Difference between revisions of "Aufgaben:Exercise 1.2Z: Lognormal Fading Revisited"

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{{quiz-Header|Buchseite=Mobile Kommunikation/Distanzabhängige Dämpfung und Abschattung
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{{quiz-Header|Buchseite=Mobile_Communications/Distance_Dependent_Attenuation_and_Shading
 
}}
 
}}
  
[[File:P_ID2123__Mob_Z_1_2.png|right|frame|Path loss model <br>with lognormal fading]]
+
[[File:P_ID2123__Mob_Z_1_2.png|right|frame|Path loss and lognormal fading]]
We assume similar conditions as in &nbsp; [[Aufgaben:Exercise_1.2:_Lognormal_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>):
+
We assume similar conditions as in&nbsp; [[Aufgaben:Exercise_1.2:_Lognormal_Channel_Model|Exercise 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 fading&nbsp; (the index "S" stands for <i>Shadowing</i>):
$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$
+
:$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$
  
 
The total path loss is then given by the equation
 
The total path loss is then given by the equation
$$V_{\rm P} = V_{\rm 1} + V_{\rm 2}(t)$$
+
:$$V_{\rm P} = V_{\rm 1} + V_{\rm 2}(t)$$
  
where&nbsp; $V_2(t)$&nbsp; describes a ''lognormal&ndash;distribution''&nbsp; with mean value zero:
+
where&nbsp; $V_2(t)$&nbsp; describes a&nbsp; ''lognormal distribution''&nbsp; with mean value zero:
$$f_{V{\rm 2}}(V_{\rm 2}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}}  \cdot {\rm exp } \left [ - \frac{ V_{\rm 2} ^2}{2 \cdot \sigma_{\rm S}^2} \right ] \hspace{0.05cm}.$$
+
:$$f_{V_{\rm S}}(V_{\rm S}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}}  \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) }\hspace{0.05cm}.$$
  
 
The path loss model shown in the graphic is suitable for the scenario described here:  
 
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 function (PDF)&nbsp; $f_{\rm z_2}(z_2)$, 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.  
+
*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 function&nbsp; $\rm (PDF)$&nbsp; $f_{\rm z_2}(z_2)$.
 +
* 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 PDF of the lognormally distributed random variable&nbsp; $z_2$&nbsp; is for&nbsp; $z_2 &#8805; 0$:
 
*The PDF of the lognormally distributed random variable&nbsp; $z_2$&nbsp; is for&nbsp; $z_2 &#8805; 0$:
$$f_{z{\rm 2}}(z_{\rm 2}) = \frac {{\rm exp } \left [ - {\rm ln}^2 (z_{\rm 2}) /({2 \cdot C^2 \cdot \sigma_{\rm S}^2}) \right ]}{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2}   \hspace{0.3cm}{\rm with}  \hspace{0.3cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$
+
:$$f_{z_{\rm 2}}(z_{\rm 2}) = \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 2})
 +
/({2 \hspace{0.05cm}\cdot \hspace{0.05cm} C^2 \hspace{0.05cm} \cdot \hspace{0.05cm} \sigma_{\rm S}^2})
 +
} } }{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2}   \hspace{0.8cm}{\rm with}  \hspace{0.8cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$
  
 
*For&nbsp; $z_2 &#8804; 0$&nbsp; this PDF is equal to zero.
 
*For&nbsp; $z_2 &#8804; 0$&nbsp; this PDF is equal to zero.
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''Notes:''
 
''Notes:''
* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distanzabhängige Dämpfung und Abschattung]].
+
* This exercise belongs to the chapter&nbsp; [[Mobile_Communications/Distance_dependent_attenuation_and_shading|Distance dependent attenuation and shading]].
 
