Aufgaben:Exercise 2.2Z: Real Two-Path Channel - 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"

Guenter at 15:11, 11 January 2021

2021-01-11T15:11:27Z

Guenter at 15:01, 11 January 2021

2021-01-11T15:01:57Z

Guenter at 15:01, 11 January 2021

2021-01-11T15:01:35Z

Javier: Text replacement - "===Sample solution===" to "===Solution==="

2020-12-16T13:16:40Z

Text replacement - "===Sample solution===" to "===Solution==="

Javier: Text replacement - "Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk" to "Mobile_Communications/Multipath_Reception_in_Mobile_Communications"

2020-12-15T14:00:21Z

Text replacement - "Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk" to "Mobile_Communications/Multipath_Reception_in_Mobile_Communications"

Javier: Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

2020-07-09T09:13:17Z

Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

Javier at 15:41, 20 May 2020

2020-05-20T15:41:53Z

Guenter at 14:10, 11 May 2020

2020-05-11T14:10:22Z

Javier at 13:18, 15 April 2020

2020-04-15T13:18:36Z

← Older revision		Revision as of 12:37, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]	+	[[Category:Mobile Communications: Exercises\|^2.2 Multi-Path Reception in Wireless Systems^]]

@@ Line 54: / Line 54: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; According to Pythagoras:
-$$d_1 = \sqrt{d^2 + (h_{\rm S}- h_{\rm E})^2} = \sqrt{10^2 + (0.5- 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10011.039\,{\rm m}}
+:$$d_1 = \sqrt{d^2 + (h_{\rm S}- h_{\rm E})^2} = \sqrt{10^2 + (0.5- 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10011.039\,{\rm m}}
    \hspace{0.05cm}.$$
 *Actually, specifying such a length with an accuracy of one millimeter is not very useful and contradicts the mentality of an engineer.
-*We have done this anyway to be able to check the accuracy of the approximation in subtask 4 (4).
+*We have done this anyway to be able to check the accuracy of the approximation in subtask&nbsp; '''(4)'''.
-'''(2)'''&nbsp; If you fold the reflected beam on the right side of $x_{\rm R}$ downwards (reflection on the ground), you get again a right triangle. From this follows:
+'''(2)'''&nbsp; If you fold the reflected beam on the right side of&nbsp; $x_{\rm R}$&nbsp; downwards&nbsp; (reflection on the ground), you get again a right triangle.&nbsp; From this follows:
-$$d_2 = \sqrt{d^2 + (h_{\rm S}+ h_{\rm E})^2} = \sqrt{10^2 + (0.5+ 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10014.035\,{\rm m}}
+:$$d_2 = \sqrt{d^2 + (h_{\rm S}+ h_{\rm E})^2} = \sqrt{10^2 + (0.5+ 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10014.035\,{\rm m}}
    \hspace{0.05cm}.$$
-'''(3)'''&nbsp; Using the results from '''(1)''' and '''(2)''', the length and delay differences are:
