Exercise 1.5: Reconstruction of the Jakes Spectrum

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

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 inverse Fourier transform of the PSD.

With the Bessel function of the first kind and zero order $({\rm J}_0)$ you get $$\varphi_z ({\rm \Delta}t) = 2 \sigma^2 \cdot {\rm J_0}(2\pi \cdot f_{\rm D, \hspace{0.05cm} max} \cdot {\rm \Delta}t)\hspace{0.05cm}.$$

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 subject area Statistische Bindungen innerhalb des Rayleigh–Prozesses.
The digital filter is treated in detail in chapter Digitale Filter of the book „Stochastic Signal Theory”.

Questionnaire

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

$\ \cdot 10^{\rm –3} \ 1/{\rm Hz}$

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

	$K$ 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 $1$ .

	Yes.
	No.

Sample solution

Musterlösung

(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) LDS ${\it \Phi}_n(f_{\rm D})$.
The PSD 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 ACF

The Jakes power spectral density (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 theory suggests.
For small values of $\delta t$, however, the approximation is very good (right graph).