Aufgaben:Exercise 1.1: Dual Slope 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^]]
	[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]		[[Category:Mobile Communications: Exercises\|^1.1 Distance-Dependent Attenuation^]]
	[[de:~~Exercises~~:~~Exercise_1~~.1:~~_Dual_Slope_Loss_Model~~]]		[[de:Aufgaben:Exercise 1.1: Dual Slope Loss Model]]

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	The graph applies to the model parameters		The graph applies to the model parameters
	:$$d_0 = 1\,{\rm m}\hspace{0.05cm},\hspace{0.2cm}d_{\rm BP} = 100\,{\rm m}\hspace{0.05cm},\hspace{0.2cm}		:$$d_0 = 1\,{\rm m}\hspace{0.05cm},\hspace{0.2cm}d_{\rm BP} = 100\,{\rm m}\hspace{0.05cm},\hspace{0.2cm}V_0 = 10\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}\gamma_0 = 2 \hspace{0.05cm},\hspace{0.2cm}\gamma_1 = 4 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}V_{\rm BP} = 50\,{\rm dB}\hspace{0.05cm}.$$
	V_0 = 10\,{\rm dB}\hspace{0.05cm},\hspace{0.2cm}\gamma_0 = 2 \hspace{0.05cm},\hspace{0.2cm}\gamma_1 = 4 \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}
	V_{\rm BP} = 50\,{\rm dB}\hspace{0.05cm}.$$

	In the questions, this piece-wise defined profile is called  $\rm A$.		In the questions, this piece-wise defined profile is called  $\rm A$.

	The second curve is the profile  $\rm B$  given by the following equation:		The second curve is the profile  $\rm B$  given by the following equation:
	:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right )		:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right )+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$
	+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$

	With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance  $d$  according to the following equation:		With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance  $d$  according to the following equation:
	:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm tot}}{V_{\rm S}(d)}		:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm tot}}{V_{\rm S}(d)}\hspace{0.05cm},\hspace{0.2cm}V_{\rm S}(d) = 10^{V_{\rm S}(d)/10}\hspace{0.05cm}.$$
	\hspace{0.05cm},\hspace{0.2cm}V_{\rm S}(d) = 10^{V_{\rm S}(d)/10}
	\hspace{0.05cm}.$$

	Here, all parameters are in natural units (not in dB).  The transmit power is assumed to be  $P_{\rm S} = 5 \ \rm W$ .  The other quantities have the following meanings and values:		Here, all parameters are in natural units (not in dB).  The transmit power is assumed to be  $P_{\rm S} = 5 \ \rm W$ .  The other quantities have the following meanings and values:
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	*This task belongs to the chapter  [[Mobile_Communications/Distance_dependent_attenuation_and_shading\|Distance dependent attenuation and shading]].		*This task belongs to the chapter  [[Mobile_Communications/Distance_dependent_attenuation_and_shading\|Distance dependent attenuation and shading]].
	*If the profile  $\rm B$  were		*If the profile  $\rm B$  were
	:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )		:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
	+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
	:then profile  $\rm A$  and profile  $\rm B$  would be identical for  $d ≥ d_{\rm BP}$.		:then profile  $\rm A$  and profile  $\rm B$  would be identical for  $d ≥ d_{\rm BP}$.
	*In this case, however, profile  $\rm B$  would be above profile  $\rm A$  for   $d < d_{\rm BP}$, suggesting clearly too good conditions.		*In this case, however, profile  $\rm B$  would be above profile  $\rm A$  for   $d < d_{\rm BP}$, suggesting clearly too good conditions.
	*For example,   $d = d_0 = 1 \ \ \rm m$  with the given numerical values gives a result that is   $40 \ \ \rm dB$  too good:		*For example,   $d = d_0 = 1 \ \ \rm m$  with the given numerical values gives a result that is   $40 \ \ \rm dB$  too good:
	:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right )		:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right )= -30\,{\rm dB} \hspace{0.05cm}. $$
	= -30\,{\rm dB} \hspace{0.05cm}. $$


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	'''(2)'''  With the profile  $\rm (B)$  on the other hand, using  $V_0 = 10 \ \rm dB$,  $\gamma_0 = 2$  and  $\gamma_1 = 4$:		'''(2)'''  With the profile  $\rm (B)$  on the other hand, using  $V_0 = 10 \ \rm dB$,  $\gamma_0 = 2$  and  $\gamma_1 = 4$:
	:$$V_{\rm S}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2)		:$$V_{\rm S}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2)\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$
	\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$



