Aufgaben:Exercise 2.8: COST Delay Models - Versionsgeschichte

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	[[Category:Exercises for Mobile Communications\|^2.3 The GWSSUS Channel Model^]]		[[Category:Exercises for Mobile Communications\|^2.3 The GWSSUS Channel Model^]]
			[[en:Aufgaben:Exercise_2.8:_COST_Delay_Models]]

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Guenter am 11. Mai 2020 um 14:20 Uhr

2020-05-11T14:20:40Z

← Nächstältere Version		Version vom 11. Mai 2020, 16:20 Uhr
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	{{quiz-Header\|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}		{{quiz-Header\|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}

	[[Datei:Mob_A_2_8_version2.png\|right\|frame\|~~COST–Verzögerungsmodelle~~]]		[[Datei:Mob_A_2_8_version2.png\|right\|frame\|COST delay models]]
	On the right, four delay power density spectra are plotted logarithmically as a function of the delay time  $\tau$		On the right, four delay power density spectra are plotted logarithmically as a function of the delay time  $\tau$
	:$$10 \cdot {\rm lg}\hspace{0.15cm} ({{\it \Phi}_{\rm V}(\tau)}/{\it \Phi}_{\rm 0}) \hspace{0.05cm},$$		:$$10 \cdot {\rm lg}\hspace{0.15cm} ({{\it \Phi}_{\rm V}(\tau)}/{\it \Phi}_{\rm 0}) \hspace{0.05cm},$$
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	*Correspondingly, the percentage of power between $0$ and $5 \ \mu\rm   s$ is		*Correspondingly, the percentage of power between $0$ and $5 \ \mu\rm   s$ is
	[[File:~~P_ID2184__Mob_A_2_8f~~.png\|right\|frame\|Delay power density of the COST profiles  ${\rm BU}$  and  ${\rm HT}$]]		[[File:EN_Mob_A_2_8.png\|right\|frame\|Delay power density of the COST profiles  ${\rm BU}$  and  ${\rm HT}$]]
	:$$\frac{P_1}{P_1+ P_2} = \frac{2}{3} \hspace{0.15cm}\underline {\approx 66.7\%}\hspace{0.05cm}.$$		:$$\frac{P_1}{P_1+ P_2} = \frac{2}{3} \hspace{0.15cm}\underline {\approx 66.7\%}\hspace{0.05cm}.$$

Javier: Javier verschob die Seite Exercises:Exercise 2.8: COST Delay Models nach Aufgaben:Exercise 2.8: COST Delay Models

2020-04-23T08:38:32Z

Javier verschob die Seite Exercises:Exercise 2.8: COST Delay Models nach Aufgaben:Exercise 2.8: COST Delay Models

← Nächstältere Version		Version vom 23. April 2020, 10:38 Uhr
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Javier am 23. April 2020 um 08:37 Uhr

2020-04-23T08:37:42Z

← Nächstältere Version		Version vom 23. April 2020, 10:37 Uhr
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	'''(2)'''  In [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth\|Exercise 2.7]], it was shown that the delay spread is $T_{\rm V} =\tau_0$ when the delay power density spectrum decreases exponentially according to ${\rm e}^{-\tau/\tau_0}$. Thus the following applies:		'''(2)'''  In [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth\|Exercise 2.7]], it was shown that the delay spread is $T_{\rm V} =\tau_0$ when the delay power density spectrum decreases exponentially according to ${\rm e}^{-\tau/\tau_0}$. Thus the following applies:
	* for &bdquo;Rural Area”: $\hspace{0.4cm} T_{\rm V} \ \underline {= 0.109 \ \rm µ s}$,		* for &bdquo;Rural Area”: $\hspace{0.4cm} T_{\rm V} \ \underline {= 0.109 \ \rm µ s}$,
	* for &bdquo;Typical Urban”: $\hspace{0.4cm} T_{\rm V} \ \ \underline {= 1 \ \rm & micro; s}$.		* for &bdquo;Typical Urban”: $\hspace{0.4cm} T_{\rm V} \ \underline {= 1 \ \rm µ s}$.


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	'''(3)'''  In [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth\|Exercise 2.7]] it was also shown that for the coherence bandwidth $B_{\rm K} \approx 0.276/\tau_0$ applies. It follows:		'''(3)'''  In [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth\|Exercise 2.7]] it was also shown that for the coherence bandwidth $B_{\rm K} \approx 0.276/\tau_0$ applies. It follows:
	*$B_{\rm K} \ \underline {\approx 2500 \ \rm kHz}$ (&bdquo;Rural Area”),		*$B_{\rm K} \ \underline {\approx 2500 \ \rm kHz}$ (&bdquo;Rural Area”),
	* $B_{\rm K} \ \underline {\approx 276 \ \ \rm kHz}$ (&bdquo;Typical Urban”)		*$B_{\rm K} \ \underline {\approx 276 \ \ \rm kHz}$ (&bdquo;Typical Urban”)