+'''(3)'''&nbsp; Using the results from&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)''', the length and delay differences are:
 :$$\Delta d = d_2 - d_1 = \hspace{0.1cm} \underline {=2.996\,{\rm m}}
    \hspace{0.05cm},\hspace{1cm}
@@ Line 73: / Line 73: @@
    \hspace{0.05cm}.$$
 '''(4)'''&nbsp; With $h_{\rm S} + h_{\rm E} \ll d$ the above equation can be expressed as follows:
 :$$d_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}- h_{\rm E})^2}{d^2}} \approx  d \cdot \left [ 1 + \frac{(h_{\rm S}- h_{\rm E})^2}{2d^2} \right ] \hspace{0.05cm},\hspace{1cm}
 d_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{d^2}} \approx  d \cdot \left [ 1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{2d^2} \right ] $$
@@ Line 81: / Line 82: @@
    \hspace{0.05cm}.$$
-*So the correct solution is the <u>solution 3</u>. With the given numerical values, we have
+*So the correct solution is the <u>solution 3</u>.&nbsp; With the given numerical values, we have
 :$$\Delta \tau  \approx  \frac {2 \cdot 500\,{\rm m}\cdot 30\,{\rm m}}{3 \cdot 10^8 \,{\rm m/s} \cdot 10000\,{\rm m}} = 10^{-8}\,{\rm s} = 10\,{\rm ns}
    \hspace{0.05cm}.$$
-*The relative error with respect to the actual value according to the subtask '''(3)'' is only $0.13\%$.
+*The relative error with respect to the actual value according to the subtask&nbsp; '''(3)'''&nbsp; is only&nbsp; $0.13\%$.
 *In solutions 1 and 2, the dimensions are wrong.
-*In solution 2, there would be no propagation delay if both antennas were the same height. This is clearly not true.
+*In solution 2, there would be no propagation delay if both antennas were the same height.&nbsp; This is clearly not true.
-'''(5)'''&nbsp; The path loss exponent $\gamma = 2$ implies that the reception power $P_{\rm E}$ decreases quadratically with distance.
+'''(5)'''&nbsp; The path loss exponent&nbsp; $\gamma = 2$&nbsp; implies that the reception power&nbsp; $P_{\rm E}$&nbsp; decreases quadratically with distance.
-*The signal amplitude thus decreases with $1/d$, so for some constant $K$ we have
+*The signal amplitude thus decreases with&nbsp; $1/d$, so for some constant&nbsp; $K$&nbsp; we have
 :$$k_1 = \frac {K}{d_1} \hspace{0.05cm},\hspace{0.2cm}|k_2| = \frac {K}{d_2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
    \frac {|k_2|}{k_1} = \frac {d_1}{d_2}= \frac {10011,039\,{\rm m}}{10014,035\,{\rm m}} \approx 0.99 \hspace{0.05cm}.$$
-*The two path weights thus only differ in magnitude by about $1\%$.
+*The two path weights thus only differ in magnitude by about&nbsp; $1\%$.
-*In addition, the coefficients $k_1$ and $k_2$ have different signs &nbsp; &#8658; &nbsp; <u>Answers 1 and 3</u> are correct.
+*In addition, the coefficients&nbsp; $k_1$&nbsp; and&nbsp; $k_2$&nbsp; have different signs &nbsp; &#8658; &nbsp; <u>Answers 1 and 3</u>&nbsp; are correct.
 {{ML-Fuß}}

