Aufgaben:Exercise 1.2: Lognormal Channel 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~~.2:~~_Lognormal_Channel_Model~~]]		[[de:Aufgaben:Exercise 1.2: Lognormal Channel Model]]

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	* You can use the following (rough) approximations for the complementary Gaussian error integral:		* 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}(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}.$$
	{\rm Q}(3) \approx 10^{-3}\hspace{0.05cm}.$$
	* Or use the interaction module  [[Applets:Complementary_Gaussian_Error_Functions\|Complementary Gaussian Error Functions]]  provided by  $\rm LNTwww$.		* Or use the interaction module  [[Applets:Complementary_Gaussian_Error_Functions\|Complementary Gaussian Error Functions]]  provided by  $\rm LNTwww$.

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	*You can also solve this problem directly with the logarithmic quantities:		*You can also solve this problem directly with the logarithmic quantities:
	:$$10 \cdot {\rm lg}\hspace{0.15cm} \frac{P_{\rm E}}{1\,\,{\rm mW}} = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{P_{\rm S}}{1\,\,{\rm mW}} - V_0 = 40\,{\rm dBm} -80\,\,{\rm dB} = -40\,\,{\rm dBm}		:$$10 \cdot {\rm lg}\hspace{0.15cm} \frac{P_{\rm E}}{1\,\,{\rm mW}} = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{P_{\rm S}}{1\,\,{\rm mW}} - V_0 = 40\,{\rm dBm} -80\,\,{\rm dB} = -40\,\,{\rm dBm}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Only the limit value  $-80 \ \rm dBm$  is required.		*Only the limit value  $-80 \ \rm dBm$  is required.
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	*The distance-dependent path loss $V_{\rm S}$ must therefore not be greater than  $20 \ \rm dB$.		*The distance-dependent path loss $V_{\rm S}$ must therefore not be greater than  $20 \ \rm dB$.
	*So it follows:		*So it follows:
	:$${\rm Pr}({\rm "System\hspace{0.15cm}does\hspace{0.15cm}not\hspace{0.15cm}work"})= {\rm Q}\left ( \frac{20\,\,{\rm dB}}{\sigma_{\rm S} = 10\,{\rm dB}}\right )		:$${\rm 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$$
	= {\rm Q}(2) \approx 0.02$$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1- 0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}{\rm Pr}({\rm "System\hspace{0.15cm}works"})= 1- 0.02 \hspace{0.15cm} \underline{\approx 98\,\%}\hspace{0.05cm}.$$
	The graphic illustrates the result.		The graphic illustrates the result.
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	*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  $V_{\rm S} ≥ 50 \ \ \rm dB$.		*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  $V_{\rm S} ≥ 50 \ \ \rm dB$.
	*This would achieve exactly the required reliability, as the following calculation shows:		*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}.$$
	{\rm Q}\left ( \frac{120-70-20}{10}\right )
	= {\rm Q}(3) \approx 0.001 \hspace{0.05cm}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

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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_1.2:_Lognormal_Channel_Model]]

Javier at 14:58, 2 July 2021

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	<quiz display=simple>		<quiz display=simple>
	{Would  $P_{\rm E}$  be sufficient ~~without consideration of~~ the ~~lognormal–fading~~?		{Would  $P_{\rm E}$  be sufficient if the loss $V_S$ due to shadowing is not present?
	\|type="()"}		\|type="()"}
	+ Yes,		+ Yes,

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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>		</quiz>

	===~~Solutions~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The correct answer is <u>YES</u>:		'''(1)'''  The correct answer is <u>YES</u>:

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

Guenter at 13:40, 16 November 2020

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	'''(3)'''  The received power is too low  $($less than $–80 \ \rm dBm)$  if the power loss due to the lognormal–term is  $40 \ \rm dB$  or more.		'''(3)'''  The received power is too low  $($less than $–80 \ \rm dBm)$  if the power loss due to the lognormal–term is  $40 \ \rm dB$  or more.
	[[File:EN_Mob_A_1_2c.png\|right\|frame\|~~loss~~ due to lognormal fading]]		[[File:EN_Mob_A_1_2c.png\|right\|frame\|Loss due to lognormal fading]]

	*The distance-dependent path loss $V_{\rm S}$ must therefore not be greater than  $20 \ \rm dB$.		*The distance-dependent path loss $V_{\rm S}$ must therefore not be greater than  $20 \ \rm dB$.

Oezdemir at 15:33, 1 November 2020

2020-11-01T15:33:55Z

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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 at 20:39, 30 July 2020

2020-07-30T20:39:57Z

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

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

	===~~Sample solution~~===		===Solutions===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The correct answer is <u>YES</u>:		'''(1)'''  The correct answer is <u>YES</u>: