Exercise 1.2Z: Lognormal Fading Revisited

Path loss plus lognormal fading

We assume similar conditions as in Task 1. 2 but now we summarize the purely distance-dependent path loss $V_0$ and the mean value $m_{\rm S}$ of the lognormal–fading (the index „S” stands for Shadowing):

$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$

The total path loss is then given by the equation $$V_{\rm P} = V_{\rm 1} + V_{\rm 2}(t)$$

where $V_2(t)$ describes a lognormal–distribution with mean value zero:

$$f_{V_{\rm S}}(V_{\rm S}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}} \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) }\hspace{0.05cm}.$$

The path loss model shown in the graphic is suitable for the scenario described here:

Multiply the transmitted signal $s(t)$ first with a constant factor $k_1$ and further with a stochastic quantity $z_2(t)$ with the probability density function (PDF) $f_{\rm z_2}(z_2)$, then the signal $r(t)$ results at the output, whose power $P_{\rm E}(t)$ is of course also time-dependent due to the stochastic component.
The PDF of the lognormally distributed random variable $z_2$ is for $z_2 ≥ 0$:

$$f_{z_{\rm 2}}(z_{\rm 2}) = \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 2}) /({2 \hspace{0.05cm}\cdot \hspace{0.05cm} C^2 \hspace{0.05cm} \cdot \hspace{0.05cm} \sigma_{\rm S}^2}) } } }{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2} \hspace{0.8cm}{\rm mit} \hspace{0.8cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$

For $z_2 ≤ 0$ this PDF is equal to zero.

Notes:

This task belongs to the chapter Distanzabhängige Dämpfung und Abschattung.
Use the following parameters:

$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$

The probability that a mean-free Gaussian random variable $z$ is greater than its standard deviation $\sigma$, is

$${\rm Pr}(z > \sigma) = {\rm Pr}(z < -\sigma) = {\rm Q}(1) \approx 0.158\hspace{0.05cm}.$$

Also, ${\rm Pr}(z > 2\sigma) = {\rm Pr}(z < -2\sigma) = {\rm Q}(2) \approx 0.023\hspace{0.05cm}.$
Again for clarification: $z_2$ is the fading coefficient in linear units, while $V_2$ is the fading coefficient in logarithmic units.
The following conversions apply:

$$z_2 = 10^{-V_{\rm 2}/20\,{\rm dB}}\hspace{0.05cm}, \hspace{0.2cm} V_{\rm 2} = -20\,{\rm dB} \cdot {\rm lg}\hspace{0.15cm}z_2\hspace{0.05cm}.$$

Questionnaire

$k_1\ = \ $

	All values between $-∞$ and $+∞$ are possible.
	The random size $z_2$ is not negative.
	The smallest possible value is $z_2 = 0.5$.
	The largest possible value is $z_2 = 2$.

$f_{\rm z2}(z_2 = 0)\ = \ $

$f_{\rm z2}(z_2 = 1)\ = \ $

$f_{\rm z2}(z_2 = 2)\ = \ $

${\rm Pr}(z_2 > 1.0)\ = \ $

${\rm Pr}(z_2 > 0.5)\ = \ $

${\rm Pr}(z_2 > 4.0)\ = \ $

	${\rm E}[P_{\rm E}(t)] = P_{\rm E}'$
	${\rm E}[P_{\rm E}(t)] < P_{\rm E}'$.
	${\rm E}[P_{\rm E}(t)] > P_{\rm E}'$.

Sample solution

Musterlösung

(1) The constant $k_1$ generates the time-independent path loss $V_1 = 60 \ \rm dB$. From this follows:

$$k_{\rm 1} = 10^{-V_{\rm 1}/(20\hspace{0.05cm} {\rm dB})} \hspace{0.15cm} \underline{= 0.001}\hspace{0.05cm}.$$

(2) Only the second statement is correct:

For the Gaussian random variable $V_2$ all values between $–∞$ and $+∞$ are (theoretically) possible.
The transformation $z_2 = 10^{{\it –V_2}\rm /20}$ results in only positive values for the linear random variable $z_2$, namely between 0 (if $V_2$ is positive and goes to infinity) and $+∞$ (very large negative values of $V_2$).

(3) The random value $z_2$ can only be positive. Therefore the PDF value $f_{\rm z2}(z_2 = 0)\hspace{0.15cm} is \underline{ = 0}$.

The PDF–value for the abscissa value $z_2 = 1$ is obtained by inserting it into the given equation:

$$f_{z{\rm 2}}(z_{\rm 2} = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 2}=1) /({2 \hspace{0.05cm}\cdot \hspace{0.05cm} C^2 \hspace{0.05cm} \cdot \hspace{0.05cm} \sigma_{\rm S}^2}) } } }{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot (z_2 = 1)} = \frac {1}{ \sqrt{2 \pi } \cdot 6\,\,{\rm dB} } \cdot \frac {20\,\,{\rm dB}}{ {\rm ln} \hspace{0.1cm}(10) } \hspace{0.15cm} \underline{\approx 0.578}\hspace{0.05cm}.$$

The first portion is equal to the WDF–value $f_{{{\it V}2}(V_2 = 0)$.
$C$ considers the amount of the derivative of the non-linear characteristic $z_2 = g(V_2)$ for $V_2 = 0 \ \rm dB$ or $z_2 = 1$.
Finally, for $z_2 = 2$:

$$f_{z{\rm 2}}(z_{\rm 2} = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac {f_{z{\rm 2}}(z_{\rm 2} = 1)}{ z_{\rm 2} = 2} \cdot {\rm exp } \left [ - \frac {{\rm ln}^2 (2)}{2 \cdot C^2 \cdot \sigma_{\rm S}^2} \right ]= \frac {0.578}{ 2} \cdot {\rm exp } \left [ - \frac {0.48}{0.952} \right ] \hspace{0.15cm} \underline{\approx 0.174}\hspace{0.05cm}. $$

(4) If you take into account the relationship between $z_2$ and $V_2$, you get

$${\rm Pr}(z_{\rm 2} > 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(V_{\rm 2} < 0\,\,{\rm dB})\hspace{0.15cm} \underline{= 0.5} \hspace{0.05cm},$$

$${\rm Pr}(z_{\rm 2} > 0.5) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(V_{\rm 2} < 6\,\,{\rm dB}) = 1- {\rm Pr}(V_{\rm 2} > 6\,\,{\rm dB})= 1- {\rm Pr}(V_{\rm 2} > \sigma_{\rm S})= 1- {\rm Q}(1)\hspace{0.15cm} \underline{= 0.842} \hspace{0.05cm},$$

$${\rm Pr}(z_{\rm 2} > 4) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(V_{\rm 2} < -12\,\,{\rm dB}) = {\rm Pr}(V_{\rm 2} > +12\,\,{\rm dB}) = {\rm Pr}(V_{\rm 2} > 2 \sigma_{\rm S}) \hspace{0.05cm}.$$

The probability that a Gaussian variable is greater than $2 \cdot \sigma$ equals ${\rm Q}(2)$:

$${\rm Pr}(z_{\rm 2} > 4) = {\rm Q}(2)\hspace{0.15cm} \underline{= 0.023} \hspace{0.05cm}.$$

(5) The statement 3 is correct:

The first statement is certainly not correct, since the mean value $m_{\rm S}$ refers to the logarithmic received power (in $\rm dBm$).
To clarify whether the second or the third statement is correct, we assume $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$ ⇒ $P_{\rm E}' = 1 \ {\rm µ W}$ and the following PDF for $V_2$:

$$f_{V{\rm 2}}(V_{\rm 2}) = 0.5 \cdot \delta (V_{\rm 2}) + 0.25 \cdot \delta (V_{\rm 2}- 10\,\,{\rm dB}) + 0.25 \cdot \delta (V_{\rm 2}+ 10\,\,{\rm dB})\hspace{0.05cm}.$$

Half of the time, $P_{\rm E} = 1 \ \rm µ W$, while each of the following has $25\%$ probability::

$$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\,}{\rm µ W}\hspace{0.05cm},$$

$$V_{\rm 2}= -10\,\,{\,}{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\,}{\rm µ W}\hspace{0.05cm}.$$

The mean value is then:

$${\rm E}[P_{\rm E}(t)] = 0.5 \cdot 1\,{\rm µ W}+ 0.25 \cdot 0.1\,{\rm µ W}+ 0.25 \cdot 10\,{\rm µ W}= 3.025\,{\rm µ W} > P_{\rm E}\hspace{0.05cm}' = 1\,{\rm µ W} \hspace{0.05cm}.$$

This simple calculation with discrete probabilities instead of a continuous PDF indicates that statement 3 is correct.