Aufgaben:Exercise 2.5Z: Multi-Path Scenario - Versionsgeschichte

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@@ Zeile 2: / Zeile 2: @@
 {{quiz-Header|Buchseite=Mobile Kommunikation/Das GWSSUS&ndash;Kanalmodell}}
-[[Datei:P_ID2169__Mob_Z_2_5.png|right|frame|Mobilfunk–Szenario mit drei Pfaden]]
+[[Datei:EN_Mob_A_2_5Z.png|right|frame|Mobile radio scenario with three paths]]
 In&nbsp; [[Aufgaben:Exercise_2.5:_Scatter_Function| Exercise 2.5]], a delay&ndash;Doppler function (or scatter function) was given. From this, we will calculate and interpret the other system functions. The given scatter function&nbsp; $s(\tau_0, f_{\rm D})$&nbsp; was
 :$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$

@@ Zeile 76: / Zeile 76: @@
 $d_2/d_1 \ = \ ${ 1,414 3% }
-{Indicate the distances&nbsp; $d_1$&nbsp; and&nbsp; $d_2$&nbsp;.
+{Find the distances&nbsp; $d_1$&nbsp; and&nbsp; $d_2$&nbsp;.
 |type="{}"}
 $d_1 \ = \ ${ 724 3% } $\ \rm m$

@@ Zeile 78: / Zeile 78: @@
 {Indicate the distances&nbsp; $d_1$&nbsp; and&nbsp; $d_2$&nbsp;.
 |type="{}"}
-$d_1 \ = \ ${724 3% } $\ \ \rm m$
+$d_1 \ = \ ${ 724 3% } $\ \rm m$
-$d_2 \ = \ ${ 1024 3% } $\ \ \rm m$
+$d_2 \ = \ ${ 1024 3% } $\ \rm m$
 </quiz>

@@ Zeile 129: / Zeile 129: @@
 *The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths.
 *From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:
-:$$\frac{\2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}
+:$$\frac{d_2}{d_1}  = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}
-   \hspace{0.15cm} \underline {= 1,414}
+   \hspace{0.15cm} \underline {= 1.414}
    \hspace{0.05cm}.$$
@@ Zeile 139: / Zeile 139: @@
 '''(8)'''&nbsp; From &nbsp; $d_2/d_1 = 2^{-0.5}$&nbsp; and&nbsp; $\Delta d = d_2 \, - d_1 = 300 \ \rm m$&nbsp; finally follows:
 :$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
-   d_1 = \frac{300\,{\rm m}}}{\sqrt{2} - 1}   \hspace{0.15cm} \underline {= 724\,{\rm m}}
+   d_1 = \frac{300\,{\rm m}}{\sqrt{2}  - 1}   \hspace{0.15cm} \underline {= 724\,{\rm m}}
    \hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}
    \hspace{0.05cm}. $$

@@ Zeile 99: / Zeile 99: @@
-'''(3)'''&nbsp; Correct are the <u>solutions 1 and 4</u>:
+'''(3)'''&nbsp; <u>Solutions 1 and 4</u> are correct:
-*The Doppler frequency $f_{\rm D} = 50 \ \rm Hz$ comes from the blue path, because the receiver somehow moves towards the virtual transmitter ${\rm S}_2$ (at the reflection point), although not in a direct direction.
+*The Doppler frequency $f_{\rm D} = 50 \ \rm Hz$ comes from the blue path, because the receiver moves towards the virtual transmitter ${\rm S}_2$ (i.e., towards the reflection point), although not directly. In other words, the movement of the receiver <b>reduces</b> the blue path's length.
 *The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 &ndash; E}$ is $60^\circ$:
-:$$\cos(\alpha_2) = \frac{f_{\rm D}}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2
+:$$\cos(\alpha_2) = \frac{f_{\rm D}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}}  = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2
    \hspace{0.1cm} \underline {= 60^{\circ} }
    \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Correct are the <u>statements 1 and 3</u>:
+'''(4)'''&nbsp; <u>Statements 1 and 3</u> are correct:
 *From $f_{\rm D} = \, &ndash;50 \ \rm Hz$ follows $\alpha_3 = \alpha_2 &plusmn; \pi$, so $\alpha_3 \ \underline {= 240^\circ}$.
@@ Zeile 113: / Zeile 113: @@
 '''(5)'''&nbsp; <u>All statements are correct</u>:
-*The two Dirac functions at $&plusmn; 50 \ \ \rm Hz$ have the same running time. For both durations $\tau_3 = \tau_2 = \tau_1 + \tau_0$ is valid.
+*The two Dirac functions at $&plusmn; 50 \ \ \rm Hz$ have the same delay. We have $\tau_3 = \tau_2 = \tau_1 + \tau_0$.
-*From the same transit time, however, also follows&nbsp; $d_3 = d_2$&nbsp; and with the same length also the same damping factors.
+*From the equality of the delays, however, also follows that&nbsp; $d_3 = d_2$. As both paths have the same length, their damping factors are also equal.
-'''(6)'''&nbsp; The runtime difference is $\tau_0 = 1 \ \rm &micro; s$, as shown in the equation for $s(\tau_0, f_{\rm D})$.
+'''(6)'''&nbsp; The delay difference is $\tau_0 = 1 \ \rm &micro; s$, as shown in the equation for $s(\tau_0, f_{\rm D})$.
 * This gives the difference in length:
 :$$\Delta d = \tau_0 \cdot c = 10^{&ndash;6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$
@@ Zeile 127: / Zeile 127: @@
 *Then $k_1 = K/d_1$ and $k_2 = K/d_2$.
-*The minus sign takes into account the $180^\circ$&ndash;phase rotation on the secondary paths.
+*The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths.
 *From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:
 :$$\frac{\2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}
@@ Zeile 137: / Zeile 137: @@
-'''(8)'''&nbsp; Aus&nbsp; $d_2/d_1 = 2^{-0.5}$&nbsp; and&nbsp; $\Delta d = d_2 \, - d_1 = 300 \ \rm m$&nbsp; finally follows:
+'''(8)'''&nbsp; From &nbsp; $d_2/d_1 = 2^{-0.5}$&nbsp; and&nbsp; $\Delta d = d_2 \, - d_1 = 300 \ \rm m$&nbsp; finally follows:
 :$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
    d_1 = \frac{300\,{\rm m}}}{\sqrt{2} - 1}   \hspace{0.15cm} \underline {= 724\,{\rm m}}