* Use the following parameters:
 
* Use the following parameters:
$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$
+
:$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$
 
   
 
   
 
* The probability that a mean-free Gaussian random variable&nbsp; $z$&nbsp; is greater than its standard deviation&nbsp; $\sigma$, is
 
* The probability that a mean-free Gaussian random variable&nbsp; $z$&nbsp; is greater than its standard deviation&nbsp; $\sigma$, is
$${\rm Pr}(z > \sigma) = {\rm Pr}(z < -\sigma) = {\rm Q}(1) \approx 0.158\hspace{0.05cm}.$$
+
:$${\rm Pr}(z > \sigma) = {\rm Pr}(z < -\sigma) = {\rm Q}(1) \approx 0.158\hspace{0.05cm}.$$
 
* Also, &nbsp; ${\rm Pr}(z > 2\sigma) = {\rm Pr}(z < -2\sigma) = {\rm Q}(2) \approx 0.023\hspace{0.05cm}.$
 
* Also, &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 capital letters&nbsp; $V_2$&nbsp; denote logarithmic units.  
+
* Again for clarification: &nbsp; $z_2$&nbsp; is the fading coefficient in linear units, while &nbsp; $V_2$&nbsp; is the fading coefficient in logarithmic units.  
 
*The following conversions apply:
 
*The following conversions apply:
$$z_2 = 10^{-V_{\rm 2}/20\,{\rm dB}}\hspace{0.05cm}, \hspace{0.2cm}
+
:$$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}.$
+
V_{\rm 2} = -20\,{\rm dB} \cdot {\rm lg}\hspace{0.15cm}z_2\hspace{0.05cm}.$$
  
  
  
===Questionnaire===
+
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
 
{How large should the constant&nbsp; $k_1$&nbsp; be?
 
{How large should the constant&nbsp; $k_1$&nbsp; be?
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- The largest possible value is&nbsp; $z_2 = 2$.
 
- The largest possible value is&nbsp; $z_2 = 2$.
  
{Calculate the WDF&nbsp; $f_{\rm z2}(z_2)$&nbsp; for some abscissa values.
+
{Calculate the PDF&nbsp; $f_{\rm z2}(z_2)$&nbsp; for some abscissa values.
 
|type="{}"}
 
|type="{}"}
 
$f_{\rm z2}(z_2 = 0)\ = \ $ { 0. }
 
$f_{\rm z2}(z_2 = 0)\ = \ $ { 0. }
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{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).
+
{What statements are valid for the average received power&nbsp; ${\rm E}\big [P_{\rm E}(t)\big]$? &nbsp; <u>Note:</u> &nbsp;$P_{\rm E}'$ is the power after multiplication by&nbsp; $k_1$&nbsp; (see diagram).
 
|type="[]"}
 
|type="[]"}
- The following applies: &nbsp; ${\rm E}[P_{\rm E}(t)] = P_{\rm E}'$
+
- ${\rm E}[P_{\rm E}(t)] = P_{\rm E}'$
- It reads: &nbsp; ${\rm E}[P_{\rm E}(t)] < P_{\rm E}'$.
+
- ${\rm E}[P_{\rm E}(t)] < P_{\rm E}'$.
+ It reads: &nbsp; ${\rm E}[P_{\rm E}(t)] > P_{\rm E}'$.
+
+ &nbsp; ${\rm E}[P_{\rm E}(t)] > P_{\rm E}'$.
 
</quiz>
 
</quiz>
  
===Sample solution===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; The constant $k_1$ generates the time-independent path loss $V_1 = 60 \ \rm dB$. From this follows:
+
'''(1)'''&nbsp; The constant&nbsp; $k_1$&nbsp; generates the time-independent path loss&nbsp; $V_1 = 60 \ \rm dB$.&nbsp; From this follows:
$$k_{\rm 1} = 10^{-V_{\rm 1}/(20\hspace{0.05cm} {\rm dB})} \hspace{0.15cm} \underline{= 0.001}\hspace{0.05cm}.$$
+
:$$k_{\rm 1} = 10^{-V_{\rm 1}/(20\hspace{0.05cm} {\rm dB})} \hspace{0.15cm} \underline{= 0.001}\hspace{0.05cm}.$$
  
