Aufgaben:Exercise 1.1: Dual Slope Loss Model - Versionsgeschichte

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	[[Category:Exercises for Mobile Communications\|^1.1 Distance-dependent attenuation^]]		[[Category:Exercises for Mobile Communications\|^1.1 Distance-dependent attenuation^]]
			[[en: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:
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right )		:$$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 (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:
	:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm zus}}{K_{\rm P}(d)}		:$$P_{\rm E}(d) = \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm zus}}{K_{\rm P}(d)}\hspace{0.05cm},\hspace{0.2cm}K_{\rm P}(d) = 10^{V_{\rm P}(d)/10} \hspace{0.05cm}.$$
	\hspace{0.05cm},\hspace{0.2cm}K_{\rm P}(d) = 10^{V_{\rm P}(d)/10}
	\hspace{0.05cm}.$$

	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:
Zeile 45:		Zeile 39:
	*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung\|Distanzabhängige Dämpfung und Abschattung]].		*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung\|Distanzabhängige Dämpfung und Abschattung]].
	*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 P}(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  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical.		:then profile  $\rm A$  and profile  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical.
	*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 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 )= -30\,{\rm dB} \hspace{0.05cm}. $$
	= -30\,{\rm dB} \hspace{0.05cm}. $$


Zeile 89:		Zeile 81:

	'''(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 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) \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
	:$$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} V_{\rm zus} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB}		:$$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 zus} = 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:
Zeile 111:		Zeile 100:

	'''(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 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}\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:
	:$$V_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 4 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (2)\hspace{0.15cm}		:$$V_{\rm P}(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 P}(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  $\Delta V_{\rm P}$  is almost symmetrical to  $d = d_{\rm BP}$  if you plot the distance  $d$  logarithmically as in the given graph.		*You can see that  $\Delta V_{\rm P}$  is almost symmetrical to  $d = d_{\rm BP}$  if you plot the distance  $d$  logarithmically as in the given graph.

Tasnad am 15. Juni 2020 um 12:51 Uhr

2020-06-15T12:51:00Z

← Nächstältere Version		Version vom 15. Juni 2020, 14:51 Uhr
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	''Notes:''		''Notes:''
	*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%~~A4mpfung_und_Abschattung_X~~\|Distanzabhängige Dämpfung und Abschattung]].		*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung\|Distanzabhängige Dämpfung und Abschattung]].
	*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 P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )

Guenter am 13. Mai 2020 um 11:49 Uhr

2020-05-13T11:49:39Z

← Nächstältere Version		Version vom 13. Mai 2020, 13:49 Uhr
Zeile 48:		Zeile 48:
	+ (\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  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical.		:then profile  $\rm A$  and profile  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical.
	*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 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 )

Guenter am 13. Mai 2020 um 11:48 Uhr

2020-05-13T11:48:53Z

← Nächstältere Version		Version vom 13. Mai 2020, 13:48 Uhr
Zeile 48:		Zeile 48:
	+ (\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  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical.		:then profile  $\rm A$  and profile  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical.
	*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 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 )

Guenter am 12. Mai 2020 um 11:59 Uhr

2020-05-12T11:59:00Z

Guenter am 11. Mai 2020 um 12:45 Uhr

2020-05-11T12:45:16Z

Guenter am 11. Mai 2020 um 12:43 Uhr

2020-05-11T12:43:08Z

Guenter am 7. Mai 2020 um 15:33 Uhr

2020-05-07T15:33:30Z

← Nächstältere Version		Version vom 7. Mai 2020, 17:33 Uhr
Zeile 2:		Zeile 2:
	}}		}}

	[[Datei:EN_Mob_A1_1.png\|right\|frame\|Path loss ~~in $\rm~~ dB$ vs. distance]]		[[Datei:EN_Mob_A1_1.png\|right\|frame\|Path loss [dB] vs. distance [m]]]
	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 (BP):		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 (BP):
	* 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:

Guenter am 7. Mai 2020 um 15:29 Uhr

2020-05-07T15:29:45Z

← Nächstältere Version		Version vom 7. Mai 2020, 17:29 Uhr
Zeile 2:		Zeile 2:
	}}		}}

