Difference between revisions of "Aufgaben:Exercise 1.5: Reconstruction of the Jakes Spectrum"

From LNTwww
m (Text replacement - "power spectral density" to "power-spectral density")
 
(28 intermediate revisions by 4 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Mobile Kommunikation/Statistische Bindungen innerhalb des Rayleigh-Prozesses}}
+
{{quiz-Header|Buchseite=Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process}}
  
[[File:P_ID2124__Mob_A_1_5.png|right|frame|Betrachtetes Jakes–Spektrum]]
+
[[File:P_ID2124__Mob_A_1_5.png|right|frame|Considered Jakes spectrum]]
Bei einem Mobilfunksystem macht sich der  [[Mobile_Kommunikation/Statistische_Bindungen_innerhalb_des_Rayleigh%E2%80%93Prozesses#Ph.C3.A4nomenologische_Beschreibung_des_Dopplereffekts|Dopplereffekt]]  auch im Leistungsdichtespektrum der Dopplerfrequenz  $f_{\rm D}$  bemerkbar.  
+
In a mobile radio system, the  [[Mobile_Communications/Statistische_Bindungen_innerhalb_des_Rayleigh%E2%80%93Prozesses#Ph.C3.A4nomenologische_Beschreibung_des_Dopplereffekts|Doppler effect]]  is also noticeable in the power-spectral density of the Doppler frequency $f_{\rm D}$.  
  
Es ergibt sich das so genannte  [[Mobile_Kommunikation/Statistische_Bindungen_innerhalb_des_Rayleigh%E2%80%93Prozesses#AKF_und_LDS_bei_Rayleigh.E2.80.93Fading|Jakes–Spektrum]], das für die maximale Dopplerfrequenz  $f_{\rm D, \ max} = 100 \ \rm Hz$  in der Grafik dargestellt ist. ${\it \Phi}_z(f_{\rm D})$  hat nur Anteile innerhalb des Bereichs  $± f_{\rm D, \ max}$, wobei gilt:
+
This results in the so-called  [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#ACF_and_PDS_with_Rayleigh.E2.80.93Fading|Jakes spectrum]], which is shown in the graph for the maximum Doppler frequency $f_{\rm D, \ max} = 100 \ \rm Hz$.  ${\it \Phi}_z(f_{\rm D})$  has only portions within the range  $± f_{\rm D, \ max}$, where
:$${\it \Phi}_z(f_{\rm D}) = \frac{2 \cdot \sigma^2}{\pi \cdot f_{\rm D, \hspace{0.05cm} max}  \cdot \sqrt { 1 - (f_{\rm D}/f_{\rm D, \hspace{0.05cm} max})^2 } }
+
:$${\it \Phi}_z(f_{\rm D}) = \frac{2 \cdot \sigma^2}{\pi \cdot f_{\rm D, \hspace{0.1cm} max}  \cdot \sqrt { 1 - (f_{\rm D}/f_{\rm D, \hspace{0.1cm} max})^2} }
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
Was im Frequenzbereich durch das Leistungsdichtespektrum (LDS) ausgedrückt wird, beschreibt man im Zeitbereich durch die Autokorrelationsfunktion (AKF). Diese ergibt sich aus  ${\it \Phi}_z(f_{\rm D})$  durch die  [[Signaldarstellung/Fouriertransformation_und_-r%C3%BCcktransformation#Das_zweite_Fourierintegral|Fourierrücktransformation]].
+
What is expressed in the frequency domain by the power-spectral density  $\rm (PSD)$  is described in the time domain by the auto-correlation function  $\rm (ACF)$.  The ACF is the   ${\it \Phi}_z(f_{\rm D})$  by the  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_Second_Fourier_Integral|inverse Fourier transform]]  of the PDS.
  
Mit der <i>Besselfunktion</i> erster Art und nullter Ordnung &nbsp;$({\rm J}_0)$&nbsp; erhält man:
+
With the&nbsp; [[Applets:Bessel_Functions_of_the_First_Kind|Bessel function]]&nbsp; of the first kind and zero order &nbsp;$({\rm J}_0)$&nbsp; you get
:$$\varphi_z ({\rm \Delta}t) = 2 \sigma^2 \cdot {\rm J_0}(2\pi \cdot f_{\rm D, \hspace{0.05cm} max} \cdot {\rm \Delta}t)\hspace{0.05cm}.$$
+
:$$\varphi_z ({\rm \Delta}t) = 2 \sigma^2 \cdot {\rm J_0}(2\pi \cdot f_{\rm D, \hspace{0.1cm} max} \cdot {\rm \Delta}t)\hspace{0.05cm}.$$
  