	'''(3)'''  The antenna gains from the transmitter  $(+17 \ \rm dB)$  and receiver  $(-3 \ \rm dB)$  and the internal losses of the base station  $(+4 \ \rm dB)$  can be combined to		'''(3)'''  The antenna gains from the transmitter  $(+17 \ \rm dB)$  and receiver  $(-3 \ \rm dB)$  and the internal losses of the base station  $(+4 \ \rm dB)$  can be combined to
	:$$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} V_{\rm tot} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB}		:$$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} V_{\rm tot} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10}\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10}
	\hspace{0.05cm}.$$

	*For the profile  $\rm (A)$  the following path loss occurred:		*For the profile  $\rm (A)$  the following path loss occurred:
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	'''(4)'''  Below the breakpoint  $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile  $\rm (B)$:		'''(4)'''  Below the breakpoint  $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile  $\rm (B)$:
	:$${\rm \Delta}V_{\rm S}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm}		:$${\rm \Delta}V_{\rm S}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm}\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$
	\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$



	'''(5)'''  Here the profile  $\rm (A)$  with  $V_{\rm BP} = 50 \ \rm dB$  gives:		'''(5)'''  Here the profile  $\rm (A)$  with  $V_{\rm BP} = 50 \ \rm dB$  gives:
	:$$V_{\rm S}(d = 200\,{\rm m}) = 50\,{\rm dB} + 4 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (2)\hspace{0.15cm}		:$$V_{\rm S}(d = 200\,{\rm m}) = 50\,{\rm dB} + 4 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (2)\hspace{0.15cm}{\approx 62\,{\rm dB}} \hspace{0.05cm}.$$
	{\approx 62\,{\rm dB}} \hspace{0.05cm}.$$

	*On the other hand, the profile  $\rm (B)$  leads to the result:		*On the other hand, the profile  $\rm (B)$  leads to the result:
	:$$V_{\rm S}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)		:$$V_{\rm S}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)+ 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB}\hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$
	+ 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB}		:$$\Rightarrow \hspace{0.3cm} {\rm \Delta}V_{\rm P}(d = 200\,{\rm m}) = 65.5\,{\rm dB} - 62\,{\rm dB}\hspace{0.15cm}\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$
	\hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$
	:$$\Rightarrow \hspace{0.3cm} {\rm \Delta}V_{\rm P}(d = 200\,{\rm m}) = 65.5\,{\rm dB} - 62\,{\rm dB}\hspace{0.15cm}
	\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$

	*You can see that  $\Delta V_{\rm S}$  is almost symmetrical to  $d = d_{\rm BP}$  if you plot the distance  $d$  logarithmically as in the given graph.		*You can see that  $\Delta V_{\rm S}$  is almost symmetrical to  $d = d_{\rm BP}$  if you plot the distance  $d$  logarithmically as in the given graph.

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

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

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

Javier at 12:35, 23 March 2021

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

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

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	</quiz>		</quiz>

	===~~Solutions~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  You can see directly from the graphic that the profile  $\rm (A)$  with the two linear sections at the breakpoint  $(d = 100 \ \rm m)$  gives the following result:		'''(1)'''  You can see directly from the graphic that the profile  $\rm (A)$  with the two linear sections at the breakpoint  $(d = 100 \ \rm m)$  gives the following result:

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

2020-12-15T13:55:07Z

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

2020-11-01T15:33:38Z

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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 20:38, 30 July 2020

2020-07-30T20:38:10Z

Oezdemir at 20:36, 30 July 2020

2020-07-30T20:36:00Z

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	With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance  $d$  according to the following equation:		With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance  $d$  according to the following equation:
	:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm tot}}{V_{\rm P}(d)}		:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm tot}}{V_{\rm S}(d)}
	\hspace{0.05cm},\hspace{0.2cm}V_{\rm P}(d) = 10^{V_{\rm P}(d)/10}		\hspace{0.05cm},\hspace{0.2cm}V_{\rm S}(d) = 10^{V_{\rm S}(d)/10}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

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	*For the profile  $\rm (A)$  the following path loss occurred:		*For the profile  $\rm (A)$  the following path loss occurred:
	:$$V_{\rm P}(d = 100\,{\rm m})\hspace{0.05cm} {= 50\,{\rm dB}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K_{\rm P} = 10^5 \hspace{0.05cm}.$$		:$$V_{\rm S}(d = 100\,{\rm m})\hspace{0.05cm} {= 50\,{\rm dB}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K_{\rm P} = 10^5 \hspace{0.05cm}.$$

