Aufgaben:Exercise 1.5: Reconstruction of the Jakes Spectrum: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process}} | ||
[[File:P_ID2124__Mob_A_1_5.png|right|frame| | [[File:P_ID2124__Mob_A_1_5.png|right|frame|Considered Jakes spectrum]] | ||
In a mobile radio system, the [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Phenomenological_description_of_the_Doppler_effect|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_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 | |||
:$${\it \Phi}_z(f_{\rm D}) = \frac{2 \cdot \sigma^2}{\pi \cdot f_{\rm D, \hspace{0. | :$${\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}.$$ | ||
What is expressed in the frequency domain by the power-spectral density $\rm (PSD)$ is described in the time domain by the auto-correlation function $\rm (ACF)$. The ACF is the ${\it \Phi}_z(f_{\rm D})$ by the [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|inverse Fourier transform]] of the PDS. | |||
With the [[Applets:Bessel_Functions_of_the_First_Kind|Bessel function]] of the first kind and zero order $({\rm J}_0)$ you get | |||
:$$\varphi_z ({\rm \Delta}t) = | :$$\varphi_z ({\rm \Delta}t) = 2 \sigma^2 \cdot {\rm J_0}(2\pi \cdot f_{\rm D, \hspace{0.1cm} max} \cdot {\rm \Delta}t)\hspace{0.05cm}.$$ | ||
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/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. | |||
* | *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/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 | ||
:$$x(t) = n(t) \star h_{\rm DF}(t) \hspace{0.05cm}.$$ | :$$x(t) = n(t) \star h_{\rm DF}(t) \hspace{0.05cm}.$$ | ||
''Notes:'' | |||
* This task belongs to the topic of [[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|Digital Filter]] of the book "Stochastic Signal Theory". | |||
=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
{ | {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 | ${\it \Phi}_x(f_{\rm D} = 0)\ = \ $ { 1.59 } $\ \cdot 10^{\rm –3} \ {\rm Hz}^{-1}$ | ||
{ | {Which dimensioning is correct, where $K$ is an appropriately chosen constant? | ||
|type="[]"} | |type="[]"} | ||
- | - It holds $H_{\rm DF}(f_{\rm D}) = K \cdot {\it \Phi}_x(f_{\rm D})$. | ||
+ | + It applies $|H_{\rm DF}(f_{\rm D})|^2 = K \cdot {\it \Phi}_x(f_{\rm D})$ | ||
{ | {From which condition can the constant $K$ be determined? | ||
|type="[]"} | |type="[]"} | ||
- $K$ | - $K$ 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 $1$ . | ||
{ | {Is $H_{\rm DF}(f)$ unambiguously defined by the two conditions according to '''(2)''' and '''(3)'''? | ||
|type="()"} | |type="()"} | ||
- | - Yes. | ||
+ | + No. | ||
</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | '''(1)''' The Jakes spectrum of the real part is half the resulting spectrum ${\it \Phi}_z(f)$: | ||
:$${\it \Phi}_x(f_{\rm D} = 0) = {\it \Phi}_y(f_{\rm D} = 0) = \frac{{\it \Phi}_z(f_{\rm D} = 0)}{2}= \frac{\sigma^2}{\pi \cdot f_{\rm D, \hspace{0.05cm} max}} = | :$${\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}.$$ | ||
'''( | '''(2)''' <u>Solution 2</u> is correct: | ||
* | *The input signal $n(t)$ has a white (constant) PDS ${\it \Phi}_n(f_{\rm D})$. | ||
*The PDS at the output is then | |||
:$${\it \Phi}_x(f_{\rm D}) = {\it \Phi}_n(f_{\rm D}) \cdot | H_{\rm DF}(f_{\rm D}|^2\hspace{0.05cm}.$$ | |||
'''(3)''' <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 $n(t)$. | |||
'''(4)''' <u>No</u>: | |||
*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: | |||
[[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 $± f_{\rm D, \ max}$. | ||
* | * 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). | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
[[Category: | [[Category:Mobile Communications: Exercises|^1.3 Rayleigh Fading with Memory^]] | ||
[[de:Aufgaben:Exercise 1.5: Reconstruction of the Jakes Spectrum]] | |||
Latest revision as of 17:55, 16 March 2026

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:
- This task belongs to the topic of Statistical bindings within the Rayleigh process.
- The digital filter is treated in detail in chapter Digital Filter of the book "Stochastic Signal Theory".
Questions
Solution
- $${\it \Phi}_x(f_{\rm D} = 0) = {\it \Phi}_y(f_{\rm D} = 0) = \frac{{\it \Phi}_z(f_{\rm D} = 0)}{2}= \frac{\sigma^2}{\pi \cdot f_{\rm D, \hspace{0.05cm} max}} =\frac{0.5}{\pi \cdot 100\,\,{\rm Hz}} \hspace{0.15cm} \underline{ = 1.59 \cdot 10^{-3}\,\,{\rm Hz^{-1}}}\hspace{0.05cm}.$$
(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:

- 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).