Um den Dopplereffekt und damit eine Relativbewegung zwischen Sender und Empfänger &ndash; bei einer Systemsimulation zu berücksichtigen, werden im&nbsp; [[Mobile_Kommunikation/Wahrscheinlichkeitsdichte_des_Rayleigh%E2%80%93Fadings#Modellierung_von_nichtfrequenzselektivem_Fading|Rayleigh&ndash;Kanalmodell]]&nbsp; zwei digitale Filter eingefügt, jeweils mit dem Frequenzgang&nbsp; $H_{\rm DF}(f_{\rm D})$.  
+
To take into account the Doppler effect and thus a relative movement between transmitter and receiver in a system simulation, two digital filters are inserted in the&nbsp; [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading|Rayleigh channel model]], each with the frequency response&nbsp; $H_{\rm DF}(f_{\rm D})$.  
  
Die Dimensionierung dieser Filter ist Inhalt dieser Aufgabe.
+
The dimensioning of these filters is part of this task.
*Wir beschränken uns hier auf den Zweig zur Generierung des Realteils&nbsp; $x(t)$. Für den Imaginärteil&nbsp; $y(t)$&nbsp; ergeben sich genau gleiche Verhältnisse.
+
*We restrict ourselves here to the branch for generating the real part&nbsp; $x(t)$.&nbsp; The ratios derived here are also valid for the imaginary part&nbsp; $y(t)$.
*Am Eingang des im&nbsp; [[Mobile_Kommunikation/Wahrscheinlichkeitsdichte_des_Rayleigh%E2%80%93Fadings#Frequenzselektives_Fading_vs._nichtfrequenzselektives_Fading|Rayleigh&ndash;Kanalmodell]]&nbsp; linken digitalen Filters liegt weißes Gaußsches Rauschen&nbsp; $n(t)$&nbsp; mit der Varianz&nbsp; $\sigma^2 = 0.5$&nbsp; an.  
+
*At the input of the left digital filter of the&nbsp; [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading#Modeling_of_non-frequency_selective_fading|Rayleigh channel model]]&nbsp;, there is white Gaussian noise&nbsp; $n(t)$&nbsp; with variance&nbsp; $\sigma^2 = 0.5$.  
*Die Realteilkomponente ergibt sich dann gemäß der Faltung zu
+
*The real component is then obtained from the  following convolution
 
:$$x(t) = n(t) \star h_{\rm DF}(t) \hspace{0.05cm}.$$
 
:$$x(t) = n(t) \star h_{\rm DF}(t) \hspace{0.05cm}.$$
  
  
  
 +
''Notes:''
 +
* This task belongs to the topic of&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process|Statistical bindings within the Rayleigh process]].
 +
* The digital filter is treated in detail in chapter&nbsp; [[Theory_of_Stochastic_Signals/Digitale_Filter|Digital Filter]]&nbsp; of the book "Stochastic Signal Theory".
  
  
 
''Hinweise:''
 
* Die Aufgabe gehört zum Themengebiet&nbsp;  [[Mobile_Kommunikation/Statistische_Bindungen_innerhalb_des_Rayleigh%E2%80%93Prozesses|Statistische Bindungen innerhalb des Rayleigh&ndash;Prozesses]].
 
* Das digitale Filter wird im Kapitel&nbsp; [[Stochastische_Signaltheorie/Digitale_Filter|Digitale Filter]]&nbsp; des Buches &bdquo;Stochastische Signaltheorie&rdquo; ausführlich behandelt.
 
 
   
 
   
  
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welchen Wert hat das Jakes&ndash;Spektrum des Realteils bei der Dopplerfrequenz&nbsp; $f_{\rm D} = 0$?
+
{What is the value of the Jakes spectrum of the real part at the Doppler frequency $f_{\rm D} = 0$?
 
|type="{}"}
 
|type="{}"}
${\it \Phi}_x(f_{\rm D} = 0)\ = \ $ { 1.59 } $\ \cdot 10^{\rm &ndash;3} \ 1/{\rm Hz}$
+
${\it \Phi}_x(f_{\rm D} = 0)\ = \ $ { 1.59 } $\ \cdot 10^{\rm &ndash;3} \ {\rm Hz}^{-1}$
  
{Welche Dimensionierung ist richtig, wobei &nbsp;$K$&nbsp; eine geeignet gewählte Konstante ist?
+
{Which dimensioning is correct, where &nbsp;$K$&nbsp; is an appropriately chosen constant?
 
|type="[]"}
 
|type="[]"}
- Es gilt&nbsp; $H_{\rm DF}(f_{\rm D}) = K \cdot {\it \Phi}_x(f_{\rm D})$.
+
- It holds&nbsp; $H_{\rm DF}(f_{\rm D}) = K \cdot {\it \Phi}_x(f_{\rm D})$.
+ Es gilt&nbsp; $|H_{\rm DF}(f_{\rm D})|^2 = K \cdot {\it \Phi}_x(f_{\rm D})$
+
+ It applies&nbsp; $|H_{\rm DF}(f_{\rm D})|^2 = K \cdot {\it \Phi}_x(f_{\rm D})$
  
{Aus welcher Bedingung lässt sich die Konstante&nbsp; $K$&nbsp; bestimmen?
+
{From which condition can the constant&nbsp; $K$&nbsp; be determined?
 
|type="[]"}
 
|type="[]"}
- $K$&nbsp; kann beliebig gewählt werden.  
+
- $K$&nbsp; can be selected as desired.  
- Das Integral über&nbsp; $|H_{\rm DF}(f_{\rm D})|$&nbsp; muss&nbsp; $1$&nbsp; ergeben.
+
- The integral over&nbsp; $|H_{\rm DF}(f_{\rm D})|$&nbsp; must equal&nbsp; $1$&nbsp;.
+ Das Integral über&nbsp; $|H_{\rm DF}(f_{\rm D})|^2$&nbsp; muss&nbsp; $1$&nbsp; ergeben.
+
+ The integral over&nbsp; $|H_{\rm DF}(f_{\rm D})|^2$&nbsp; must be&nbsp; $1$&nbsp;.
  
{Ist&nbsp; $H_{\rm DF}(f)$&nbsp; durch die beiden Bedingungen  gemäß '''(2)''' und '''(3)''' eindeutig festgelegt?
+
{Is&nbsp; $H_{\rm DF}(f)$&nbsp; unambiguously defined by the two conditions according to&nbsp; '''(2)'''&nbsp; and&nbsp; '''(3)'''?
 
|type="()"}
 
|type="()"}
- Ja.
+
- Yes.
+ Nein.
+
+ No.
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Das Jakes&ndash;Spektrum des Realteils ist halb so groß wie das resultierende Spektrum ${\it \Phi}_z(f)$:
+
'''(1)'''&nbsp; The Jakes spectrum of the real part is half the resulting spectrum&nbsp; ${\it \Phi}_z(f)$:
 
:$${\it \Phi}_x(f_{\rm D} = 0) = {\it \Phi}_y(f_{\rm D} = 0) = \frac{{\it \Phi}_z(f_{\rm D} = 0)}{2}= \frac{\sigma^2}{\pi \cdot f_{\rm D, \hspace{0.05cm} max}} =  
 
:$${\it \Phi}_x(f_{\rm D} = 0) = {\it \Phi}_y(f_{\rm D} = 0) = \frac{{\it \Phi}_z(f_{\rm D} = 0)}{2}= \frac{\sigma^2}{\pi \cdot f_{\rm D, \hspace{0.05cm} max}} =  
 
  \frac{0.5}{\pi \cdot 100\,\,{\rm Hz}} \hspace{0.15cm} \underline{ = 1.59 \cdot 10^{-3}\,\,{\rm Hz^{-1}}}
 
  \frac{0.5}{\pi \cdot 100\,\,{\rm Hz}} \hspace{0.15cm} \underline{ = 1.59 \cdot 10^{-3}\,\,{\rm Hz^{-1}}}
Line 66: Line 65:
  
  
'''(2)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 2</u>:  
+
'''(2)'''&nbsp; <u>Solution 2</u> is correct:  
*Das Eingangssignal $n(t)$ besitzt ein weißes (konstantes) LDS ${\it \Phi}_n(f_{\rm D})$.  
+
*The input signal&nbsp; $n(t)$&nbsp; has a white (constant) PDS&nbsp; ${\it \Phi}_n(f_{\rm D})$.  
*Für das LDS am Ausgang gilt dann:
+
*The PDS at the output is then
 
:$${\it \Phi}_x(f_{\rm D}) = {\it \Phi}_n(f_{\rm D}) \cdot | H_{\rm DF}(f_{\rm D}|^2
 
:$${\it \Phi}_x(f_{\rm D}) = {\it \Phi}_n(f_{\rm D}) \cdot | H_{\rm DF}(f_{\rm D}|^2
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
  
'''(3)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>.  
+
'''(3)'''&nbsp; <u>Solution 3</u> is correct.  
*Nur wenn diese Bedingung erfüllt ist, hat das Signal $x(t)$ die gleiche Varianz $\sigma^2$ wie das Rauschsignal $n(t)$.
+
*Only if this condition is fulfilled, the signal&nbsp; $x(t)$&nbsp; has the same variance&nbsp; $\sigma^2$&nbsp; as the noise signal&nbsp; $n(t)$.
 
 
 
 
  
'''(4)'''&nbsp; <u>Richtig ist NEIN</u>:
 
*Die beiden Bedingungen nach den Teilaufgaben (2) und (3) beziehen sich nur auf die Betragsfunktion.
 
*Für die Phase des digitalen Filters gibt es keine Vorschrift.
 
*Diese ist frei wählbar. Meist wählt man diese so, dass sich ein minimalphasiges Netzwerk ergibt.
 
*In diesem Fall hat dann die Impulsantwort $h_{\rm DF}(t)$ die geringst mögliche Ausdehnung.
 
  
 +
'''(4)'''&nbsp; <u>No</u>:
 +
*The two conditions after subtasks&nbsp; '''(2)'''&nbsp; and&nbsp; '''(3)'''&nbsp; only refer to the magnitude of the digital filter.
 +
*There is no constraint for the phase of the digital filter.
 +
*This phase can be chosen arbitrarily.&nbsp; Usually it is chosen in such a way that a minimum phase network results.
 +
*In this case, the impulse response&nbsp; $h_{\rm DF}(t)$&nbsp; then has the lowest possible duration.
  
Die Grafik zeigt das Ergebnis der Approximation. Die roten Kurven wurden simulativ über $100\hspace{0.05cm}000$ Abtastwerte ermittelt. Man erkennt:
 
[[File:P_ID2125__Mob_A_1_5d.png|right|frame|Approximation des Jakes–Spektrums und der AKF]]
 
  
 +
The graph shows the result of the approximation.&nbsp; The red curves were determined simulatively over $100\hspace{0.05cm}000$ samples.&nbsp; You can see:
 +
[[File:EN_Mob_A_1_5d.png|right|frame|Approximation of the Jakes spectrum and the auto-correlation function]]
  
* Das Jakes&ndash;Leistungsdichtespektrum (linke Grafik) lässt sich aufgrund des senkrechten Abfalls bei $&plusmn; f_{\rm D, \ max}$ nur sehr ungenau nachbilden.
+
* The Jakes PDS (left graph) can only be reproduced very inaccurately due to the vertical drop at&nbsp; $&plusmn; f_{\rm D, \ max}$.
* Für den Zeitbereich bedeutet dies, dass die AKF sehr viel schneller abfällt, als es die Theorie besagt.  
+
* For the time domain, this means that the ACF decreases much faster than the theory suggests.  
*Für kleine $\Delta t$&ndash;Werte ist die Approximation aber sehr gut (rechte Grafik).
+
*For small values of&nbsp;  $\Delta t$, however, the approximation is very good (right graph).
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Exercises for Mobile Communications|^1.3 Rayleigh Fading with Memory^]]
+
[[Category:Mobile Communications: Exercises|^1.3 Rayleigh Fading with Memory^]]

Latest revision as of 13:41, 17 February 2022

Considered Jakes spectrum

In a mobile radio system, the  Doppler effect  is also noticeable in the power-spectral density of the Doppler frequency $f_{\rm D}$.

This results in the so-called  Jakes spectrum, which is shown in the graph for the maximum Doppler frequency $f_{\rm D, \ max} = 100 \ \rm Hz$.  ${\it \Phi}_z(f_{\rm D})$  has only portions within the range  $± f_{\rm D, \ max}$, where

$${\it \Phi}_z(f_{\rm D}) = \frac{2 \cdot \sigma^2}{\pi \cdot f_{\rm D, \hspace{0.1cm} max} \cdot \sqrt { 1 - (f_{\rm D}/f_{\rm D, \hspace{0.1cm} max})^2} } \hspace{0.05cm}.$$

What is expressed in the frequency domain by the power-spectral density  $\rm (PSD)$  is described in the time domain by the auto-correlation function  $\rm (ACF)$.  The ACF is the   ${\it \Phi}_z(f_{\rm D})$  by the  inverse Fourier transform  of the PDS.