	:This gives you for the received power after  $d = 100\ \rm m$:		:This gives you for the received power after  $d = 100\ \rm m$:
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	'''(4)'''  Below the breakpoint  $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile  $\rm (B)$:		'''(4)'''  Below the breakpoint  $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile  $\rm (B)$:
	:$${\rm \Delta}V_{\rm P}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm}		:$${\rm \Delta}V_{\rm S}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm}
	\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$		\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$

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	'''(5)'''  Here the profile  $\rm (A)$  with  $V_{\rm BP} = 50 \ \rm dB$  gives:		'''(5)'''  Here the profile  $\rm (A)$  with  $V_{\rm BP} = 50 \ \rm dB$  gives:
	:$$V_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 4 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (2)\hspace{0.15cm}		:$$V_{\rm S}(d = 200\,{\rm m}) = 50\,{\rm dB} + 4 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (2)\hspace{0.15cm}
	{\approx 62\,{\rm dB}} \hspace{0.05cm}.$$		{\approx 62\,{\rm dB}} \hspace{0.05cm}.$$

	*On the other hand, the profile  $\rm (B)$  leads to the result:		*On the other hand, the profile  $\rm (B)$  leads to the result:
	:$$V_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)		:$$V_{\rm S}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)
	+ 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB}		+ 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB}
	\hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$		\hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$
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	\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$		\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$

	*You can see that  $\Delta V_{\rm P}$  is almost symmetrical to  $d = d_{\rm BP}$  if you plot the distance  $d$  logarithmically as in the given graph.		*You can see that  $\Delta V_{\rm S}$  is almost symmetrical to  $d = d_{\rm BP}$  if you plot the distance  $d$  logarithmically as in the given graph.
	{{ML-Fuß}}		{{ML-Fuß}}

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	To simulate path loss in an urban environment, the asymptotic dual-slope model is often used, which is shown as a red curve in the diagram.  This simple model is characterized by two linear sections separated by the so-called breakpoint  $\rm (BP)$. The variable name "$V$" stands for "Verlust", which is the German word for "loss":		To simulate path loss in an urban environment, the asymptotic dual-slope model is often used, which is shown as a red curve in the diagram.  This simple model is characterized by two linear sections separated by the so-called breakpoint  $\rm (BP)$. The variable name "$V$" stands for "Verlust", which is the German word for "loss":
	* For  $d \le d_{\rm BP}$  and the exponent   $\gamma_0$  we have:		* For  $d \le d_{\rm BP}$  and the exponent   $\gamma_0$  we have:
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$		:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$

	* For  $d > d_{\rm BP}$  we must apply the path loss exponent  $\gamma_1$  where  $\gamma_1 > \gamma_0$:		* For  $d > d_{\rm BP}$  we must apply the path loss exponent  $\gamma_1$  where  $\gamma_1 > \gamma_0$:
	:$$V_{\rm P}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$		:$$V_{\rm S}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$

	In these equations, the variables are:		In these equations, the variables are:
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	The second curve is the profile  $\rm B$  given by the following equation:		The second curve is the profile  $\rm B$  given by the following equation:
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right )		:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right )
	+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$		+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$

	With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance  $d$  according to the following equation:		With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance  $d$  according to the following equation:
	:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm t} \cdot G_{\rm r} /V_{\rm tot}}{V_{\rm P}(d)}		:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm tot}}{V_{\rm P}(d)}
	\hspace{0.05cm},\hspace{0.2cm}V_{\rm P}(d) = 10^{V_{\rm P}(d)/10}		\hspace{0.05cm},\hspace{0.2cm}V_{\rm P}(d) = 10^{V_{\rm P}(d)/10}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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	*This task belongs to the chapter  [[Mobile_Communications/Distance_dependent_attenuation_and_shading\|Distance dependent attenuation and shading]].		*This task belongs to the chapter  [[Mobile_Communications/Distance_dependent_attenuation_and_shading\|Distance dependent attenuation and shading]].
	*If the profile  $\rm B$  were		*If the profile  $\rm B$  were
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )		:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )
	+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$		+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
	:then profile  $\rm A$  and profile  $\rm B$  would be identical for  $d ≥ d_{\rm BP}$.		:then profile  $\rm A$  and profile  $\rm B$  would be identical for  $d ≥ d_{\rm BP}$.
	*In this case, however, profile  $\rm B$  would be above profile  $\rm A$  for   $d < d_{\rm BP}$, suggesting clearly too good conditions.		*In this case, however, profile  $\rm B$  would be above profile  $\rm A$  for   $d < d_{\rm BP}$, suggesting clearly too good conditions.
	*For example,   $d = d_0 = 1 \ \ \rm m$  with the given numerical values gives a result that is   $40 \ \ \rm dB$  too good:		*For example,   $d = d_0 = 1 \ \ \rm m$  with the given numerical values gives a result that is   $40 \ \ \rm dB$  too good:
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right )		:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right )
	= -30\,{\rm dB} \hspace{0.05cm}. $$		= -30\,{\rm dB} \hspace{0.05cm}. $$

Line 56:		Line 56:


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

	<quiz display=simple>		<quiz display=simple>
	{How large is the path loss $($in  $\rm dB)$  after  $d= 100 \ \rm m$  according to profile  $\rm A$?		{How large is the path loss $($in  $\rm dB)$  after  $d= 100 \ \rm m$  according to profile  $\rm A$?
	\|type="{}"}		\|type="{}"}
	$V_{\rm P}(d = 100 \ \rm m) \ = \ $ { 50 3% } $\ \rm dB$		$V_{\rm S}(d = 100 \ \rm m) \ = \ $ { 50 3% } $\ \rm dB$

	{How large is the path loss $($in  $\rm dB)$  after  $d= 100 \ \rm m$  according to profile  $\rm B$?		{How large is the path loss $($in  $\rm dB)$  after  $d= 100 \ \rm m$  according to profile  $\rm B$?
	\|type="{}"}		\|type="{}"}
	$V_{\rm P}(d = 100 \ \rm m) \ = \ $ { 56 3% } $\ \rm dB$		$V_{\rm S}(d = 100 \ \rm m) \ = \ $ { 56 3% } $\ \rm dB$

	{What is the received power after  $100 \ \ \rm m$  with both profiles?		{What is the received power after  $100 \ \ \rm m$  with both profiles?
Line 72:		Line 72:
	Profile $\text{B:} \hspace{0.2cm} P_{\rm E}(d = 100 \ \rm m) \ = \ $ { 0.125 3% } $\ \ \rm mW$		Profile $\text{B:} \hspace{0.2cm} P_{\rm E}(d = 100 \ \rm m) \ = \ $ { 0.125 3% } $\ \ \rm mW$

	{How big is the deviation  $ΔV_{\rm P}$  between profile  $\rm A$  and  $\rm B$  at  $d = 50 \ \rm m$?		{How big is the deviation  $ΔV_{\rm S}$  between profile  $\rm A$  and  $\rm B$  at  $d = 50 \ \rm m$?
	\|type="{}"}		\|type="{}"}
	$ΔV_{\rm P}(d = 50 \ \rm m) \ = \ $ { 3.5 3% } $\ \rm dB$		$ΔV_{\rm P}(d = 50 \ \rm m) \ = \ $ { 3.5 3% } $\ \rm dB$

	{How big is the deviation  $ΔV_{\rm P}$  between profile  $\rm A$  and  $\rm B$  at  $d = 200 \ \rm m$?		{How big is the deviation  $ΔV_{\rm S}$  between profile  $\rm A$  and  $\rm B$  at  $d = 200 \ \rm m$?
	\|type="{}"}		\|type="{}"}
	$ΔV_{\rm P}(d = 200 \ \rm m)\ = \ $ { 3.5 3% } $\ \rm dB$		$ΔV_{\rm P}(d = 200 \ \rm m)\ = \ $ { 3.5 3% } $\ \rm dB$
	</quiz>		</quiz>

	===~~Sample solution~~===		===Solutions===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  You can see directly from the graphic that the profile  $\rm (A)$  with the two linear sections at the breakpoint  $(d = 100 \ \rm m)$  gives the following result:		'''(1)'''  You can see directly from the graphic that the profile  $\rm (A)$  with the two linear sections at the breakpoint  $(d = 100 \ \rm m)$  gives the following result:
	:$$V_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$		:$$V_{\rm S}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$



	'''(2)'''  With the profile  $\rm (B)$  on the other hand, using  $V_0 = 10 \ \rm dB$,  $\gamma_0 = 2$  and  $\gamma_1 = 4$:		'''(2)'''  With the profile  $\rm (B)$  on the other hand, using  $V_0 = 10 \ \rm dB$,  $\gamma_0 = 2$  and  $\gamma_1 = 4$:
	:$$V_{\rm P}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2)		:$$V_{\rm S}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2)
	\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$		\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$