Aufgaben:Exercise 2.2Z: Real Two-Path Channel - Revision history

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	[[Category:Mobile Communications: Exercises\|^2.2 Multi-Path Reception in Wireless Systems^]]		[[Category:Mobile Communications: Exercises\|^2.2 Multi-Path Reception in Wireless Systems^]]
	[[de:~~Exercises~~:~~Exercise_2~~.2Z:~~_Real_Two~~-~~Path_Channel~~]]		[[de:Aufgaben:Exercise 2.2Z: Real Two-Path Channel]]

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	[[File:EN_Mob_A_2_2Z.png\|right\|frame\|Two-path scenario]]		[[File:EN_Mob_A_2_2Z.png\|right\|frame\|Two-path scenario]]
	The sketched scenario is considered in which the transmitted signal  $s(t)$  reaches the antenna of the receiver via two paths:		The sketched scenario is considered in which the transmitted signal  $s(t)$  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}.$$
	\hspace{0.05cm}.$$

	Note the following:		Note the following:
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  According to Pythagoras:		'''(1)'''  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}.$$
	\hspace{0.05cm}.$$

	*Actually, specifying such a length with an accuracy of one millimeter is not very useful and contradicts the mentality of an engineer.		*Actually, specifying such a length with an accuracy of one millimeter is not very useful and contradicts the mentality of an engineer.
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	'''(2)'''  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)'''  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:
	:$$d_2 = \sqrt{d^2 + (h_{\rm S}+ h_{\rm E})^2} = \sqrt{10^2 + (0.5+ 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10014.035\,{\rm m}}		:$$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}.$$
	\hspace{0.05cm}.$$


	'''(3)'''  Using the results from  '''(1)'''  and  '''(2)''', the length and delay differences are:		'''(3)'''  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}}\hspace{0.05cm}.$$
	\hspace{0.05cm},\hspace{1cm}
	\Delta \tau = \frac{\Delta d}{c} = \frac{2.996\,{\rm m}}{3 \cdot 10^8 \,{\rm m/s}} \hspace{0.1cm} \underline {=9.987\,{\rm ns}}
	\hspace{0.05cm}.$$


	'''(4)'''  With  $h_{\rm S} + h_{\rm E} \ll d$  the above equation can be expressed as follows:		'''(4)'''  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 ] $$
	d_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{d^2}} \approx d \cdot \left [ 1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{2d^2} \right ] $$		:$$\Rightarrow \hspace{0.3cm} \Delta d = d_2 - d_1 \approx \frac {1}{2d} \cdot \left [ (h_{\rm S}+ h_{\rm E})^2 - (h_{\rm S}- h_{\rm E})^2 \right ]= \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{d}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \Delta \tau = \frac{\Delta d}{c} \approx \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{c \cdot d}\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} \Delta d = d_2 - d_1 \approx \frac {1}{2d} \cdot \left [ (h_{\rm S}+ h_{\rm E})^2 - (h_{\rm S}- h_{\rm E})^2 \right ]
	= \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{d}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} \Delta \tau = \frac{\Delta d}{c} \approx \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{c \cdot d}
	\hspace{0.05cm}.$$

	*So the correct solution is the <u>solution 3</u>.  With the given numerical values, we have		*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}.$$
	\hspace{0.05cm}.$$

	*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  '''(3)'''  is only  $0.13\%$.
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	'''(5)'''  The path loss exponent  $\gamma = 2$  implies that the reception power  $P_{\rm E}$  decreases quadratically with distance.		'''(5)'''  The path loss exponent  $\gamma = 2$  implies that the reception power  $P_{\rm E}$  decreases quadratically with distance.
	*The signal amplitude thus decreases with  $1/d$, so for some constant  $K$  we have		*The signal amplitude thus decreases with  $1/d$, so for some constant  $K$  we have
	:$$k_1 = \frac {K}{d_1} \hspace{0.05cm},\hspace{0.2cm}\|k_2\| = \frac {K}{d_2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$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}.$$
	\frac {\|k_2\|}{k_1} = \frac {d_1}{d_2}= \frac {10011,039\,{\rm m}}{10014,035\,{\rm m}} \approx 0.99 \hspace{0.05cm}.$$

	*The two path weights thus only differ in magnitude by about  $1\%$.		*The two path weights thus only differ in magnitude by about  $1\%$.

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	[[Category:Mobile Communications: Exercises\|^2.2 Multi-Path Reception in Wireless Systems^]]		[[Category:Mobile Communications: Exercises\|^2.2 Multi-Path Reception in Wireless Systems^]]
			[[de:Exercises:Exercise_2.2Z:_Real_Two-Path_Channel]]

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

2021-03-23T12:37:21Z

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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^]]

Guenter at 15:11, 11 January 2021

2021-01-11T15:11:27Z

← Older revision		Revision as of 17:11, 11 January 2021
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  According to Pythagoras:		'''(1)'''  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}.$$		\hspace{0.05cm}.$$

	*Actually, specifying such a length with an accuracy of one millimeter is not very useful and contradicts the mentality of an engineer.		*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  '''(4)'''.



	'''(2)'''  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)'''  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:
	$$d_2 = \sqrt{d^2 + (h_{\rm S}+ h_{\rm E})^2} = \sqrt{10^2 + (0.5+ 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10014.035\,{\rm m}}		:$$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}.$$		\hspace{0.05cm}.$$


	'''(3)'''  Using the results from '''(1)''' and '''(2)''', the length and delay differences are:		'''(3)'''  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}		\hspace{0.05cm},\hspace{1cm}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	'''(4)'''  With $h_{\rm S} + h_{\rm E} \ll d$ the above equation can be expressed as follows:
			'''(4)'''  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 ] $$		d_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{d^2}} \approx d \cdot \left [ 1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{2d^2} \right ] $$
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*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>.  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}.$$		\hspace{0.05cm}.$$

	*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  '''(3)'''  is only  $0.13\%$.
	*In solutions 1 and 2, 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.		*In solution 2, there would be no propagation delay if both antennas were the same height.  This is clearly not true.



	'''(5)'''  The path loss exponent $\gamma = 2$ implies that the reception power $P_{\rm E}$ decreases quadratically with distance.		'''(5)'''  The path loss exponent  $\gamma = 2$  implies that the reception power  $P_{\rm E}$  decreases quadratically with distance.
	*The signal amplitude thus decreases with $1/d$, so for some constant $K$ we have		*The signal amplitude thus decreases with  $1/d$, so for some constant  $K$  we have
	:$$k_1 = \frac {K}{d_1} \hspace{0.05cm},\hspace{0.2cm}\|k_2\| = \frac {K}{d_2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$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}.$$		\frac {\|k_2\|}{k_1} = \frac {d_1}{d_2}= \frac {10011,039\,{\rm m}}{10014,035\,{\rm m}} \approx 0.99 \hspace{0.05cm}.$$

	*The two path weights thus only differ in magnitude by about $1\%$.		*The two path weights thus only differ in magnitude by about  $1\%$.
	*In addition, the coefficients $k_1$ and $k_2$ have different signs   ⇒   <u>Answers 1 and 3</u> are correct.		*In addition, the coefficients  $k_1$  and  $k_2$  have different signs   ⇒   <u>Answers 1 and 3</u>  are correct.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:01, 11 January 2021

2021-01-11T15:01:57Z

← Older revision		Revision as of 17:01, 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]].

Guenter at 15:01, 11 January 2021

2021-01-11T15:01:35Z

← Older revision		Revision as of 17:01, 11 January 2021
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	[[File:EN_Mob_A_2_2Z.png\|right\|frame\|Two-path scenario]]		[[File:EN_Mob_A_2_2Z.png\|right\|frame\|Two-path scenario]]
	The sketched scenario is considered in which the transmitted signal  $s(t)$  reaches the antenna of the receiver via two paths:		The sketched scenario is considered in which the transmitted signal  $s(t)$  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}.$$		\hspace{0.05cm}.$$

	Note the following:		Note the following:
	* The delays  $\tau_1$  and  $\tau_2$  of the main and secondary paths can be calculated from the path lengths  $d_1$  and  $d_2$  using the speed of light  $c = 3 \cdot 10^8 \ \rm m/s$ .		* The delays  $\tau_1$  and  $\tau_2$  of the main and the secondary paths can be calculated from the path lengths  $d_1$  and  $d_2$  using the speed of light  $c = 3 \cdot 10^8 \ \rm m/s$ .
	* The amplitude factors  $k_1$  and  $k_2$  are obtained according to the path loss model with path loss exponent  $\gamma = 2$  (free-space attenuation).		* The amplitude factors  $k_1$  and  $k_2$  are obtained according to the path loss model with path loss exponent  $\gamma = 2$  (free–space attenuation).
	* The height of the transmit antenna is  $h_{\rm S} = 500 \ \rm m$. The height of the receiving antenna is  $h_{\rm E} = 30 \ \rm m$. The antennas are separated by a distance of  $d = 10 \ \ \rm km$.		* The height of the transmit antenna is  $h_{\rm S} = 500 \ \rm m$.  The height of the receiving antenna is  $h_{\rm E} = 30 \ \rm m$.  The antennas are separated by a distance of  $d = 10 \ \ \rm km$.
	* The reflection on the secondary path causes a phase change of  $\pi$, so that the partial signals must be subtracted. This is taken into account by a negative  $k_2$ value.		* The reflection on the secondary path causes a phase change of  $\pi$, so that the partial signals must be subtracted.  This is taken into account by a negative  $k_2$ value.


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	''Note:''		''Note:''
	* This task belongs to the chapter  [[Mobile_Communications/~~Mehrwegeempfang_beim_Mobilfunk~~\| ~~Mehrwegeempfang beim Mobilfunk~~]].		* This task belongs to the chapter    [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications\| Multi–Path Reception in Mobile Communications]].


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	===Questionnaire===		===Questionnaire===
	<quiz display=simple>		<quiz display=simple>
	{Calculate the length  $d_1$  of the direct path		{Calculate the length  $d_1$  of the direct path.
	\|type="{}"}		\|type="{}"}
	$d_1 \ = \ ${ 10011 1% } $\ \ \rm m$		$d_1 \ = \ ${ 10011 1% } $\ \ \rm m$

	{Calculate the length  $d_2$  of the reflected path		{Calculate the length  $d_2$  of the reflected path.
	\|type="{}"}		\|type="{}"}
	$d_2 \ = \ ${ 10014 1% } $\ \ \rm m$		$d_2 \ = \ ${ 10014 1% } $\ \ \rm m$
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	$\Delta \tau \ = \ ${ 9,987 3% } $\ \ \rm ns$		$\Delta \tau \ = \ ${ 9,987 3% } $\ \ \rm ns$

	{What equation results for the path delay difference  $\~~delta~~ \tau$  with the approximation $\sqrt{(1 + \varepsilon)} \approx 1 + \varepsilon/2$ valid for small  $\varepsilon$ ?		{What equation results for the path delay difference  $\Delta \tau$  with the approximation $\sqrt{(1 + \varepsilon)} \approx 1 + \varepsilon/2$  valid for small  $\varepsilon$ ?
	\|type="[]"}		\|type="[]"}
	- $\Delta \tau = (h_{\rm S} \ - h_{\rm E})/d$,		- $\Delta \tau = (h_{\rm S} \ - h_{\rm E})/d$,

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

2020-12-16T13:16:40Z

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)'''  According to Pythagoras:		'''(1)'''  According to Pythagoras:

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"

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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]]

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

2020-07-09T09:13:17Z

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

← Older revision		Revision as of 11:13, 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]].