← Older revision		Revision as of 15:01, 11 January 2021
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	''Note:''		''Note:''
−	* This task belongs to the chapter~~ ~~   [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications\| Multi–Path Reception in Mobile Communications]].	+	* This task belongs to the chapter   [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications\| Multi–Path Reception in Mobile Communications]].

@@ Line 4: / Line 4: @@
 [[File:EN_Mob_A_2_2Z.png|right|frame|Two-path scenario]]
 The sketched scenario is considered in which the transmitted signal&nbsp; $s(t)$&nbsp; reaches the antenna of the receiver via two paths:
-$$r(t) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} r_1(t) + r_2(t) =k_1 \cdot s( t - \tau_1) + k_2 \cdot s( t - \tau_2)
+:$$r(t) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} r_1(t) + r_2(t) =k_1 \cdot s( t - \tau_1) + k_2 \cdot s( t - \tau_2)
    \hspace{0.05cm}.$$
 Note the following:
-* The delays&nbsp; $\tau_1$&nbsp; and&nbsp; $\tau_2$&nbsp; of the main and secondary paths can be calculated from the path lengths&nbsp; $d_1$&nbsp; and&nbsp; $d_2$&nbsp; using the speed of light&nbsp; $c = 3 \cdot 10^8 \ \rm m/s$&nbsp;.
+* The delays&nbsp; $\tau_1$&nbsp; and&nbsp; $\tau_2$&nbsp; of the main and the secondary paths can be calculated from the path lengths&nbsp; $d_1$&nbsp; and&nbsp; $d_2$&nbsp; using the speed of light&nbsp; $c = 3 \cdot 10^8 \ \rm m/s$&nbsp;.
-* The amplitude factors&nbsp; $k_1$&nbsp; and&nbsp; $k_2$&nbsp; are obtained according to the path loss model with path loss exponent&nbsp; $\gamma = 2$&nbsp; (free-space attenuation).
+* The amplitude factors&nbsp; $k_1$&nbsp; and&nbsp; $k_2$&nbsp; are obtained according to the path loss model with path loss exponent&nbsp; $\gamma = 2$&nbsp; (free&ndash;space attenuation).
-* The height of the transmit antenna is&nbsp; $h_{\rm S} = 500 \ \rm m$. The height of the receiving antenna is&nbsp; $h_{\rm E} = 30 \ \rm m$. The antennas are separated by a distance of&nbsp; $d = 10 \ \ \rm km$.
+* The height of the transmit antenna is&nbsp; $h_{\rm S} = 500 \ \rm m$.&nbsp; The height of the receiving antenna is&nbsp; $h_{\rm E} = 30 \ \rm m$.&nbsp; The antennas are separated by a distance of&nbsp; $d = 10 \ \ \rm km$.
-* The reflection on the secondary path causes a phase change of&nbsp; $\pi$, so that the partial signals must be subtracted. This is taken into account by a negative&nbsp; $k_2$ value.
+* The reflection on the secondary path causes a phase change of&nbsp; $\pi$, so that the partial signals must be subtracted.&nbsp; This is taken into account by a negative&nbsp; $k_2$ value.
@@ Line 18: / Line 18: @@
 ''Note:''
-* This task belongs to the chapter&nbsp; [[Mobile_Communications/Mehrwegeempfang_beim_Mobilfunk| Mehrwegeempfang beim Mobilfunk]].
+* This task belongs to the chapter&nbsp; &nbsp; [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications| Multi&ndash;Path Reception in Mobile Communications]].
@@ Line 25: / Line 25: @@
 ===Questionnaire===
 <quiz display=simple>
-{Calculate the length&nbsp; $d_1$&nbsp; of the direct path
+{Calculate the length&nbsp; $d_1$&nbsp; of the direct path.
 |type="{}"}
 $d_1 \ = \ ${ 10011 1% } $\ \ \rm m$
-{Calculate the length&nbsp; $d_2$&nbsp; of the reflected path
+{Calculate the length&nbsp; $d_2$&nbsp; of the reflected path.
 |type="{}"}
 $d_2 \ = \ ${ 10014 1% } $\ \ \rm m$
@@ Line 38: / Line 38: @@
 $\Delta \tau \ = \ ${ 9,987 3% } $\ \ \rm ns$
-{What equation results for the path delay difference&nbsp; $\delta \tau$&nbsp; with the approximation $\sqrt{(1 + \varepsilon)} \approx 1 + \varepsilon/2$ valid for small&nbsp; $\varepsilon$&nbsp;?
+{What equation results for the path delay difference&nbsp; $\Delta \tau$&nbsp; with the approximation $\sqrt{(1 + \varepsilon)} \approx 1 + \varepsilon/2$&nbsp; valid for small&nbsp; $\varepsilon$&nbsp;?
 |type="[]"}
 - $\Delta \tau = (h_{\rm S} \ - h_{\rm E})/d$,

@@ Line 51: / Line 51: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; According to Pythagoras:

← Older revision		Revision as of 14:00, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Mehrwegeempfang beim Mobilfunk~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/Multipath_Reception_in_Mobile_Communications}}

	[[File:EN_Mob_A_2_2Z.png\|right\|frame\|Two-path scenario]]		[[File:EN_Mob_A_2_2Z.png\|right\|frame\|Two-path scenario]]

← Older revision		Revision as of 09:13, 9 July 2020
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	''Note:''		''Note:''
−	* This task belongs to the chapter  [[~~Mobile_Kommunikation~~/Mehrwegeempfang_beim_Mobilfunk\| Mehrwegeempfang beim Mobilfunk]].	+	* This task belongs to the chapter  [[Mobile_Communications/Mehrwegeempfang_beim_Mobilfunk\| Mehrwegeempfang beim Mobilfunk]].

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}
-[[File:P_ID2159__Mob_Z_2_2.png|right|frame|Zweiwege–Szenario]]
+[[File:EN_Mob_A_2_2Z.png|right|frame|Two-way scenario]]
 The sketched scenario is considered in which the transmitted signal&nbsp; $s(t)$&nbsp; reaches the antenna of the receiver via two paths:
 $$r(t) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} r_1(t) + r_2(t) =k_1 \cdot s( t - \tau_1) + k_2 \cdot s( t - \tau_2)