Aufgaben:Exercise 1.2Z: Lognormal Fading Revisited - Revision history

Javier: Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

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Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

Javier: Text replacement - "===Solutions===" to "===Solution==="

2020-12-15T14:10:56Z

Text replacement - "===Solutions===" to "===Solution==="

Javier: Text replacement - "Mobile Kommunikation/Distanzabhängige Dämpfung und Abschattung" to "Mobile_Communications/Distance_Dependent_Attenuation_and_Shading"

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Text replacement - "Mobile Kommunikation/Distanzabhängige Dämpfung und Abschattung" to "Mobile_Communications/Distance_Dependent_Attenuation_and_Shading"

Oezdemir at 15:34, 1 November 2020

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Oezdemir at 20:41, 30 July 2020

2020-07-30T20:41:09Z

Oezdemir at 20:40, 30 July 2020

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Guenter at 09:15, 26 July 2020

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Guenter at 08:43, 26 July 2020

2020-07-26T08:43:32Z

Javier: Text replacement - "receive power" to "received power"

2020-07-24T15:27:12Z

Text replacement - "receive power" to "received power"

Javier: Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

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Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^1.1 Distance-Dependent Attenuation^]]	+	[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]

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 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; The constant&nbsp; $k_1$&nbsp; generates the time-independent path loss&nbsp; $V_1 = 60 \ \rm dB$.&nbsp; From this follows:

← Older revision		Revision as of 13:55, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Distanzabhängige Dämpfung und Abschattung~~	+	{{quiz-Header\|Buchseite=Mobile_Communications/Distance_Dependent_Attenuation_and_Shading
	}}		}}

← Older revision		Revision as of 15:34, 1 November 2020
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−	[[Category:Exercises for Mobile Communications\|^1.1 Distance-~~dependent attenuation~~^]]	+	[[Category:Exercises for Mobile Communications\|^1.1 Distance-Dependent Attenuation^]]

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 }}
-[[File:P_ID2123__Mob_Z_1_2.png|right|frame|Path loss plus 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|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>):
 :$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$

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-===Questionnaire===
+===Questions===
 <quiz display=simple>
 {How large should the constant&nbsp; $k_1$&nbsp; be?
@@ Line 77: / Line 77: @@
 </quiz>
-===Sample solution===
+===Solutions===
 {{ML-Kopf}}
 '''(1)'''&nbsp; The constant&nbsp; $k_1$&nbsp; generates the time-independent path loss&nbsp; $V_1 = 60 \ \rm dB$.&nbsp; From this follows:

@@ Line 79: / Line 79: @@
 ===Sample solution===
 {{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:
 :$$k_{\rm 1} = 10^{-V_{\rm 1}/(20\hspace{0.05cm} {\rm dB})} \hspace{0.15cm} \underline{= 0.001}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; Only the <u>second statement</u> is correct:
+'''(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 goes 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 PDF 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; $f_{\rm z2}(z_2 = 0)\hspace{0.15cm}\underline{= 0}$.
-*The PDF&ndash;value for the abscissa value $z_2 = 1$ is obtained by inserting it into the given equation:
+*The PDF&ndash;value for the abscissa value&nbsp; $z_2 = 1$&nbsp; is obtained by inserting it into the given equation:
 :$$f_{z{\rm 2}}(z_{\rm 2} = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 2}=1)
 /({2 \hspace{0.05cm}\cdot \hspace{0.05cm} C^2 \hspace{0.05cm} \cdot \hspace{0.05cm} \sigma_{\rm S}^2})
@@ Line 98: / Line 98: @@
   \hspace{0.15cm} \underline{\approx 0.578}\hspace{0.05cm}.$$
-*The first portion is equal to the WDF&ndash;value $f_{{{\it V}2}(V_2 = 0)$.
+*The first portion is equal to the PDF&ndash;value&nbsp; $f_{V2}(V_2 = 0)$.
-*$C$ considers the amount of the derivative of the non-linear characteristic $z_2 = g(V_2)$ for $V_2 = 0 \ \rm dB$ or $z_2 = 1$.
+*$C$&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; $z_2 = 1$.
-*Finally, for $z_2 = 2$:
+*Finally, for&nbsp; $z_2 = 2$:
 :$$f_{z{\rm 2}}(z_{\rm 2} = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \frac {f_{z{\rm 2}}(z_{\rm 2} = 1)}{ z_{\rm 2} = 2} \cdot
   {\rm exp } \left [ - \frac {{\rm ln}^2 (2)}{2 \cdot C^2 \cdot \sigma_{\rm S}^2} \right ]= \frac {0.578}{ 2} \cdot
@@ Line 106: / Line 106: @@
-'''(4)'''&nbsp; If you take into account the relationship between $z_2$ and $V_2$, you get
+'''(4)'''&nbsp; If you take into account the relationship between&nbsp; $z_2$&nbsp; and&nbsp; $V_2$, you get
 :$${\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}
   \hspace{0.05cm},$$
@@ Line 114: / Line 114: @@
   \hspace{0.05cm}.$$
-*The probability that a Gaussian variable is greater than $2 \cdot \sigma$ equals ${\rm Q}(2)$:
+*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}.$$
 '''(5)'''&nbsp; The <u>statement 3</u> is correct:
-*The first statement is certainly not correct, since the mean value $m_{\rm S}$ refers to the logarithmic received power (in $\rm dBm$).
+*The first statement is certainly not correct, since the mean value&nbsp; $m_{\rm S}$&nbsp; refers to the logarithmic received power&nbsp; $($in&nbsp; $\rm dBm)$.
-*To clarify whether the second or the third statement is correct, we assume $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$ and the following PDF for  $V_2$:
+*To clarify whether the second or the third statement is correct, we assume&nbsp; $P_{\rm S} = 1 \ \rm W$, $V_1 = 60 \ \rm dB$ &nbsp;&#8658;&nbsp; $P_{\rm E}' = 1 \ {\rm &micro; W}$&nbsp; and the following PDF for&nbsp;  $V_2$:
 :$$f_{V{\rm 2}}(V_{\rm 2}) =  0.5 \cdot \delta (V_{\rm 2}) + 0.25 \cdot \delta (V_{\rm 2}- 10\,\,{\rm dB})
   + 0.25 \cdot \delta (V_{\rm 2}+ 10\,\,{\rm dB})\hspace{0.05cm}.$$
-*Half of the time, $P_{\rm E} = 1 \ \rm &micro; W$, while each of the following has $25\%$ probability::
+*Half of the time,&nbsp; $P_{\rm E} = 1 \ \rm &micro; W$, while each of the following has&nbsp; $25\%$ probability::
 :$$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}}{10^5} = 10\,\,{\,}{\rm &micro; W}\hspace{0.05cm}.$$

@@ Line 4: / Line 4: @@
 [[File: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>):
+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>):
 :$$V_{\rm 1} = V_{\rm 0} + m_{\rm S} \hspace{0.05cm}.$$
 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:
 :$$f_{V_{\rm S}}(V_{\rm S}) =  \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}}  \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) }\hspace{0.05cm}.$$
 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.
+*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; $f_{\rm z_2}(z_2)$.
 *The PDF of the lognormally distributed random variable&nbsp; $z_2$&nbsp; is for&nbsp; $z_2 &#8805; 0$:
 :$$f_{z_{\rm 2}}(z_{\rm 2}) =  \frac {{\rm e^{- {\rm ln}^2 (z_{\rm 2})
 /({2 \hspace{0.05cm}\cdot \hspace{0.05cm} C^2 \hspace{0.05cm} \cdot \hspace{0.05cm} \sigma_{\rm S}^2})
-} } }{ \sqrt{2 \pi }\cdot C \cdot \sigma_{\rm S} \cdot z_2}   \hspace{0.8cm}{\rm mit}  \hspace{0.8cm} C = \frac{{\rm ln} \hspace{0.1cm}(10)}{20\,\,{\rm dB}}\hspace{0.05cm}.$$
+} } }{ \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}.$$
 *For&nbsp; $z_2 &#8804; 0$&nbsp; this PDF is equal to zero.
@@ Line 29: / Line 30: @@
 ''Notes:''
-* This task belongs to the chapter&nbsp; [[Mobile_Communications/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X|Distanzabhängige Dämpfung und Abschattung]].
+* This exercise belongs to the chapter&nbsp; [[Mobile_Communications/Distance_dependent_attenuation_and_shading|Distance dependent attenuation and shading]].
 * Use the following parameters:
 :$$V_{\rm 1} = 60\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 6\,{\rm dB}\hspace{0.05cm}.$$
@@ Line 69: / Line 70: @@
-{What statements are valid for the average received 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="[]"}
 - ${\rm E}[P_{\rm E}(t)] = P_{\rm E}'$