Aufgaben:Exercise 2.1: Two-Dimensional Impulse Response - Revision history

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2026-03-16T13:30:34Z

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	In this task the following quantities are also referred to:		In this task the following quantities are also referred to:
	* the "time-variant transfer function"  according to the definition		* the "time-variant transfer function"  according to the definition
	:$$H(f,\hspace{0.05cm} t)		:$$H(f,\hspace{0.05cm} t)\hspace{0.2cm} \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)\hspace{0.05cm},$$
	\hspace{0.2cm} \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)
	\hspace{0.05cm},$$

	* the approximation of the "coherence bandwidth"  as the reciprocal of the maximal duration of the delay profile  $h(\tau, t)$:		* the approximation of the "coherence bandwidth"  as the reciprocal of the maximal duration of the delay profile  $h(\tau, t)$:
	:$$B_{\rm K} \hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}}		:$$B_{\rm K} \hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	'''(4)'''  At time  $t = 3T$  the two Dirac functions occur at  $\tau_{\rm min} = 0$  and  $\tau_{\rm max} = 4 \ \rm µ s$.		'''(4)'''  At time  $t = 3T$  the two Dirac functions occur at  $\tau_{\rm min} = 0$  and  $\tau_{\rm max} = 4 \ \rm µ s$.
	*The (simple approximation for the) coherence bandwidth is the reciprocal of the delay span of these Dirac functions:		*The (simple approximation for the) coherence bandwidth is the reciprocal of the delay span of these Dirac functions:
	:$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{4\,\,{\rm µ s} } \hspace{0.25cm} \underline{ = 250\,\,{\rm kHz}}		:$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{4\,\,{\rm µ s} } \hspace{0.25cm} \underline{ = 250\,\,{\rm kHz}}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	*The same as at  $t = 4T$  the time between the Dirac functions is  $4 \ \rm µ s$:  $B_{\rm K} \hspace{0.01cm}' = \underline {250 \ \rm kHz}$.		*The same as at  $t = 4T$  the time between the Dirac functions is  $4 \ \rm µ s$:  $B_{\rm K} \hspace{0.01cm}' = \underline {250 \ \rm kHz}$.
	*At  $t = 5T$  the impulse response has a duration of  $6 \ \ \rm µ s \ \ \Rightarrow \ {\it B}_{\rm K} \hspace{0.01cm}' \ \underline {\approx 166.7 \ \rm kHz}$.		*At  $t = 5T$  the impulse response has a duration of  $6 \ \ \rm µ s \ \ \Rightarrow \ {\it B}_{\rm K} \hspace{0.01cm}' \ \underline {\approx 166.7 \ \rm kHz}$.

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2026-03-16T12:27:04Z

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	[[Category:Mobile Communications: Exercises\|^2.1 Description of Time-Variant Systems^]]		[[Category:Mobile Communications: Exercises\|^2.1 Description of Time-Variant Systems^]]
			[[de:Exercises:Exercise_2.1:_Two-Dimensional_Impulse_Response]]

Javier: Text replacement - "time-discrete" to "discrete-time"

2021-10-11T09:01:00Z

Text replacement - "time-discrete" to "discrete-time"

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	:$$h(\tau,\hspace{0.05cm}t) = \sum_{m = 1}^{M} z_m(t) \cdot {\rm \delta} (\tau - \tau_m)$$		:$$h(\tau,\hspace{0.05cm}t) = \sum_{m = 1}^{M} z_m(t) \cdot {\rm \delta} (\tau - \tau_m)$$

	is to be analyzed according to the adjoining diagram.  The two axes are time~~-discrete~~:		is to be analyzed according to the adjoining diagram.  The two axes are discrete-time:
	* $\tau$  is the  <i>delay</i>  and can take values between  $0$  and  $6 \ {\rm µ s}$  in the example.		* $\tau$  is the  <i>delay</i>  and can take values between  $0$  and  $6 \ {\rm µ s}$  in the example.
	* The (absolute) time  $t$  is related to the frequency of snapshots and characterizes the variation of the channel over time. We have  $t = n \cdot T$, where  $T \gg \tau_{\rm max}$ .		* The (absolute) time  $t$  is related to the frequency of snapshots and characterizes the variation of the channel over time. We have  $t = n \cdot T$, where  $T \gg \tau_{\rm max}$ .

Javier: Text replacement - "band pass" to "band-pass"

2021-05-03T15:13:21Z

Text replacement - "band pass" to "band-pass"

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The signal described in the equivalent low-pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\Delta \tau \ \underline {= 500 \ \rm kHz}$.		'''(1)'''  The signal described in the equivalent low-pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\Delta \tau \ \underline {= 500 \ \rm kHz}$.
	*This mathematical (two-sided) bandwidth of the low-pass signal is also the maximum physical (one-sided) bandwidth of the corresponding band pass signal.		*This mathematical (two-sided) bandwidth of the low-pass signal is also the maximum physical (one-sided) bandwidth of the corresponding band-pass signal.

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

2021-03-23T12:37:21Z

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

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	[[Category:~~Exercises for~~ Mobile Communications\|^2.1 Description of Time-Variant Systems^]]		[[Category:Mobile Communications: Exercises\|^2.1 Description of Time-Variant Systems^]]

Guenter at 10:09, 16 February 2021

2021-02-16T10:09:17Z

← Older revision		Revision as of 12:09, 16 February 2021
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	In this task the following quantities are also referred to:		In this task the following quantities are also referred to:
	* the ~~<i>~~time-variant transfer function~~</i>~~  according to the definition		* the "time-variant transfer function"  according to the definition
	:$$H(f,\hspace{0.05cm} t)		:$$H(f,\hspace{0.05cm} t)
	\hspace{0.2cm} \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)		\hspace{0.2cm} \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)
	\hspace{0.05cm},$$		\hspace{0.05cm},$$

	* the approximation of the ~~<i>~~coherence bandwidth~~</i>~~  as the reciprocal of the maximal duration of the delay profile  $h(\tau, t)$:		* the approximation of the "coherence bandwidth"  as the reciprocal of the maximal duration of the delay profile  $h(\tau, t)$:
	:$$B_{\rm K} \hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}}		:$$B_{\rm K} \hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The signal described in the equivalent low-pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\~~delta~~ \tau \ \underline {= 500 \ \rm kHz}$.		'''(1)'''  The signal described in the equivalent low-pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\Delta \tau \ \underline {= 500 \ \rm kHz}$.
	*This mathematical (two-sided) bandwidth of the low-pass signal is also the maximum physical (one-sided) bandwidth of the corresponding band pass signal.		*This mathematical (two-sided) bandwidth of the low-pass signal is also the maximum physical (one-sided) bandwidth of the corresponding band pass signal.

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	'''(5)'''  The impulse responses are identical at the times  $5T$,  $6T$  and  $7T$&;nbsp; each consists of three Diracs.		'''(5)'''  The impulse responses are identical at the times  $5T$,  $6T$  and  $7T$  each consists of three Diracs.
	* Assuming that nothing changes in this respect for  $t ≥ 8T$:   $t_{\rm 5} \ \ \underline {= 5T}$.		* Assuming that nothing changes in this respect for  $t ≥ 8T$:   $t_{\rm 5} \ \ \underline {= 5T}$.

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

2020-12-16T13:16:19Z

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

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	</quiz>		</quiz>

	===~~Sample solution~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The signal described in the equivalent low-pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\delta \tau \ \underline {= 500 \ \rm kHz}$.		'''(1)'''  The signal described in the equivalent low-pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\delta \tau \ \underline {= 500 \ \rm kHz}$.

Javier: Text replacement - "low pass" to "low-pass"

2020-12-15T14:19:22Z

Text replacement - "low pass" to "low-pass"

← Older revision		Revision as of 16:19, 15 December 2020
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	===Sample solution===		===Sample solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The signal described in the equivalent low pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\delta \tau \ \underline {= 500 \ \rm kHz}$.		'''(1)'''  The signal described in the equivalent low-pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\delta \tau \ \underline {= 500 \ \rm kHz}$.
	*This mathematical (two-sided) bandwidth of the low pass signal is also the maximum physical (one-sided) bandwidth of the corresponding band pass signal.		*This mathematical (two-sided) bandwidth of the low-pass signal is also the maximum physical (one-sided) bandwidth of the corresponding band pass signal.

Javier: Text replacement - "Mobile Kommunikation/Allgemeine Beschreibung zeitvarianter Systeme" to "Mobile_Communications/General_Description_of_Time_Variant_Systems"

2020-12-15T13:59:18Z

Text replacement - "Mobile Kommunikation/Allgemeine Beschreibung zeitvarianter Systeme" to "Mobile_Communications/General_Description_of_Time_Variant_Systems"

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	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Allgemeine Beschreibung zeitvarianter Systeme~~}}		{{quiz-Header\|Buchseite=Mobile_Communications/General_Description_of_Time_Variant_Systems}}

	[[File:P_ID2144__Mob_A_2_1.png\|right\|frame\|Two-dimensional impulse response]]		[[File:P_ID2144__Mob_A_2_1.png\|right\|frame\|Two-dimensional impulse response]]

Guenter at 16:25, 14 December 2020

2020-12-14T16:25:20Z

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	$B_{\rm max} \ = \ ${ 500 3% } $\ \ \rm kHz$		$B_{\rm max} \ = \ ${ 500 3% } $\ \ \rm kHz$

	{At what time  $t_2$  is the channel ideal, characterized by  $H(f, t_{\rm 2}) = 1$?		{At what time  $t_2$  the channel is ideal, characterized by  $H(f, t_{\rm 2}) = 1$?
	\|type="{}"}		\|type="{}"}
	$t_{\rm 2} \ = \ ${ 0. } $\ \cdot T$		$t_{\rm 2} \ = \ ${ 0. } $\ \cdot T$

	{From what time  $t_{\rm 3}$  ~~does~~ this channel cause distortion?		{From what time  $t_{\rm 3}$  this channel does cause distortion?
	\|type="{}"}		\|type="{}"}
	$t_{\rm 3} \ = \ ${ 3 3% } $\ \cdot T$		$t_{\rm 3} \ = \ ${ 3 3% } $\ \cdot T$
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	$t = 5T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ ${ 166.7 3% } $\ \ \rm kHz$		$t = 5T \text{:} \hspace{0.4cm} B_{\rm K} \hspace{0.01cm}' \ = \ ${ 166.7 3% } $\ \ \rm kHz$

	{From what time  $t_{\rm 5}$  ~~could~~ this channel be considered as time–invariant?		{From what time  $t_{\rm 5}$  this channel could be considered as time–invariant?
	\|type="{}"}		\|type="{}"}
	$t_{\rm 5} \ = \ ${ 5 3% } $\ \cdot T$		$t_{\rm 5} \ = \ ${ 5 3% } $\ \cdot T$
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	===Sample solution===		===Sample solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The ~~message~~ signal described in the equivalent low-pass ~~band shall~~ not have a bandwidth greater than $B_{\rm max} = 1/\delta \tau \ \underline {= 500 \ \rm kHz}$.		'''(1)'''  The signal described in the equivalent low pass range should not have a bandwidth greater than  $B_{\rm max} = 1/\delta \tau \ \underline {= 500 \ \rm kHz}$.
	*This mathematical (two-sided) bandwidth of the low pass~~–~~signal is also the maximum physical (one-sided) bandwidth of the corresponding ~~bandpass–~~signal.		*This mathematical (two-sided) bandwidth of the low pass signal is also the maximum physical (one-sided) bandwidth of the corresponding band pass signal.


			'''(2)'''  $H(f, t_{\rm 2}) = 1$  means in the time domain  $h(\tau, t_{\rm 2}) = \delta(\tau)$.
	'''(2)'''  $H(f, t_{\rm 2}) = 1$ means in the time domain $h(\tau, t_{\rm 2}) = \delta(\tau)$.
	*Only then the channel is ideal.		*Only then the channel is ideal.
	*You can see from the graph that this only applies to the time $t_{\rm 2} \ \underline {= 0}$.		*You can see from the graph that this only applies to the time  $t_{\rm 2} \ \underline {= 0}$.

	~~'''(3)'''  Distortions occur if at time $t$ the impulse response is composed of two or more Dirac functions   ⇒   $t ≥ t_{\rm 3} \ \underline {= 3T}$.~~
	*At time $t = T$ the signal $s(t)$ is delayed only by $2 \ \rm µ s$.
	*At $t = 2T$ the amplitude is additionally reduced by $50 \%$ ($6 \ \ \rm dB$ loss).

			'''(3)'''  Distortions occur if at time  $t$  the impulse response is composed of two or more Dirac functions   ⇒   $t ≥ t_{\rm 3} \ \underline {= 3T}$.
			*At time  $t = T$  the signal  $s(t)$  is delayed only by  $2 \ \rm µ s$.
			*At  $t = 2T$  the amplitude is additionally reduced by  $50 \%$  $(6 \ \ \rm dB$ loss$)$.


	'''(4)'''  At time $t = 3T$ the two Dirac functions occur at $\tau_{\rm min} = 0$ and $\tau_{\rm max} = 4 \ \rm µ s$.		'''(4)'''  At time  $t = 3T$  the two Dirac functions occur at  $\tau_{\rm min} = 0$  and  $\tau_{\rm max} = 4 \ \rm µ s$.
	*The (simple approximation for the) coherence bandwidth is the reciprocal of the delay span of these Dirac functions		*The (simple approximation for the) coherence bandwidth is the reciprocal of the delay span of these Dirac functions:
	:$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{4\,\,{\rm µ s} } \hspace{0.25cm} \underline{ = 250\,\,{\rm kHz}}		:$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{4\,\,{\rm µ s} } \hspace{0.25cm} \underline{ = 250\,\,{\rm kHz}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
			*The same as at  $t = 4T$  the time between the Dirac functions is  $4 \ \rm µ s$:  $B_{\rm K} \hspace{0.01cm}' = \underline {250 \ \rm kHz}$.
			*At  $t = 5T$  the impulse response has a duration of  $6 \ \ \rm µ s \ \ \Rightarrow \ {\it B}_{\rm K} \hspace{0.01cm}' \ \underline {\approx 166.7 \ \rm kHz}$.
	*~~As also~~ at $t = 4T$ the Dirac functions ~~are~~ $4 \ \rm µ s$ ~~apart, you also get~~ $B_{\rm K} \hspace{0.01cm}' = \underline {250 \ \rm kHz}$ ~~here~~ .
	*At $t = 5T$ the impulse response has an duration of $6 \ \ \rm µ s \ \ \Rightarrow \ {\it B}_{\rm K} \hspace{0.01cm}' \ \underline {\approx 166.7 \ \rm kHz}$.



	'''(5)'''  The impulse responses are identical at the times $5T$, $6T$ and $7T$ ~~and consist~~ of 3 Diracs ~~each~~.		'''(5)'''  The impulse responses are identical at the times  $5T$,  $6T$  and  $7T$&;nbsp; each consists of three Diracs.
	* Assuming that nothing changes in this respect for $t ≥ 8T$~~, you get~~ $t_{\rm 5} \ \ \underline {= 5T}$.		* Assuming that nothing changes in this respect for  $t ≥ 8T$:   $t_{\rm 5} \ \ \underline {= 5T}$.



	'''(6)'''  <u>Solution 2</u> is correct:		'''(6)'''  <u>Solution 2</u> is correct:
	*The temporal change of the impulse response, whose dynamics is expressed by the parameter $T$, must be slow in comparison to the maximum delay span of $h(\tau, t)$, which in this task equals $\tau_{\rm max} = 6 \ \rm µ s$: &~~e.g~~;		*The temporal change of the impulse response, whose dynamics is expressed by the parameter  $T$, must be slow in comparison to the maximum delay span of  $h(\tau, t)$, which is in this task equals $\tau_{\rm max} = 6 \ \rm µ s$   ⇒   $T \gg \tau_{\rm max}.$
	$$T \gg \tau_{\rm max}.$$
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Exercises for Mobile Communications\|^2.1 Description of Time-Variant Systems^]]		[[Category:Exercises for Mobile Communications\|^2.1 Description of Time-Variant Systems^]]