Aufgaben:Exercise 1.1Z: Simple Path Loss Model - Versionsgeschichte

Tasnad am 15. Juni 2020 um 12:52 Uhr

2020-06-15T12:52:10Z

Guenter am 11. Mai 2020 um 12:47 Uhr

2020-05-11T12:47:53Z

Guenter am 11. Mai 2020 um 12:47 Uhr

2020-05-11T12:47:09Z

Guenter am 7. Mai 2020 um 15:34 Uhr

2020-05-07T15:34:25Z

Guenter am 7. Mai 2020 um 15:30 Uhr

2020-05-07T15:30:57Z

Javier am 25. März 2020 um 14:57 Uhr

2020-03-25T14:57:09Z

Javier am 25. März 2020 um 14:56 Uhr

2020-03-25T14:56:53Z

Javier am 25. März 2020 um 14:55 Uhr

2020-03-25T14:55:08Z

Javier: /* Sample solution */

2020-03-25T14:53:33Z

Sample solution

Javier am 25. März 2020 um 14:52 Uhr

2020-03-25T14:52:25Z

@@ Zeile 31: / Zeile 31: @@
 ''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]].
 * The speed of light is&nbsp; $c = 3 \cdot 10^8 \ {\rm m/s}$.

@@ Zeile 31: / Zeile 31: @@
 ''Notes:''
-* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distanzabhängige Dämpfung und Abschattung_X]].
+* This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X|Distanzabhängige Dämpfung und Abschattung]].
 * The speed of light is&nbsp; $c = 3 \cdot 10^8 \ {\rm m/s}$.

@@ Zeile 2: / Zeile 2: @@
 }}
-[[Datei:EN_Mob_Z1_1.png|right|frame|Path loss [dB] vs. distance [m]]]
+[[Datei:EN_Mob_Z1_1.png|right|frame|Simplest path loss diagram]]
 Radio transmission with line-of-sight can be described by the so-called path loss model, which is given by the following equations:
 $$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm},$$
@@ Zeile 31: / Zeile 31: @@
 ''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|Distanzabhängige Dämpfung und Abschattung_X]].
 * The speed of light is&nbsp; $c = 3 \cdot 10^8 \ {\rm m/s}$.

@@ Zeile 2: / Zeile 2: @@
 }}
-[[Datei:EN_Mob_Z1_1.png|right|frame|Path loss model]]
+[[Datei:EN_Mob_Z1_1.png|right|frame|Path loss [dB] vs. distance [m]]]
 Radio transmission with line-of-sight can be described by the so-called path loss model, which is given by the following equations:
 $$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm},$$

@@ Zeile 2: / Zeile 2: @@
 }}
-[[Datei:P_ID2121__Mob_Z_1_1.png|right|frame|Path loss model]]
+[[Datei:EN_Mob_Z1_1.png|right|frame|Path loss model]]
 Radio transmission with line-of-sight can be described by the so-called path loss model, which is given by the following equations:
 $$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm},$$

@@ Zeile 100: / Zeile 100: @@
 '''(4)'''&nbsp; The <u>first suggested solution</u> is correct:
 *In the free-space scenario (A), the Fraunhofer distance&nbsp; $d_{\rm F} = \lambda_{\rm A}/2 \approx 63 \ \rm cm$. Thus, &nbsp; $d > d_{\rm F}$ always holds.
-*Also in scenario (B), the entire path loss curve is correct because of&nbsp; $\lambda_{\rm B} \approx 2 \ \rm m$&nbsp; or &nbsp;$d_{\rm F} \approx 1 \ \rm m$&nbsp; .
+*Also in scenario (B), the entire path loss curve is correct because &nbsp; $\lambda_{\rm B} \approx 2 \ \rm m$&nbsp; or &nbsp;$d_{\rm F} \approx 1 \ \rm m$&nbsp; .
 {{ML-Fuß}}

@@ Zeile 84: / Zeile 84: @@
 *The frequency $f_{\rm A}$ is related to the wavelength $\lambda_{\rm A}$ over the speed of light $c$:
-$$f_{\rm A} = \frac{c}{\lambda_{\rm A}} = \frac{3 \cdot 10^8\,{\rm m/s}}}{1.257\,{\rm m}} = 2.39 \cdot 10^8\,{\rm Hz}
+:$$f_{\rm A} =  \frac{c}{\lambda_{\rm A}} = \frac{3 \cdot 10^8\,{\rm m/s}}{1.257\,{\rm m}}  = 2.39 \cdot 10^8\,{\rm Hz}
-  \hspace{0.15cm} \underline{\approx 240 \,\,{\rm MHz}}
+  \hspace{0.15cm} \underline{\approx  240 \,\,{\rm MHz}}
    \hspace{0.05cm}.$$
-*On the other hand, the scenario (B)
+*On the other hand, for scenario (B),
 $$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ]^{2.5} = 20\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 25 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ] = 20\,{\rm dB}$$
-$$\Rightarrow \hspace{0.3cm}  \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}} = 10^{0.8} \approx 6.31
+:$$\Rightarrow \hspace{0.3cm}  \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}} = 10^{0.8} \approx 6.31
-   \hspace {0.3cm} \Rightarrow \hspace{0.3cm}
+   \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
   {\lambda_{\rm B}} = \frac{10}{6.31} \cdot {\lambda_{\rm A}}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}
-  {f_{\rm B}}} = \frac{6.31}{10} \cdot {f_{\rm A}}} = 0.631 \cdot 240 \,{\rm MHz}\hspace{0.15cm} \underline {\approx 151.4 \,\,{\rm MHz}}
+  {f_{\rm B}} = \frac{6.31}{10} \cdot {f_{\rm A}} = 0.631 \cdot 240 \,{\rm MHz}\hspace{0.15cm} \underline{\approx  151.4 \,\,{\rm MHz}}
    \hspace{0.05cm}.$$
-'''(4)'''&nbsp; <u>first suggested solution</u> is correct:
+'''(4)'''&nbsp; The <u>first suggested solution</u> is correct:
-*In free space&ndash;scenario (A) the Fraunhofer&ndash;distance&nbsp; $d_{\rm F} = \lambda_{\rm A}/2 \approx 63 \ \rm cm$. Thus, the following always applies&nbsp; $d > d_{\rm F}$.
+*In the free-space scenario (A), the Fraunhofer distance&nbsp; $d_{\rm F} = \lambda_{\rm A}/2 \approx 63 \ \rm cm$. Thus, the following always applies&nbsp; $d > d_{\rm F}$.
 *Also in scenario (B) is because of&nbsp; $\lambda_{\rm B} \approx 2 \ \rm m$&nbsp; or &nbsp;$d_{\rm F} \approx 1 \ \rm m$&nbsp; the entire displayed course correct.
 {{ML-Fuß}}

@@ Zeile 64: / Zeile 64: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)''&nbsp; The (simplest) path loss equation is
+'''(1)'''&nbsp; The (simplest) path loss equation is
 $$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm}.$$
-*In scenario (A), the waste per decade (for example, between $d_0 = 1 \ \rm m$ and $d = 10 \ \rm m$) is exactly $20 \ \rm dB$ and in scenario (B) $25 \ \rm dB$.
+*In scenario (A), the decay per decade (for example, between $d_0 = 1 \ \rm m$ and $d = 10 \ \rm m$) is exactly $20 \ \rm dB$ and in scenario (B) $25 \ \rm dB$.
 *It follows:
 $$\gamma_{\rm A} \hspace{0.15cm} \underline{= 2}\hspace{0.05cm},\hspace{0.2cm}\gamma_{\rm B} \hspace{0.15cm} \underline{= 2.5}\hspace{0.05cm}.$$
@@ Zeile 73: / Zeile 73: @@
-'''(2)'''&nbsp; Correct is <u>solution 1</u>, since the free space attenuation is characterized by the path loss exponent $\gamma = 2$.
+'''(2)'''&nbsp; <u>Solution 1</u> is correct, since the free space attenuation is characterized by the path loss exponent $\gamma = 2$.

@@ Zeile 56: / Zeile 56: @@
 $f_{\rm B} \ = \ $ { 151.4 3% } $\ \ \rm MHz$
-{Does the free-space&ndash scenario apply to all distances between&nbsp; $1 \ \rm m$&nbsp; and&nbsp; $10 \ \rm km$?
+{Does the free-space scenario apply to all distances between&nbsp; $1 \ \rm m$&nbsp; and&nbsp; $10 \ \rm km$?
 |type="()"}
 + Yes,