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

Javier am 20. Mai 2020 um 15:41 Uhr

2020-05-20T15:41:53Z

Guenter am 11. Mai 2020 um 14:10 Uhr

2020-05-11T14:10:22Z

Javier am 15. April 2020 um 13:18 Uhr

2020-04-15T13:18:36Z

Javier am 15. April 2020 um 13:17 Uhr

2020-04-15T13:17:41Z

Javier am 15. April 2020 um 13:12 Uhr

2020-04-15T13:12:45Z

Javier am 15. April 2020 um 13:10 Uhr

2020-04-15T13:10:21Z

Javier am 15. April 2020 um 13:09 Uhr

2020-04-15T13:09:56Z

Javier am 15. April 2020 um 13:06 Uhr

2020-04-15T13:06:23Z

Javier am 15. April 2020 um 13:05 Uhr

2020-04-15T13:05:43Z

Javier am 15. April 2020 um 13:05 Uhr

2020-04-15T13:05:21Z

Änderungen zeigen

@@ Zeile 2: / Zeile 2: @@
 {{quiz-Header|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}
-[[Datei:EN_Mob_A_2_2Z.png|right|frame|Two-way scenario]]
+[[Datei: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)

@@ Zeile 2: / Zeile 2: @@
 {{quiz-Header|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}
-[[Datei:P_ID2159__Mob_Z_2_2.png|right|frame|Zweiwege–Szenario]]
+[[Datei: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)

@@ Zeile 86: / Zeile 86: @@
 *The relative error with respect to the actual value according to the subtask '''(3)'' is only $0.13\%$.
-*In solution 1, the dimensions are wrong.
+*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.

@@ Zeile 67: / Zeile 67: @@
-'''(3)'''&nbsp; With the results from '''(1)''' and '''(2)''' you get for the lengths&ndash; and the runtime difference:
+'''(3)'''&nbsp; Using the results from '''(1)''' and '''(2)''', the length and delay differences are:
-:$$\Delta d = d_2 - d_1 = \hspace{0.1cm} \underline {=2,996\,{\rm m}
+:$$\Delta d = d_2 - d_1 = \hspace{0.1cm} \underline {=2.996\,{\rm m}}
    \hspace{0.05cm},\hspace{1cm}
-\delta \tau = \frac{\delta d}{c} = \frac{2,996\,{\rm m}}}{3 \cdot 10^8 \,{\rm m/s}}  \hspace{0.1cm} \underline {=9,987\,{\rm ns}
+\Delta \tau = \frac{\Delta d}{c} =  \frac{2.996\,{\rm m}}{3 \cdot 10^8 \,{\rm m/s}}  \hspace{0.1cm} \underline {=9.987\,{\rm ns}}
    \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}
+:$$d_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}- h_{\rm E})^2}{d^2}} \approx  d \cdot \left [ 1 + \frac{(h_{\rm S}- h_{\rm E})^2}{2d^2} \right ] \hspace{0.05cm},\hspace{1cm}
-d_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{d^2} \approx d \cdot \left [ 1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{2d^2} \right ] $$
+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 ] $$
-$$\Rightarrow \hspace{0.3cm} \delta d = d_2 - d_1 \approx \frac {1}{2d} \cdot \left [ (h_{\rm S}+ h_{\rm E})^2 - (h_{\rm S}- h_{\rm E})^2 \right ]
+:$$\Rightarrow \hspace{0.3cm} \Delta d = d_2 - d_1 \approx \frac {1}{2d} \cdot \left [ (h_{\rm S}+ h_{\rm E})^2 - (h_{\rm S}- h_{\rm E})^2 \right ]
-  = \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{d}\hspace{0.3cm}
+  =  \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{d}\hspace{0.3cm}
-\Rightarrow \hspace{0.3cm} \delta \tau = \frac{\delta d}{c} \approx \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{c \cdot d}
+\Rightarrow \hspace{0.3cm} \Delta \tau = \frac{\Delta d}{c} \approx  \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{c \cdot d}
    \hspace{0.05cm}.$$
-*So the correct solution is the <u>solution 3</u>. With the given numerical values you get for this:
+*So the correct solution is the <u>solution 3</u>. With the given numerical values, we have
-$$\Delta \tau \approx \frac {2 \cdot 500\,{\rm m}\cdot 30\,{\rm m}}{3 \cdot 10^8 \,{\rm m/s} \cdot 10000\,{\rm m}} = 10^{-8}\,{\rm s} = 10\,{\rm ns}
+:$$\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}.$$
-*The relative falsification to the actual value according to the subtask '''(3)'' is only $0.13\%$.
+*The relative error with respect to the actual value according to the subtask '''(3)'' is only $0.13\%$.
-*In solution 1 the unit is already wrong.
+*In solution 1, the dimensions are wrong.
-*In solution 2, there would be no propagation delay if both antennas were the same height. This is certainly not true.
+*In solution 2, there would be no propagation delay if both antennas were the same height. This is clearly not true.
-'''(5)'''&nbsp; The path loss exponent $\gamma = 2$ says that the reception power $P_{\rm E}$ decreases quadratically with distance.
+'''(5)'''&nbsp; The path loss exponent $\gamma = 2$ implies that the reception power $P_{\rm E}$ decreases quadratically with distance.
-*The signal amplitude thus decreases with $1/d$, and with a constant $K$ applies:
+*The signal amplitude thus decreases with $1/d$, so for some constant $K$ we have
 :$$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 amount by about $1\%$.
+*The two path weights thus only differ in magnitude by about $1\%$.
-*However, the coefficients $k_1$ and $k_2$ have different signs &nbsp; &#8658; &nbsp; Correct are the <u>answers 1 and 3</u>.
+*In addition, the coefficients $k_1$ and $k_2$ have different signs &nbsp; &#8658; &nbsp; <u>Answers 1 and 3</u> are correct.
 {{ML-Fuß}}

@@ Zeile 53: / Zeile 53: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; According to &bdquo;Pythagoras&rdquo;:
+'''(1)'''&nbsp; According to Pythagoras:
 $$d_1 = \sqrt{d^2 + (h_{\rm S}- h_{\rm E})^2} = \sqrt{10^2 + (0.5- 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10011.039\,{\rm m}}
    \hspace{0.05cm}.$$
 *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 searched for in the subtask (4).
+*We have done this anyway to be able to check the accuracy of the approximation in subtask 4 (4).
-'''(2)'''&nbsp; If you fold the reflected beam right vpn $x_{\rm R}$ downwards (reflection on the ground), you get again a right-angled triangle. From this follows:
+'''(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:
 $$d_2 = \sqrt{d^2 + (h_{\rm S}+ h_{\rm E})^2} = \sqrt{10^2 + (0.5+ 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10014.035\,{\rm m}}
    \hspace{0.05cm}.$$

@@ Zeile 36: / Zeile 36: @@
 |type="{}"}
 $\Delta d \ = \ ${ 2,996 3% } $\ \ \rm m$
-$\ Delta \tau \ = \ ${ 9,987 3% } $\ \ \rm ns$
+$\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;?

@@ Zeile 9: / Zeile 9: @@
 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 amplitude factors&nbsp; $k_1$&nbsp; and&nbsp; $k_2$&nbsp; are to be assumed in a simplified way according to the path loss model with the 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-space attenuation).
-* The height of the transmitting antenna is&nbsp; $h_{\rm S} = 500 \ \rm m$, that of the receiving antenna&nbsp; $h_{\rm E} = 30 \ \rm m$. The antennas are at a distance of&nbsp; $d = 10 \ \ \rm km$.
+* 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 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; $k_2$&ndash;value.
+* 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.
@@ Zeile 29: / Zeile 29: @@
 $d_1 \ = \ ${ 10011 1% } $\ \ \rm m$
-{Calculate the length&nbsp; $d_2$&nbsp; of the detour path
+{Calculate the length&nbsp; $d_2$&nbsp; of the reflected path
 |type="{}"}
 $d_2 \ = \ ${ 10014 1% } $\ \ \rm m$
@@ Zeile 38: / Zeile 38: @@
 $\ Delta \tau \ = \ ${ 9,987 3% } $\ \ \rm ns$
-{What equation results for the transit time 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 $\sqrt{(1 + \varepsilon)} \approx 1 + \varepsilon/2$ valid for small&nbsp; $\varepsilon$&nbsp;?
 |type="[]"}
 - $\Delta \tau = (h_{\rm S} \ - h_{\rm E})/d$,
@@ Zeile 46: / Zeile 46: @@
 {Which statements apply for the amplitude coefficients&nbsp; $k_1$&nbsp; and&nbsp; $k_2$&nbsp;?
 |type="[]"}
-+ The coefficients&nbsp; $k_1$&nbsp; and&nbsp; $k_2$&nbsp; are almost equal in amount.
++ The coefficients&nbsp; $k_1$&nbsp; and&nbsp; $k_2$&nbsp; are almost equal in magnitude.
-- The amounts&nbsp; $|k_1|$&nbsp; and&nbsp; $|k_2|$&nbsp; differ significantly.
+- The magnitudes&nbsp; $|k_1|$&nbsp; and&nbsp; $|k_2|$&nbsp; differ significantly.
 + The coefficients&nbsp; $|k_1|$&nbsp; and&nbsp; $|k_2|$&nbsp; differ in sign.
 </quiz>