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Difference between revisions of "Aufgaben:Exercise 2.1: ACF and PSD with Coding"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Grundlagen der codierten Übertragung
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Basics_of_Coded_Transmission
 
}}
 
}}
  
  
  
[[File:P_ID1308__Dig_A_2_1.png|right|frame|Leistungsdichtespektrum bei Codierung]]
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[[File:P_ID1308__Dig_A_2_1.png|right|frame|Power-spectral density with coding]]
Wir betrachten das Digitalsignal
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We consider the digital signal  $s(t)$,   using the following descriptive quantities:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm},$$
+
*aν  are the amplitude coefficients,
wobei wir folgende Beschreibungsgrößen verwenden:
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*gs(t)  indicates the basic transmission pulse,
*aν sind die Amplitudenkoeffizienten,
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*T  is the symbol duration  (spacing of the pulses).
*gs(t) gibt den Sendegrundimpuls an,
 
*T ist die Symboldauer (Abstand der Impulse).
 
  
  
Zur Charakterisierung der spektralen Eigenschaften, die sich aufgrund der Codierung und der Impulsformung ergeben, verwendet man unter anderem
+
Then holds:
*die Autokorrelationsfunktion (AKF)
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:s(t)=+ν=aνgs(tνT).
 +
 
 +
To characterize the spectral properties resulting from the coding and pulse shaping,  one uses,  among other things
 +
*the auto-correlation function  $\rm (ACF)$
 
:φs(τ)=+λ=1/Tφa(λ)φgs(τλT),
 
:φs(τ)=+λ=1/Tφa(λ)φgs(τλT),
*das Leistungsdichtespektrum (LDS)
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*the power-spectral density   $\rm (PSD)$
 
:Φs(f)=1/TΦa(f)Φgs(f).
 
:Φs(f)=1/TΦa(f)Φgs(f).
  
Hierbei bezeichnet φa(λ) die diskrete Autokorrelationsfunktion der Amplitudenkoeffizienten, die mit der spektralen Leistungsdichte $\Phi_{a}(f)$ über die Fouriertransformation zusammenhängt. Für diese gilt somit:
+
Here,  φa(λ)  denotes the discrete ACF of the amplitude coefficients related to the power-spectral density  ${\it \Phi}_{a}(f)$  via the Fourier transform.  Thus,  for this holds:
 
:Φa(f)=+λ=φa(λ)ej2πfλT.
 
:Φa(f)=+λ=φa(λ)ej2πfλT.
Weiterhin sind in obigen Gleichungen die Energie–AKF und das Energiespektrum verwendet:
+
Furthermore,  the energy ACF and energy spectrum are used in above equations:
 
:φgs(τ)=+gs(t)gs(t+τ)dtΦgs(f)=|Gs(f)|2.
 
:φgs(τ)=+gs(t)gs(t+τ)dtΦgs(f)=|Gs(f)|2.
In der vorliegenden Aufgabe soll für die spektrale Leistungsdichte der Amplitudenkoeffizienten folgender Funktionsverlauf angenommen werden (siehe Grafik):
+
 
 +
 
 +
In the present exercise,  the following function is to be assumed for the power-spectral density of the amplitude coefficients  (see graph):
 
:Φa(f)=1/21/2cos(4πfT).
 
:Φa(f)=1/21/2cos(4πfT).
Für den Sendegrundimpuls werden folgende Annahmen getroffen:
+
The following assumptions are made for the basic transmission pulse:
*n der Teilfrage (2) sei gs(t) ein NRZ–Rechteckimpuls, so dass eine dreieckförmige Energie–AKF vorliegt, die auf den Bereich |τ|T beschränkt ist. Das Maximum ist dabei
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*In question  '''(2)''',  let  gs(t)  be an NRZ rectangular pulse,  so that there is a triangular energy ACF confined to the range  |τ|T.  The maximum value here is
 
:φgs(τ=0)=s20T.
 
