Aufgaben:Exercise 2.5Z: Multi-Path Scenario - Revision history

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

2021-03-23T12:38:15Z

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

Guenter at 10:39, 16 February 2021

2021-02-16T10:39:11Z

Guenter at 17:26, 13 January 2021

2021-01-13T17:26:13Z

Guenter at 17:06, 13 January 2021

2021-01-13T17:06:54Z

Guenter at 17:05, 13 January 2021

2021-01-13T17:05:51Z

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

2020-12-16T13:16:30Z

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

Javier: Text replacement - "Mobile Kommunikation/Das GWSSUS–Kanalmodell" to "Mobile_Communications/The_GWSSUS_Channel_Model"

2020-12-15T14:01:18Z

Text replacement - "Mobile Kommunikation/Das GWSSUS–Kanalmodell" to "Mobile_Communications/The_GWSSUS_Channel_Model"

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

2020-07-09T09:13:26Z

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

Guenter at 14:12, 11 May 2020

2020-05-11T14:12:48Z

Javier: Javier verschob die Seite Exercises:Exercise 2.5Z: Multi-Path Scenario nach Aufgaben:Exercise 2.5Z: Multi-Path Scenario

2020-04-22T12:40:47Z

Javier verschob die Seite Exercises:Exercise 2.5Z: Multi-Path Scenario nach Exercise 2.5Z: Multi-Path Scenario

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^2.3 The GWSSUS Channel Model^]]	+	[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]

@@ Line 32: / Line 32: @@
 ''Notes:''
 * This task belongs to chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model| The GWSSUS Channel Model]].
-*We also refer to the pages&nbsp; [[Mobile_Communications/Distance_Dependent_Attenuation_and_Shading#Common_path_loss_model| Common path-loss model]]&nbsp; and&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution| Doppler effect and its distribution]].
+*We also refer to the pages&nbsp; [[Mobile_Communications/Distance_Dependent_Attenuation_and_Shading#Common_path_loss_model| Common path-loss model]]&nbsp; and&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution| Doppler frequency and its distribution]].

@@ Line 84: / Line 84: @@
 ===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>solution 2</u> is correct.
+'''(1)'''&nbsp; The Doppler frequency is positive for&nbsp; $\tau_0$.&nbsp; 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
 :$$f_{\rm D}= \frac{v}{c} \cdot f_{\rm S} \cdot \cos(\alpha)
-  \hspace{0.05cm},$$
+  \hspace{0.05cm}.$$
- If the angle of incidence is $\alpha=0$, the Doppler frequency is
+*If the angle of incidence is&nbsp; $\alpha=0$, then the Doppler frequency is
-:$$f_d=\frac{v}{c}\cdot f_S$$
+:$$f_{\rm D}=\frac{v}{c}\cdot f_{\rm S}.$$
-*The speed of the receiver is then
+*In this case the speed of the receiver is
 :$$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}}
@@ Line 100: / Line 100: @@
 '''(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 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 Doppler frequency $f_{\rm D} = 50 \ \rm Hz$&nbsp; comes from the blue path, because the receiver moves towards the virtual transmitter&nbsp; ${\rm S}_2$ (i.e., towards the reflection point), although not directly.&nbsp;
-*The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 &ndash; E}$ is $60^\circ$:
+*In other words:&nbsp; The movement of the receiver <b>reduces</b> the blue path's length.
 :$$\cos(\alpha_2) = \frac{f_{\rm D}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}}  = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2
    \hspace{0.1cm} \underline {= 60^{\circ} }
@@ Line 107: / Line 108: @@
-'''(4)'''&nbsp; <u>Statements 1 and 3</u> are correct:
+'''(4)'''&nbsp; <u>Statements 1 and 3</u> are correct:&nbsp; From&nbsp; $f_{\rm D} = \, &ndash;50 \ \rm Hz$&nbsp; follows&nbsp; $\alpha_3 = \alpha_2 &plusmn; \pi$ &nbsp; &rArr; &nbsp; $\alpha_3 \ \underline {= 240^\circ}$.
 '''(5)'''&nbsp; <u>All statements are correct</u>:
-*The two Dirac functions at $&plusmn; 50 \ \ \rm Hz$ have the same delay. We have $\tau_3 = \tau_2 = \tau_1 + \tau_0$.
+*The two Dirac functions at&nbsp; $&plusmn; 50 \ \ \rm Hz$&nbsp; have the same delay.&nbsp; We have&nbsp; $\tau_3 = \tau_2 = \tau_1 + \tau_0$.
-*From the equality of the delays, however, also follows that&nbsp; $d_3 = d_2$. As both paths have the same length, their damping factors are also equal.
+*From the equality of the delays, however, it follows also&nbsp; $d_3 = d_2$.
-'''(6)'''&nbsp; The delay difference is $\tau_0 = 1 \ \rm &micro; s$, as shown in the equation for $s(\tau_0, f_{\rm D})$.
+'''(6)'''&nbsp; The delay difference is&nbsp; $\tau_0 = 1 \ \rm &micro; s$, as shown in the equation for&nbsp; $s(\tau_0, f_{\rm D})$.&nbsp; This gives the difference in length:
 :$$\Delta d = \tau_0 \cdot c = 10^{&ndash;6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$
-'''(7)'''&nbsp; The path loss exponent was assumed to be $\gamma = 2$ for this task.
+*The minus sign takes into account the&nbsp; $180^\circ$&nbsp; phase rotation on the secondary paths.
-*Then $k_1 = K/d_1$ and $k_2 = K/d_2$.
+*From the weights of the Dirac functions one can read&nbsp; $k_1 = \sqrt{0.5}$&nbsp; and&nbsp; $k_2 = -0.5$.&nbsp; Therefore:
 :$$\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{0.05cm}.$$

