Exercise 1.3: Rayleigh Fading

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Time evolution of Rayleigh fading

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

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

The multiplicative fading coefficient

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

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:

$$f_a(a) = \left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^ { -a^2/(2\sigma^2)} \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} a \ge 0 \\ {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\ \end{array} \hspace{0.05cm}.$$

• The absolute square  $p(t) = a(t)^2 = |z(t)|^2$  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}$$ \right.\quad \begin{array}{*{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} p \ge 0 \\ {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\\ \\ \end{array} \hspace{0.05cm}.$$

Measurements have shown that the time intervals with  $a(t) ≤ 1$  (highlighted in yellow in the graphic) add up to  $\text{59 ms}$  (areas highlighted in red). With the total measurement time of  $\text{150 ms}$  the probability that the amount of the Rayleigh–fading is less than or equal to  $1$  results in $${\rm Pr}(a(t) \le 1) = \frac{59\,\,{\,{\rm ms}}}{150\,\,{\rm ms}} = 39.4 \% \hspace{0.05cm}.$$

In the lower graphic the value range between  $\text{-3 dB}$  and  $\text{+3 dB}$  regarding the logarithmic Rayleigh–Size  $20 \cdot {\rm lg} is highlighted in green. \ a(t)$. The subtask '(4) refers to this.

Notes:

• The task belongs to chapter  Probability Density of Rayleigh–Fadings  of this book.
• A similar topic is treated with a different approach in chapter  Weitere Verteilungen  of the book „Stochastic Signal Theory”.
• To check your results you can use the interactive applet  WDF, VTF and Moments  of the book „Stochastic Signal Theory”.

Questionnaire

1

For the entire range, the amount function  $a(t) ≤ 2$ applies. What is the maximum value for the logarithmic quantity in this range?

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

2

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

${\rm Max}\big[p(t)\big] \ = \ $ {\ $ 4 3% }

${\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 of the logarithmic Rayleigh–size   ⇒   $10 \cdot {\rm lg} \ p(t)$  in the range between between  $\text{-3 dB}$  and  $\text{+3 dB}$?

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

Sample 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 square of the amount $p(t)$ is identical to the logarithmic representation of the amount $a(t)$. Since $p(t)$ is a power quantity

$$10 \cdot {\rm lg}\hspace{0.15cm} p(t) = 10 \cdot {\rm lg}\hspace{0.15cm}a(t)^2 = 20 \cdot {\rm lg}\hspace{0.15cm} a(t) \hspace{0.05cm}.$$

• 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 $p ≥ 0$ applies accordingly:

$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}] \hspace{0.05cm}.$$

WDF and probability regions

• It follows:

$${\rm Pr}(p(t) \le 1) = \frac{1}{2\sigma^2} \cdot \int_{0}^{1}{\rm exp} [ -\frac{p}{2\sigma^2}] \hspace{0.15cm}{\rm d}p = 1 - {\rm exp} [ -\frac{1}{2\sigma^2}] = 0.394$$ $$\Rightarrow \hspace{0.3cm} {\rm exp} [ -\frac{1}{2\sigma^2}] = 0.606 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sigma^2 = \frac{1}{2 \cdot {\rm ln}\hspace{0.1cm}(0.606)} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{\sigma = 1} \hspace{0.05cm}.$$

The graphic shows

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

'(4)  From $10 \cdot {\rm lg} \ p_1 = \ –3 \ \ \rm dB$ follows $p_1 = 0.5$ and the upper limit of the integration range results from the condition $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$ to $p_2 = 2$.

• This gives you, according to the above graphic:

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