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

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	[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]		[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]
	[[de:~~Exercises~~:~~Exercise_2~~.5Z:~~_Multi~~-~~Path_Scenario~~]]		[[de:Aufgaben:Exercise 2.5Z: Multi-Path Scenario]]

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	In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given.  From this, you should calculate and interpret the other system functions.  The given scatter function  $s(\tau_0, f_{\rm D})$  was		In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given.  From this, you should calculate and interpret the other system functions.  The given scatter function  $s(\tau_0, f_{\rm D})$  was
	:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$		:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$
	:$$\hspace{1.5cm} \ - \ \hspace{-0.2cm} \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D} \hspace{-0.05cm}- \hspace{-0.05cm}50\,{\rm Hz}) \ - \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D}\hspace{-0.05cm} + \hspace{-0.05cm}50\,{\rm Hz})		:$$\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})\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	* ${\rm S}_2$  and  ${\rm S}_3$  are to be understood here as fictitious transmitters from whose position the angles of incidence  $\alpha_2$  and  $\alpha_3$  of the secondary paths can be determined.		* ${\rm S}_2$  and  ${\rm S}_3$  are to be understood here as fictitious transmitters from whose position the angles of incidence  $\alpha_2$  and  $\alpha_3$  of the secondary paths can be determined.
	* Let the signal frequency be  $f_{\rm S}$,  the angle of incidence  $\alpha$, the velocity  $v$  and the velocity of light  $c = 3 \cdot 10^8 \ \rm m/s$.  Then, the Doppler frequency is		* Let the signal frequency be  $f_{\rm S}$,  the angle of incidence  $\alpha$, the velocity  $v$  and the velocity of light  $c = 3 \cdot 10^8 \ \rm m/s$.  Then, the Doppler frequency is
	:$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)		:$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	* The damping factors  $k_1$,  $k_2$  and  $k_3$  are inversely proportional to the path lengths  $d_1$,  $d_2$  and  $d_3$. This corresponds to the path loss exponent  $\gamma = 2$.		* The damping factors  $k_1$,  $k_2$  and  $k_3$  are inversely proportional to the path lengths  $d_1$,  $d_2$  and  $d_3$. This corresponds to the path loss exponent  $\gamma = 2$.
	*This means:   The signal power decreases quadratically with distance  $d$  and accordingly the signal amplitude decreases linearly with  $d$.		*This means:   The signal power decreases quadratically with distance  $d$  and accordingly the signal amplitude decreases linearly with  $d$.
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	'''(2)'''  The equation for the Doppler frequency is		'''(2)'''  The equation for the Doppler frequency is
	:$$f_{\rm D}= \frac{v}{c} \cdot f_{\rm S} \cdot \cos(\alpha)		:$$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  $\alpha=0$, then the Doppler frequency is		*If the angle of incidence is  $\alpha=0$, then the Doppler frequency is
	:$$f_{\rm D}=\frac{v}{c}\cdot f_{\rm S}.$$		:$$f_{\rm D}=\frac{v}{c}\cdot f_{\rm S}.$$

	*In this case the speed of the receiver is		*In this case the speed of the receiver is
	:$$v = \frac{f_{\rm D}}{f_{\rm S}} \cdot c = \frac{10^2\,{\rm Hz}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}		:$$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{0.05cm}.$$
	\hspace{0.1cm} \underline {= 54 \,{\rm km/h}}
	\hspace{0.05cm}.$$


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	*In other words:  The movement of the receiver <b>reduces</b> the blue path's length.		*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  ${\rm S_2 – E}$  is  $60^\circ$:		*The angle  $\alpha_2$  between the direction of movement and the connecting line  ${\rm S_2 – E}$  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}.$$
	\hspace{0.1cm} \underline {= 60^{\circ} }
	\hspace{0.05cm}.$$


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	*From the weights of the Dirac functions one can read  $k_1 = \sqrt{0.5}$  and  $k_2 = -0.5$.  Therefore:		*From the weights of the Dirac functions one can read  $k_1 = \sqrt{0.5}$  and  $k_2 = -0.5$.  Therefore:

	:$$\frac{d_2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}		:$$\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{0.05cm}.$$
	\hspace{0.15cm} \underline {= 1.414}
	\hspace{0.05cm}.$$


	'''(8)'''  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:		'''(8)'''  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:
	:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$\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}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}\hspace{0.05cm}. $$
	d_1 = \frac{300\,{\rm m}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}}
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}
	\hspace{0.05cm}. $$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]		[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]
			[[de:Exercises:Exercise_2.5Z:_Multi-Path_Scenario]]

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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^]]

Guenter at 10:39, 16 February 2021

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

Guenter at 17:26, 13 January 2021

2021-01-13T17:26:13Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The Doppler frequency is positive for $\tau_0$. This means that the receiver is moving towards the transmitter   ⇒   <u>solution 2</u> is correct.		'''(1)'''  The Doppler frequency is positive for  $\tau_0$.  This means that the receiver is moving towards the transmitter   ⇒   <u>solution 2</u> is correct.


