Aufgaben:Exercise 1.2Z: Lognormal Fading Revisited - Versionsgeschichte

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@@ Zeile 29: / Zeile 29: @@
 ''Notes:''
-* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X|Distanzabhängige Dämpfung und Abschattung]].
+* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distanzabhängige Dämpfung und Abschattung]].
 * Use the following parameters:
 :$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$

@@ Zeile 3: / Zeile 3: @@
 }}
-[[Datei:P_ID2123__Mob_Z_1_2.png|right|frame|Path loss model <br>with lognormal fading]]
+[[Datei:P_ID2123__Mob_Z_1_2.png|right|frame|Path loss plus 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>):
-$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$
+:$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$
 The total path loss is then given by the equation
@@ Zeile 11: / Zeile 11: @@
 where&nbsp; $V_2(t)$&nbsp; describes a ''lognormal&ndash;distribution''&nbsp; with mean value zero:
-$$f_{V{\rm 2}}(V_{\rm 2}) = \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}}  \cdot {\rm exp } \left [ - \frac{ V_{\rm 2} ^2}{2 \cdot \sigma_{\rm S}^2} \right ] \hspace{0.05cm}.$$
+:$$f_{V_{\rm S}}(V_{\rm S}) =  \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}}  \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) }\hspace{0.05cm}.$$
 The path loss model shown in the graphic is suitable for the scenario described here:
 *Multiply the transmitted signal&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.
 *The PDF of the lognormally distributed random variable&nbsp; $z_2$&nbsp; is for&nbsp; $z_2 &#8805; 0$:
-$$f_{z{\rm 2}}(z_{\rm 2}) = \frac {{\rm exp } \left [ - {\rm ln}^2 (z_{\rm 2}) /({2 \cdot C^2 \cdot \sigma_{\rm S}^2}) \right ]}{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2}    \hspace{0.3cm}{\rm with}  \hspace{0.3cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$
+:$$f_{z_{\rm 2}}(z_{\rm 2}) =  \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 2})
 *For&nbsp; $z_2 &#8804; 0$&nbsp; this PDF is equal to zero.
@@ Zeile 27: / Zeile 29: @@
 ''Notes:''
-* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distanzabhängige Dämpfung und Abschattung]].
+* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X|Distanzabhängige Dämpfung und Abschattung]].
 * Use the following parameters:
-$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$
+:$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$
 * The probability that a mean-free Gaussian random variable&nbsp; $z$&nbsp; is greater than its standard deviation&nbsp; $\sigma$, is
-$${\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}.$$
 * Also, &nbsp; ${\rm Pr}(z > 2\sigma) = {\rm Pr}(z < -2\sigma) = {\rm Q}(2) \approx 0.023\hspace{0.05cm}.$
 * Again for clarification: &nbsp; $z_2$&nbsp; is the fading coefficient in linear units, while &nbsp; $V_2$&nbsp; is the fading coefficient in logarithmic units.
 *The following conversions apply:
-$$z_2 = 10^{-V_{\rm 2}/20\,{\rm dB}}\hspace{0.05cm}, \hspace{0.2cm}
+:$$z_2 = 10^{-V_{\rm 2}/20\,{\rm dB}}\hspace{0.05cm}, \hspace{0.2cm}
 V_{\rm 2} = -20\,{\rm dB} \cdot {\rm lg}\hspace{0.15cm}z_2\hspace{0.05cm}.$$
@@ Zeile 77: / Zeile 79: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The constant $k_1$ generates the time-independent path loss $V_1 = 60 \ \rm dB$. From this follows:
-$$k_{\rm 1} = 10^{-V_{\rm 1}/(20\hspace{0.05cm} {\rm dB})} \hspace{0.15cm} \underline{= 0.001}\hspace{0.05cm}.$$
+:$$k_{\rm 1} = 10^{-V_{\rm 1}/(20\hspace{0.05cm} {\rm dB})} \hspace{0.15cm} \underline{= 0.001}\hspace{0.05cm}.$$
@@ Zeile 89: / Zeile 91: @@
 *The PDF&ndash;value for the abscissa value $z_2 = 1$ is obtained by inserting it into the given equation:
-:$$f_{z{\rm 2}}(z_{\rm 2} = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \frac {{\rm exp } \left [ - {\rm ln}^2 (z_2 = 1) /({2 \cdot C^2 \cdot \sigma_{\rm S}^2}) \right ]}{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot (z_2 = 1)}=\frac {1}{ \sqrt{2 \pi } \cdot \sigma_{\rm S} }  \cdot \frac {1}{  C  } =
+:$$f_{z{\rm 2}}(z_{\rm 2} = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 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}.$$
@@ Zeile 102: / Zeile 106: @@
 '''(4)'''&nbsp; If you take into account the relationship between $z_2$ and $V_2$, you get
-$${\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}
+:$${\rm Pr}(z_{\rm 2} > 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(V_{\rm 2} < 0\,\,{\rm dB})\hspace{0.15cm} \underline{= 0.5}
   \hspace{0.05cm},$$
-$${\rm Pr}(z_{\rm 2} > 0.5) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(V_{\rm 2} < 6\,\,{\rm dB}) = 1- {\rm Pr}(V_{\rm 2} > 6\,\,{\rm dB})= 1- {\rm Pr}(V_{\rm 2} > \sigma_{\rm S})= 1- {\rm Q}(1)\hspace{0.15cm} \underline{= 0.842}
+:$${\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})
+:$${\rm Pr}(z_{\rm 2} > 4) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(V_{\rm 2} < -12\,\,{\rm dB}) = {\rm Pr}(V_{\rm 2} > +12\,\,{\rm dB}) = {\rm Pr}(V_{\rm 2} > 2 \sigma_{\rm S})
   \hspace{0.05cm}.$$
@@ Zeile 121: / Zeile 125: @@
 *Half of the time, $P_{\rm E} = 1 \ \rm &micro; W$, while each of the following has $25\%$ probability::
-$$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\,}{\rm &micro; W}\hspace{0.05cm},$$
+:$$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\,}{\rm &micro; W}\hspace{0.05cm},$$
-$$V_{\rm 2}= -10\,\,{\,}{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\,}{\rm &micro; W}\hspace{0.05cm}.$$
+:$$V_{\rm 2}= -10\,\,{\,}{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\,}{\rm &micro; W}\hspace{0.05cm}.$$
 *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}.$$

@@ Zeile 116: / Zeile 116: @@
 '''(5)'''&nbsp; The <u>statement 3</u> 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 $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$ and the following $V_2$&ndash;WDF
+*To clarify whether the second or the third statement 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 PDF for  $V_2$:
 :$$f_{V{\rm 2}}(V_{\rm 2}) =  0.5 \cdot \delta (V_{\rm 2}) + 0.25 \cdot \delta (V_{\rm 2}- 10\,\,{\rm dB})
   + 0.25 \cdot \delta (V_{\rm 2}+ 10\,\,{\rm dB})\hspace{0.05cm}.$$
-*Halfway through the time, $P_{\rm E} = 1 \ \ \rm &micro; W$, while in the other two quarters, each is valid:
+*Half of the time, $P_{\rm E} = 1 \ \rm &micro; W$, while each of the following has $25\%$ probability::
 $$V_{\rm 2}= +10\,\,{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^7} = 0.1\,\,{\,}{\rm &micro; W}\hspace{0.05cm},$$
 $$V_{\rm 2}= -10\,\,{\,}{\rm dB}\text{:} \hspace{0.3cm} P_{\rm E}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1\,\,{\rm W}}{10^5} = 10\,\,{\,}{\rm &micro; W}\hspace{0.05cm}.$$
-*The mean value thus gives:
+*The mean value is then:
 $${\rm E}[P_{\rm E}(t)] = 0.5 \cdot 1\,{\rm &micro; W}+ 0.25 \cdot 0.1\,{\rm &micro; W}+ 0.25 \cdot 10\,{\rm &micro; W}= 3.025\,{\rm &micro; W} > P_{\rm E}\hspace{0.05cm}' = 1\,{\rm &micro; W}
   \hspace{0.05cm}.$$
-*This simple calculation with discrete probabilities instead of a continuous WDF indicates that the <u>solution 3</u> will be correct.
+*This simple calculation with discrete probabilities instead of a continuous PDF indicates that <u>statement 3</u> is correct.
 {{ML-Fuß}}

@@ Zeile 109: / Zeile 109: @@
   \hspace{0.05cm}.$$
-*The probability that a Gaussian variable is greater than $2 \cdot \sigma$, but equals ${\rm Q}(2)$:
+*The probability that a Gaussian variable is greater than $2 \cdot \sigma$ equals ${\rm Q}(2)$:
 $${\rm Pr}(z_{\rm 2} > 4) = {\rm Q}(2)\hspace{0.15cm} \underline{= 0.023}
   \hspace{0.05cm}.$$
-'''(5)'''&nbsp; Correct is the <u>solution 3</u>:
+'''(5)'''&nbsp; The <u>statement 3</u> is correct:
 *The first statement is certainly not correct, since the mean value $m_{\rm S}$ refers to the logarithmic received power (in $\rm dBm$).
-*To clarify whether the second or the third solution alternative is correct, we assume $P_{\rm S} = 1 \ \ \rm W$, $V_1 = 60 \ \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$ and the following $V_2$&ndash;WDF
+*To clarify whether the second or the third statement is correct, we assume $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$ and the following $V_2$&ndash;WDF
-$$f_{V{\rm 2}}(V_{\rm 2}) = 0.5 \cdot \delta (V_{\rm 2}) + 0.25 \cdot \delta (V_{\rm 2}- 10\,\,{\rm dB})
+:$$f_{V{\rm 2}}(V_{\rm 2}) =  0.5 \cdot \delta (V_{\rm 2}) + 0.25 \cdot \delta (V_{\rm 2}- 10\,\,{\rm dB})
-  + 0.25 \cdot \delta (V_{\rm 2}+ 10\,\,{\rm dB})\hspace{0.05cm}.$
+  + 0.25 \cdot \delta (V_{\rm 2}+ 10\,\,{\rm dB})\hspace{0.05cm}.$$
 *Halfway through the time, $P_{\rm E} = 1 \ \ \rm &micro; W$, while in the other two quarters, each is valid:

@@ Zeile 86: / Zeile 86: @@
-'''(3)''&nbsp; The random value $z_2$ can only be positive. Therefore the PDF value $f_{\rm z2}(z_2 = 0)\hspace{0.15cm} is \underline{ = 0}$.
+'''(3)'''&nbsp; 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&ndash;value for the abscissa value $z_2 = 1$ is obtained by inserting it into the given equation:
@@ Zeile 96: / Zeile 96: @@
 *$C$ 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 $z_2 = 2$:
-$$f_{z{\rm 2}}(z_{\rm 2} = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \frac {f_{z{\rm 2}}}(z_{\rm 2} = 1)}{ z_{\rm 2} = 2} \cdot
+:$$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
   {\rm exp } \left [ - \frac {0.48}{0.952} \right ] \hspace{0.15cm} \underline{\approx 0.174}\hspace{0.05cm}. $$

← Nächstältere Version		Version vom 25. März 2020, 15:30 Uhr
Zeile 87:		Zeile 87:

	'''(3)''  The random value $z_2$ can only be positive. Therefore the PDF–value $f_{\rm z2}(z_2 = 0)\hspace{0.15cm} is \underline{ = 0}$.		'''(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 <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 $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$  ⇒  $P_{\rm E}' = 1 \ {\rm µ W}$ 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 <u>Lösungsvorschlag 3</u> richtig sein wird.

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

@@ Zeile 80: / Zeile 80: @@
-'''(2)'''&nbsp; Correct is only the <u>second solution suggestion</u>:
+'''(2)'''&nbsp; Only the <u>second statement</u> is correct:
 *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 $V_2$).
+*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 goes 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 $f_{\rm z2}(z_2 = 0)\hspace{0.15cm} is \underline{ = 0}$.
+'''(3)''&nbsp; The random value $z_2$ can only be positive. Therefore the PDF&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:

@@ Zeile 54: / Zeile 54: @@
 - 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="{}"}
 $f_{\rm z2}(z_2 = 0)\ = \ $ { 0. }
@@ Zeile 67: / Zeile 67: @@
-{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 receive 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; ${\rm E}[P_{\rm E}(t)] = P_{\rm E}'$
+- ${\rm E}[P_{\rm E}(t)] = P_{\rm E}'$
-- 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>