Aufgaben:Exercise 1.1: Dual Slope Loss Model - Revision history

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

2021-03-23T12:37:21Z

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

Javier at 12:35, 23 March 2021

2021-03-23T12:35:26Z

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

2020-12-15T14:11:32Z

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

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

2020-12-15T13:55:07Z

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

Oezdemir at 15:33, 1 November 2020

2020-11-01T15:33:38Z

Oezdemir at 20:38, 30 July 2020

2020-07-30T20:38:10Z

Oezdemir at 20:36, 30 July 2020

2020-07-30T20:36:00Z

Javier: Text replacement - "receive power" to "received power"

2020-07-24T15:27:15Z

Text replacement - "receive power" to "received power"

Guenter at 14:07, 24 July 2020

2020-07-24T14:07:37Z

Javier at 11:49, 23 July 2020

2020-07-23T11:49:34Z

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

← Older revision		Revision as of 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^]]

@@ Line 81: / Line 81: @@
 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; You can see directly from the graphic that the profile&nbsp; $\rm (A)$&nbsp; with the two linear sections at the breakpoint&nbsp; $(d = 100 \ \rm m)$&nbsp; gives the following result:

← 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^]]

@@ Line 28: / Line 28: @@
 With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance&nbsp; $d$&nbsp; 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}.$$
@@ Line 100: / Line 100: @@
 *For the profile&nbsp; $\rm (A)$&nbsp; the following path loss occurred:
-:$$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}.$$
 :This gives you for the received power after&nbsp; $d = 100\ \rm m$:
@@ Line 111: / Line 111: @@
 '''(4)'''&nbsp; Below the breakpoint&nbsp; $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile&nbsp; $\rm (B)$:
-:$${\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}
+:$${\rm \Delta}V_{\rm S}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm}
   \underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$
@@ Line 117: / Line 117: @@
 '''(5)'''&nbsp; Here the profile&nbsp; $\rm (A)$&nbsp; with&nbsp; $V_{\rm BP} = 50 \ \rm dB$&nbsp; 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}.$$
 *On the other hand, the profile&nbsp; $\rm (B)$&nbsp; leads to the result:
-:$$V_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)
+:$$V_{\rm S}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)
   + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB}
   \hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$
@@ Line 127: / Line 127: @@
   \underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$
-*You can see that&nbsp; $\Delta V_{\rm P}$&nbsp; is almost symmetrical to&nbsp; $d = d_{\rm BP}$&nbsp; if you plot the distance&nbsp; $d$&nbsp; logarithmically as in the given graph.
+*You can see that&nbsp; $\Delta V_{\rm S}$&nbsp; is almost symmetrical to&nbsp; $d = d_{\rm BP}$&nbsp; if you plot the distance&nbsp; $d$&nbsp; logarithmically as in the given graph.
 {{ML-Fuß}}

@@ Line 5: / Line 5: @@
 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.&nbsp; This simple model is characterized by two linear sections separated by the so-called breakpoint&nbsp; $\rm (BP)$. The variable name "$V$" stands for "Verlust", which is the German word for "loss":
 * For&nbsp; $d \le d_{\rm BP}$&nbsp; and the exponent &nbsp; $\gamma_0$&nbsp; 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&nbsp; $d > d_{\rm BP}$&nbsp; we must apply the path loss exponent&nbsp; $\gamma_1$&nbsp; where&nbsp; $\gamma_1 > \gamma_0$:
-:$$V_{\rm P}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$
+:$$V_{\rm S}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$
 In these equations, the variables are:
@@ Line 24: / Line 24: @@
 The second curve is the profile &nbsp;$\rm B$&nbsp; 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}.$$
 With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance&nbsp; $d$&nbsp; according to the following equation:
-:$$P_{\rm E}(d) =  \frac{P_{\rm S} \cdot G_{\rm t} \cdot G_{\rm r} /V_{\rm tot}}{V_{\rm P}(d)}
+:$$P_{\rm E}(d) =  \frac{P_{\rm S} \cdot G_{\rm S} \cdot G_{\rm E} /V_{\rm tot}}{V_{\rm P}(d)}
   \hspace{0.05cm},\hspace{0.2cm}V_{\rm P}(d) = 10^{V_{\rm P}(d)/10}
   \hspace{0.05cm}.$$
@@ Line 45: / Line 45: @@
 *This task belongs to the chapter&nbsp; [[Mobile_Communications/Distance_dependent_attenuation_and_shading|Distance dependent attenuation and shading]].
 *If the profile &nbsp;$\rm B$&nbsp; 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 )$$
 :then profile &nbsp;$\rm A$&nbsp; and profile &nbsp;$\rm B$&nbsp; would be identical  for&nbsp; $d &#8805; d_{\rm BP}$.
 *In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $d &lt; d_{\rm BP}$, suggesting clearly too good conditions.
 *For example, &nbsp; $d = d_0 = 1 \ \ \rm m$&nbsp; with the given numerical values gives a result that is &nbsp; $40 \ \ \rm dB$&nbsp; too good:
-:$$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 )
+:$$V_{\rm S}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ( {d}/{d_0} \right ) + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right ) =10\,{\rm dB} + 2 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({1}/{100} \right )
   = -30\,{\rm dB} \hspace{0.05cm}. $$
@@ Line 56: / Line 56: @@
-===Questionnaire===
+===Questions===
 <quiz display=simple>
 {How large is the path loss $($in &nbsp;$\rm dB)$&nbsp; after&nbsp; $d= 100 \ \rm m$&nbsp; according to profile &nbsp;$\rm A$?
 |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 &nbsp;$\rm dB)$&nbsp; after&nbsp; $d= 100 \ \rm m$&nbsp; according to profile &nbsp;$\rm B$?
 |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&nbsp; $100 \ \ \rm m$&nbsp; with both profiles?
@@ Line 72: / Line 72: @@
 Profile $\text{B:}   \hspace{0.2cm} P_{\rm E}(d = 100 \ \rm m) \ = \ $ { 0.125 3% } $\ \ \rm mW$
-{How big is the deviation&nbsp; $ΔV_{\rm P}$&nbsp; between profile &nbsp;$\rm A$&nbsp; and &nbsp;$\rm B$&nbsp; at&nbsp; $d = 50 \ \rm m$?
+{How big is the deviation&nbsp; $ΔV_{\rm S}$&nbsp; between profile &nbsp;$\rm A$&nbsp; and &nbsp;$\rm B$&nbsp; at&nbsp; $d = 50 \ \rm m$?
 |type="{}"}
 $ΔV_{\rm P}(d = 50 \ \rm m) \ = \ $ { 3.5 3% } $\ \rm dB$
-{How big is the deviation&nbsp; $ΔV_{\rm P}$&nbsp; between profile &nbsp;$\rm A$&nbsp; and &nbsp;$\rm B$&nbsp; at&nbsp; $d = 200 \ \rm m$?
+{How big is the deviation&nbsp; $ΔV_{\rm S}$&nbsp; between profile &nbsp;$\rm A$&nbsp; and &nbsp;$\rm B$&nbsp; at&nbsp; $d = 200 \ \rm m$?
 |type="{}"}
 $ΔV_{\rm P}(d = 200 \ \rm m)\ = \ $ { 3.5 3% } $\ \rm dB$
 </quiz>
-===Sample solution===
+===Solutions===
 {{ML-Kopf}}
 '''(1)'''&nbsp; You can see directly from the graphic that the profile&nbsp; $\rm (A)$&nbsp; with the two linear sections at the breakpoint&nbsp; $(d = 100 \ \rm m)$&nbsp; 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)'''&nbsp; With the profile&nbsp; $\rm (B)$&nbsp; on the other hand, using&nbsp; $V_0 = 10 \ \rm dB$,&nbsp; $\gamma_0 = 2$&nbsp; and&nbsp; $\gamma_1 = 4$:
-:$$V_{\rm P}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2)
+:$$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}.$$

