Aufgaben:Exercise 1.2: Lognormal Channel Model - Versionsgeschichte

Tasnad am 15. Juni 2020 um 12:52 Uhr

2020-06-15T12:52:56Z

Tasnad: Textersetzung - „Komplementäre_Gaußsche_Fehlerfunktionen_(neues_Applet)“ durch „Komplementäre_Gaußsche_Fehlerfunktionen“

2020-06-10T15:50:36Z

Textersetzung - „Komplementäre_Gaußsche_Fehlerfunktionen_(neues_Applet)“ durch „Komplementäre_Gaußsche_Fehlerfunktionen“

Guenter am 13. Mai 2020 um 11:56 Uhr

2020-05-13T11:56:44Z

Guenter am 11. Mai 2020 um 12:53 Uhr

2020-05-11T12:53:43Z

Javier: Javier verschob die Seite Exercises:Exercise 1.2: Lognormal Channel Model nach Aufgaben:Exercise 1.2: Lognormal Channel Model

2020-03-25T15:12:02Z

Javier verschob die Seite Exercises:Exercise 1.2: Lognormal Channel Model nach Exercise 1.2: Lognormal Channel Model

Javier am 25. März 2020 um 15:11 Uhr

2020-03-25T15:11:38Z

Javier am 25. März 2020 um 15:10 Uhr

2020-03-25T15:10:58Z

Javier am 25. März 2020 um 15:09 Uhr

2020-03-25T15:09:45Z

Javier am 25. März 2020 um 15:08 Uhr

2020-03-25T15:08:42Z

Javier am 25. März 2020 um 15:08 Uhr

2020-03-25T15:08:06Z

@@ Zeile 25: / Zeile 25: @@
 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]].
 * You can use the following (rough) approximations for the complementary Gaussian error integral:
 :$${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm}
   {\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$
-* Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen_X|Komplementäre Gaußsche Fehlerfunktionen]].
+* Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Komplementäre Gaußsche Fehlerfunktionen]].

@@ Zeile 30: / Zeile 30: @@
 :$${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm}
   {\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$
-* Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen_(neues_Applet)_X|Komplementäre Gaußsche Fehlerfunktionen]].
+* Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen_X|Komplementäre Gaußsche Fehlerfunktionen]].

@@ Zeile 6: / Zeile 6: @@
 Thus the total path loss can be described by the following equation:
-$$V_{\rm P} = V_{\rm 0} + V_{\rm S}  \hspace{0.05cm}.$$
+:$$V_{\rm P} = V_{\rm 0} + V_{\rm S}  \hspace{0.05cm}.$$
 *$V_0$&nbsp; takes into account the distance-dependent path loss which is assumed to be constant: $V_0 = 80 \ \rm dB$&nbsp;.
 *The loss&nbsp; $V_{\rm S}$&nbsp; is due to shadowing caused by the lognormal&ndash;distribution with the probability density function (PDF)
-:$$f_{V{\rm S}}(V_{\rm S}) =  \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}}  \cdot {\rm exp } \left [ - \frac{ (V_{\rm S}-  m_{\rm S})^2}{2 \cdot \sigma_{\rm S}^2} \right ] \hspace{0.05cm}$$
+:$$f_{V_{\rm S}}(V_{\rm S}) =  \frac {1}{ \sqrt{2 \pi }\cdot \sigma_{\rm S}}  \cdot {\rm e }^{ - { (V_{\rm S}\hspace{0.05cm}- \hspace{0.05cm}m_{\rm S})^2}/(2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma_{\rm S}^2) }$$
 : see diagram. The following numerical values apply:
-$$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0.15cm}{\rm or }\hspace{0.15cm}\sigma_{\rm S} = 0\,\,{\rm dB}\hspace{0.15cm}{\rm (subtask\hspace{0.15cm} 2)}\hspace{0.05cm}.$$
+:$$m_{\rm S} = 20\,\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}  \sigma_{\rm S} = 10\,\,{\rm dB}\hspace{0.15cm}{\rm or }\hspace{0.15cm}\sigma_{\rm S} = 0\,\,{\rm dB}\hspace{0.15cm}{\rm (subtask\hspace{0.15cm} 2)}\hspace{0.05cm}.$$
 Also make the following simple assumptions:
@@ Zeile 25: / Zeile 25: @@
 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]].
 * You can use the following (rough) approximations for the complementary Gaussian error integral:
 :$${\rm Q}(1) \approx 0.16\hspace{0.05cm},\hspace{0.2cm} {\rm Q}(2) \approx 0.02\hspace{0.05cm},\hspace{0.2cm}
   {\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$
-* Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen_(neues_Applet)|Komplementäre Gaußsche Fehlerfunktionen]].
+* Or use the interaction module provided by $\rm LNTwww$ [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen_(neues_Applet)_X|Komplementäre Gaußsche Fehlerfunktionen]].
@@ Zeile 59: / Zeile 59: @@
 '''(1)'''&nbsp; The correct answer is <u>YES</u>:
 *From the $\rm dB$&ndash;value $V_0 = 80 \ \rm dB$ follows the absolute (linear) value $K_0 = 10^8$. Thus the received power is
-$$P_{\rm E} = P_{\rm S}/K_0 = 10 \ {\rm W}/10^8 = 100 \ {\rm nW} > 10 \ \ \rm pW.$$
+:$$P_{\rm E} = P_{\rm S}/K_0 = 10 \ {\rm W}/10^8 = 100 \ {\rm nW} > 10 \ \ \rm pW.$$
 *You can also solve this problem directly with the logarithmic quantities:
@@ Zeile 86: / Zeile 86: @@
 *The probability density $f_{\rm VS}(V_{\rm S})$ of the path loss due to shadowing (Longnormal&ndash;Fading) is shown here.
-*The probability that the system will fail is marked in red:
+*The probability that the system will fail is marked in red.
 <br clear=all>
 '''(4)'''&nbsp; From the availability probability $99.9 \%$ follows the failure probability $10^{\rm &ndash;3} \approx \ {\rm Q}(3)$.
 *If the distance-dependent path loss $V_0$ is reduced by $10 \ \ \rm dB$ to $\underline {70 \ \rm dB}$, a failure will only occur when $V_{\rm S} &#8805; 50 \ \ \rm dB$.
 *This would achieve exactly the required reliability, as the following calculation shows:
-$${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})=
+:$${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})=
    {\rm Q}\left ( \frac{120-70-20}{10}\right )
    = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$

