+
*From the given equation for the integrator, it follows that&nbsp; $d(\nu T)$&nbsp; can only take the values $±1$:
 
:$$d (\nu T) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t.$$  
 
:$$d (\nu T) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t.$$  
  
  
  
'''(2)'''&nbsp; Richtig ist wieder der <u>letzte Lösungsvorschlag</u>:  
+
'''(2)'''&nbsp; Correct again is the <u>last proposed solution</u>:  
*Im rauschfreien Fall &nbsp; &rArr; &nbsp; $n(t) = 0$ kann auf die zweifache Multiplikation mit&nbsp; $c(t) ∈ \{+1, –1\} \ \Rightarrow \ c(t)^{2} = 1$&nbsp; verzichtet werden,  
+
*In the noise-free case &nbsp; &rArr; &nbsp; $n(t) = 0$, the twofold multiplication with&nbsp; $c(t) ∈ \{+1, -1\} \ \Rightarrow \ c(t)^{2} = 1$&nbsp; can be omitted,  
*so dass das obere Modell mit dem unteren Modell identisch ist.
+
*so that the upper model is identical to the lower model.
  
  
  
'''(3)'''&nbsp; Zutreffend ist nur der <u>Lösungsvorschlag 1</u>:
+
'''(3)'''&nbsp; Only the <u>proposed solution 1</u> is applicable:
*Da beide Modelle im rauschfreien Fall identisch sind, muss nur das Rauschsignal angepasst werden: &nbsp; $n\hspace{0.05cm}'(t) = n(t) \cdot c(t)$.  
+
*Since both models are identical in the noise-free case, only the noise signal needs to be adjusted: &nbsp; $n\hspace{0.05cm}'(t) = n(t) \cdot c(t)$.  
*Die Vorschläge 2 und 3 sind dagegen nicht zutreffend: &nbsp; Die Integration muss weiterhin über&nbsp; $T = J \cdot T_{\rm c}$&nbsp; erfolgen und die PN–Modulation verringert das AWGN–Rauschen nicht.  
+
*Suggestions 2 and 3, on the other hand, are not applicable: &nbsp; Integration must still be done over&nbsp; $T = J \cdot T_{\rm c}$&nbsp; and PN modulation does not reduce AWGN noise.  
  
  
  
'''(4)'''&nbsp; Richtig ist der <u>letzte Lösungsvorschlag</u>.:
+
'''(4)'''&nbsp; Correct is the <u>last proposed solution</u>:
*Multipliziert man das AWGN–Rauschen mit dem hochfrequenten&nbsp; $±1$–Signal&nbsp; $c(t)$, so ist das Rauschen ebenfalls gaußförmig und weiß.  
+
*Multiplying the AWGN noise by the high frequency&nbsp; $±1$ signal&nbsp; $c(t)$, the noise is also Gaussian and white.  
*Wegen&nbsp; $E[c^{2}(t)] = 1$&nbsp; wird auch die Rauschvarianz nicht verändert. Die für BPSK gültige Gleichung
+
*Because&nbsp; $E[c^{2}(t)] = 1$&nbsp; the noise variance is also not changed. The equation valid for BPSK
:$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { {2 \cdot E_{\rm B}}/{N_{\rm 0}} } \hspace{0.05cm} \right )$$
+
:$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { {2 \cdot E_{\rm B}}/{N_{\rm 0}} } \hspace{-0.05cm} \right )$$
:ist somit auch bei der PN&ndash;Modulation anwendbar und zwar unabhängig vom Spreizfaktor&nbsp; $J$&nbsp; und von der spezifischen Spreizfolge.  
+
:is thus also applicable to PN&ndash;modulation and is independent of the spreading factor&nbsp; $J$&nbsp; and of the specific spreading sequence.  
  