With the  Bessel function  of the first kind and zero order  $({\rm J}_0)$  you get

$$\varphi_z ({\rm \Delta}t) = 2 \sigma^2 \cdot {\rm J_0}(2\pi \cdot f_{\rm D, \hspace{0.1cm} max} \cdot {\rm \Delta}t)\hspace{0.05cm}.$$

To take into account the Doppler effect and thus a relative movement between transmitter and receiver in a system simulation, two digital filters are inserted in the  Rayleigh channel model, each with the frequency response  $H_{\rm DF}(f_{\rm D})$.

The dimensioning of these filters is part of this task.

  • We restrict ourselves here to the branch for generating the real part  $x(t)$.  The ratios derived here are also valid for the imaginary part  $y(t)$.
  • At the input of the left digital filter of the  Rayleigh channel model , there is white Gaussian noise  $n(t)$  with variance  $\sigma^2 = 0.5$.
  • The real component is then obtained from the following convolution
$$x(t) = n(t) \star h_{\rm DF}(t) \hspace{0.05cm}.$$


Notes:




Questions

1

What is the value of the Jakes spectrum of the real part at the Doppler frequency $f_{\rm D} = 0$?

${\it \Phi}_x(f_{\rm D} = 0)\ = \ $

$\ \cdot 10^{\rm –3} \ {\rm Hz}^{-1}$

2

Which dimensioning is correct, where  $K$  is an appropriately chosen constant?

It holds  $H_{\rm DF}(f_{\rm D}) = K \cdot {\it \Phi}_x(f_{\rm D})$.
It applies  $|H_{\rm DF}(f_{\rm D})|^2 = K \cdot {\it \Phi}_x(f_{\rm D})$

3

From which condition can the constant  $K$  be determined?

$K$  can be selected as desired.
The integral over  $|H_{\rm DF}(f_{\rm D})|$  must equal  $1$ .
The integral over  $|H_{\rm DF}(f_{\rm D})|^2$  must be  $1$ .

4

Is  $H_{\rm DF}(f)$  unambiguously defined by the two conditions according to  (2)  and  (3)?

Yes.
No.


Solution

(1)  The Jakes spectrum of the real part is half the resulting spectrum  ${\it \Phi}_z(f)$:

$${\it \Phi}_x(f_{\rm D} = 0) = {\it \Phi}_y(f_{\rm D} = 0) = \frac{{\it \Phi}_z(f_{\rm D} = 0)}{2}= \frac{\sigma^2}{\pi \cdot f_{\rm D, \hspace{0.05cm} max}} = \frac{0.5}{\pi \cdot 100\,\,{\rm Hz}} \hspace{0.15cm} \underline{ = 1.59 \cdot 10^{-3}\,\,{\rm Hz^{-1}}} \hspace{0.05cm}.$$


(2)  Solution 2 is correct:

  • The input signal  $n(t)$  has a white (constant) PDS  ${\it \Phi}_n(f_{\rm D})$.
  • The PDS at the output is then
$${\it \Phi}_x(f_{\rm D}) = {\it \Phi}_n(f_{\rm D}) \cdot | H_{\rm DF}(f_{\rm D}|^2 \hspace{0.05cm}.$$


(3)  Solution 3 is correct.

  • Only if this condition is fulfilled, the signal  $x(t)$  has the same variance  $\sigma^2$  as the noise signal  $n(t)$.


(4)  No:

  • The two conditions after subtasks  (2)  and  (3)  only refer to the magnitude of the digital filter.
  • There is no constraint for the phase of the digital filter.
  • This phase can be chosen arbitrarily.  Usually it is chosen in such a way that a minimum phase network results.
  • In this case, the impulse response  $h_{\rm DF}(t)$  then has the lowest possible duration.


The graph shows the result of the approximation.  The red curves were determined simulatively over $100\hspace{0.05cm}000$ samples.  You can see:

Approximation of the Jakes spectrum and the auto-correlation function
  • The Jakes PDS (left graph) can only be reproduced very inaccurately due to the vertical drop at  $± f_{\rm D, \ max}$.
  • For the time domain, this means that the ACF decreases much faster than the theory suggests.
  • For small values of  $\Delta t$, however, the approximation is very good (right graph).