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 ''Notes:''
 * This task belongs to chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model| The GWSSUS Channel Model]].
-*We also refer to the pages&nbsp; [[Mobile_Communications/Distance_Dependent_Attenuation_and_Shading#Common_path_loss_model| Common path-loss model]]&nbsp; and&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution| Doppler effect]].
+*We also refer to the pages&nbsp; [[Mobile_Communications/Distance_Dependent_Attenuation_and_Shading#Common_path_loss_model| Common path-loss model]]&nbsp; and&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution| Doppler effect and its distribution]].

@@ Line 3: / Line 3: @@
 [[File: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
+In&nbsp; [[Aufgaben:Exercise_2.5:_Scatter_Function| Exercise 2.5]], a delay&ndash;Doppler function (or scatter function) was given.&nbsp; From this, you should calculate and interpret the other system functions.&nbsp; The given scatter function&nbsp; $s(\tau_0, f_{\rm D})$&nbsp; was
 :$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$
 :$$\hspace{1.5cm} \ - \ \hspace{-0.2cm} \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D} \hspace{-0.05cm}- \hspace{-0.05cm}50\,{\rm Hz}) \ - \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D}\hspace{-0.05cm} + \hspace{-0.05cm}50\,{\rm Hz})
@@ Line 9: / Line 9: @@
-''Note:'' &nbsp; In our learning tutorial,&nbsp; $s(\tau_0, \hspace{0.05cm} f_{\rm D})$&nbsp; is also identified with&nbsp; $\eta_{\rm VD}(\tau_0, \hspace{0.05cm}f_{\rm D})$&nbsp;.
+''Note:'' &nbsp; In our learning tutorial,&nbsp; $s(\tau_0, \hspace{0.05cm} f_{\rm D})$&nbsp; is also identified with&nbsp; $\eta_{\rm VD}(\tau_0, \hspace{0.05cm}f_{\rm D})$.
-Here we have replaced the delay variable&nbsp; $\tau$&nbsp; with&nbsp; $\tau_0$&nbsp;. The new variable&nbsp; $\tau_0$&nbsp; describes the difference between the delay of a path and the delay&nbsp; $\tau_1$&nbsp; of the main path. The main path is thus identified in the above equation by&nbsp; $\tau_0 = 0$&nbsp;.
+Here we have replaced the delay variable&nbsp; $\tau$&nbsp; with&nbsp; $\tau_0$&nbsp;.&nbsp; The new variable&nbsp; $\tau_0$&nbsp; describes the difference between the delay of a path and the delay&nbsp; $\tau_1$&nbsp; of the main path.&nbsp; The main path is thus identified in the above equation by&nbsp; $\tau_0 = 0$.
-Now, we try to find a mobile radio scenario in which this scatter function would actually occur. The basic structure is sketched above as a top view, and the following hold:
+Now, we try to find a mobile radio scenario in which this scatter function would actually occur.&nbsp; The basic structure is sketched above as a top view, and the following hold:
-* A single frequency is transmitted&nbsp; $f_{\rm S} = 2 \ \rm GHz$.
+* A single frequency is transmitted:&nbsp; $f_{\rm S} = 2 \ \rm GHz$.
-* The mobile receiver&nbsp; $\rm (E)$&nbsp; is represented here by a yellow dot. It is not known whether the vehicle is stationary, moving towards the transmitter&nbsp; $\rm (S)$&nbsp; or moving away from it.
+* The mobile receiver&nbsp; $\rm (E)$&nbsp; is represented here by a yellow dot.&nbsp; It is not known whether the vehicle is stationary, moving towards the transmitter&nbsp; $\rm (S)$&nbsp; or moving away.
-* The signal reaches the receiver via a main path (red) and two secondary paths (blue and green). Reflections from the obstacles cause phase shifts of&nbsp; $\pi$.
+* The signal reaches the receiver via a main path (red) and two secondary paths (blue and green).&nbsp; Reflections from the obstacles cause phase shifts of&nbsp; $\pi$.
 * ${\rm S}_2$&nbsp; and&nbsp; ${\rm S}_3$&nbsp; are to be understood here as fictitious transmitters from whose position the angles of incidence&nbsp; $\alpha_2$&nbsp; and&nbsp; $\alpha_3$&nbsp; of the secondary paths can be determined.
-* Let the signal frequency be &nbsp; $f_{\rm S}$, the angle of incidence &nbsp; $\alpha$, the velocity&nbsp; $v$&nbsp; and the velocity of light&nbsp; $c = 3 \cdot 10^8 \ \rm m/s$. Then, the Doppler frequency is
+* Let the signal frequency be&nbsp; $f_{\rm S}$,&nbsp; the angle of incidence&nbsp; $\alpha$, the velocity&nbsp; $v$&nbsp; and the velocity of light&nbsp; $c = 3 \cdot 10^8 \ \rm m/s$.&nbsp; Then, the Doppler frequency is
 :$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)
   \hspace{0.05cm}.$$
@@ Line 31: / Line 31: @@
 ''Notes:''
-* This task belongs to chapter&nbsp; [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell| Das GWSSUS&ndash;Kanalmodell]].
+* This task belongs to chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model| The GWSSUS Channel Model]].
-*We focus especially on the &nbsp; [[Mobile_Communications/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung#Gebr.C3.A4uchliches_Pfadverlustmodell| path-loss model]]&nbsp; and the &nbsp; [[Mobile_Communications/Statistische_Bindungen_innerhalb_des_Rayleigh-Prozesses#Dopplerfrequenz_und_deren_Verteilung| Doppler effect]].
+*We also refer to the pages&nbsp; [[Mobile_Communications/Distance_Dependent_Attenuation_and_Shading#Common_path_loss_model| Common path-loss model]]&nbsp; and&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution| Doppler effect]].
@@ Line 39: / Line 39: @@
 ===Questionnaire===
 <quiz display=simple>
-{At first, consider only the Dirac function at&nbsp; $\tau = 0$&nbsp; and&nbsp; $f_{\rm D} = 100 \ \rm Hz$. Which statements apply to the receiver?
+{At first, consider only the Dirac function at&nbsp; $\tau = 0$&nbsp; and&nbsp; $f_{\rm D} = 100 \ \rm Hz$.&nbsp; Which statements apply to the receiver?
 |type="()"}
 - The receiver is standing.
@@ Line 58: / Line 58: @@
 {What statements apply to the green path?
 |type="[]"}
-+ We have $\tau_0 = 1 \ \ \rm &micro; s$&nbsp; and&nbsp; $f_{\rm D} = \, &ndash;50 \ \ \rm Hz$.
++ We have $\tau_0 = 1 \  \rm &micro; s$&nbsp; and&nbsp; $f_{\rm D} =  -50 \  \rm Hz$.
 - The angle&nbsp; $\alpha_3$&nbsp; (see graph) is&nbsp; $60^\circ$.
 + The angle&nbsp; $\alpha_3$&nbsp; is&nbsp; $240^\circ$.
@@ Line 68: / Line 68: @@
 + $\tau_3 = \tau_2$.
-{What is the difference in time&nbsp; $\Delta d = d_2 - d_1$?
+{What is the difference&nbsp; $\Delta d = d_2 - d_1$&nbsp;  in time?
 |type="{}"}
 $\Delta d \ = \ ${ 300 3% } $\ \ \rm m$

← Older revision		Revision as of 14:01, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Das GWSSUS–Kanalmodell~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}

	[[File:EN_Mob_A_2_5Z.png\|right\|frame\|Mobile radio scenario with three paths]]		[[File:EN_Mob_A_2_5Z.png\|right\|frame\|Mobile radio scenario with three paths]]

← Older revision	Revision as of 12:40, 22 April 2020
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