@@ Line 28: / Line 28: @@
 With this dual model, the entire distance course can be written in closed form, and the received power depends on the distance&nbsp; $d$&nbsp; according to the following equation:
-:$$P_{\rm t}(d) =  \frac{P_{\rm t} \cdot G_{\rm t} \cdot G_{\rm r} /V_{\rm tot}}{V_{\rm P}(d)}
+:$$P_{\rm E}(d) =  \frac{P_{\rm S} \cdot G_{\rm t} \cdot G_{\rm r} /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}.$$
-Here, all parameters are in natural units (not in dB).&nbsp; The transmit power is assumed to be&nbsp; $P_{\rm t} = 5 \ \rm W$&nbsp;.&nbsp; The other quantities have the following meanings and values:
+Here, all parameters are in natural units (not in dB).&nbsp; The transmit power is assumed to be&nbsp; $P_{\rm S} = 5 \ \rm W$&nbsp;.&nbsp; The other quantities have the following meanings and values:
-* $10 \cdot \lg \ G_{\rm t} = 17 \ \rm dB$&nbsp; (gain of the transmit antenna),
+* $10 \cdot \lg \ G_{\rm S} = 17 \ \rm dB$&nbsp; (gain of the transmit antenna),
-* $10 \cdot \lg \ G_{\rm r} = -3 \ \ \rm dB$&nbsp; (gain of receiving antenna &ndash; so actually a loss),
+* $10 \cdot \lg \ G_{\rm E} = -3 \ \ \rm dB$&nbsp; (gain of receiving antenna &ndash; so actually a loss),
 * $10 \cdot \lg \ V_{\rm tot} = 4 \ \ \rm dB$&nbsp; (loss through feeds).
@@ Line 43: / Line 43: @@
 ''Notes:''
-*This task belongs to the chapter&nbsp; [[Mobile_Communications/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_|Distanzabhängige Dämpfung und Abschattung]].
+*This task belongs to the chapter&nbsp; [[Mobile_Communications/Distance_dependent_attenuation_and_shading|Distance dependent attenuation and shading]].
 *If the profile &nbsp;$\rm B$&nbsp; were
 :$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )
   + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
-:then profile &nbsp;$\rm A$&nbsp; and profile &nbsp;$\rm B$&nbsp; for&nbsp; $d &#8805; d_{\rm BP}$&nbsp; would be identical.
+:then profile &nbsp;$\rm A$&nbsp; and profile &nbsp;$\rm B$&nbsp; would be identical  for&nbsp; $d &#8805; d_{\rm BP}$.
-*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $(d &lt; d_{\rm BP})$&nbsp; , suggesting clearly too good conditions.
+*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $d &lt; d_{\rm BP}$, suggesting clearly too good conditions.
 *For example, &nbsp; $d = d_0 = 1 \ \ \rm m$&nbsp; with the given numerical values gives a result that is &nbsp; $40 \ \ \rm dB$&nbsp; too good:
 :$$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 )
@@ Line 69: / Line 69: @@
 {What is the receive power after&nbsp; $100 \ \ \rm m$&nbsp; with both profiles?
 |type="{}"}
-Profile $\text{A:}   \hspace{0.2cm} P_{\rm r}(d = 100 \ \rm m) \ = \ $ { 0.5 3% } $\ \ \rm mW$
+Profile $\text{A:}   \hspace{0.2cm} P_{\rm E}(d = 100 \ \rm m) \ = \ $ { 0.5 3% } $\ \ \rm mW$
-Profile $\text{B:}   \hspace{0.2cm} P_{\rm r}(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&nbsp; $ΔV_{\rm P}$&nbsp; between profile &nbsp;$\rm A$&nbsp; and &nbsp;$\rm B$&nbsp; at&nbsp; $d = 50 \ \rm m$?