Exercise 1.2Z: Lognormal Fading Revisited

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Path loss model
with 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: $$f_{V{\rm 2}}(V_{\rm 2}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}} \cdot {\rm exp } \left [ - \frac{ V_{\rm 2} ^2}{2 \cdot \sigma_{\rm S}^2} \right ] \hspace{0.05cm}.$$

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  $z_2 ≥ 0$:

$$f_{z{\rm 2}}(z_{\rm 2}) = \frac {{\rm exp } \left [ - {\rm ln}^2 (z_{\rm 2}) /({2 \cdot C^2 \cdot \sigma_{\rm S}^2}) \right ]}{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2} \hspace{0.3cm}{\rm with} \hspace{0.3cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$

• 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

1

How large should the constant  $k_1$  be?

$k_1\ = \ $

2

Which value range applies to the random variable  $z_2$?

	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  $z_2 = 2$.

3

Calculate the PDF  $f_{\rm z2}(z_2)$  for some abscissa values.

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

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

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

4

Calculate the following probabilities.

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

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

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

5

What statements are valid for the average receive power  ${\rm E}\big [P_{\rm E}(t)\big]$?
Note:  $P_{\rm E}'$ is the power after multiplication by  $k_1$  (see diagram).

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

Sample solution

Solution

(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 $f_{\rm z2}(z_2 = 0)\hspace{0.15cm} is \underline{ = 0}$.

• Der WDF–Wert für den Abszissenwert $z_2 = 1$ erhält man durch Einsetzen in die gegebene Gleichung:

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

• Der erste Anteil ist gleich dem WDF–Wert $f_{{\it V}2}(V_2 = 0)$.
• $C$ berücksichtigt den Betrag der Ableitung der nichtlinearen Kennlinie $z_2 = g(V_2)$ für $V_2 = 0 \ \rm dB$ bzw. $z_2 = 1$.
• Schließlich erhält man für $z_2 = 2$:

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

(4)  Berücksichtigt man den Zusammenhang zwischen $z_2$ und $V_2$, so erhält man:

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

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

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

• Die Wahrscheinlichkeit, dass eine Gaußvariable größer ist als $2 \cdot \sigma$, ist aber gleich ${\rm Q}(2)$:

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

(5)  Richtig ist der Lösungsvorschlag 3:

• Die erste Aussage ist mit Sicherheit nicht zutreffend, da sich der Mittelwert $m_{\rm S}$ auf die logarithmierte Empfangsleistung (in $\rm dBm$) bezieht.
• Um zu klären, ob nun die zweite oder die dritte Lösungsalternative zutrifft, gehen wir von $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$  ⇒  $P_{\rm E}' = 1 \ {\rm µ W}$ und folgender $V_2$–WDF aus:

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

• In der Hälfte der Zeit ist dann $P_{\rm E} = 1 \ \rm µ W$, während in den beiden anderen Vierteln jeweils gilt:

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

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

• Der Mittelwert ergibt somit:

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

• Diese einfache Rechnung mit diskreten Wahrscheinlichkeiten anstelle einer kontinuierlichen WDF deutet darauf hin, dass der Lösungsvorschlag 3 richtig sein wird.

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

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

• 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 $2 \cdot \sigma$, but equals ${\rm Q}(2)$:

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

(5)  Correct is the solution 3:

• 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 solution alternative 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 $V_2$–WDF

$$f_{V{\rm 2}}(V_{\rm 2}) = 0.5 \cdot \delta (V_{\rm 2}) + 0.25 \cdot \delta (V_{\rm 2}- 10\,\,{\rm dB}) + 0.25 \cdot \delta (V_{\rm 2}+ 10\,\,{\rm dB})\hspace{0.05cm}.$ *Halfway through the time, $P_{\rm E} = 1 \ \ \rm µ W$, while in the other two quarters, each is valid:$$ 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}. $$*The mean value thus gives:$$ {\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 WDF indicates that the solution 3 will be correct.