Aufgaben:Exercise 1.1Z: Simple Path Loss Model - Revision history

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

2021-03-23T12:37:43Z

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

Javier: Text replacement - "free space attenuation" to "free-space attenuation"

2021-03-22T10:47:37Z

Text replacement - "free space attenuation" to "free-space attenuation"

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

2020-12-15T14:11:11Z

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

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

2020-12-15T13:55:17Z

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

Oezdemir at 15:33, 1 November 2020

2020-11-01T15:33:12Z

Oezdemir: /* Questionnaire */

2020-07-30T20:39:24Z

Questionnaire

Oezdemir: /* Sample solution */

2020-07-30T20:39:14Z

Sample solution

Guenter at 14:22, 24 July 2020

2020-07-24T14:22:24Z

Guenter at 14:16, 24 July 2020

2020-07-24T14:16:16Z

Javier at 11:50, 23 July 2020

2020-07-23T11:50:25Z

← Older revision		Revision as of 12:37, 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^]]

@@ Line 19: / Line 19: @@
 :$$V_{\rm 0} = V_{\rm P}(d = d_0) = 20\,{\rm dB}  \hspace{0.05cm}.$$
-One of these two scenarios describes the so-called <i>free space attenuation</i>, characterized by the path loss exponent&nbsp; $\gamma = 2$. However, the equation for the free space attenuation only applies in the <i>far-field</i>, i.e. when the distance&nbsp; $d$&nbsp; between transmitter and receiver is greater than the <i>Fraunhofer distance</i>,
+One of these two scenarios describes the so-called <i>free-space attenuation</i>, characterized by the path loss exponent&nbsp; $\gamma = 2$. However, the equation for the free-space attenuation only applies in the <i>far-field</i>, i.e. when the distance&nbsp; $d$&nbsp; between transmitter and receiver is greater than the <i>Fraunhofer distance</i>,
 :$$d_{\rm F} = {2 D^2}/{\lambda} \hspace{0.05cm}.$$
@@ Line 72: / Line 72: @@
-'''(2)'''&nbsp; <u>Solution 1</u> is correct, 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 $\gamma = 2$.

@@ Line 62: / Line 62: @@
 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; The (simplest) path loss equation is

← 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 15:33, 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^]]

← Older revision		Revision as of 20:39, 30 July 2020
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−	===~~Questionnaire~~===	+	===Questions===

	<quiz display=simple>		<quiz display=simple>

@@ Line 4: / Line 4: @@
 [[file: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:
-$$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm},$$
+:$$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm},$$
-$$V_{\rm 0} = \gamma \cdot 10\,{\rm dB}  \cdot {\rm lg} \hspace{0.1cm} \frac{4 \cdot \pi \cdot d_0}{\lambda} \hspace{0.05cm}.$$
+:$$V_{\rm 0} = \gamma \cdot 10\,{\rm dB}  \cdot {\rm lg} \hspace{0.1cm} \frac{4 \cdot \pi \cdot d_0}{\lambda} \hspace{0.05cm}.$$
 The graphic shows the path loss&nbsp; $V_{\rm P}(d)$&nbsp; in&nbsp; $\rm dB$. The abscissa&nbsp; $d$&nbsp; is also displayed logarithmically.
@@ Line 17: / Line 17: @@
 Two scenarios are shown &nbsp;$\rm (A)$&nbsp; and &nbsp;$\rm (B)$&nbsp; with the same path loss at distance&nbsp; $d_0 = 1 \ \rm m$:
-$$V_{\rm 0} = V_{\rm P}(d = d_0) = 20\,{\rm dB}  \hspace{0.05cm}.$$
+:$$V_{\rm 0} = V_{\rm P}(d = d_0) = 20\,{\rm dB}  \hspace{0.05cm}.$$
 One of these two scenarios describes the so-called <i>free space attenuation</i>, characterized by the path loss exponent&nbsp; $\gamma = 2$. However, the equation for the free space attenuation only applies in the <i>far-field</i>, i.e. when the distance&nbsp; $d$&nbsp; between transmitter and receiver is greater than the <i>Fraunhofer distance</i>,
-$$d_{\rm F} = {2 D^2}/{\lambda} \hspace{0.05cm}.$$
+:$$d_{\rm F} = {2 D^2}/{\lambda} \hspace{0.05cm}.$$
 Here,&nbsp; $D$&nbsp; is the largest physical dimension of the transmitting antenna. With an&nbsp; $\lambda/2$&ndash;antenna, the Fraunhofer distance has a simple expression:
-$$d_{\rm F} = \frac{2 \cdot (\lambda/2)^2}{\lambda} = {\lambda}/{2}\hspace{0.05cm}.$$
+:$$d_{\rm F} = \frac{2 \cdot (\lambda/2)^2}{\lambda} = {\lambda}/{2}\hspace{0.05cm}.$$
@@ Line 65: / Line 65: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The (simplest) path loss equation is
-$$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm}.$$
+:$$V_{\rm P}(d) = V_{\rm 0} + \gamma \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (d/d_0)\hspace{0.05cm}.$$
 *In scenario (A), the 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:
-$$\gamma_{\rm A} \hspace{0.15cm} \underline{= 2}\hspace{0.05cm},\hspace{0.2cm}\gamma_{\rm B} \hspace{0.15cm} \underline{= 2.5}\hspace{0.05cm}.$$
+:$$\gamma_{\rm A} \hspace{0.15cm} \underline{= 2}\hspace{0.05cm},\hspace{0.2cm}\gamma_{\rm B} \hspace{0.15cm} \underline{= 2.5}\hspace{0.05cm}.$$
 '''(2)'''&nbsp; <u>Solution 1</u> is correct, since the free space attenuation is characterized by the path loss exponent $\gamma = 2$.
 '''(3)'''&nbsp; The path loss at $d_0 = 1 \ \rm m$ is in both cases $V_0 = 20 \ \rm dB$. For scenario (A) the same applies:
-$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm A}}\right ]^2 = 20\,{\rm dB} \hspace{0.2cm} \Rightarrow \hspace{0.2cm}
+:$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm A}}\right ]^2 = 20\,{\rm dB} \hspace{0.2cm} \Rightarrow \hspace{0.2cm}
   \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm A}} = 10 \hspace{0.2cm} \Rightarrow \hspace{0.2cm}
   \lambda_{\rm A} = 4 \pi \cdot 0.1\,{\rm m} = 1,257\,{\rm m}
    \hspace{0.05cm}.$$
-*The frequency $f_{\rm A}$ is related to the wavelength $\lambda_{\rm A}$ over the speed of light $c$:
+*The frequency $f_{\rm A}$ is related to the wavelength $\lambda_{\rm A}$ over the speed of light $(c)$:
 :$$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}
   \hspace{0.15cm} \underline{\approx  240 \,\,{\rm MHz}}
@@ Line 89: / Line 87: @@
 *On the other hand, for scenario (B),
-$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ]^{2.5} = 20\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 25 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ] = 20\,{\rm dB}$$
+:$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ]^{2.5} = 20\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 25 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ] = 20\,{\rm dB}$$
 :$$\Rightarrow \hspace{0.3cm}  \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}} = 10^{0.8} \approx 6.31
    \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
@@ Line 96: / Line 94: @@
   {f_{\rm B}} = \frac{6.31}{10} \cdot {f_{\rm A}} = 0.631 \cdot 240 \,{\rm MHz}\hspace{0.15cm} \underline{\approx  151.4 \,\,{\rm MHz}}
    \hspace{0.05cm}.$$
 '''(4)'''&nbsp; The <u>first suggested solution</u> is correct:

@@ Line 19: / Line 19: @@
 $$V_{\rm 0} = V_{\rm P}(d = d_0) = 20\,{\rm dB}  \hspace{0.05cm}.$$
-One of these two scenarios describes the so-called <i>free space attenuation</i>, characterized by the path loss exponent&nbsp; $\gamma = 2$. However, the equation for the free space attenuation only applies in the <i>far-field</i>, i.e. when the distance&nbsp; $d$&nbsp; between transmitter and receiver is greater than the <i>Fraunhofer distance</i>;
+One of these two scenarios describes the so-called <i>free space attenuation</i>, characterized by the path loss exponent&nbsp; $\gamma = 2$. However, the equation for the free space attenuation only applies in the <i>far-field</i>, i.e. when the distance&nbsp; $d$&nbsp; between transmitter and receiver is greater than the <i>Fraunhofer distance</i>,
 $$d_{\rm F} = {2 D^2}/{\lambda} \hspace{0.05cm}.$$
@@ Line 31: / Line 31: @@
 ''Notes:''
-* This task belongs to the chapter&nbsp; [[Mobile_Communications/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distanzabhängige Dämpfung und Abschattung]].
+* This task belongs to the chapter&nbsp; [[Mobile_Communications/Distance_dependent_attenuation_and_shading|Distance dependent attenuation and shading]].
 * The speed of light is&nbsp; $c = 3 \cdot 10^8 \ {\rm m/s}$.