	'''(2)'''  The equation for the Doppler frequency is		'''(2)'''  The equation for the Doppler frequency is
	:$$f_{\rm D}= \frac{v}{c} \cdot f_{\rm S} \cdot \cos(\alpha)		:$$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  $\alpha=0$, 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
	:$$v = \frac{f_{\rm D}}{f_{\rm S}} \cdot c = \frac{10^2\,{\rm Hz}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}		:$$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}}
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	'''(3)'''  <u>Solutions 1 and 4</u> are correct:		'''(3)'''  <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 $f_{\rm D} = 50 \ \rm Hz$  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.
	*The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 – E}$ is $60^\circ$:		*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  ${\rm S_2 – E}$  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{0.1cm} \underline {= 60^{\circ} }
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	'''(4)'''  <u>Statements 1 and 3</u> are correct:		'''(4)'''  <u>Statements 1 and 3</u> are correct:  From  $f_{\rm D} = \, –50 \ \rm Hz$  follows  $\alpha_3 = \alpha_2 ± \pi$   ⇒   $\alpha_3 \ \underline {= 240^\circ}$.
	*From $f_{\rm D} = \, –50 \ \rm Hz$ follows $\alpha_3 = \alpha_2 ± \pi$~~, so~~ $\alpha_3 \ \underline {= 240^\circ}$.



	'''(5)'''  <u>All statements are correct</u>:		'''(5)'''  <u>All statements are correct</u>:
	*The two Dirac functions at $± 50 \ \ \rm Hz$ have the same delay. We have $\tau_3 = \tau_2 = \tau_1 + \tau_0$.		*The two Dirac functions at  $± 50 \ \ \rm Hz$  have the same delay.  We have  $\tau_3 = \tau_2 = \tau_1 + \tau_0$.
	*From the equality of the delays, however, also ~~follows that~~  $d_3 = d_2$. As both paths have the same length, their damping factors are ~~also~~ equal.		*From the equality of the delays, however, it follows also  $d_3 = d_2$.
			*As both paths have the same length, their damping factors are equal, too.



	'''(6)'''  The delay difference is $\tau_0 = 1 \ \rm µ s$, as shown in the equation for $s(\tau_0, f_{\rm D})$.		'''(6)'''  The delay difference is  $\tau_0 = 1 \ \rm µ s$, as shown in the equation for  $s(\tau_0, f_{\rm D})$.  This gives the difference in length:
	* This gives the difference in length:
	:$$\Delta d = \tau_0 \cdot c = 10^{–6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$		:$$\Delta d = \tau_0 \cdot c = 10^{–6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$


			'''(7)'''  The path loss exponent was assumed to be  $\gamma = 2$  for this task.
			*Then  $k_1 = K/d_1$  and  $k_2 = K/d_2$.  The constant  $K$  is only an auxiliary variable that does not need to be considered further.

	~~'''(7)'''~~  ~~The path loss exponent was assumed to be~~ $\~~gamma = 2~~$ ~~for this task~~.		*The minus sign takes into account the  $180^\circ$  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  $k_1 = \sqrt{0.5}$  and  $k_2 = -0.5$.  Therefore:

	*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:
	:$$\frac{d_2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}		:$$\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}.$$		\hspace{0.05cm}.$$

	*The constant $K$ is only an auxiliary variable that does not need to be considered further.

Guenter at 17:06, 13 January 2021

2021-01-13T17:06:54Z

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

Guenter at 17:05, 13 January 2021

2021-01-13T17:05:51Z

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	[[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]]
	In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given. From this, ~~we will~~ calculate and interpret the other system functions. The given scatter function  $s(\tau_0, f_{\rm D})$  was		In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given.  From this, you should calculate and interpret the other system functions.  The given scatter function  $s(\tau_0, f_{\rm D})$  was
	:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$		:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$
	:$$\hspace{1.5cm} \ - \ \hspace{-0.2cm} \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D} \hspace{-0.05cm}- \hspace{-0.05cm}50\,{\rm Hz}) \ - \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D}\hspace{-0.05cm} + \hspace{-0.05cm}50\,{\rm Hz})		:$$\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})
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	''Note:''   In our learning tutorial,  $s(\tau_0, \hspace{0.05cm} f_{\rm D})$  is also identified with  $\eta_{\rm VD}(\tau_0, \hspace{0.05cm}f_{\rm D})$~~ ~~.		''Note:''   In our learning tutorial,  $s(\tau_0, \hspace{0.05cm} f_{\rm D})$  is also identified with  $\eta_{\rm VD}(\tau_0, \hspace{0.05cm}f_{\rm D})$.

	Here we have replaced the delay variable  $\tau$  with  $\tau_0$ . The new variable  $\tau_0$  describes the difference between the delay of a path and the delay  $\tau_1$  of the main path. The main path is thus identified in the above equation by  $\tau_0 = 0$~~ ~~.		Here we have replaced the delay variable  $\tau$  with  $\tau_0$ .  The new variable  $\tau_0$  describes the difference between the delay of a path and the delay  $\tau_1$  of the main path.  The main path is thus identified in the above equation by  $\tau_0 = 0$.