:&nbsp; &rArr; &nbsp; '''Bei AWGN&ndash;Rauschen wird also die Fehlerwahrscheinlichkeit durch Bandspreizung weder vergrößert noch verkleinert'''.  
+
:&nbsp; &rArr; &nbsp; '''Thus, in the case of AWGN noise, the error probability is neither increased nor decreased by band spreading'''.  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

Revision as of 21:55, 28 February 2023

Equivalent circuit of PN modulation and BPSK

The graph above shows the equivalent circuit of "pseudo-noise" modulation  (  Direct Sequence Spread Spectrum, abbreviated  DS-SS)  in the equivalent low-pass region. $n(t)$  denotes AWGN noise.

Below is sketched the low pass model of  "Binary Phase Shift Keying".

  • The low-pass transmit signal  $s(t)$  is set equal to the rectangular source signal  $q(t) ∈ \{+1, -1\}$  with rectangular duration  $T$  only for reasons of uniform representation.
  • The integrator function can be written as follows:
$$d (\nu T) = \frac{1}{T} \cdot \hspace{0.03cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t \hspace{0.05cm}.$$
  • The two models differ by multiplication by the  $±1$-spread signal  $c(t)$  at the transmitter and receiver, where only the degree of spread  $J$  is known from  $c(t)$ .
  • For the solution of this exercise, the specification of the specific spreading sequence  (M sequence or Walsh function)  is not important.


What needs to be investigated is whether the lower BPSK model can also be applied in 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 to modify the given equation.






Hints:

  • This exercise belongs to the chapter  "Telecommunications Aspects of UMTS".
  • The CDMA method used in UMTS also goes by the name "PN modulation".
  • The nomenclature used in this exercise is partly based on the chapter  "PN Modulation"  in the book "Modulation Methods".


Questions

1

What detection signal values are possible with BPSK in the noise-free case?

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

2

What values are possible with PN modulation in the noise-free case?

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

3

What modification must be made to the BPSK model to make it applicable to 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 over  $J \cdot T$  .
The noise power must be reduced by a factor of  $J$ .

4

What is the bit error probability  $p_{\rm B}$  for  $10 \cdot {\rm lg} \ (E_{\rm B}/N_{0}) = 6 \ \rm dB$  for PN modulation?
Note:   For BPSK, in this case:   $p_{\rm B} \approx 2.3 \cdot 10^{-3}$.

The larger  $J$  is chosen, the smaller  $p_{\rm B}$ is.
The larger  $J$  is chosen, the larger  $p_{\rm B}$.
It always results in the value  $2.3 \cdot 10^{-3}$ regardless of  $J$ .


Solution

(1)  Correct is the last proposed solution:

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


(2)  Correct again is the last proposed solution:

  • In the noise-free case   ⇒   $n(t) = 0$, the twofold multiplication with  $c(t) ∈ \{+1, -1\} \ \Rightarrow \ c(t)^{2} = 1$  can be omitted,
  • so that the upper model is identical to the lower model.


(3)  Only the proposed solution 1 is applicable:

  • Since both models are identical in the noise-free case, only the noise signal needs to be adjusted:   $n\hspace{0.05cm}'(t) = n(t) \cdot c(t)$.
  • Suggestions 2 and 3, on the other hand, are not applicable:   Integration must still be done over  $T = J \cdot T_{\rm c}$  and PN modulation does not reduce AWGN noise.


(4)  Correct is the last proposed solution:

  • Multiplying the AWGN noise by the high frequency  $±1$ signal  $c(t)$, the noise is also Gaussian and white.
  • Because  $E[c^{2}(t)] = 1$  the noise variance is also not changed. The equation valid for BPSK
$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { {2 \cdot E_{\rm B}}/{N_{\rm 0}} } \hspace{-0.05cm} \right )$$
is thus also applicable to PN–modulation and is independent of the spreading factor  $J$  and of the specific spreading sequence.
  ⇒   Thus, in the case of AWGN noise, the error probability is neither increased nor decreased by band spreading.