Exercise 1.5: Reconstruction of the Jakes Spectrum

From LNTwww

Considered Jakes spectrum

In a mobile radio system, the  Doppler effect  is also noticeable in the power density spectrum 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 density spectrum  $\rm (PDS)$  is described in the time domain by the autocorrelation 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 autocorrelation 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).