:φgs(τ=0)=s20T.
*Für die Teilaufgabe (3) soll von einer Wurzel–Nyquist–Charakteristik mit Rolloff–Faktor r=0 ausgegangen werden. In diesem Fall gilt:
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*For question  '''(3)''',  assume a root-Nyquist characteristic with rolloff factor  r=0.  In this case holds:
:$$|G_s(f)|^2 = \left\{ s20T20 \right.\quad \begin{array}{*{1}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array} |f|<1/(2T),|f|>1/(2T).$$
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:$$|G_s(f)|^2 = \left\{ s20T20 \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array} |f|<1/(2T),|f|>1/(2T).$$
*Für numerische Berechnungen ist stets s20=10 mW zu verwenden.
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*For numerical calculations,&nbsp; use always &nbsp;s20=10 mW.&nbsp;
  
  
''Hinweis:''
 
  
Die Aufgabe gehört zum [[Digitalsignalübertragung/Grundlagen_der_codierten_Übertragung|Grundlagen der codierten Übertragung]] des vorliegenden Buches. Berücksichtigen Sie, dass die Sendeleistung PS gleich der AKF ϕs(τ) an der Stelle τ=0 ist, aber auch als Integral über das LDS Φs(f) berechnet werden kann.
 
  
===Fragebogen===
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Notes:
 +
*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|"Basics of Coded Transmission"]].
 +
 +
*Consider that the transmit power &nbsp;PS&nbsp; is equal to the ACF &nbsp;φs(τ)&nbsp; at the point &nbsp;$\tau = 0,&nbsp; but can also be calculated as an integral over the PSD &nbsp;{\it \Phi}_{s}(f)$.&nbsp;
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
 
|type="[]"}
 
- Falsch
 
+ Richtig
 
  
 +
{What are the discrete ACF values &nbsp;φa(λ)&nbsp; of the amplitude coefficients? Enter the numerical values for &nbsp;λ=0, &nbsp;λ=1&nbsp; and &nbsp;λ=2.&nbsp;
 +
|type="{}"}
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φa(λ=0) =  { 0.5 3% }
 +
φa(λ=1) =  { 0. }
 +
φa(λ=2) =  { -0.2575--0.2425 }
 +
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{What is the transmit power with the &nbsp;<u>NRZ basic transmission pulse</u>?
 +
|type="{}"}
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PS =  { 5 3% }  mW
  
{Input-Box Frage
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{What is the transmit power with &nbsp;<u>root-Nyquist characteristic</u> &nbsp;(r=0)?
 
|type="{}"}
 
|type="{}"}
$\alpha$ = { 0.3 }
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$P_{\rm S} \ = \ $ { 5 3% }  mW
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;
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'''(1)'''&nbsp; Since&nbsp; Φa(f)&nbsp; as a power-spectral density is always real &nbsp;(plus even and positive,&nbsp; but that does not matter here)&nbsp; and the ACF values&nbsp; φa(λ)&nbsp; are symmetric about&nbsp; λ=0,&nbsp; the given equation can be transformed as follows:
'''(2)'''&nbsp;
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:Φa(f)=+λ=φa(λ)ej2πfλT=φa(0)+λ=12φa(λ)cos(2πfλT).
'''(3)'''&nbsp;
+
*By comparison with the sketched function
'''(4)'''&nbsp;
+
:Φa(f)=1/21/2cos(4πfT).
'''(5)'''&nbsp;
+
:one obtains:
'''(6)'''&nbsp;
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:φa(λ=0)=0.5_,φa(λ=2)=φa(λ=2)=0.25_.
 +
*All other ACF values result to zero,&nbsp; so also&nbsp; φa(λ=±1)=0_.
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; For the rectangular NRZ basic pulse,&nbsp; due to the limitation of the energy ACF to the range |τ|T,&nbsp; we obtain:
 +
:PS=φs(τ=0)=1Tφa(λ=0)φgs(τ=0)=1T12s20T=s202=5mW_.
 +
 
 +
 
 +
'''(3)'''&nbsp; For rectangular spectral function,&nbsp; it is more convenient to calculate the transmit power by integration over the power-spectral density:
 +
:PS= +1/(2T)1/(2T)Φs(f)df=1T+1/(2T)1/(2T)Φa(f)Φgs(f)df
 +
:$$\Rightarrow\hspace{0.3cm}P_{\rm S} = \ \frac{1}{T} \cdot \left [ s_0^2 \cdot T^2 \right ] \cdot \int_{-1/(2T)}^{+1/(2T)} \left( {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\right ) \,{\rm d} f\hspace{0.05cm} = {s_0^2}/{2}\hspace{0.15cm}\underline { = 5\,\,{\rm mW}} .$$
 +
*Here it is considered that the energy PSD&nbsp; |Gs(f)|2&nbsp; is given as constant&nbsp; (within the integration interval)&nbsp; and thus can be drawn in front of the integral.
 +
*In spite of a completely different signal form&nbsp; $s(t)$,&nbsp; the same transmit power results here,&nbsp; since the integral yields the value 1/(2T).
 +
*It should be noted that this simple calculation is only possible for the rolloff factor r=0.
 +
 