	[[Datei:~~P_ID2120__Mob_A_1_1~~.png\|right\|frame\|~~Dual-Slope-Pfadverlustmodell~~]]		[[Datei:EN_Mob_A1_1.png\|right\|frame\|Path loss in $\rm dB$ vs. distance]]
	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 (BP):		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 (BP):
	* 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:

← Nächstältere Version		Version vom 12. Mai 2020, 13:59 Uhr
Zeile 3:		Zeile 3:

	[[Datei:EN_Mob_A1_1.png\|right\|frame\|Dual-slope path loss diagram]]		[[Datei:EN_Mob_A1_1.png\|right\|frame\|Dual-slope path loss diagram]]
	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  $\rm (BP)$:		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  $\rm (BP)$:
	* 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:
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$		:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$

	* 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  $\gamma_1 > \gamma_0$:
	:$$V_{\rm P}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$		:$$V_{\rm P}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$

Zeile 32:		Zeile 32:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	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:
	* $10 \cdot \lg \ G_{\rm S} = 17 \ \rm dB$  (gain of the transmit antenna),		* $10 \cdot \lg \ G_{\rm S} = 17 \ \rm dB$  (gain of the transmit antenna),
	* $10 \cdot \lg \ G_{\rm E} = -3 \ \ \rm dB$  (gain of receiving antenna – so actually a loss),		* $10 \cdot \lg \ G_{\rm E} = -3 \ \ \rm dB$  (gain of receiving antenna – so actually a loss),
Zeile 47:		Zeile 47:
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )		:$$V_{\rm P}(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  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical		:then profile  $\rm A$  and profile  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical.
	*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:		*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:
	:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right )		:$$V_{\rm 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 )
	= -30\,{\rm dB} \hspace{0.05cm}. $$		= -30\,{\rm dB} \hspace{0.05cm}. $$
Zeile 82:		Zeile 83:
	===Sample solution===		===Sample solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  You can see directly from the graphic that the profile ~~'''~~(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:
	:$$V_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$		:$$V_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$



	'''(2)'''  With the profile ~~<b>~~(B)~~</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 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)
	\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$		\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$
Zeile 93:		Zeile 94:


	'''(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
	:$$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} V_{\rm zus} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB}		:$$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 zus} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB}
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

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

	:This gives you for the received power after $d = 100\ \rm m$:		:This gives you for the received power after  $d = 100\ \rm m$:
	:$$P_{\rm E}(d = 100\,{\rm m}) = \frac{P_{\rm S} \cdot G}{K_{\rm P}} = \frac{5\,{\,} \cdot 10}{10^5}\hspace{0.15cm} \underline{= 0.5\,{\rm mW}} \hspace{0.05cm}.$$		:$$P_{\rm E}(d = 100\,{\rm m}) = \frac{P_{\rm S} \cdot G}{K_{\rm P}} = \frac{5\,{\,} \cdot 10}{10^5}\hspace{0.15cm} \underline{= 0.5\,{\rm mW}} \hspace{0.05cm}.$$

	*For profile ~~'''~~(B)~~'''~~ the received power is about $4$ times smaller:		*For profile  $\rm (B)$  the received power is about  $4$  times smaller:
	:$$P_{\rm E}(d = 100\,{\rm m}) = \frac{5\,{\rm W} \cdot 10}{10^{5.6}}\approx \frac{5\,{\rm W} \cdot 10}{4 \cdot 10^{5}}\hspace{0.15cm} \underline{= 0.125\,{\rm mW}} \hspace{0.05cm}.$$		:$$P_{\rm E}(d = 100\,{\rm m}) = \frac{5\,{\rm W} \cdot 10}{10^{5.6}}\approx \frac{5\,{\rm W} \cdot 10}{4 \cdot 10^{5}}\hspace{0.15cm} \underline{= 0.125\,{\rm mW}} \hspace{0.05cm}.$$



	'''(4)'''  Below the breakpoint $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile ~~'''~~(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 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}
	\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$		\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$



	'''(5)'''  Here the profile ~~'''~~(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 P}(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 ~~'''~~(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 P}(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}}$$		\hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$
	$$\Rightarrow \hspace{0.3cm} {\rm \~~delta~~}V_{\rm P}(d = 200\,{\rm m}) = 65.5\,{\rm dB} - 62\,{\rm dB}\hspace{0.15cm}		:$$\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}.$$		\underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$