  
'''(2)'''&nbsp; Correct is only the <u>second solution suggestion</u>:  
+
'''(2)'''&nbsp; Only the&nbsp; <u>second statement</u>&nbsp; is correct:  
*For the Gaussian random variable $V_2$ all values between $&ndash;&#8734;$ and $+&#8734;$ are (theoretically) possible.  
+
*For the Gaussian random variable&nbsp; $V_2$&nbsp; all values between&nbsp; $-&#8734;$&nbsp; and&nbsp; $+&#8734;$&nbsp; 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 $V_2$).
+
*The transformation&nbsp; $z_2 = 10^{{\it -V_2}\rm /20}$&nbsp; results in only positive values for the linear random variable&nbsp; $z_2$, <br>namely between&nbsp; 0&nbsp; $($if&nbsp; $V_2$&nbsp; is positive and goes to infinity$)$&nbsp; and&nbsp; $+&#8734;$&nbsp; $($very large negative values&nbsp; of $V_2)$.
  
  
  
'''(3)''&nbsp; The random value $z_2$ can only be positive. Therefore the WDF&ndash;value $f_{\rm z2}(z_2 = 0)\hspace{0.15cm} is \underline{ = 0}$.  
+
'''(3)'''&nbsp; The random value&nbsp; $z_2$&nbsp; can only be positive.&nbsp; Therefore the PDF value&nbsp; $f_{\rm z2}(z_2 = 0)\hspace{0.15cm}\underline{= 0}$.  
  
*Der WDF&ndash;Wert für den Abszissenwert $z_2 = 1$ erhält man durch Einsetzen in die gegebene Gleichung:
+
*The PDF&ndash;value for the abscissa value&nbsp; $z_2 = 1$&nbsp; is obtained by inserting it into the given equation:
:$$f_{z{\rm 2}}(z_{\rm 2} = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac {{\rm exp } \left [ - {\rm ln}^2 (z_2 = 1) /({2 \cdot C^2 \cdot \sigma_{\rm S}^2}) \right ]}{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot (z_2 = 1)}=\frac {1}{ \sqrt{2 \pi } \cdot \sigma_{\rm S} }  \cdot \frac {1}{ C  } =
+
:$$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)  }   
 
  \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}.$$
 
  \hspace{0.15cm} \underline{\approx 0.578}\hspace{0.05cm}.$$
  
*Der erste Anteil ist gleich dem WDF&ndash;Wert $f_{{\it V}2}(V_2 = 0)$.
+
*The first portion is equal to the PDF&ndash;value&nbsp; $f_{V2}(V_2 = 0)$.
*$C$ berücksichtigt den Betrag der Ableitung der nichtlinearen Kennlinie $z_2 = g(V_2)$ für $V_2 = 0 \ \rm dB$ bzw. $z_2 = 1$.  
+
*$C$&nbsp; considers the magnitude of the derivative of the non-linear characteristic&nbsp; $z_2 = g(V_2)$&nbsp; for&nbsp; $V_2 = 0 \ \rm dB$&nbsp; or&nbsp; $z_2 = 1$.  
*Schließlich erhält man für $z_2 = 2$:
+
*Finally, for&nbsp; $z_2 = 2$:
 
:$$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  
 
:$$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 {{\rm ln}^2 (2)}{2 \cdot C^2 \cdot \sigma_{\rm S}^2} \right ]= \frac {0.578}{ 2} \cdot  
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'''(4)'''&nbsp; Berücksichtigt man den Zusammenhang zwischen $z_2$ und $V_2$, so erhält man:
+
'''(4)'''&nbsp; If you take into account the relationship between&nbsp; $z_2$&nbsp; and&nbsp; $V_2$, 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}
 
:$${\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},$$
 
  \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}
 
:$${\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},$$
 
  \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})
+
:$${\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}.$$
 