	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.  The basic structure is sketched above as a top view, and the following hold:
	* A single frequency is transmitted  $f_{\rm S} = 2 \ \rm GHz$.		* A single frequency is transmitted:  $f_{\rm S} = 2 \ \rm GHz$.
	* The mobile receiver  $\rm (E)$  is represented here by a yellow dot. It is not known whether the vehicle is stationary, moving towards the transmitter  $\rm (S)$  or moving away ~~from it~~.		* The mobile receiver  $\rm (E)$  is represented here by a yellow dot.  It is not known whether the vehicle is stationary, moving towards the transmitter  $\rm (S)$  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  $\pi$.		* 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  $\pi$.
	* ${\rm S}_2$  and  ${\rm S}_3$  are to be understood here as fictitious transmitters from whose position the angles of incidence  $\alpha_2$  and  $\alpha_3$  of the secondary paths can be determined.		* ${\rm S}_2$  and  ${\rm S}_3$  are to be understood here as fictitious transmitters from whose position the angles of incidence  $\alpha_2$  and  $\alpha_3$  of the secondary paths can be determined.
	* Let the signal frequency be   $f_{\rm S}$, the angle of incidence   $\alpha$, the velocity  $v$  and the velocity of light  $c = 3 \cdot 10^8 \ \rm m/s$. Then, the Doppler frequency is		* Let the signal frequency be  $f_{\rm S}$,  the angle of incidence  $\alpha$, the velocity  $v$  and the velocity of light  $c = 3 \cdot 10^8 \ \rm m/s$.  Then, the Doppler frequency is
	:$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)		:$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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	''Notes:''		''Notes:''
	* This task belongs to chapter  [[Mobile_Communications/~~Das_GWSSUS%E2%80%93Kanalmodell~~\| ~~Das~~ GWSSUS~~–Kanalmodell~~]].		* This task belongs to chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\| The GWSSUS Channel Model]].
	*We ~~focus especially on~~ the   [[Mobile_Communications/~~Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung~~#~~Gebr.C3.A4uchliches_Pfadverlustmodell~~\| path-loss model]]  and ~~the~~   [[Mobile_Communications/~~Statistische_Bindungen_innerhalb_des_Rayleigh-Prozesses~~#~~Dopplerfrequenz_und_deren_Verteilung~~\| Doppler effect]].		*We also refer to the pages  [[Mobile_Communications/Distance_Dependent_Attenuation_and_Shading#Common_path_loss_model\| Common path-loss model]]  and  [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution\| Doppler effect]].


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	===Questionnaire===		===Questionnaire===
	<quiz display=simple>		<quiz display=simple>
	{At first, consider only the Dirac function at  $\tau = 0$  and  $f_{\rm D} = 100 \ \rm Hz$. Which statements apply to the receiver?		{At first, consider only the Dirac function at  $\tau = 0$  and  $f_{\rm D} = 100 \ \rm Hz$.  Which statements apply to the receiver?
	\|type="()"}		\|type="()"}
	- The receiver is standing.		- The receiver is standing.
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	{What statements apply to the green path?		{What statements apply to the green path?
	\|type="[]"}		\|type="[]"}
	+ We have $\tau_0 = 1 \ \ \rm µ s$  and  $f_{\rm D} = ~~\, –~~50 \ \ \rm Hz$.		+ We have $\tau_0 = 1 \ \rm µ s$  and  $f_{\rm D} = -50 \ \rm Hz$.
	- The angle  $\alpha_3$  (see graph) is  $60^\circ$.		- The angle  $\alpha_3$  (see graph) is  $60^\circ$.
	+ The angle  $\alpha_3$  is  $240^\circ$.		+ The angle  $\alpha_3$  is  $240^\circ$.
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	+ $\tau_3 = \tau_2$.		+ $\tau_3 = \tau_2$.

	{What is the difference ~~in time~~  $\Delta d = d_2 - d_1$?		{What is the difference  $\Delta d = d_2 - d_1$  in time?
	\|type="{}"}		\|type="{}"}
	$\Delta d \ = \ ${ 300 3% } $\ \ \rm m$		$\Delta d \ = \ ${ 300 3% } $\ \ \rm m$

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

2020-12-16T13:16:30Z

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

← Older revision		Revision as of 15:16, 16 December 2020
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	</quiz>		</quiz>

	===~~Sample solution~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The Doppler frequency is positive for $\tau_0$. This means that the receiver is moving towards the transmitter   ⇒   <u>solution 2</u> is correct.		'''(1)'''  The Doppler frequency is positive for $\tau_0$. This means that the receiver is moving towards the transmitter   ⇒   <u>solution 2</u> is correct.

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"

← Older revision		Revision as of 16:01, 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]]