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

← Nächstältere Version		Version vom 11. Mai 2020, 14:45 Uhr
Zeile 43:		Zeile 43:

	''Notes:''		''Notes:''
	*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%~~A4mpfung_und_Abschattung_~~\|Distanzabhängige Dämpfung und Abschattung]].		*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X\|Distanzabhängige Dämpfung und Abschattung]].
	*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 P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )
Zeile 83:		Zeile 83:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  You can see directly from the graphic that the profile '''(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 '''(A)''' with the two linear sections at the breakpoint $(d = 100 \ \rm m)$ gives the following result:
	$$V_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$		:$$V_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}} \hspace{0.05cm}.$$



	'''(2)'''  With the profile <b>(B)</b> on the other hand, using $V_0 = 10 \ \rm dB$, $\gamma_0 = 2$ and $\gamma_1 = 4$:		'''(2)'''  With the profile <b>(B)</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 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)
	\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$		\hspace{0.15cm} \underline{\approx 56\,{\rm dB}} \hspace{0.05cm}.$$

Zeile 94:		Zeile 94:

	'''(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
	$$10 \cdot {\rm lg}\hspace{0.1cm} G = 10 \cdot {\rm lg}\hspace{0.1cm} G_{\rm S} + 10 \cdot {\rm lg} \hspace{0.1cm} G_{\rm E} - 10 \cdot {\rm lg}\hspace{0.1cm} V_{\rm zus} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB}		:$$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 zus} = 17\,{\rm dB} -3\,{\rm dB} - 4\,{\rm dB} = 10\,{\rm dB}
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

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

	:This gives you for the received power after $d = 100\ \rm m$:		:This gives you for the received power after $d = 100\ \rm m$:
	$$P_{\rm E}(d = 100\,{\rm m}) = \frac{P_{\rm S} \cdot G}{K_{\rm P}} = \frac{5\,{\,} \cdot 10}{10^5}\hspace{0.15cm} \underline{= 0.5\,{\rm mW}} \hspace{0.05cm}.$$		:$$P_{\rm E}(d = 100\,{\rm m}) = \frac{P_{\rm S} \cdot G}{K_{\rm P}} = \frac{5\,{\,} \cdot 10}{10^5}\hspace{0.15cm} \underline{= 0.5\,{\rm mW}} \hspace{0.05cm}.$$

	*For profile '''(B)''' the received power is about $4$ times smaller:		*For profile '''(B)''' the received power is about $4$ times smaller:

← Nächstältere Version		Version vom 11. Mai 2020, 14:43 Uhr
Zeile 2:		Zeile 2:
	}}		}}

	[[Datei:EN_Mob_A1_1.png\|right\|frame\|~~Path~~ loss ~~[dB] vs. distance [m]~~]]		[[Datei:EN_Mob_A1_1.png\|right\|frame\|Dual-slope path loss diagram]]
	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 (BP):		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  $\rm (BP)$:
	* 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:
	$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$		:$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$

	* 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  $\gamma_1 > \gamma_0$ :
	$$V_{\rm P}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$		:$$V_{\rm P}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$

	In these equations, the variables are:		In these equations, the variables are:
Zeile 16:		Zeile 16:

	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}		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}		\Rightarrow \hspace{0.3cm}
Zeile 24:		Zeile 24:

	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 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 (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$		+ (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )\hspace{0.05cm}.$$

Zeile 43:		Zeile 43:

	''Notes:''		''Notes:''
	*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%~~A4mpfung_und_Abschattung~~\|Distanzabhängige Dämpfung und Abschattung]].		*This task belongs to the chapter  [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_\|Distanzabhängige Dämpfung und Abschattung]].
	*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 P}(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  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical		:then profile  $\rm A$  and profile  $\rm B$  for  $d ≥ d_{\rm BP}$  would be identical
	*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:		*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:
	$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right )		:$$V_{\rm 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 )
	= -30\,{\rm dB} \hspace{0.05cm}. $$		= -30\,{\rm dB} \hspace{0.05cm}. $$