@@ Zeile 70: / Zeile 70: @@
 {What is the difference in time&nbsp; $\Delta d = d_2 - d_1$?
 |type="{}"}
-$\ Delta d \ = \ ${ 300 3% } $\ \ \rm m$
+$\Delta d \ = \ ${ 300 3% } $\ \ \rm m$
 {What is the relationship between&nbsp; $d_2$&nbsp; and&nbsp; $d_1$?

@@ Zeile 58: / Zeile 58: @@
 {What statements apply to the green path?
 |type="[]"}
-+ For this,&nbsp; $\tau_0 = 1 \ \ \rm &micro; s$&nbsp; and&nbsp; $f_{\rm D} = \, &ndash;50 \ \ \rm Hz$.
++ We have $\tau_0 = 1 \ \ \rm &micro; s$&nbsp; and&nbsp; $f_{\rm D} = \, &ndash;50 \ \ \rm Hz$.
-- The angle&nbsp; $\alpha_3$&nbsp; (see graphic) is&nbsp; $60^\circ$.
+- The angle&nbsp; $\alpha_3$&nbsp; (see graph) is&nbsp; $60^\circ$.
 + The angle&nbsp; $\alpha_3$&nbsp; is&nbsp; $240^\circ$.

@@ Zeile 62: / Zeile 62: @@
 + The angle&nbsp; $\alpha_3$&nbsp; is&nbsp; $240^\circ$.
-{What are the relations between the two side paths?
+{Which of the following relations hold between the two side paths?
 |type="[]"}
-+ It applies&nbsp; $d_3 = d_2$.
++ $d_3 = d_2$.
-+ It is&nbsp; $k_3 = k_2$.
++ $k_3 = k_2$.
-+ It is&nbsp; $\tau_3 = \tau_2$.
++ $\tau_3 = \tau_2$.
 {What is the difference in time&nbsp; $\Delta d = d_2 - d_1$?
@@ Zeile 94: / Zeile 94: @@
 *The speed of the receiver is then
-:$$v = \frac{f_{\rm D}}}{f_{\rm S} \cdot c = \frac{10^2\,{\rm Hz}}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}
+:$$v = \frac{f_{\rm D}}{f_{\rm S}} \cdot c = \frac{10^2\,{\rm Hz}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}
-   \hspace{0.1cm} \underline {= 54 \,{\rm km/h}
+   \hspace{0.1cm} \underline {= 54 \,{\rm km/h}}
    \hspace{0.05cm}.$$

@@ Zeile 84: / Zeile 84: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The Doppler frequency is positive for $\tau_0$. This means that the receiver is moving towards the transmitter &nbsp; &#8658; &nbsp; <u>statement 2</u>.
+'''(1)'''&nbsp; The Doppler frequency is positive for $\tau_0$. This means that the receiver is moving towards the transmitter &nbsp; &#8658; &nbsp; <u>solution 2</u> is correct.
-'''(2)'''&nbsp; The equation for the Doppler frequency is general or for the angle $\alpha = 0$.
+'''(2)'''&nbsp; The equation for the Doppler frequency is
 :$$f_{\rm D}= \frac{v}{c} \cdot f_{\rm S} \cdot \cos(\alpha)
-  \hspace{0.05cm},\hspace{0.3cm}\alpha = 0 \hspace{0.05cm}{\rm :} \hspace{0.15cm}f_{\rm D}= \frac{v}{c} \cdot f_{\rm S}\hspace{0.05cm}.$$
+  \hspace{0.05cm},$$
-*This is what you get for speed:
+*The speed of the receiver is then
 :$$v = \frac{f_{\rm D}}}{f_{\rm S} \cdot c = \frac{10^2\,{\rm Hz}}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}
    \hspace{0.1cm} \underline {= 54 \,{\rm km/h}