Aufgaben:Exercise 1.6: Autocorrelation Function and PDS with Rice Fading

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Rice PDF for different values of   $z_0^2$

One speaks of  Rice fading if the complex factor describing the mobile radio channel contains   $z(t)$  besides the purely stochastic component  $x(t) +{\rm j} \cdot y(t)$  a deterministic part of the form  $x_0 + {\rm j} \cdot y_0$ .

The equations of Rice fading can be summarized briefly as follows:

$$r(t) = z(t) \cdot s(t) ,$$

$$z(t) = x(t) + {\rm j} \cdot y(t) ,$$

$$x(t) = u(t) + x_0 ,$$

$$y(t) = v(t) + y_0 .$$

The following applies:

• The direct path is defined by the complex constant  $z_0 = x_0 + {\rm j} \cdot y_0$ . The magnitude of this time-invariant component is

$$|z_0| = \sqrt{x_0^2 + y_0^2}\hspace{0.05cm}.$$

• $u(t)$  and  $v(t)$  are zero-mean Gaussian random processes, both with variance  $\sigma^2$  and uncorrelated with each other. They model scattering, refraction; and diffraction effects on a variety of indirect paths.
• The magnitude  $a(t) = |z(t)|$  has a Rice PDF, which gives this channel model its name. For   $a ≥ 0$, the PDF is

$$f_a(a) = \frac{a}{\sigma^2} \cdot {\rm exp} [ -\frac{a^2 + |z_0|^2}{2\sigma^2}] \cdot {\rm I}_0 \left [ \frac{a \cdot |z_0|}{\sigma^2} \right ]\hspace{0.05cm}, \hspace{0.2cm}{\rm I }_0 (u) = \sum_{k = 0}^{\infty} \frac{ (u/2)^{2k}}{k! \cdot \Gamma (k+1)} \hspace{0.05cm}.$$

The graph shows the Rice PDF for  $|z_0|^2 = 0,\ 2, \ 4, \ 10$  and  $20$. For all curves, we have   $\sigma = 1$   ⇒   $\sigma^2 = 1$.

In this task, however, we will not consider the PDF of the magnitude, but the ACF of the complex factor  $z(t)$, $$\varphi_z ({\rm \Delta}t) = {\rm E}\big [ z(t) \cdot z^{\star}(t + {\rm \Delta}t)\big ] \hspace{0.05cm},$$

and the corresponding power spectral density

$${\it \Phi}_z (f_{\rm D}) \hspace{0.3cm} \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.3cm} \varphi_z ({\rm \Delta}t) \hspace{0.05cm}.$$

Notes:

• This task belongs to chapter  Nichtfrequenzselektives Fading mit Direktkomponente.
• Reference is also made to the chapters  Autokorrelationsfunktion (AKF)  and  Leistungsdichtespektrum (LDS)  in the book „Stochastic Signal Theory”.

=Questionnaire

1

Which value of   $|z_0|^2$ corresponds to Rayleigh fading?

$|z_0|^2 \ = \ $ $\ \ $rm $

2

Let $|z_0|^2 \ne 0$. Which of the following functions depend only on  $|z_0|^2 = x_0^2$ + $y_0^2$  but not on its components  $x_0^2$  and  $y_0^2$  alone?

	PDF  $f_x(x)$  of the real part,
	PDF  $f_y(y)$  of the imaginary part,
	PDF  $f_a(a)$  of the amount
	PDF  $f_{\rm \phi}(\phi)$  of the phase,
	ACF  $\varphi_z(\Delta t)$  the complex quantity  $z(t)$,
	PSD  ${\it \Phi}_z(f_{\rm D})$  the complex quantity  $z(t)$.

3

Calculate the root mean square value  ${\rm E}\big[|z(t)|^2\big]$ for different values of  $|z_0|^2$. Assume   $\sigma^2 = 1$.

$|z_0|^2 = 0\text{:} \hspace{0.52cm} {\rm E}\big[|z(t)|^2\big] \ = \ $ $\ \ \rm $

$|z_0|^2 = 2\text{:} \hspace{0.52cm} {\rm E}\big[|z(t)|^2\big] \ = \ $ $\ \ \rm $

$|z_0|^2 = 10\text{:} \hspace{0.3cm} {\rm E}\big[|z(t)|^2\big] \ = \ $ $\ \ \rm $

4

How do the autocorrelation functions (ACFs) of the black, the blue and the green channel differ?

	The „blue” ACF is above the „black” ACF by about  $4$  units.
	The „blue” ACF is below the „black” ACF by about  $2$  units.
	The „green” ACF is wider than the „blue” by the factor  $2.5$ .

5

How do the power spectral densities (PSDs) differ among the black, blue, and green mobile radio channels?

	The „black” PSD is purely continuous (no Dirac).
	The „blue” and „green” PSDs contain one Dirac each.
	The „green” Dirac has a greater weight than the „blue” one.

Sample solution

Solution

(1) Rayleigh fading results from the Rice fading with $|z_0|^2 \ \underline {= \ 0}$.

(2) Options 3, 5 and 6 are correct:

It is obvious that

• $f_x(x)$ depends on $x_0$
• $f_y(y)$ depends on $y_0$
• $f_{\rm \phi}(\phi)$ depends on the ratio $y_0/x_0$.

The given equation for the PDF $f_a(a)$ shows that the magnitude $a$ depends only on $|z_0|$.

For the ACF, using $z(t) = x(t) + {\rm j} \cdot y(t)$ we have

$$\varphi_z ({\rm \Delta}t) = {\rm E}\left [ z(t) \cdot z^{\star}(t + {\rm \Delta}t)\right] = {\rm E}\left [ \left ( x(t) + {\rm j} \cdot y(t) \right )\cdot (x(t + {\rm \Delta}t) - {\rm j} \cdot (y(t+ {\rm \Delta}t)\right ] \hspace{0.05cm}.$$

Because of the statistical independence between real and imaginary parts, the equation can be simplified as follows:

$$\varphi_z ({\rm \Delta}t) = {\rm E}\left [ x(t) \cdot x(t + {\rm \Delta}t)\right ] + {\rm E}\left [ y(t) \cdot y(t + {\rm \Delta}t)\right ] \hspace{0.05cm}.$$

With $x(t) = u(t) + x_0$ and $t' = t + \Delta t$, the first part results in $x(t) = u(t) + x_0$:

$${\rm E}\left [ x(t) \cdot x(t')\right ] = {\rm E}\left [ u(t) \cdot u(t')\right ] + x_0 \cdot {\rm E}\left [ u(t) \right ] + x_0 \cdot {\rm E}\left [ u(t') \right ] + x_0^2\hspace{0.05cm},$$

$$\Rightarrow \hspace{0.3cm} {\rm E}\left [ x(t) \cdot x(t + {\rm \Delta}t)\right ] = {\rm E}\left [ u(t) \cdot u(t + {\rm \Delta}t)\right ] + x_0^2 = \varphi_u ({\rm \Delta}t) + x_0^2 \hspace{0.05cm}.$$

This takes into account that the Gaussian random variable $u(t)$ has zero mean and has the variance $\sigma^2$.

In the same way with $y(t) = \upsilon (t) + y_0$ is obtained:

$${\rm E}\left [ y(t) \cdot y(t + {\rm \Delta}t)\right ] = \ ... \ = \varphi_v ({\rm \Delta}t) + y_0^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \varphi_z ({\rm \Delta}t) = \varphi_u ({\rm \Delta}t) + \varphi_v ({\rm \Delta}t) + x_0^2 + y_0^2 = 2 \cdot \varphi_u ({\rm \Delta}t) + |z_0|^2 \hspace{0.05cm}.$$

But if the ACF $\varphi_z(\Delta t)$ only depends on $|z_0^2|$, then this also applies to the Fourier transform „LDS”.

(3) The root mean square can be calculated from the PDF of the magnitude:

$${\rm E}\left [ |z(t)|^2 \right ] = {\rm E}\left [ a^2 \right ] = \int_{0}^{\infty}a^2 \cdot f_a(a)\hspace{0.15cm}{\rm d}a \hspace{0.05cm}.$$

At the same time, the root mean square value – i.e. the power – is also determined by the AKF: $${\rm E}\left [ |z(t)|^2 \right ] = \varphi_z ({\rm \delta}t = 0) = 2 \cdot \varphi_u ({\rm \delta}t = 0) + |z_0|^2 = 2 \cdot \sigma^2 + |z_0|^2 \hspace{0.05cm}.$$

With $\sigma = 1$ you get the following numerical results:

$$ \ |z_0|^2 = 0\text{:} \ \hspace{0.3cm}{\rm E}\left [ |z(t)|^2 \right ] = 2 + 0 \hspace{0.15cm} \underline{ = 2} \hspace{0.05cm},$$

$$ \ |z_0|^2 = 2\text{:} \ \hspace{0.3cm}{\rm E}\left [ |z(t)|^2 \right ] = 2 + 2 \hspace{0.15cm} \underline{ = 4} \hspace{0.05cm},$$

$$|z_0|^2 = 10\text{:} \ \hspace{0.3cm}{\rm E}\left [ |z(t)|^2 \right ] = 2 + 10 \hspace{0.15cm} \underline{ = 12} \hspace{0.05cm}.$$

(4) Correct is the solution 1, as already derived in the sample solution for (2).

The following statements would also be correct:

• The „blue” AKF is 4 over the „black”.
• The „green” AKF is 6 over the „blue”.

(5) All solution suggestions apply.

• The „black” LDS is a Jakes–Spectrum and therefore continuous, i.e. all frequencies are present within an interval.
• In the autocorrelation function (AKF) of the blue or green channel, the constant $|z_0|^2$ also occurs.
• In the power density spectrum (LDS), there are Dirac functions in the GCF at the Doppler frequency $f_{\rm D} = 0$ with the weight $|z_0|^2$ because of these constants.