Aufgaben:Exercise 2.5: Scatter Function - Revision history

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2026-03-16T13:47:52Z

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	The delay–Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$  is shown in the plot:		The delay–Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$  is shown in the plot:
	:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})-$$		:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})-$$
	:$$\hspace{1.75cm} \ - \ \hspace{-0.1cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})-		:$$\hspace{1.75cm} \ - \ \hspace{-0.1cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})-\frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz})\hspace{0.05cm}.$$
	\frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz})
	\hspace{0.05cm}.$$

	In the literature,  $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$  is often also called  '''scatter function'''  and denoted with  $s(\tau, \hspace{0.05cm}f_{\rm D})$ .		In the literature,  $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$  is often also called  '''scatter function'''  and denoted with  $s(\tau, \hspace{0.05cm}f_{\rm D})$ .
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The time-variant impulse response  $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$  is the inverse Fourier transform of the delay–Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:		'''(1)'''  The time-variant impulse response  $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$  is the inverse Fourier transform of the delay–Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:
	:$$\eta_{\rm VZ}(\tau, \hspace{0.05cm} t)		:$$\eta_{\rm VZ}(\tau, \hspace{0.05cm} t)\hspace{0.2cm} \stackrel{t, \hspace{0.02cm}f_{\rm D}}{\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.2cm} \eta_{\rm VD}(\tau, f_{\rm D})\hspace{0.05cm}.$$
	\hspace{0.2cm} \stackrel{t, \hspace{0.02cm}f_{\rm D}}{\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.2cm} \eta_{\rm VD}(\tau, f_{\rm D})\hspace{0.05cm}.$$

	*Accordingly,  $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)=0$  for the values of  $\tau$  that make  $\eta_{\rm VD}(\tau, f_{\rm D})=0$.		*Accordingly,  $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)=0$  for the values of  $\tau$  that make  $\eta_{\rm VD}(\tau, f_{\rm D})=0$.
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	:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$$		:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$$
	:$$\hspace{2cm}+\hspace{0.22cm}\frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})+ $$		:$$\hspace{2cm}+\hspace{0.22cm}\frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})+ $$
	:$$\hspace{2cm}+\hspace{0.22cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz})		:$$\hspace{2cm}+\hspace{0.22cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz})\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Comparison with the equation in this task shows that only the signs of the Diracs have changed at  $\tau = 1 \ \rm µ s$.		*Comparison with the equation in this task shows that only the signs of the Diracs have changed at  $\tau = 1 \ \rm µ s$.

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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_2.5:_Scatter_Function]]

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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 12:57, 14 January 2021

2021-01-14T12:57:58Z

← Older revision		Revision as of 14:57, 14 January 2021
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	''Notes:''		''Notes:''
	* This ~~task~~ should clarify the subject matter of the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\| The GWSSUS Channel Model]].		* This exercise should clarify the subject matter of the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\| The GWSSUS Channel Model]].
	* The relationship between the individual system functions is given in the  [[Mobile_Communications/The_GWSSUS_Channel_Model#Generalized_system_functions_of_time_variant_systems\|graph on the first page]]  of this chapter.		* The relationship between the individual system functions is given in the  [[Mobile_Communications/The_GWSSUS_Channel_Model#Generalized_system_functions_of_time_variant_systems\|graph on the first page]]  of this chapter.
	*Note that the magnitude function  $\|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})\|$  is shown above, so negative weights of the Dirac functions cannot be recognized.		*Note that the magnitude function  $\|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})\|$  is shown above, so negative weights of the Dirac functions cannot be recognized.

Guenter at 16:42, 13 January 2021

2021-01-13T16:42:26Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The time-variant impulse response $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$ is the inverse Fourier transform of the delay–Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:		'''(1)'''  The time-variant impulse response  $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$  is the inverse Fourier transform of the delay–Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:
	:$$\eta_{\rm VZ}(\tau, \hspace{0.05cm} t)		:$$\eta_{\rm VZ}(\tau, \hspace{0.05cm} t)
	\hspace{0.2cm} \stackrel{t, \hspace{0.02cm}f_{\rm D}}{\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.2cm} \eta_{\rm VD}(\tau, f_{\rm D})\hspace{0.05cm}.$$		\hspace{0.2cm} \stackrel{t, \hspace{0.02cm}f_{\rm D}}{\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.2cm} \eta_{\rm VD}(\tau, f_{\rm D})\hspace{0.05cm}.$$

			*Accordingly,  $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)=0$  for the values of  $\tau$  that make  $\eta_{\rm VD}(\tau, f_{\rm D})=0$.
			*Correct are therefore the <u>solutions 1 and 2</u>:  <br>Only for  $\tau = 0$  and  $\tau = 1 \ \ \rm \mu s$  does the time-variant impulse response have non-zero values.

	*Accordingly, $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)=0$ for the values of $\tau$ that make $\eta_{\rm VD}(\tau, f_{\rm D})=0$.
	*The <u>solutions 1 and 2</u> are therefore correct: Only for $\tau = 0$ and $\tau = 1 \ \ \rm \mu s$ does the time variant impulse response have nonzero values.


			'''(2)'''  For the delay  $\tau = 0$, the scatter function  $\eta_{\rm VD}$  consists of a single Dirac at $f_{\rm D} = 100 \ \rm Hz$.
	'''(2)'''  For the delay $\tau = 0$, the scatter function ($\eta_{\rm VD}$) consists of a single Dirac at $f_{\rm D} = 100 \ \rm Hz$.
	*According to the second Fourier integral, the desired time-domain function satisfies:		*According to the second Fourier integral, the desired time-domain function satisfies:
	:$$\eta_{\rm VZ}(\tau = 0, t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \int\limits_{-\infty}^{+\infty} \delta (f_{\rm D} - 100\,{\rm Hz}) \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi f_{\rm D} t}\hspace{0.15cm}{\rm d}f_{\rm D} =\frac{1}{\sqrt{2}} \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi t \hspace{0.05cm}\cdot \hspace{0.05cm}100\,{\rm Hz}} .$$		:$$\eta_{\rm VZ}(\tau = 0, t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \int\limits_{-\infty}^{+\infty} \delta (f_{\rm D} - 100\,{\rm Hz}) \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi f_{\rm D} t}\hspace{0.15cm}{\rm d}f_{\rm D} =\frac{1}{\sqrt{2}} \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi t \hspace{0.05cm}\cdot \hspace{0.05cm}100\,{\rm Hz}} .$$

	*~~The correct solution~~ is ~~therefore~~ <u>solution 1</u>.		*Correct is <u>solution 1</u>.



	'''(3)'''  For the delay $\tau = 1 \ \ \rm µ s$ the delay–Doppler function consists of two Dirac functions at $±50 \ \rm Hz$, each with weight $-0.5$.		'''(3)'''  For the delay  $\tau = 1 \ \ \rm µ s$  the delay–Doppler function consists of two Dirac functions at  $±50 \ \rm Hz$, each with weight  $-0.5$.
	*The time function is ~~then~~		*The time function is  $\eta_{\rm VZ}(\tau = 1\,{\rm \mu s}, t) = - \cos( 2 \pi t \cdot 50\,{\rm Hz})\hspace{0.05cm}.$
	:$$\eta_{\rm VZ}(\tau = 1\,{\rm \mu s}, t) = - \cos( 2 \pi t \cdot 50\,{\rm Hz})\hspace{0.05cm}.$$

	*This function can be represented with $A = -1$ and $f_0 = 50 \ \rm Hz$ according to <u>solution 2</u>.		*This function can be represented with  $A = -1$  and  $f_0 = 50 \ \rm Hz$  according to <u>solution 2</u>.



	'''(4)'''  The three Dirac functions $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ are at the Doppler frequencies $+100 \ \rm Hz$, $+50 \ \rm Hz$ and $-50 \ \rm Hz$.		'''(4)'''  The three Dirac functions  $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$  are at the Doppler frequencies  $+100 \ \rm Hz$, $+50 \ \rm Hz$  and  $-50 \ \rm Hz$.
	*For all other Doppler frequencies~~, therefore~~, we must have $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}) \equiv 0$.		*For all other Doppler frequencies, we must have  $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}) \equiv 0$.
	*<u>Solution 2</u> is ~~therefore~~ correct.		*<u>Solution 2</u> is correct.



	'''(5)'''  If ~~one looks~~ at the scatter function $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ in the direction of the $\tau$–axis, there is one Dirac function at each of the Doppler frequencies $100 \ \rm Hz$ and $±50 \ \rm Hz$.		'''(5)'''  If you look at the scatter function  $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$  in the direction of the  $\tau$–axis, there is one Dirac function at each of the Doppler frequencies  $100 \ \rm Hz$  and  $±50 \ \rm Hz$.
	*Here, depending on $f$, complex exponential oscillations with constant magnitude result in each case (from which it follows that ~~the~~ <u>solution 1</u> is correct):		*Here, depending on $f$,  complex exponential oscillations with constant magnitude result in each case  (from which it follows that <u>solution 1</u> is correct):
	:$$\|\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 100\,{\rm Hz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{\sqrt{2}} = {\rm const.}$$		:$$\|\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 100\,{\rm Hz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{\sqrt{2}} = {\rm const.}$$
	:$$\| \eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}= \pm 50\,{\rm Hz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5 = {\rm const.}$$		:$$\| \eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}= \pm 50\,{\rm Hz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5 = {\rm const.}$$
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	[[File:P_ID2168__Mob_A_2_5e_neu.png\|right\|frame\|Relationships between all system functions]]		[[File:P_ID2168__Mob_A_2_5e_neu.png\|right\|frame\|Relationships between all system functions]]
	'''(6)'''  As can be seen from the given [[Mobile_Communications/~~Das_GWSSUS%E2%80%93Kanalmodell~~#~~Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme~~\|graph]], <u>solutions 2 and 3</u> are correct.		'''(6)'''  As can be seen from the given [[Mobile_Communications/The_GWSSUS_Channel_Model#Generalized_system_functions_of_time_variant_systems\|graph]], <u>solutions 2 and 3</u> are correct.

	*The graph shows all system functions.		*The graph shows all system functions.
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	''Note:''		''Note:''

	Compare the time-variant transfer function $\|\eta_{\rm FZ}(f, \hspace{0.05cm} t)\|$ in the bottom right figure with the corresponding graph ~~for~~ [[Aufgaben:Exercise_2.4:_2-D_Transfer_Function\| Exercise 2.4]]:		Compare the time-variant transfer function  $\|\eta_{\rm FZ}(f, \hspace{0.05cm} t)\|$  in the bottom right figure with the corresponding graph in  [[Aufgaben:Exercise_2.4:_2-D_Transfer_Function\| Exercise 2.4]]:
	*The respective magnitude functions differ significantly, although $\|\eta_{\rm VZ}(\tau, t)\|$ is the same in both cases.		*The respective magnitude functions differ significantly, although  $\|\eta_{\rm VZ}(\tau, t)\|$  is the same in both cases.
	*In Exercise 2.4, a cosine was implicitly assumed for $\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, t)$; here we have a negative cosine function.		*In Exercise 2.4, a cosine was implicitly assumed for  $\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, t)$;  here we have a negative cosine function.
	*The (not explicitly) specified delay–Doppler function for Exercise 2.4 was		*The (not explicitly) specified delay–Doppler function for Exercise 2.4 was
	:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$$		:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$$
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*Comparison with the equation in this task shows that only the signs of the Diracs have changed at $\tau = 1 \ \rm µ s$.		*Comparison with the equation in this task shows that only the signs of the Diracs have changed at  $\tau = 1 \ \rm µ s$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 16:18, 13 January 2021

2021-01-13T16:18:48Z

← Older revision		Revision as of 18:18, 13 January 2021
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	[[File:P_ID2164__Mob_A_2_5.png\|right\|frame\|Delay-Doppler profile]]		[[File:P_ID2164__Mob_A_2_5.png\|right\|frame\|Delay-Doppler profile]]
	For the mobile radio channel as a time-variant system, there are a total of four system functions that are linked with each other via the Fourier transform. With the nomenclature from our ~~learning~~ tutorial, these are:		For the mobile radio channel as a time-variant system, there are a total of four system functions that are linked with each other via the Fourier transform.  With the nomenclature from our tutorial, these are:
	* the time-variant impulse response  $h(\tau, \hspace{0.05cm}t)$, which we also denote here as  $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$~~ ~~,		* the time-variant impulse response  $h(\tau, \hspace{0.05cm}t)$, which we also denote here as  $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$,
	* the delay-Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$,		* the delay-Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$,
	* the frequency-Doppler function  $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$,		* the frequency-Doppler function  $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$,
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	The ~~indices represent the delay (<b>V</b>) $\tau$, the time (<b>Z</b>)~~   $~~t$, the frequency (<b>F</b>)  $f~~$  ~~and the Doppler frequency (~~<b>~~D</b>)  $f_{\rm D}$.~~		The four possible system functions are uniformly denoted by  $\boldsymbol{\eta}_{12}$ .<br>

	The delay–Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$  is shown in the ~~top~~ plot:		*The first subindex is either a  $\boldsymbol{\rm V}$  $($because of German  $\rm V\hspace{-0.05cm}$erzögerung   ⇒   delay time  $\tau)$  or  a  $\boldsymbol{\rm F}$  $($frequency  $f)$.<br>

			*Either a  $\boldsymbol{\rm Z}$  $($because of German  $\rm Z\hspace{-0.05cm}$eit   ⇒   time  $t)$  or a  $\boldsymbol{\rm D}$  $($Doppler frequency  $f_{\rm D})$  is possible as the second subindex.


			The delay–Doppler function  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$  is shown in the plot:
	:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})-$$		:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})-$$
	:$$\hspace{1.75cm} \ - \ \hspace{-0.1cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})-		:$$\hspace{1.75cm} \ - \ \hspace{-0.1cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})-
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In the literature,  $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$  is often also called ~~<i>~~scatter function~~</i>~~ and denoted with  $s(\tau, \hspace{0.05cm}f_{\rm D})$ .		In the literature,  $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$  is often also called  '''scatter function'''  and denoted with  $s(\tau, \hspace{0.05cm}f_{\rm D})$ .

	In this task, the associated delay–time function  $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$  and the frequency–Doppler function  $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$  are to be determined.		In this task, the associated delay–time function  $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$  and the frequency–Doppler function  $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$  are to be determined.






	''Notes:''		''Notes:''
	* This task should clarify the subject matter of the chapter  [[Mobile_Communications/~~Das_GWSSUS%E2%80%93Kanalmodell~~\| ~~Das~~ GWSSUS~~–Kanalmodell~~]].		* This task should clarify the subject matter of the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\| The GWSSUS Channel Model]].
	* The relationship between the individual system functions is given in the  [[Mobile_Communications/~~Das_GWSSUS%E2%80%93Kanalmodell~~#~~Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme~~\|graph on the first page]]  of this chapter.		* The relationship between the individual system functions is given in the  [[Mobile_Communications/The_GWSSUS_Channel_Model#Generalized_system_functions_of_time_variant_systems\|graph on the first page]]  of this chapter.
	*Note that the magnitude function  $\|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})\|$  is shown above, so negative weights of the Dirac functions cannot be recognized.		*Note that the magnitude function  $\|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})\|$  is shown above, so negative weights of the Dirac functions cannot be recognized.

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	===Questionnaire===		===Questionnaire===
	<quiz display=simple>		<quiz display=simple>
	{At which values of   $\tau$ are the components of 2D impulse response  $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$ ?		{At which values of  $\tau$  there are the components of 2D impulse response  $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$ ?
	\|type="[]"}		\|type="[]"}
	+ $\tau = 0$,		+ $\tau = 0$,
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	- other $\tau$–values.		- other $\tau$–values.

	{Calculate  $\|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm}t)\|$. Which of the following statements are true?		{Calculate  $\|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm}t)\|$.  Which of the following statements are true?
	\|type="()"}		\|type="()"}
	+ $\|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm} t)\|$  is independent of  $t$.		+ $\|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm} t)\|$  is independent of  $t$.
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	-   $\eta_{\rm VZ}(\tau = 0, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.		-   $\eta_{\rm VZ}(\tau = 0, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.

	{Calculate  $\|\eta_{\rm VZ}(\tau = 1 \ {\rm µ s},\hspace{0.05cm} t)\|$. Which of the following statements are true?		{Calculate  $\|\eta_{\rm VZ}(\tau = 1 \ {\rm µ s},\hspace{0.05cm} t)\|$.  Which of the following statements are true?
	\|type="()"}		\|type="()"}
	- $\|\eta_{\rm VZ}(\tau = 1 \ {\rm µ s},\hspace{0.05cm} t)\|$  is independent of  $t$.		- $\|\eta_{\rm VZ}(\tau = 1 \ {\rm µ s},\hspace{0.05cm} t)\|$  is independent of  $t$.
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	-   $\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.		-   $\eta_{\rm VZ}(\tau = 1 \ {\rm µ s}, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.

	{Consider ~~now~~ the frequency–Doppler representation  $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$. For which values of   $f_{\rm D}$ is this function <b>not</b> equal to zero?		{Consider the frequency–Doppler representation  $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$.  For which values of  $f_{\rm D}$ is this function <b>not</b> equal to zero?
	\|type="[]"}		\|type="[]"}
	- $f_{\rm D} = 0$,		- $f_{\rm D} = 0$,
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	{Which of the following statements are true for  $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$?		{Which of the following statements are true for  $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$?
	\|type="()"}		\|type="()"}
	+ $\|\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D} = 100 \ \rm Hz)\|$  is independent of~~ ~~ $f_{\rm D}$.		+ $\|\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D} = 100 \ \rm Hz)\|$  is independent of $f_{\rm D}$.
	-   $\eta_{\rm FD}(f, \hspace{0.05cm} f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \cos {(2\pi t_0 f)}$.		-   $\eta_{\rm FD}(f, \hspace{0.05cm} f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \cos {(2\pi t_0 f)}$.
	-   $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \sin {(2\pi t_0 f)}$.		-   $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \sin {(2\pi t_0 f)}$.
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	{How do you get the time-variant transfer function  $\eta_{\rm FZ}(f, \hspace{0.05cm}t)$?		{How do you get the time-variant transfer function  $\eta_{\rm FZ}(f, \hspace{0.05cm}t)$?
	\|type="[]"}		\|type="[]"}
	- By Fourier transformation of  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$  with respect to   $\tau$.		- By Fourier transformation of  $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$  with respect to  $\tau$.
	+ By Fourier transformation of  $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$  with respect to  $\tau$.		+ By Fourier transformation of  $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$  with respect to  $\tau$.
	+ By Fourier ~~inverse~~ transformation of  $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$  with respect to~~ ~~ $f_{\rm D}$.		+ By inverse Fourier transformation of  $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$  with respect to $f_{\rm D}$.
	</quiz>		</quiz>

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

2020-12-16T13:16:19Z

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 time-variant impulse response $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$ is the inverse Fourier transform of the delay–Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:		'''(1)'''  The time-variant impulse response $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$ is the inverse Fourier transform of the delay–Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:

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

2020-12-15T14:01:17Z

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:P_ID2164__Mob_A_2_5.png\|right\|frame\|Delay-Doppler profile]]		[[File:P_ID2164__Mob_A_2_5.png\|right\|frame\|Delay-Doppler profile]]

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

2020-07-09T09:13:27Z

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

← Older revision		Revision as of 11:13, 9 July 2020
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	''Notes:''		''Notes:''
	* This task should clarify the subject matter of the chapter  [[~~Mobile_Kommunikation~~/Das_GWSSUS%E2%80%93Kanalmodell\| Das GWSSUS–Kanalmodell]].		* This task should clarify the subject matter of the chapter  [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell\| Das GWSSUS–Kanalmodell]].
	* The relationship between the individual system functions is given in the  [[~~Mobile_Kommunikation~~/Das_GWSSUS%E2%80%93Kanalmodell#Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme\|graph on the first page]]  of this chapter.		* The relationship between the individual system functions is given in the  [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell#Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme\|graph on the first page]]  of this chapter.
	*Note that the magnitude function  $\|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})\|$  is shown above, so negative weights of the Dirac functions cannot be recognized.		*Note that the magnitude function  $\|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})\|$  is shown above, so negative weights of the Dirac functions cannot be recognized.

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	[[File:P_ID2168__Mob_A_2_5e_neu.png\|right\|frame\|Relationships between all system functions]]		[[File:P_ID2168__Mob_A_2_5e_neu.png\|right\|frame\|Relationships between all system functions]]
	'''(6)'''  As can be seen from the given [[~~Mobile_Kommunikation~~/Das_GWSSUS%E2%80%93Kanalmodell#Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme\|graph]], <u>solutions 2 and 3</u> are correct.		'''(6)'''  As can be seen from the given [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell#Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme\|graph]], <u>solutions 2 and 3</u> are correct.

	*The graph shows all system functions.		*The graph shows all system functions.

Javier at 15:43, 20 May 2020

2020-05-20T15:43:40Z

← Older revision		Revision as of 17:43, 20 May 2020
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	{{quiz-Header\|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}		{{quiz-Header\|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}

	[[File:P_ID2164__Mob_A_2_5.png\|right\|frame\|~~Verzögerungs–Doppler–Funktion~~]]		[[File:P_ID2164__Mob_A_2_5.png\|right\|frame\|Delay-Doppler profile]]
	For the mobile radio channel as a time-variant system, there are a total of four system functions that are linked with each other via the Fourier transform. With the nomenclature from our learning tutorial, these are:		For the mobile radio channel as a time-variant system, there are a total of four system functions that are linked with each other via the Fourier transform. With the nomenclature from our learning tutorial, these are:
	* the time-variant impulse response  $h(\tau, \hspace{0.05cm}t)$, which we also denote here as  $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$ ,		* the time-variant impulse response  $h(\tau, \hspace{0.05cm}t)$, which we also denote here as  $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$ ,