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

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m (Text replacement - "power spectral density" to "power-spectral density")
 
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{{quiz-Header|Buchseite=Mobile Kommunikation/Statistische Bindungen innerhalb des Rayleigh-Prozesses}}
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{{quiz-Header|Buchseite=Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process}}
  
 
[[File:P_ID2124__Mob_A_1_5.png|right|frame|Considered Jakes spectrum]]
 
[[File:P_ID2124__Mob_A_1_5.png|right|frame|Considered Jakes spectrum]]
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 density spectrum of the Doppler frequency  $f_{\rm D}$ .  
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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 $f_{\rm D}$.  
  
This results in the so-called  [[Mobile_Communications/Statistical_bindings_within_the_Rayleigh_process#ACF_und_PSD_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
+
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}.$$
  
What is expressed in the frequency domain by the power spectral density (PSD) is described in the time domain by the autocorrelation function (ACF).  The ACF is the   ${\it \Phi}_z(f_{\rm D})$  by the  [[Signal_Representation/Fouriertransformation_und_-r%C3%BCcktransformation#Das_zweite_Fourierintegral|''' checkLink:_Buch_1 ⇒ ''' inverse Fourier transform]] of the PSD.
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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.
  
 
With the  [[Applets:Bessel_Functions_of_the_First_Kind|Bessel function]]  of the first kind and zero order  $({\rm J}_0)$  you get
 
With the  [[Applets:Bessel_Functions_of_the_First_Kind|Bessel function]]  of the first kind and zero order  $({\rm J}_0)$  you get
:$$\varphi_z ({\rm \Delta}t) = 2 \sigma^2 \cdot {\rm J_0}(2\pi \cdot f_{\rm D, \hspace{0.05cm} max} \cdot {\rm \Delta}t)\hspace{0.05cm}.$$
+
:$$\varphi_z ({\rm \Delta}t) = 2 \sigma^2 \cdot {\rm J_0}(2\pi \cdot f_{\rm D, \hspace{0.1cm} max} \cdot {\rm \Delta}t)\hspace{0.05cm}.$$
  
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  [[Mobile_Communications/Wahrscheinlichkeitsdichte_des_Rayleigh%E2%80%93Fadings#Modellierung_von_nichtfrequenzselektivem_Fading|Rayleigh channel model]], each with the frequency response  $H_{\rm DF}(f_{\rm D})$.  
+
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  [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading|Rayleigh channel model]], each with the frequency response  $H_{\rm DF}(f_{\rm D})$.  
  
 
The dimensioning of these filters is part of this task.
 
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  $y(t)$.
 
*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  [[Mobile_Communications/Wahrscheinlichkeitsdichte_des_Rayleigh%E2%80%93Fadings#Frequenzselektives_Fading_vs._nichtfrequenzselektives_Fading|Rayleigh channel model]] , there is white Gaussian noise  $n(t)$  with variance  $\sigma^2 = 0.5$.  
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*At the input of the left digital filter of the  [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading#Modeling_of_non-frequency_selective_fading|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
 
*The real component is then obtained from the  following convolution
 
:$$x(t) = n(t) \star h_{\rm DF}(t) \hspace{0.05cm}.$$
 
:$$x(t) = n(t) \star h_{\rm DF}(t) \hspace{0.05cm}.$$
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''Notes:''  
 
''Notes:''  
* This task belongs to the topic of  [[Mobile_Communications/Statistical_bindings_within_the_Rayleigh_process|Statistical bindings within the Rayleigh process]].  
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* This task belongs to the topic of  [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process|Statistical bindings within the Rayleigh process]].  
* The digital filter is treated in detail in chapter  [[Theory_of_Stochastic_Signals/Digitale_Filter|''' checkLink:_Buch_3 ⇒ ''' Digital Filter]]  of the book „Stochastic Signal Theory”.
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* The digital filter is treated in detail in chapter  [[Theory_of_Stochastic_Signals/Digitale_Filter|Digital Filter]]  of the book "Stochastic Signal Theory".
  