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Digitalsignalübertragung|^2.1 Codierte Übertragung - Grundlagen^]]
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[[Category:Digital Signal Transmission: Exercises|^2.1 Basics of Coded Transmission^]]

Latest revision as of 16:23, 23 May 2022



Power-spectral density with coding

We consider the digital signal  s(t),  using the following descriptive quantities:

  • aν  are the amplitude coefficients,
  • gs(t)  indicates the basic transmission pulse,
  • T  is the symbol duration  (spacing of the pulses).


Then holds:

s(t)=+ν=aνgs(tνT).

To characterize the spectral properties resulting from the coding and pulse shaping,  one uses,  among other things

  • the auto-correlation function  (ACF)
φs(τ)=+λ=1/Tφa(λ)φgs(τλT),
  • the power-spectral density  (PSD)
Φs(f)=1/TΦa(f)Φgs(f).

Here,  φa(λ)  denotes the discrete ACF of the amplitude coefficients related to the power-spectral density  Φa(f)  via the Fourier transform.  Thus,  for this holds:

Φa(f)=+λ=φa(λ)ej2πfλT.

Furthermore,  the energy ACF and energy spectrum are used in above equations:

φgs(τ)=+gs(t)gs(t+τ)dtΦgs(f)=|Gs(f)|2.


In the present exercise,  the following function is to be assumed for the power-spectral density of the amplitude coefficients  (see graph):

Φa(f)=1/21/2cos(4πfT).

The following assumptions are made for the basic transmission pulse:

  • In question  (2),  let  gs(t)  be an NRZ rectangular pulse,  so that there is a triangular energy ACF confined to the range  |τ|T.  The maximum value here is
φgs(τ=0)=s20T.
  • For question  (3),  assume a root-Nyquist characteristic with rolloff factor  r=0.  In this case holds:
|Gs(f)|2={s20T20forfor|f|<1/(2T),|f|>1/(2T).
  • For numerical calculations,  use always  s20=10 mW



Notes:

  • Consider that the transmit power  PS  is equal to the ACF  φs(τ)  at the point  τ=0,  but can also be calculated as an integral over the PSD  Φs(f)

Questions

1

What are the discrete ACF values  φa(λ)  of the amplitude coefficients? Enter the numerical values for  λ=0,  λ=1  and  λ=2

φa(λ=0) = 

φa(λ=1) = 

φa(λ=2) = 

2

What is the transmit power with the  NRZ basic transmission pulse?

PS = 

 mW

3

What is the transmit power with  root-Nyquist characteristic  (r=0)?

PS = 

 mW


Solution

(1)  Since  Φa(f)  as a power-spectral density is always real  (plus even and positive,  but that does not matter here)  and the ACF values  φa(λ)  are symmetric about  λ=0,  the given equation can be transformed as follows:

Φa(f)=+λ=φa(λ)ej2πfλT=φa(0)+λ=12φa(λ)cos(2πfλT).
  • By comparison with the sketched function
Φa(f)=1/21/2cos(4πfT).
one obtains:
φa(λ=0)=0.5_,φa(λ=2)=φa(λ=2)=0.25_.
  • All other ACF values result to zero,  so also  φa(λ=±1)=0_.


(2)  For the rectangular NRZ basic pulse,  due to the limitation of the energy ACF to the range |τ|T,  we obtain:

PS=φs(τ=0)=1Tφa(λ=0)φgs(τ=0)=1T12s20T=s202=5mW_.


(3)  For rectangular spectral function,  it is more convenient to calculate the transmit power by integration over the power-spectral density:

PS= +1/(2T)1/(2T)Φs(f)df=1T+1/(2T)1/(2T)Φa(f)Φgs(f)df
PS= 1T[s20T2]+1/(2T)1/(2T)(1/21/2cos(4πfT))df=s20/2=5mW_.
  • Here it is considered that the energy PSD  |Gs(f)|2  is given as constant  (within the integration interval)  and thus can be drawn in front of the integral.
  • In spite of a completely different signal form  s(t),  the same transmit power results here,  since the integral yields the value 1/(2T).
  • It should be noted that this simple calculation is only possible for the rolloff factor r=0.