  \hspace{0.05cm}.$$
  
*Die Wahrscheinlichkeit, dass eine Gaußvariable größer ist als $2 \cdot \sigma$, ist aber gleich ${\rm Q}(2)$:
+
*The probability that a Gaussian variable is greater than&nbsp; $2 \cdot \sigma$&nbsp; equals&nbsp; ${\rm Q}(2)$:
:$${\rm Pr}(z_{\rm 2} > 4) = {\rm Q}(2)\hspace{0.15cm} \underline{= 0.023}
+
:$${\rm Pr}(z_{\rm 2} > 4) = {\rm Q}(2)\hspace{0.15cm} \underline{= 0.023}
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
  
'''(5)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:
+
'''(5)'''&nbsp; The <u>statement 3</u> is correct:
*Die erste Aussage ist mit Sicherheit nicht zutreffend, da sich der Mittelwert $m_{\rm S}$ auf die logarithmierte Empfangsleistung (in $\rm dBm$) bezieht.  
+
*The first statement is certainly not correct, since the mean value&nbsp; $m_{\rm S}$&nbsp; refers to the logarithmic received power&nbsp; $($in&nbsp; $\rm dBm)$.  
*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:
+
*To clarify whether the second or the third statement is correct, we assume&nbsp; $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$&nbsp; and the following PDF for&nbsp;  $V_2$:
 
:$$f_{V{\rm 2}}(V_{\rm 2}) =  0.5 \cdot \delta (V_{\rm 2}) + 0.25 \cdot \delta (V_{\rm 2}- 10\,\,{\rm dB})
 
:$$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}.$$
 
  + 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:
+
*Half of the time,&nbsp; $P_{\rm E} = 1 \ \rm &micro; W$, while each of the following has&nbsp; $25\%$ probability::
:$$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\rm &micro; W}\hspace{0.05cm},$$
+
:$$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\,}{\rm &micro; W}\hspace{0.05cm},$$
:$$V_{\rm 2}= -10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\rm &micro; W}\hspace{0.05cm}.$$
+
:$$V_{\rm 2}= -10\,\,{\,}{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\,}{\rm &micro; W}\hspace{0.05cm}.$$
  
*Der Mittelwert ergibt somit:
+
*The mean value is then:
:$${\rm E}[P_{\rm E}(t)] = 0.5 \cdot 1\,{\rm &micro; W}+ 0.25 \cdot 0.1\,{\rm &micro; W}+0.25 \cdot 10\,{\rm &micro; W}= 3.025\,{\rm &micro; W} > P_{\rm E}\hspace{0.05cm}' = 1\,{\rm &micro; W}
+
:$${\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}.$$
 
  \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.
+
*This simple calculation with discrete probabilities instead of a continuous PDF indicates that <u>statement 3</u> is correct.
 
 
*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.
 
 
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Latest revision as of 13:38, 23 March 2021

Path loss and lognormal fading

We assume similar conditions as in  Exercise 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  $\rm (PDF)$  $f_{\rm z_2}(z_2)$.
  • 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  $z_2 ≥ 0$:
$$f_{z_{\rm 2}}(z_{\rm 2}) = \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 2}) /({2 \hspace{0.05cm}\cdot \hspace{0.05cm} C^2 \hspace{0.05cm} \cdot \hspace{0.05cm} \sigma_{\rm S}^2}) } } }{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2} \hspace{0.8cm}{\rm with} \hspace{0.8cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$
  • For  $z_2 ≤ 0$  this PDF is equal to zero.




Notes:

$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm} \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$
  • 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:
$$z_2 = 10^{-V_{\rm 2}/20\,{\rm dB}}\hspace{0.05cm}, \hspace{0.2cm} V_{\rm 2} = -20\,{\rm dB} \cdot {\rm lg}\hspace{0.15cm}z_2\hspace{0.05cm}.$$


Questions

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 received power  ${\rm E}\big [P_{\rm E}(t)\big]$?   Note:  $P_{\rm E}'$ is the power after multiplication by  $k_1$  (see diagram).

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


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

  • 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 PDF–value  $f_{V2}(V_2 = 0)$.
  • $C$  considers the magnitude 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.