Exercise 1.3: Rayleigh Fading

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 falsification

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

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 bindings, but this is not relevant for the solution of the present task. There are no bonds between $x(t)$ and $y(t)$; their cross-correlation function is identical to zero.

The amount $a(t) = |z(t)|$ has a Rayleigh–WDF, 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

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

$\ \rm dB$

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

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

$ \ \rm dB$

$\sigma \ = \ $

${\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:

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