Aufgaben:Exercise 5.1: Sampling Theorem - Revision history

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	[[Category:Signal Representation: Exercises\|^5.1 Discrete-Time Signal Representation^]]		[[Category:Signal Representation: Exercises\|^5.1 Discrete-Time Signal Representation^]]
	[[de:Aufgaben:5.~~1_Abtasttheorem~~]]		[[de:Aufgaben:Aufgabe 5.1: Zum Abtasttheorem]]

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	*For signal reconstruction a low-pass filter is used, for whose frequency response applies:		*For signal reconstruction a low-pass filter is used, for whose frequency response applies:

	:$$H(f) = \left\{ \begin{array}{c} 1 \\		:$$H(f) = \left\{ \begin{array}{c} 1 \\0 \\ \end{array} \right.\quad\begin{array}{{5}c} {\rm{{\rm{f\ddot{u}r}}}}\\ {\rm{{\rm{f\ddot{u}r}}}} \\ \end{array}\begin{array}{{5}c}\|f\| < f_1 \hspace{0.05cm}, \\\|f\| > f_2 \hspace{0.05cm} \\\end{array}$$
	0 \\ \end{array} \right.\quad
	\begin{array}{*{5}c} {\rm{{\rm{f\ddot{u}r}}}}
	\\ {\rm{{\rm{f\ddot{u}r}}}} \\ \end{array}\begin{array}{*{5}c}
	\|f\| < f_1 \hspace{0.05cm}, \\
	\|f\| > f_2 \hspace{0.05cm} \\
	\end{array}$$

	The range between the frequencies  $f_1$  and  $f_2 > f_1$  is not relevant for the solution of this task.		The range between the frequencies  $f_1$  and  $f_2 > f_1$  is not relevant for the solution of this task.

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	[[Category:Signal Representation: Exercises\|^5.1 Discrete-Time Signal Representation^]]		[[Category:Signal Representation: Exercises\|^5.1 Discrete-Time Signal Representation^]]
			[[de:Aufgaben:5.1_Abtasttheorem]]

Javier: Text replacement - "Time Discrete" to "Discrete-Time"

2021-10-11T09:03:34Z

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	[[Category:Signal Representation: Exercises\|^5.1 Time ~~Discrete~~ Signal Representation^]]		[[Category:Signal Representation: Exercises\|^5.1 Discrete-Time Signal Representation^]]

Javier: Text replacement - "Time-Discrete" to "Discrete-Time"

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	''Hints:''		''Hints:''
	*This task belongs to the chapter  [[Signal_Representation/Time_Discrete_Signal_Representation\|Time~~-Discrete~~ Signal Representation]].		*This task belongs to the chapter  [[Signal_Representation/Time_Discrete_Signal_Representation\|Discrete-Time Signal Representation]].

	*There is an interactive applet for the topic dealt with here:  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction\|Sampling of Analog Signals and Signal Reconstruction]]		*There is an interactive applet for the topic dealt with here:  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction\|Sampling of Analog Signals and Signal Reconstruction]]

Guenter at 12:47, 13 May 2021

2021-05-13T12:47:40Z

Guenter at 12:43, 13 May 2021

2021-05-13T12:43:06Z

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	''Hints:''		''Hints:''
	*This task belongs to the chapter  [[Signal_Representation/Time_Discrete_Signal_Representation\|Time Discrete Signal Representation]].		*This task belongs to the chapter  [[Signal_Representation/Time_Discrete_Signal_Representation\|Time-Discrete Signal Representation]].

	*There is an interactive applet for the topic dealt with here:  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction\|Sampling of Analog Signals and Signal Reconstruction]]		*There is an interactive applet for the topic dealt with here:  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction\|Sampling of Analog Signals and Signal Reconstruction]]

Guenter at 12:41, 13 May 2021

2021-05-13T12:41:16Z

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:38:57Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

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	[[Category:~~Exercises for~~ Signal Representation\|^5.1 Time Discrete Signal Representation^]]		[[Category:Signal Representation: Exercises\|^5.1 Time Discrete Signal Representation^]]

Noah at 11:56, 1 March 2021

2021-03-01T11:56:08Z

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	[[File:P_ID1126__Sig_A_5_1.png\|right\|frame\|Zur Abtastung eines analogen Signals  $x(t)$]]		[[File:P_ID1126__Sig_A_5_1.png\|right\|frame\|Zur Abtastung eines analogen Signals  $x(t)$]]

	~~Gegeben ist ein Analogsignal~~  $x(t)$  ~~entsprechend der Skizze~~:		Given is an analogue signal  $x(t)$  according to the sketch:
	*~~Bekannt ist, dass dieses Signal keine höheren Frequenzen als~~  $B_{\rm NF} = 4 \ \text{kHz}$  ~~beinhaltet~~.		*It is known that this signal does not contain any frequencies higher than  $B_{\rm NF} = 4 \ \text{kHz}$ .
	*~~Durch Abtastung mit der Abtastrate~~  $f_{\rm A}$  ~~erhält man das in der Grafik rot skizzierte Signal~~  $x_{\rm A}(t)$.		*By sampling with the sampling rate  $f_{\rm A}$ , the signal  $x_{\rm A}(t)$ sketched in red in the diagram is obtained.
	*~~Zur Signalrekonstruktion wird ein Tiefpass verwendet~~, ~~für dessen Frequenzgang gilt~~:		*For signal reconstruction a low-pass filter is used, for whose frequency response applies:

	:$$H(f) = \left\{ \begin{array}{c} 1 \\		:$$H(f) = \left\{ \begin{array}{c} 1 \\
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	\end{array}$$		\end{array}$$

	~~Der Bereich zwischen den Frequenzen~~  $f_1$  ~~und~~  $f_2 > f_1$  ~~ist für die Lösung dieser Aufgabe nicht~~ relevant.		The range between the frequencies  $f_1$  and  $f_2 > f_1$  is not relevant for the solution of this task.

	~~Die Eckfrequenzen~~  $f_1$  ~~und~~  $f_2$  ~~sind so zu bestimmen, dass das Ausgangssignal~~  $y(t)$  ~~des Tiefpasses mit dem Signal~~  $x(t)$  ~~exakt übereinstimmt~~.		The corner frequencies  $f_1$  and  $f_2$  are to be determined in such a way that the output signal  $y(t)$  of the low-pass filter exactly matches the signal  $x(t)$ .


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	''~~Hinweise~~:''		''Hints:''
	*~~Die Aufgabe gehört zum Kapitel~~  [[Signal_Representation/Time_Discrete_Signal_Representation\|~~Zeitdiskrete Signaldarstellung~~]].		*This task belongs to the chapter  [[Signal_Representation/Time_Discrete_Signal_Representation\|Time Discrete Signal Representation]].

	*~~Zu der hier behandelten Thematik gibt es ein interaktives Applet~~:  [[Applets:Abtastung_periodischer_Signale_und_Signalrekonstruktion_(Applet)\|Abtastung periodischer Signale & Signalrekonstruktion]]		*There is an interactive applet for the topic dealt with here:  [[Applets:Abtastung_periodischer_Signale_und_Signalrekonstruktion_(Applet)\|Abtastung periodischer Signale & Signalrekonstruktion]]


	===~~Fragebogen~~===		===Questions===

	<quiz display=simple>		<quiz display=simple>
	{~~Ermitteln Sie aus der Grafik die zugrundeliegende Abtastrate~~.		{Determine the underlying sampling rate from the graph.
	\|type="{}"}		\|type="{}"}
	$f_{\rm A}\ = \ $ { 10 3% }  $\text{kHz}$		$f_{\rm A}\ = \ $ { 10 3% }  $\text{kHz}$

	{~~Bei welchen Frequenzen besitzt die Spektralfunktion~~  $X_{\rm A}(f)$  ~~mit Sicherheit~~ <u>~~keine Anteile~~</u>?		{At which frequencies does the spectral function  $X_{\rm A}(f)$  have <u>no components</u> with certainty?
	\|type="[]"}		\|type="[]"}
	- $f = 2.5 \ \text{kHz},$		- $f = 2.5 \ \text{kHz},$
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	+ $f= 34.5 \ \text{kHz}.$		+ $f= 34.5 \ \text{kHz}.$

	{~~Wie groß muss die untere Eckfrequenz~~  $f_1$  ~~mindestens sein, damit das Signal perfekt rekonstruiert wird~~?		{What is the minimum size of the lower cut-off frequency  $f_1$  that the signal is perfectly reconstructed?
	\|type="{}"}		\|type="{}"}
	$f_{1,\ \text{min}}\ = \ ${ 4 3% }  $\text{kHz}$		$f_{1,\ \text{min}}\ = \ ${ 4 3% }  $\text{kHz}$

	{~~Wie groß darf die obere Eckfrequenz~~  $f_2$  ~~höchstens sein, damit das Signal perfekt rekonstruiert wird~~?		{What is the maximum size of the upper corner frequency  $f_2$  for the signal to be perfectly reconstructed?
	\|type="{}"}		\|type="{}"}
	$f_{2,\ \text{max}}\ = \ ${ 6 3% }  $\text{kHz}$		$f_{2,\ \text{max}}\ = \ ${ 6 3% }  $\text{kHz}$
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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Der Abstand zweier benachbarter Abtastwerte beträgt~~  $T_{\rm A} = 0.1 \ \text{ms}$. ~~Somit erhält man für die Abtastrate~~  $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$.		'''(1)'''  The distance between two adjacent samples is  $T_{\rm A} = 0.1 \ \text{ms}$. Thus, for the sampling rate  $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$is obtained.


	[[File:P_ID1127__Sig_A_5_1_b.png\|450px\|right\|frame\|~~Spektrum~~  $X_{\rm A}(f)$  ~~des abgetasteten Signals~~ (~~schematische Darstellung~~)]]		[[File:P_ID1127__Sig_A_5_1_b.png\|450px\|right\|frame\|Spectrum  $X_{\rm A}(f)$  of the sampled signal (schematic representation)]]
	'''(2)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 2 ~~und~~ 4</u>:		'''(2)'''  Proposed <u>solutions 2 and 4</u> are correct:
	*~~Das Spektrum~~  $X_{\rm A}(f)$  ~~des abgetasteten Signals erhält man aus~~  $X(f)$  ~~durch periodische Fortsetzung im Abstand~~  $f_{\rm A} = 10 \ \text{kHz}$.		*The spectrum  $X_{\rm A}(f)$  of the sampled signal is obtained from  $X(f)$  by periodic continuation at a distance of  $f_{\rm A} = 10 \ \text{kHz}$.
	*~~Aus der Skizze erkennt man, dass~~  $X_{\rm A}(f)$  ~~durchaus Anteile bei~~  $f = 2.5 \ \text{kHz}$  ~~und~~  $f = 6.5 \ \text{kHz}$  ~~besitzen kann~~.		*From the sketch you can see that  $X_{\rm A}(f)$  can have parts at  $f = 2.5 \ \text{kHz}$  and  $f = 6.5 \ \text{kHz}$ .
	* ~~Dagegen gibt es bei~~  $f = 5.5 \ \text{kHz}$  ~~keine Anteile~~.		*In contrast, there are no components at  $f = 5.5 \ \text{kHz}$ .
	*~~Auch bei~~  $f = 34.5 \ \text{kHz}$  ~~wird auf jeden Fall~~  $X_{\rm A}(f) = 0$  ~~gelten~~.		*Also at  $f = 34.5 \ \text{kHz}$  will be valid in any case.  $X_{\rm A}(f) = 0$ .
	<br clear=all>		<br clear=all>
	'''(3)'''  ~~Es muss sichergestellt sein, dass alle Frequenzen des Analogsignals mit~~  $H(f) = 1$  ~~bewertet werden~~.		'''(3)'''  It must be ensured that all frequencies of the analogue signal are weighted with  $H(f) = 1$ .
	*~~Daraus folgt entsprechend der Skizze~~:		*From this follows according to the sketch:

	:$$f_{1, \ \text{min}} = B_{\rm NF} \;\underline{= 4 \ \text{kHz}}.$$		:$$f_{1, \ \text{min}} = B_{\rm NF} \;\underline{= 4 \ \text{kHz}}.$$


	'''(4)'''  ~~Ebenso muss garantiert werden~~, ~~dass alle Spektralanteile von~~  $X_{\rm A}(f)$, ~~die~~ in  $X(f)$  ~~nicht enthalten sind, durch den Tiefpass entfernt werden~~.		'''(4)'''  Likewise, it must be guaranteed that all spectral components of  $X_{\rm A}(f)$, that are not contained in  $X(f)$  are removed by the low-pass filter.
	*~~Entsprechend der Skizze muss also gelten~~:		*According to the sketch, the following must therefore apply:

	:$$f_{2, \ \text{max}} = f_{\rm A} – B_{\rm NF} \;\underline{= 6 \ \text{kHz}}.$$		:$$f_{2, \ \text{max}} = f_{\rm A} – B_{\rm NF} \;\underline{= 6 \ \text{kHz}}.$$

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The distance between two adjacent samples is  $T_{\rm A} = 0.1 \ \text{ms}$. Thus, for the sampling rate  $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$is obtained.		'''(1)'''  The distance between two adjacent samples is  $T_{\rm A} = 0.1 \ \text{ms}$.  Thus, for the sampling rate  $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$is obtained.


