Rayleigh fading should be used when

• there is no direct connection between transmitter and receiver, and
• the signal reaches the receiver through many paths, but their transit times are approximately the same.

An example of such a Rayleigh channel occurs in urban mobile communications when narrow-band signals are used with ranges between  $50$  and  $100$  meters.

Looking at the radio signals  $s(t)$  and  $r(t)$  in the equivalent low-pass range  $($that is, around the frequency  $f = 0)$,  the signal transmission is described completely by the equation

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

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

is always complex and has the following characteristics:

• The real part  $x(t)$  and the imaginary part  $y(t)$  are Gaussian mean-free random variables, both with equal variance  $\sigma^2$.  Within the components  $x(t)$  and  $y(t)$  there may be statistical dependence, but this is not relevant for the solution of the present task.  We assume that  $x(t)$  and  $y(t)$  are uncorrelated.
• The magnitude  $a(t) = |z(t)|$  has a Rayleigh PDF, from which the name "Rayleigh Fading" is derived:
$$f_a(a) = \left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^ { -a^2/(2\sigma^2)} \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm for}\hspace{0.15cm} a \ge 0 \\ {\rm for}\hspace{0.15cm} a < 0 \\ \end{array} \hspace{0.05cm}.$$
• The squared magnitude  $p(t) = a(t)^2 = |z(t)|^2$  is exponentially distributed according to the equation
$$f_p(p) = \left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^ { -p/(2\sigma^2)} \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm for}\hspace{0.15cm} p \ge 0 \\ {\rm for}\hspace{0.15cm} p < 0 \\ \end{array} \hspace{0.05cm}.$$

Measurements have shown that the time intervals with  $a(t) ≤ 1$  $($highlighted in yellow in the graphic$)$ add up to  $\text{59 ms}$  $($intervals highlighted in red$).$  Being the total measurement time  $\text{150 ms}$,  the probability that the magnitude of the Rayleigh fading is less than or equal to  $1$  is

$${\rm Pr}(a(t) \le 1) = \frac{59\,\,{\,{\rm ms}}}{150\,\,{\rm ms}} = 39.4 \% \hspace{0.05cm}.$$

In the lower graph, the value range between  $\text{-3 dB}$  and  $\text{+3 dB}$  of the logarithmic Rayleigh coefficient  $20 \cdot {\rm lg} \ a(t)$  is highlighted in green.  The subtask (4)  refers to this.

Notes:

### Questions

1

For the entire range,  we have  $a(t) ≤ 2$.  What is the maximum value of the logarithmic quantity in this range?

 ${\rm Max}\big [20 \cdot {\rm lg} \ {a(t)}\big] \ = \$ $\ \rm dB$

2

What is the maximum value of  $p(t) = |z(t)|^2$,  both in linear and logarithmic representation?

 ${\rm Max}\big[p(t)\big] \ = \$ ${\rm Max}\big[10 \cdot {\rm lg} \ p(t)\big] \ = \$ $\ \rm dB$

3

Let   ${\rm Pr}\big[a(t) ≤ 1\big] = 0.394$.  Determine the Rayleigh parameter  $\sigma$.

 $\sigma \ = \$

4

What is the probability that the logarithmic Rayleigh coefficient  $10 \cdot {\rm lg} \ p(t)$  is between   $\text{-3 dB}$  and  $\text{+3 dB}$?

 ${\rm Pr}(|10 \cdot {\rm lg} \ p(t)| < 3 \ \rm dB) \ = \$

### Solution

#### Solution

(1)  From  ${\rm Max}[a(t)] = 2$  follows directly:

$${\rm Max} \left [ 20 \cdot {\rm lg}\hspace{0.15cm}a(t) \right ] = 20 \cdot {\rm lg}\hspace{0.15cm}(2) \hspace{0.15cm} \underline{\approx 6\,\,{\rm dB}} \hspace{0.05cm}.$$

(2)  The maximum value of the square  $p(t) = a(t)^2$  is

$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{=4} \hspace{0.05cm}.$$
• The logarithmic representation of the squared magnitude  $p(t)$  is identical to the logarithmic representation of the magnitude  $a(t)$.
• Since  $p(t)$  is a power quantity:
$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{= 4} \hspace{0.05cm}.$$
• The maximum value is thus also  $\underline{\approx 6\,\,{\rm dB}}$.

(3)  The condition  $a(t) ≤ 1$  is equivalent to the requirement  $p(t) = a(t)^2 ≤ 1$.

• The absolute square is known to be exponentially distributed, and for  $p ≥ 0$  we have
$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm e}^{ -p/(2\sigma^2)} \hspace{0.05cm}.$$
PDF and probability regions
• It follows:
$${\rm Pr}(p(t) \le 1) = \frac{1}{2\sigma^2} \cdot \int_{0}^{1}{\rm e}^{ -p/(2\sigma^2)} \hspace{0.15cm}{\rm d}p = 1 - {\rm e}^{ -1/(2\sigma^2)} = 0.394$$
$$\Rightarrow \hspace{0.3cm} {\rm e}^{ -1/(2\sigma^2)} =0.606 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sigma^2 = \frac{1}{2 \cdot {\rm ln}\hspace{0.1cm}(0.606)} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{\sigma = 1} \hspace{0.05cm}.$$

The graph shows

• on the left side the probability  ${\rm Pr}(p(t) ≤ 1)$,
• on the right side the probability  ${\rm Pr}(0.5 \le p(t) ≤ 2)$.

(4)  From  $10 \cdot {\rm lg} \ p_1 = \ -3 \ \ \rm dB$  follows  $p_1 = 0.5$.  The upper limit of the integration range results from the condition  $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$,  so  $p_2 = 2$.

• This gives, according to the above graph:

$${\rm Pr}(-3\,\,{\rm dB}\le 10 \cdot {\rm lg}\hspace{0.15cm}p(t) \le +3\,\,{\rm dB}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \int_{0.5}^{2}f_p(p)\hspace{0.15cm}{\rm d}p = \left [ - {\rm e}^{ -{p}/(2\sigma^2)}\hspace{0.15cm} \right ]_{0.5}^{2} ={\rm e}^{-0.25}- {\rm e}^{-1} \approx 0.779 - 0.368 \hspace{0.15cm} \underline{ = 0.411} \hspace{0.05cm}.$$