  
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<quiz display=simple>
 
<quiz display=simple>
{What is the value of the Jakes, spectrum of the real part at the Doppler frequency&nbsp; $f_{\rm D} = 0$?
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{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 } $\ \cdot 10^{\rm &ndash;3} \ 1/{\rm Hz}$
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${\it \Phi}_x(f_{\rm D} = 0)\ = \ $ { 1.59 } $\ \cdot 10^{\rm &ndash;3} \ {\rm Hz}^{-1}$
  
 
{Which dimensioning is correct, where &nbsp;$K$&nbsp; is an appropriately chosen constant?
 
{Which dimensioning is correct, where &nbsp;$K$&nbsp; is an appropriately chosen constant?
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+ The integral over&nbsp; $|H_{\rm DF}(f_{\rm D})|^2$&nbsp; must be&nbsp; $1$&nbsp;.
 
+ The integral over&nbsp; $|H_{\rm DF}(f_{\rm D})|^2$&nbsp; must be&nbsp; $1$&nbsp;.
  
{Is&nbsp; $H_{\rm DF}(f)$&nbsp; unambiguously defined by the two conditions according to '''(2)'' and '''(3)'''?
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{Is&nbsp; $H_{\rm DF}(f)$&nbsp; unambiguously defined by the two conditions according to&nbsp; '''(2)'''&nbsp; and&nbsp; '''(3)'''?
 
|type="()"}
 
|type="()"}
 
- Yes.
 
- Yes.
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</quiz>
 
</quiz>
  
===Solutions===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; The Jakes spectrum of the real part is half the resulting spectrum ${\it \Phi}_z(f)$:
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'''(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}}}
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'''(2)'''&nbsp; <u>Solution 2</u> is correct:  
 
'''(2)'''&nbsp; <u>Solution 2</u> is correct:  
*The input signal $n(t)$ has a white (constant) LDS ${\it \Phi}_n(f_{\rm D})$.  
+
*The input signal&nbsp; $n(t)$&nbsp; has a white (constant) PDS&nbsp; ${\it \Phi}_n(f_{\rm D})$.  
*The PSD at the output is then
+
*The PDS at the output is then
$${\it \Phi}_x(f_{\rm D}) = {\it \Phi}_n(f_{\rm D}) \cdot | H_{\rm DF}(f_{\rm D}|^2
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:$${\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; <u>Solution 3</u> is correct.  
 
'''(3)'''&nbsp; <u>Solution 3</u> is correct.  
*Only if this condition is fulfilled, the signal $x(t)$ has the same variance $\sigma^2$ as the noise signal $n(t)$.
+
*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>No</u>:  
 
'''(4)'''&nbsp; <u>No</u>:  
*The two conditions after subtasks (2) and (3) only refer to the magnitude of the digital filter.  
+
*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.  
 
*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.  
+
*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 $h_{\rm DF}(t)$ then has the lowest possible duration.
+
*In this case, the impulse response&nbsp; $h_{\rm DF}(t)$&nbsp; 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:
+
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 ACF]]
+
[[File:EN_Mob_A_1_5d.png|right|frame|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&nbsp; $&plusmn; f_{\rm D, \ max}$.
* The Jakes power spectral density (left graph) can only be reproduced very inaccurately due to the vertical drop at $&plusmn; f_{\rm D, \ max}$.
+
* For the time domain, this means that the ACF decreases much faster than the theory suggests.  
* For the time domain, this means that the ACF decreases much faster than theory suggests.  
+
*For small values of&nbsp; $\Delta t$, however, the approximation is very good (right graph).
*For small values of  $\delta t$, however, the approximation is very good (right graph).
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Exercises for Mobile Communications|^1.3 Rayleigh Fading with Memory^]]
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[[Category:Mobile Communications: Exercises|^1.3 Rayleigh Fading with Memory^]]

Latest revision as of 12:41, 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).