@@ Zeile 2: / Zeile 2: @@
 }}
-[[Datei:P_ID2122__Mob_A_1_2.png|right|frame|PDF of lognormal fading]]
+[[Datei:EN_Mob_A1_2.png|right|frame|PDF of lognormal fading]]
 We consider a mobile radio cell in an urban area and a vehicle that is approximately at a fixed distance&nbsp; $d_0$&nbsp; from the base station. For example, it moves on an arc around the base station.
@@ Zeile 82: / Zeile 82: @@
    = {\rm Q}(2) \approx 0.02\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1-  0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
-[[File:P_ID2187__Mob_A_1_2c_v1.png|right|frame|loss due to lognormal fading]]
+[[File:EN_Mob_A_1_2c.png|right|frame|loss due to lognormal fading]]
 The graphic illustrates the result.
 *The probability density $f_{\rm VS}(V_{\rm S})$ of the path loss due to shadowing (Longnormal&ndash;Fading) is shown here.

@@ Zeile 91: / Zeile 91: @@
 *If the distance-dependent path loss $V_0$ is reduced by $10 \ \ \rm dB$ to $\underline {70 \ \rm dB}$, a failure will only occur when $V_{\rm S} &#8805; 50 \ \ \rm dB$.
 *This would achieve exactly the required reliability, as the following calculation shows:
-$${\rm Pr}({\rm "System\hspace{0.15cm} does not work\hspace{0.15cm}"})=
+$${\rm Pr}({\rm "System\hspace{0.15cm} does\hspace{0.15cm} not\hspace{0.15cm} work\hspace{0.15cm}"})=
    {\rm Q}\left ( \frac{120-70-20}{10}\right )
    = {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$

@@ Zeile 79: / Zeile 79: @@
 *The variable portion $V_{\rm S}$ must therefore not be greater than $20 \ \rm dB$.
 *So it follows:
-:$${\rm Pr}({\rm "System\hspace{0.15cm}funktioniert\hspace{0.15cm}nicht"})= {\rm Q}\left ( \frac{20\,\,{\rm dB}}{\sigma_{\rm S} = 10\,{\rm dB}}\right )
+:$${\rm Pr}({\rm "System\hspace{0.15cm}does\hspace{0.15cm}not\hspace{0.15cm}work"})= {\rm Q}\left ( \frac{20\,\,{\rm dB}}{\sigma_{\rm S} = 10\,{\rm dB}}\right )
    = {\rm Q}(2) \approx 0.02\hspace{0.3cm}
-\Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}funktioniert"})= 1-  0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
+\Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1-  0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
 [[File:P_ID2187__Mob_A_1_2c_v1.png|right|frame|loss due to lognormal fading]]
 The graphic illustrates the result.

@@ Zeile 72: / Zeile 72: @@
 *Compared to the subtask '''(1)'' this is $m_{\rm S} = 20 \ \ \rm dB$ smaller &nbsp; &#8658; &nbsp; $P_{\rm E} = \ &ndash;60 \ \ \rm dBm$.
 *But it is still greater than the specified limit value ($&ndash;80 \ \rm dBm$).
-*It follows: &nbsp; The system is (almost) <u>100% functional</u>. &bdquo;Fast&rdquo; because with a Gaussian random quantity there is always a (small) residual uncertainty.
+*It follows: &nbsp; The system is (almost) <u>100% functional</u>. &bdquo;Almost&rdquo; because with a Gaussian random quantity there is always a (small) residual uncertainty.
@@ Zeile 85: / Zeile 85: @@
 The graphic illustrates the result.
 *The probability density $f_{\rm VS}(V_{\rm S})$ of the path loss due to shadowing (Longnormal&ndash;Fading) is shown here.
 *The probability that the system will fail is marked in red: