Exercise 1.6: Autocorrelation Function and PDS with Rice Fading

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:

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 differ the autocorrelation functions  $\rm (ACFs)$  of the black, the blue and the green channel?

 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 differ the power density spectra  $\rm (PDSs)$  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.

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 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  $y(t) = v(t) + y_0$  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 ACF:

$${\rm E}\left [ |z(t)|^2 \right ] = \varphi_z ({\rm \delta}t = 0) = 2 \cdot \varphi_u ({\rm \delta}t = 0) + |z_0|^2 = 2 \cdot \sigma^2 + |z_0|^2 \hspace{0.05cm}.$$

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 solution  (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.