Aufgaben:Exercise 1.1: Dual Slope Loss Model - Versionsgeschichte

Tasnad am 15. Juni 2020 um 12:51 Uhr

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Guenter am 13. Mai 2020 um 11:49 Uhr

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Guenter am 11. Mai 2020 um 12:43 Uhr

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Guenter am 7. Mai 2020 um 15:33 Uhr

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Guenter am 7. Mai 2020 um 15:29 Uhr

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Javier am 25. März 2020 um 14:13 Uhr

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Javier am 25. März 2020 um 14:11 Uhr

2020-03-25T14:11:50Z

@@ Zeile 43: / Zeile 43: @@
 ''Notes:''
-*This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X|Distanzabhängige Dämpfung und Abschattung]].
+*This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung|Distanzabhängige Dämpfung und Abschattung]].
 *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 )

@@ Zeile 48: / Zeile 48: @@
   + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
 :then profile &nbsp;$\rm A$&nbsp; and profile &nbsp;$\rm B$&nbsp; for&nbsp; $d &#8805; d_{\rm BP}$&nbsp; would be identical.
-*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $(d &lt; d_{\rm BP})$&nbsp;, suggesting clearly too good conditions.
+*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $(d &lt; d_{\rm BP})$&nbsp; , suggesting clearly too good conditions.
 *For example, &nbsp; $d = d_0 = 1 \ \ \rm m$&nbsp; with the given numerical values gives a result that is &nbsp; $40 \ \ \rm dB$&nbsp; too good:
 :$$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 )

@@ Zeile 48: / Zeile 48: @@
   + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
 :then profile &nbsp;$\rm A$&nbsp; and profile &nbsp;$\rm B$&nbsp; for&nbsp; $d &#8805; d_{\rm BP}$&nbsp; would be identical.
-*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $(d &lt; d_{\rm BP})$&nbsp; , suggesting clearly too good conditions.
+*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $(d &lt; d_{\rm BP})$&nbsp;, suggesting clearly too good conditions.
 *For example, &nbsp; $d = d_0 = 1 \ \ \rm m$&nbsp; with the given numerical values gives a result that is &nbsp; $40 \ \ \rm dB$&nbsp; too good:
 :$$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 )

@@ Zeile 3: / Zeile 3: @@
 [[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&nbsp; $\rm (BP)$:
+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.&nbsp; This simple model is characterized by two linear sections separated by the so-called breakpoint&nbsp; $\rm (BP)$:
-* For&nbsp; $d \le d_{\rm BP}$&nbsp; and the exponent &nbsp; $\gamma_0$ we have:
+* For&nbsp; $d \le d_{\rm BP}$&nbsp; and the exponent &nbsp; $\gamma_0$&nbsp; we have:
 :$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_0})\hspace{0.05cm}.$$
-* For&nbsp; $d > d_{\rm BP}$&nbsp; we must apply the path loss exponent&nbsp; $\gamma_1$&nbsp; where&nbsp; $\gamma_1 > \gamma_0$&nbsp;:
+* For&nbsp; $d > d_{\rm BP}$&nbsp; we must apply the path loss exponent&nbsp; $\gamma_1$&nbsp; where&nbsp; $\gamma_1 > \gamma_0$:
 :$$V_{\rm P}(d) = V_{\rm BP} + \gamma_1 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} ({d}/{d_{\rm BP}})\hspace{0.05cm}.$$
@@ Zeile 32: / Zeile 32: @@
   \hspace{0.05cm}.$$
-Here, all parameters are in natural units (not in dB). The transmit power is assumed to be&nbsp; $P_{\rm S} = 5 \ \rm W$&nbsp;. The other quantities have the following meanings and values:
+Here, all parameters are in natural units (not in dB).&nbsp; The transmit power is assumed to be&nbsp; $P_{\rm S} = 5 \ \rm W$&nbsp;.&nbsp; The other quantities have the following meanings and values:
 * $10 \cdot \lg \ G_{\rm S} = 17 \ \rm dB$&nbsp; (gain of the transmit antenna),
 * $10 \cdot \lg \ G_{\rm E} = -3 \ \ \rm dB$&nbsp; (gain of receiving antenna &ndash; so actually a loss),
@@ Zeile 47: / Zeile 47: @@
 :$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )
   + (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left ({d}/{d_{\rm BP}} \right )$$
-:then profile &nbsp;$\rm A$&nbsp; and profile &nbsp;$\rm B$&nbsp; for&nbsp; $d &#8805; d_{\rm BP}$&nbsp; would be identical
+:then profile &nbsp;$\rm A$&nbsp; and profile &nbsp;$\rm B$&nbsp; for&nbsp; $d &#8805; d_{\rm BP}$&nbsp; would be identical.
-*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $(d &lt; d_{\rm BP})$&nbsp; , suggesting clearly too good conditions. For example, &nbsp; $d = d_0 = 1 \ \ \rm m$&nbsp; with the given numerical values gives a result that is &nbsp; $40 \ \ \rm dB$&nbsp; too good:
+*In this case, however, profile &nbsp;$\rm B$&nbsp; would be above profile &nbsp;$\rm A$&nbsp; for &nbsp; $(d &lt; d_{\rm BP})$&nbsp; , suggesting clearly too good conditions.
 :$$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}. $$
@@ Zeile 82: / Zeile 83: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; You can see directly from the graphic that the profile '''(A)''' with the two linear sections at the breakpoint $(d = 100 \ \rm m)$ gives the following result:
+'''(1)'''&nbsp; You can see directly from the graphic that the profile&nbsp; $\rm (A)$&nbsp; with the two linear sections at the breakpoint&nbsp; $(d = 100 \ \rm m)$&nbsp; gives the following result:
 :$$V_{\rm P}(d = 100\,{\rm m})\hspace{0.15cm} \underline{= 50\,{\rm dB}}  \hspace{0.05cm}.$$
-'''(2)'''&nbsp; With the profile <b>(B)</b> on the other hand, using $V_0 = 10 \ \rm dB$, $\gamma_0 = 2$ and $\gamma_1 = 4$:
+'''(2)'''&nbsp; With the profile&nbsp; $\rm (B)$&nbsp; on the other hand, using&nbsp; $V_0 = 10 \ \rm dB$,&nbsp; $\gamma_0 = 2$&nbsp; and&nbsp; $\gamma_1 = 4$:
 :$$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)
    \hspace{0.15cm} \underline{\approx 56\,{\rm dB}}  \hspace{0.05cm}.$$
@@ Zeile 93: / Zeile 94: @@
-'''(3)'''&nbsp; The antenna gains from the transmitter $(+17 \ \rm dB)$ and receiver $(-3 \ \rm dB)$ and the internal losses of the base station $(+4 \ \rm dB)$ can be combined to
+'''(3)'''&nbsp; The antenna gains from the transmitter&nbsp; $(+17 \ \rm dB)$&nbsp; and receiver&nbsp; $(-3 \ \rm dB)$&nbsp; and the internal losses of the base station&nbsp; $(+4 \ \rm dB)$&nbsp; can be combined to
 :$$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}
   \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {G = 10}
    \hspace{0.05cm}.$$
-*For the profile '''(A)''' the following path loss occurred:
+*For the profile&nbsp; $\rm (A)$&nbsp; the following path loss occurred:
 :$$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}.$$
-:This gives you for the received power after $d = 100\ \rm m$:
+:This gives you for the received power after&nbsp; $d = 100\ \rm m$:
 :$$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}.$$
-*For profile '''(B)''' the received power is about $4$ times smaller:
+*For profile&nbsp; $\rm (B)$&nbsp; the received power is about&nbsp; $4$&nbsp; times smaller:
 :$$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}.$$
-'''(4)'''&nbsp; Below the breakpoint $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile '''(B)''':
+'''(4)'''&nbsp; Below the breakpoint&nbsp; $(d < 100 \ \rm m)$, the deviation is determined by the last summand of profile&nbsp; $\rm (B)$:
-$${\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}
+:$${\rm \Delta}V_{\rm P}(d = 50\,{\rm m}) = (\gamma_1 - \gamma_0) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} \left (1 + {d}/{d_{\rm BP}} \right )= (4-2) \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (1.5)\hspace{0.15cm}
   \underline{\approx 3.5\,{\rm dB}} \hspace{0.05cm}.$$
-'''(5)'''&nbsp; Here the profile '''(A)''' with $V_{\rm BP} = 50 \ \rm dB$ gives:
+'''(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 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}.$$
-*On the other hand, the profile '''(B)''' leads to the result:
+*On the other hand, the profile&nbsp; $\rm (B)$&nbsp; leads to the result:
-$$V_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)
+:$$V_{\rm P}(d = 200\,{\rm m}) = 50\,{\rm dB} + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (200)
   + 20\,{\rm dB} \cdot {\rm lg} \hspace{0.1cm} (3) = 10\,{\rm dB} + 46\,{\rm dB} + 9.5\,{\rm dB}
   \hspace{0.15cm} {\approx 65.5\,{\rm dB}}$$
-$$\Rightarrow \hspace{0.3cm} {\rm \delta}V_{\rm P}(d = 200\,{\rm m}) = 65.5\,{\rm dB} - 62\,{\rm dB}\hspace{0.15cm}
+:$$\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}.$$
-*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&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.
 {{ML-Fuß}}

@@ Zeile 43: / Zeile 43: @@
 ''Notes:''
-*This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_|Distanzabhängige Dämpfung und Abschattung]].
+*This task belongs to the chapter&nbsp; [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung_X|Distanzabhängige Dämpfung und Abschattung]].
 *If the profile &nbsp;$\rm B$&nbsp; were
 :$$V_{\rm P}(d) = V_{\rm 0} + \gamma_0 \cdot 10\,{\rm dB} \cdot {\rm lg} \hspace{0.15cm} \left ( {d}/{d_0} \right )
@@ Zeile 83: / Zeile 83: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; You can see directly from the graphic that the profile '''(A)''' with the two linear sections at the breakpoint $(d = 100 \ \rm m)$ gives the following result:
-$$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}.$$
 '''(2)'''&nbsp; With the profile <b>(B)</b> on the other hand, using $V_0 = 10 \ \rm dB$, $\gamma_0 = 2$ and $\gamma_1 = 4$:
-$$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)
+:$$V_{\rm P}(d = 100\,{\rm m}) = 10\,{\rm dB} + 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(100)+ 20\,{\rm dB}\cdot {\rm lg} \hspace{0.1cm}(2)
    \hspace{0.15cm} \underline{\approx 56\,{\rm dB}}  \hspace{0.05cm}.$$
@@ Zeile 94: / Zeile 94: @@
 '''(3)'''&nbsp; 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}.$$
 *For the profile '''(A)''' the following path loss occurred:
-$$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}.$$
 :This gives you for the received power after $d = 100\ \rm m$:
-$$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:

@@ 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):
 * For&nbsp; $d \le d_{\rm BP}$&nbsp; and the exponent &nbsp; $\gamma_0$ we have: