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

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	[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]		[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]
	[[de:~~Exercises~~:~~Exercise_1~~.1Z:~~_Simple_Path_Loss_Model~~]]		[[de:Aufgaben:Exercise 1.1Z: Simple Path Loss Model]]

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	'''(3)'''  The path loss at $d_0 = 1 \ \rm m$ is in both cases $V_0 = 20 \ \rm dB$. For scenario (A) the same applies:		'''(3)'''  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}.$$
	\frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm A}} = 10 \hspace{0.2cm} \Rightarrow \hspace{0.2cm}
	\lambda_{\rm A} = 4 \pi \cdot 0.1\,{\rm m} = 1,257\,{\rm m}
	\hspace{0.05cm}.$$

	*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 $(c)$:
	:$$f_{\rm A} = \frac{c}{\lambda_{\rm A}} = \frac{3 \cdot 10^8\,{\rm m/s}}{1.257\,{\rm m}} = 2.39 \cdot 10^8\,{\rm Hz}		:$$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{0.05cm}.$$
	\hspace{0.15cm} \underline{\approx 240 \,\,{\rm MHz}}
	\hspace{0.05cm}.$$

	*On the other hand, for scenario (B),		*On the other hand, for scenario (B),
	:$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ]^{2.5} = 20\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 25 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ] = 20\,{\rm dB}$$		:$$10 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ]^{2.5} = 20\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 25 \cdot {\rm lg}\hspace{0.1cm} \left [ \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}}\right ] = 20\,{\rm dB}$$
	:$$\Rightarrow \hspace{0.3cm} \frac{4 \cdot \pi \cdot d_0}{\lambda_{\rm B}} = 10^{0.8} \approx 6.31		:$$\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}{\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}}\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	{\lambda_{\rm B}} = \frac{10}{6.31} \cdot {\lambda_{\rm A}}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}
	{f_{\rm B}} = \frac{6.31}{10} \cdot {f_{\rm A}} = 0.631 \cdot 240 \,{\rm MHz}\hspace{0.15cm} \underline{\approx 151.4 \,\,{\rm MHz}}
	\hspace{0.05cm}.$$

	'''(4)'''  The <u>first suggested solution</u> is correct:		'''(4)'''  The <u>first suggested solution</u> is correct:

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	[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]		[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]
			[[de:Exercises:Exercise_1.1Z:_Simple_Path_Loss_Model]]

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	[[Category:~~Exercises for~~ Mobile Communications\|^1.1 Distance-Dependent Attenuation^]]		[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]

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

2021-03-22T10:47:37Z

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

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	:$$V_{\rm 0} = V_{\rm P}(d = d_0) = 20\,{\rm dB} \hspace{0.05cm}.$$		:$$V_{\rm 0} = V_{\rm P}(d = d_0) = 20\,{\rm dB} \hspace{0.05cm}.$$

	One of these two scenarios describes the so-called <i>free space attenuation</i>, characterized by the path loss exponent  $\gamma = 2$. However, the equation for the free space attenuation only applies in the <i>far-field</i>, i.e. when the distance  $d$  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  $\gamma = 2$. However, the equation for the free-space attenuation only applies in the <i>far-field</i>, i.e. when the distance  $d$  between transmitter and receiver is greater than the <i>Fraunhofer distance</i>,
	:$$d_{\rm F} = {2 D^2}/{\lambda} \hspace{0.05cm}.$$		:$$d_{\rm F} = {2 D^2}/{\lambda} \hspace{0.05cm}.$$

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	'''(2)'''  <u>Solution 1</u> is correct, since the free space attenuation is characterized by the path loss exponent $\gamma = 2$.		'''(2)'''  <u>Solution 1</u> is correct, since the free-space attenuation is characterized by the path loss exponent $\gamma = 2$.

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

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	</quiz>		</quiz>

	===~~Solutions~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The (simplest) path loss equation is		'''(1)'''  The (simplest) path loss equation is

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"

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	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Distanzabhängige Dämpfung und Abschattung~~		{{quiz-Header\|Buchseite=Mobile_Communications/Distance_Dependent_Attenuation_and_Shading
	}}		}}

Oezdemir at 15:33, 1 November 2020

2020-11-01T15:33:12Z

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	[[Category:Exercises for Mobile Communications\|^1.1 Distance-~~dependent attenuation~~^]]		[[Category:Exercises for Mobile Communications\|^1.1 Distance-Dependent Attenuation^]]

Oezdemir: /* Questionnaire */

2020-07-30T20:39:24Z

Questionnaire

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	===~~Questionnaire~~===		===Questions===

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

Oezdemir: /* Sample solution */

2020-07-30T20:39:14Z

Sample solution

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	</quiz>		</quiz>

	===~~Sample solution~~===		===Solutions===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The (simplest) path loss equation is		'''(1)'''  The (simplest) path loss equation is