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  $z(t)$  describing the mobile radio channel contains 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 probability density function  $\rm (PDF)$, 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 autocorrelation function  $\rm (ACF)$  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 density spectrum  $\rm (PDS)$

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

Notes:

• This task belongs to chapter  Non-frequency selective fading with direct component.
• Reference is also made to the chapters  Autocorrelation function (ACF)  and  Power Density Spectrum (PDS)  in the book „Stochastic Signal Theory”.

Questions

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 magnitude,
	PDF  $f_{\rm \phi}(\phi)$  of the phase,
	ACF  $\varphi_z(\Delta t)$  the complex quantity  $z(t)$,
	PDS  ${\it \Phi}_z(f_{\rm D})$  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  $\rm (ACFs)$  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 density spectranbsp; $\rm (PDSs)$  differ among the black, blue, and green mobile radio channels?

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

Solutions

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 only on  $x_0$,
• $f_y(y)$  depends only on  $y_0$,
• $f_{\rm \phi}(\phi)$ depends only 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   ⇒   power density spectrum  $\rm PDS$.

(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 ACF is 4 over the black one.
• The green ACF is 6 over the blue one.

(5) All solution suggestions apply.

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