Javier am 23. April 2020 um 08:36 Uhr

2020-04-23T08:36:36Z

← Nächstältere Version		Version vom 23. April 2020, 10:36 Uhr
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	${\it \Phi}_{\rm V}(\tau = 4,999 \ \rm µ s) \ = \ ${ 0.00674 3% } $\ \cdot {\it \Phi}_0$		${\it \Phi}_{\rm V}(\tau = 4,999 \ \rm µ s) \ = \ ${ 0.00674 3% } $\ \cdot {\it \Phi}_0$

	{We consider  ${\rm BU}$ again. Let $P_1$ be the power of the signal between $0$  and  $5 \ \rm µ s$, and let $P_2$ be the remaining signal power. What percentage of the total signal power is $0$  and  $5 \ \rm µ s$? \|type="{}"}		{We consider  ${\rm BU}$ again. Let $P_1$ be the power of the signal between $0$  and  $5 \ \rm µ s$, and let $P_2$ be the remaining signal power. What percentage of the total signal power is $0$  and  $5 \ \rm µ s$?
	$P_1/(P_1 + P_2) \ = \ ${ 66.7 3% } $\ \ \rm \%$		\|type="{}"}
			$P_1/(P_1 + P_2) \ = \ ${ 66.7 3% } $\ \rm \%$

	{Calculate the delay spread  $T_{\rm V}$  of the profile  ${\rm BU}$. ''Note'': The average delay is  $m_{\rm V} = E[\tau] = 2.667 \ \rm µ s$.		{Calculate the delay spread  $T_{\rm V}$  of the profile  ${\rm BU}$. ''Note'': The average delay is  $m_{\rm V} = E[\tau] = 2.667 \ \rm µ s$.

Javier am 23. April 2020 um 08:35 Uhr

2020-04-23T08:35:40Z

← Nächstältere Version		Version vom 23. April 2020, 10:35 Uhr
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	${\it \Phi}_{\rm V}(\tau = 4,999 \ \rm µ s) \ = \ ${ 0.00674 3% } $\ \cdot {\it \Phi}_0$		${\it \Phi}_{\rm V}(\tau = 4,999 \ \rm µ s) \ = \ ${ 0.00674 3% } $\ \cdot {\it \Phi}_0$

	{We consider  ${\rm BU}$ again. Let $P_1$ be the power of the signal between $0$  and  $5 \ \rm µ s$, and let $P_2$ be the remaining signal power. What percentage of the total signal power is $0$  and  $5 \ \rm µ s$? \|type="{}" }		{We consider  ${\rm BU}$ again. Let $P_1$ be the power of the signal between $0$  and  $5 \ \rm µ s$, and let $P_2$ be the remaining signal power. What percentage of the total signal power is $0$  and  $5 \ \rm µ s$? \|type="{}"}
	$P_1/(P_1 + P_2) \ = \ ${ 66.7 3% } $\ \ \rm \%$		$P_1/(P_1 + P_2) \ = \ ${ 66.7 3% } $\ \ \rm \%$

Javier am 23. April 2020 um 08:34 Uhr

2020-04-23T08:34:52Z

← Nächstältere Version		Version vom 23. April 2020, 10:34 Uhr
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	{Specify the parameter  $\tau_0$  of the delay power density spectrum for the profiles  ${\rm RA}$  and  ${\rm TU}$ .		{Specify the parameter  $\tau_0$  of the delay power density spectrum for the profiles  ${\rm RA}$  and  ${\rm TU}$ .
	\|type="{}"}		\|type="{}"}
	${\rm RA} \text \ \hspace{0.4cm} \tau_0 \ = \ ${ 0.109 3% } $\ \rm µ s$		${\rm RA} \text{:} \ \hspace{0.4cm} \tau_0 \ = \ ${ 0.109 3% } $\ \rm µ s$
	${\rm TU} \text \ \hspace{0.4cm} \tau_0 \ = \ ${ 1 3% } $\ \rm µ s$		${\rm TU} \text{:} \ \hspace{0.4cm} \tau_0 \ = \ ${ 1 3% } $\ \rm µ s$

	{How large is the delay spread  $T_{\rm V}$  of these channels?		{How large is the delay spread  $T_{\rm V}$  of these channels?
	\|type="{}"}		\|type="{}"}
	${\rm RA} \text \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 0.109 3% } $\ \rm µ s$		${\rm RA} \text{:} \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 0.109 3% } $\ \rm µ s$
	${\rm TU} \text \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 1 3% } $\ \rm µ s$		${\rm TU} \text{:} \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 1 3% } $\ \rm µ s$

	{What is the coherence bandwidth  $B_{\rm K}$  of these channels?		{What is the coherence bandwidth  $B_{\rm K}$  of these channels?