	[[File:P_ID1127__Sig_A_5_1_b.png\|450px\|right\|frame\|Spectrum  $X_{\rm A}(f)$  of the sampled signal (schematic representation)]]		[[File:P_ID1127__Sig_A_5_1_b.png\|450px\|right\|frame\|Spectrum  $X_{\rm A}(f)$  of the sampled signal <br>(schematic representation)]]
	'''(2)'''  Proposed <u>solutions 2 and 4</u> are correct:		'''(2)'''  Proposed <u>solutions 2 and 4</u> are correct:
	*The spectrum  $X_{\rm A}(f)$  of the sampled signal is obtained from  $X(f)$  by periodic continuation at a distance of  $f_{\rm A} = 10 \ \text{kHz}$.		*The spectrum  $X_{\rm A}(f)$  of the sampled signal is obtained from  $X(f)$  by periodic continuation at a distance of  $f_{\rm A} = 10 \ \text{kHz}$.
	*From the sketch you can see that  $X_{\rm A}(f)$  can have parts at  $f = 2.5 \ \text{kHz}$  and  $f = 6.5 \ \text{kHz}$~~&nbsp~~;.		*From the sketch you can see that  $X_{\rm A}(f)$  can have signal parts at  $f = 2.5 \ \text{kHz}$  and  $f = 6.5 \ \text{kHz}$;.
	*In contrast, there are no components at  $f = 5.5 \ \text{kHz}$~~ ~~.		*In contrast, there are no components at  $f = 5.5 \ \text{kHz}$.
	*Also at  $f = 34.5 \ \text{kHz}$  will be valid ~~in any case.~~  $X_{\rm A}(f) = 0$~~ ~~.		*Also at  $f = 34.5 \ \text{kHz}$  will be valid  $X_{\rm A}(f) = 0$.
	<br clear=all>		<br clear=all>
	'''(3)'''  It must be ensured that all frequencies of the ~~analogue~~ signal are weighted with  $H(f) = 1$~~ ~~.		'''(3)'''  It must be ensured that all frequencies of the analog signal are weighted with  $H(f) = 1$.
	*From this follows according to the sketch:		*From this follows according to the sketch:

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	'''(4)'''  Likewise, it must be guaranteed that all spectral components of  $X_{\rm A}(f)$, that are not contained in  $X(f)$  are removed by the low-pass filter.		'''(4)'''  Likewise, it must be guaranteed that all spectral components of  $X_{\rm A}(f)$, that are not contained in  $X(f)$  are removed by the low-pass filter.
	*According to the sketch, the following must ~~therefore~~ apply:		*According to the sketch, the following must apply:

	:$$f_{2, \ \text{max}} = f_{\rm A} – B_{\rm NF} \;\underline{= 6 \ \text{kHz}}.$$		:$$f_{2, \ \text{max}} = f_{\rm A} – B_{\rm NF} \;\underline{= 6 \ \text{kHz}}.$$

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	}}		}}

	[[File:P_ID1126__Sig_A_5_1.png\|right\|frame\|~~Zur Abtastung eines analogen Signals~~  $x(t)$]]		[[File:P_ID1126__Sig_A_5_1.png\|right\|frame\|Sampling of an analog signal  $x(t)$]]

	Given is an ~~analogue~~ signal  $x(t)$  according to the sketch:		Given is an analog signal  $x(t)$  according to the sketch:
	*It is known that this signal does not contain any frequencies higher than  $B_{\rm NF} = 4 \ \text{kHz}$~~ ~~.		*It is known that this signal does not contain any frequencies higher than  $B_{\rm NF} = 4 \ \text{kHz}$.
	*By sampling with the sampling rate  $f_{\rm A}$ , the signal  $x_{\rm A}(t)$ sketched in red in the diagram is obtained.		*By sampling with the sampling rate  $f_{\rm A}$ , the signal  $x_{\rm A}(t)$ sketched in red in the diagram is obtained.
	*For signal reconstruction a low-pass filter is used, for whose frequency response applies:		*For signal reconstruction a low-pass filter is used, for whose frequency response applies:
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	*This task belongs to the chapter  [[Signal_Representation/Time_Discrete_Signal_Representation\|Time Discrete Signal Representation]].		*This task belongs to the chapter  [[Signal_Representation/Time_Discrete_Signal_Representation\|Time Discrete Signal Representation]].

	*There is an interactive applet for the topic dealt with here:  [[Applets:~~Abtastung_periodischer_Signale_und_Signalrekonstruktion_(Applet)~~\|~~Abtastung periodischer Signale & Signalrekonstruktion~~]]		*There is an interactive applet for the topic dealt with here:  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction\|Sampling of Analog Signals and Signal Reconstruction]]


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	$f_{1,\ \text{min}}\ = \ ${ 4 3% }  $\text{kHz}$		$f_{1,\ \text{min}}\ = \ ${ 4 3% }  $\text{kHz}$

	{What is the maximum size of the upper corner frequency  $f_2$  ~~for~~ the signal ~~to be~~ perfectly reconstructed?		{What is the maximum size of the upper corner frequency  $f_2$  that the signal is perfectly reconstructed?
	\|type="{}"}		\|type="{}"}
	$f_{2,\ \text{max}}\ = \ ${ 6 3% }  $\text{kHz}$		$f_{2,\ \text{max}}\ = \ ${ 6 3% }  $\text{kHz}$