Quelltext der Seite Aufgaben:Exercise 1.2Z: Lognormal Fading Revisited

 
{{quiz-Header|Buchseite=Mobile Kommunikation/Distanzabhängige Dämpfung und Abschattung
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[[Datei:P_ID2123__Mob_Z_1_2.png|right|frame|Pfadverlustmodell <br>mit Lognormal-Fading]]
We assume similar conditions as in the&nbsp; [[Tasks:1.2_Lognormal_%E2%80%93_channel model|Task 1. 2]]&nbsp; but now we summarize the purely distance-dependent path loss&nbsp; $V_0$&nbsp; and the mean value&nbsp; $m_{\rm S}$&nbsp; of the lognormal&ndash;fading (the index &bdquo;S&rdquo; stands for <i>Shadowing</i>):
$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$

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

where&nbsp; $V_2(t)$&nbsp; describes a ''lognormal&ndash;distribution''&nbsp; 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&nbsp; $s(t)$&nbsp; first with a constant factor&nbsp; $k_1$&nbsp; and further with a stochastic quantity&nbsp; $z_2(t)$&nbsp; with the probability density (WDF)&nbsp; $f_{\rm z2}(z_2)$, then the signal&nbsp; $r(t)$ results at the output, whose power&nbsp; $P_{\rm E}(t)$&nbsp; is of course also time-dependent due to the stochastic component. 
*The WDF of the lognormally distributed random variable&nbsp; $z_2$&nbsp; is for&nbsp; $z_2 &#8805; 0$:
$$f_{z{\rm 2}}(z_{\rm 2}) = \frac {{\rm exp } \left [ - {\rm ln}^2 (z_{\rm 2}) /({2 \cdot C^2 \cdot \sigma_{\rm S}^2}) \right ]}{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2}    \hspace{0.3cm}{\rm with}  \hspace{0.3cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$

*For&nbsp; $z_2 &#8804; 0$&nbsp; this WDF is equal to zero.







''Notes:''
* The task belongs to the chapter&nbsp; [[Mobile_Communication/Distancedependent%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distance-dependent attenuation and shading]].
* 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&nbsp; $z$&nbsp; is greater than its dispersion&nbsp; $\sigma$, is
$${\rm Pr}(z > \sigma) = {\rm Pr}(z < -\sigma) = {\rm Q}(1) \approx 0.158\hspace{0.05cm}.$$
* Also applies: &nbsp; ${\rm Pr}(z > 2\sigma) = {\rm Pr}(z < -2\sigma) = {\rm Q}(2) \approx 0.023\hspace{0.05cm}.$
* Again for clarification: &nbsp; $z_2$&nbsp; is the linear fading&ndash;size, while the description size&nbsp; $V_2$&nbsp; is based on the tenner&ndash;logarithm. 
*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===
<quiz display=simple>
{How large should the constant&nbsp; $k_1$&nbsp; be?
|type="{}"}
$k_1\ = \ $ { 0.001 3% } 

{Which value range applies to the random variable&nbsp; $z_2$?
|type="[]"}
- All values between&nbsp; $-&#8734;$ and $+&#8734;$&nbsp; are possible.
+ The random size&nbsp; $z_2$&nbsp; is not negative.
- The smallest possible value is&nbsp; $z_2 = 0.5$.
- The largest possible value is&nbsp; $z_2 = 2$.

{Calculate the WDF&nbsp; $f_{\rm z2}(z_2)$&nbsp; for some abscissa values.
|type="{}"}
$f_{\rm z2}(z_2 = 0)\ = \ $ { 0. }
$f_{\rm z2}(z_2 = 1)\ = \ $ { 0.578 3% } 
$f_{\rm z2}(z_2 = 2)\ = \ $ { 0.174 3% } 

{Calculate the following probabilities.
|type="{}"}
${\rm Pr}(z_2 > 1.0)\ = \ $ { 0.5 3% }
${\rm Pr}(z_2 > 0.5)\ = \ $ { 0.842 3% }
${\rm Pr}(z_2 > 4.0)\ = \ $ { 0.023 3% }


{What statements are valid for the average receiving power&nbsp; ${\rm E}\big [P_{\rm E}(t)\big]$? <br><u>Note:</u> &nbsp;$P_{\rm E}'$ is the power after multiplication by&nbsp; $k_1$&nbsp; (see diagram).
|type="[]"}
- The following applies: &nbsp; ${\rm E}[P_{\rm E}(t)] = P_{\rm E}'$
- It reads: &nbsp; ${\rm E}[P_{\rm E}(t)] < P_{\rm E}'$.
+ It reads: &nbsp; ${\rm E}[P_{\rm E}(t)] > P_{\rm E}'$.
</quiz>

===Sample solution===
{{ML-Kopf}}
'''(1)'''&nbsp; 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)'''&nbsp; Correct is only the <u>second solution suggestion</u>: 
*For the Gaussian random variable $V_2$ all values between $&ndash;&#8734;$ and $+&#8734;$ are (theoretically) possible. 
*The transformation $z_2 = 10^{{\it &ndash;V_2}\rm /20}$ results in only positive values for the linear random variable $z_2$, namely between 0 (if $V_2$ is positive and reaches to infinity) and $+&#8734;$ (very large negative values of $V_2$).



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

*Der WDF&ndash;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&ndash;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)'''&nbsp; 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)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:
*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 $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$ und folgender $V_2$&ndash;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 $P_{\rm E} = 1 \ \rm &micro; W$, während in den beiden anderen Vierteln jeweils gilt:
:$$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm}  P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\rm &micro; W}\hspace{0.05cm},$$
:$$V_{\rm 2}= -10\,\,{\rm dB}\text{:} \hspace{0.3cm}  P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\rm &micro; W}\hspace{0.05cm}.$$

*Der Mittelwert ergibt somit:
:$${\rm E}[P_{\rm E}(t)] =  0.5 \cdot 1\,{\rm &micro; W}+ 0.25 \cdot 0.1\,{\rm &micro; W}+0.25 \cdot 10\,{\rm &micro; W}= 3.025\,{\rm &micro; W} > P_{\rm E}\hspace{0.05cm}' = 1\,{\rm &micro; W}
 \hspace{0.05cm}.$$

*Diese einfache Rechnung mit diskreten Wahrscheinlichkeiten anstelle einer kontinuierlichen WDF deutet darauf hin, dass der <u>Lösungsvorschlag 3</u> richtig sein wird.

*The WDF&ndash;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&ndash;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)'''&nbsp; 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)'''&nbsp; Correct is the <u>solution 3</u>:
*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 $P_{\rm S} = 1 \ \ \rm W$, $V_1 = 60 \ \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$ and the following $V_2$&ndash;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, $P_{\rm E} = 1 \ \ \rm &micro; W$, while in the other two quarters, each is valid:
$$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\,}{\rm &micro; W}\hspace{0.05cm},$$
$$V_{\rm 2}= -10\,\,{\,}{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\,}{\rm &micro; W}\hspace{0.05cm}.$$

*The mean value thus gives:
$${\rm E}[P_{\rm E}(t)] = 0.5 \cdot 1\,{\rm &micro; W}+ 0.25 \cdot 0.1\,{\rm &micro; W}+ 0.25 \cdot 10\,{\rm &micro; W}= 3.025\,{\rm &micro; W} > P_{\rm E}\hspace{0.05cm}' = 1\,{\rm &micro; W}
 \hspace{0.05cm}.$$

*This simple calculation with discrete probabilities instead of a continuous WDF indicates that the <u>solution 3</u> will be correct.
{{ML-Fuß}}



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[[Category:Exercises for Mobile Communications|^1.1 Distance-dependent attenuation^]]