Javier am 23. April 2020 um 08:33 Uhr

2020-04-23T08:33:52Z

← Nächstältere Version		Version vom 23. April 2020, 10:33 Uhr
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	===Questionnaire==		===Questionnaire===
	<quiz display=simple>		<quiz display=simple>
	{Specify the parameter  $\tau_0$  of the delay power density spectrum for the profiles  ${\rm RA}$  and  ${\rm TU}$ .		{Specify the parameter  $\tau_0$  of the delay power density spectrum for the profiles  ${\rm RA}$  and  ${\rm TU}$ .
Zeile 65:		Zeile 65:
	{What is the coherence bandwidth  $B_{\rm K}$  of these channels?		{What is the coherence bandwidth  $B_{\rm K}$  of these channels?
	\|type="{}"}		\|type="{}"}
	${\rm RA} \text \ \hspace{0.4cm} B_{\rm K} \ = \ ${ 2500 3% } $\ \ \rm kHz$		${\rm RA} \text{:} \ \hspace{0.4cm} B_{\rm K} \ = \ ${ 2500 3% } $\ \ \rm kHz$
	${\rm TU} \text \ \hspace{0.4cm} B_{\rm K} \ = \ ${ 276 3% } $\ \ \rm kHz$		${\rm TU} \text{:} \ \hspace{0.4cm} B_{\rm K} \ = \ ${ 276 3% } $\ \ \rm kHz$

	{For which channel does frequency selectivity play a greater role?		{For which channel does frequency selectivity play a greater role?
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	$P_1/(P_1 + P_2) \ = \ ${ 66.7 3% } $\ \ \rm \%$		$P_1/(P_1 + P_2) \ = \ ${ 66.7 3% } $\ \ \rm \%$

	{Calculate the delay spread  $T_{\rm V}$  of the profile  ${\rm BU}$. ''Note'': The average delay is  $m_{\rm V} = E[\~~rm V} = E[\rm BU~~] = 2,667 \ \rm µ s$.		{Calculate the delay spread  $T_{\rm V}$  of the profile  ${\rm BU}$. ''Note'': The average delay is  $m_{\rm V} = E[\tau] = 2.667 \ \rm µ s$.
	\|type="{}"}		\|type="{}"}
	$T_{\rm V} \ = \ ${ 2.56 3% } $\ \rm µ s$		$T_{\rm V} \ = \ ${ 2.56 3% } $\ \rm µ s$

Javier am 23. April 2020 um 08:32 Uhr

2020-04-23T08:32:51Z

← Nächstältere Version		Version vom 23. April 2020, 10:32 Uhr
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	The upper sketch contains the two profiles  ${\rm RA}$  (<i>Rural Area</i>) and  ${\rm TU}$  (<i>Typical Urban</i>). Both of these are exponential:		The upper sketch contains the two profiles  ${\rm RA}$  (<i>Rural Area</i>) and  ${\rm TU}$  (<i>Typical Urban</i>). Both of these are exponential:
	$${{{\it \Phi}_{\rm V}(\tau)}/{\it \Phi}_{\rm 0} = {\rm e}^{ -\tau / \tau_0} \hspace{0.05cm}.$$		:$${{\it \Phi}_{\rm V}(\tau)}/{\it \Phi}_{\rm 0} = {\rm e}^{ -\tau / \tau_0} \hspace{0.05cm}.$$

	The value of the parameter  $\tau_0$  (time constant of the ACF) should be determined from the graphic in subtask '''(1)'''. Note the specified values of   $\tau_{-30}$ for  ${\it \Phi}_{\rm V}(\tau_{-30})=-30 \ \rm dB$:		The value of the parameter  $\tau_0$  (time constant of the ACF) should be determined from the graphic in subtask '''(1)'''. Note the specified values of   $\tau_{-30}$ for  ${\it \Phi}_{\rm V}(\tau_{-30})=-30 \ \rm dB$:
Zeile 45:		Zeile 45:
	:$$\frac{1}{\tau_0} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\rm e}^{ -\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = 1		:$$\frac{1}{\tau_0} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\rm e}^{ -\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = 1
	\hspace{0.05cm},\hspace{0.6cm}		\hspace{0.05cm},\hspace{0.6cm}
	\frac {1}{\~~frac_0~~} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\tau} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = \tau_0		\frac{1}{\tau_0} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\tau} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = \tau_0
	\hspace{0.05cm},\hspace{0.6cm}		\hspace{0.05cm},\hspace{0.6cm}
	\frac {1}{\~~frac_0~~} \cdot \int_{0}^{\infty} \hspace{-0.15cm}{\tau^2} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = 2\tau_0^2\hspace{0.05cm}.$$		\frac{1}{\tau_0} \cdot \int_{0}^{\infty} \hspace{-0.15cm}{\tau^2} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = 2\tau_0^2\hspace{0.05cm}.$$