Aufgaben:Exercise 5.1: Sampling Theorem - Revision history

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

2021-10-11T09:03:34Z

Text replacement - "Time Discrete" to "Discrete-Time"

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

2021-10-11T09:02:23Z

Text replacement - "Time-Discrete" to "Discrete-Time"

Guenter at 12:47, 13 May 2021

2021-05-13T12:47:40Z

Guenter at 12:43, 13 May 2021

2021-05-13T12:43:06Z

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"

Noah at 11:56, 1 March 2021

2021-03-01T11:56:08Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 5.1: Zum Abtasttheorem to Aufgaben:Exercise 5.1: Sampling Theorem

2020-11-29T18:47:07Z

Oezdemir moved page Aufgabe 5.1: Zum Abtasttheorem to Exercise 5.1: Sampling Theorem

Oezdemir at 18:46, 29 November 2020

2020-11-29T18:46:38Z

Javier: Text replacement - "Signaldarstellung/Zeitdiskrete Signaldarstellung" to "Signal_Representation/Time_Discrete_Signal_Representation"

2020-09-01T13:27:15Z

Text replacement - "Signaldarstellung/Zeitdiskrete Signaldarstellung" to "Signal_Representation/Time_Discrete_Signal_Representation"

← Older revision		Revision as of 09:03, 11 October 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:Signal Representation: Exercises\|^5.1 Time ~~Discrete~~ Signal Representation^]]	+	[[Category:Signal Representation: Exercises\|^5.1 Discrete-Time Signal Representation^]]

@@ Line 31: / Line 31: @@
 ''Hints:''
-*This task belongs to the chapter&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation|Time-Discrete Signal Representation]].
+*This task belongs to the chapter&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation|Discrete-Time Signal Representation]].
 *There is an interactive applet for the topic dealt with here: &nbsp;[[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]

@@ Line 63: / Line 63: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  The distance between two adjacent samples is&nbsp; $T_{\rm A} = 0.1 \ \text{ms}$. Thus, for the sampling rate&nbsp; $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$is obtained.
+'''(1)'''&nbsp;  The distance between two adjacent samples is&nbsp; $T_{\rm A} = 0.1 \ \text{ms}$.&nbsp; Thus, for the sampling rate&nbsp; $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$is obtained.
-[[File:P_ID1127__Sig_A_5_1_b.png|450px|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal (schematic representation)]]
+[[File:P_ID1127__Sig_A_5_1_b.png|450px|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal <br>(schematic representation)]]
 '''(2)'''&nbsp;   Proposed <u>solutions 2 and 4</u> are correct:
 *The spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal is obtained from&nbsp; $X(f)$&nbsp; by periodic continuation at a distance of&nbsp; $f_{\rm A} =  10 \ \text{kHz}$.
-*From the sketch you can see that&nbsp; $X_{\rm A}(f)$&nbsp; can have parts at&nbsp; $f =  2.5 \ \text{kHz}$&nbsp; and&nbsp;  $f =  6.5 \ \text{kHz}$&nbsp;.
+*From the sketch you can see that&nbsp; $X_{\rm A}(f)$&nbsp; can have signal parts at&nbsp; $f =  2.5 \ \text{kHz}$&nbsp; and&nbsp;  $f =  6.5 \ \text{kHz}$;.
-*In contrast, there are no components at&nbsp;  $f =  5.5 \ \text{kHz}$&nbsp;.
+*In contrast, there are no components at&nbsp;  $f =  5.5 \ \text{kHz}$.
-*Also at&nbsp;  $f =  34.5 \ \text{kHz}$&nbsp; will be valid in any case.&nbsp; $X_{\rm A}(f) = 0$&nbsp;.
+*Also at&nbsp;  $f =  34.5 \ \text{kHz}$&nbsp; will be valid&nbsp; $X_{\rm A}(f) = 0$.
 <br clear=all>
-'''(3)'''&nbsp; It must be ensured that all frequencies of the analogue signal are weighted with&nbsp; $H(f) = 1$&nbsp;.
+'''(3)'''&nbsp; It must be ensured that all frequencies of the analog signal are weighted with&nbsp; $H(f) = 1$.
 *From this follows according to the sketch:
@@ Line 80: / Line 80: @@
 '''(4)'''&nbsp; Likewise, it must be guaranteed that all spectral components of&nbsp; $X_{\rm A}(f)$, that are not contained in&nbsp; $X(f)$&nbsp; are removed by the low-pass filter.
-*According to the sketch, the following must therefore apply:
+*According to the sketch, the following must apply:
 :$$f_{2, \ \text{max}} = f_{\rm A} – B_{\rm NF} \;\underline{= 6 \ \text{kHz}}.$$

@@ Line 31: / Line 31: @@
 ''Hints:''
-*This task belongs to the chapter&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation|Time Discrete Signal Representation]].
+*This task belongs to the chapter&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation|Time-Discrete Signal Representation]].
 *There is an interactive applet for the topic dealt with here: &nbsp;[[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1126__Sig_A_5_1.png|right|frame|Zur Abtastung eines analogen Signals&nbsp; $x(t)$]]
+[[File:P_ID1126__Sig_A_5_1.png|right|frame|Sampling of an analog signal&nbsp; $x(t)$]]
-Given is an analogue signal&nbsp; $x(t)$&nbsp; according to the sketch:
+Given is an analog signal&nbsp; $x(t)$&nbsp; according to the sketch:
-*It is known that this signal does not contain any frequencies higher than&nbsp; $B_{\rm NF} = 4 \ \text{kHz}$&nbsp;.
+*It is known that this signal does not contain any frequencies higher than&nbsp; $B_{\rm NF} = 4 \ \text{kHz}$.
 *By sampling with the sampling rate&nbsp; $f_{\rm A}$&nbsp;, the signal&nbsp; $x_{\rm A}(t)$ sketched in red in the diagram is obtained.
 *For signal reconstruction a low-pass filter is used, for whose frequency response applies:
@@ Line 33: / Line 33: @@
 *This task belongs to the chapter&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation|Time Discrete Signal Representation]].
-*There is an interactive applet for the topic dealt with here: &nbsp;[[Applets:Abtastung_periodischer_Signale_und_Signalrekonstruktion_(Applet)|Abtastung periodischer Signale & Signalrekonstruktion]]
+*There is an interactive applet for the topic dealt with here: &nbsp;[[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]
@@ Line 54: / Line 54: @@
 $f_{1,\ \text{min}}\ = \ ${ 4 3% } &nbsp;$\text{kHz}$
-{What is the maximum size of the upper corner frequency&nbsp; $f_2$&nbsp; for the signal to be perfectly reconstructed?
+{What is the maximum size of the upper corner frequency&nbsp; $f_2$&nbsp; that the signal is perfectly reconstructed?
 |type="{}"}
 $f_{2,\ \text{max}}\ = \ ${ 6 3% } &nbsp;$\text{kHz}$

← Older revision		Revision as of 12:38, 23 March 2021
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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^]]

@@ Line 5: / Line 5: @@
 [[File:P_ID1126__Sig_A_5_1.png|right|frame|Zur Abtastung eines analogen Signals&nbsp; $x(t)$]]
-Gegeben ist ein Analogsignal&nbsp; $x(t)$&nbsp; entsprechend der Skizze:
+Given is an analogue signal&nbsp; $x(t)$&nbsp; according to the sketch:
-*Bekannt ist, dass dieses Signal keine höheren Frequenzen als&nbsp; $B_{\rm NF} = 4 \ \text{kHz}$&nbsp; beinhaltet.
+*It is known that this signal does not contain any frequencies higher than&nbsp; $B_{\rm NF} = 4 \ \text{kHz}$&nbsp;.
-*Durch Abtastung mit der Abtastrate&nbsp; $f_{\rm A}$&nbsp; erhält man das in der Grafik rot skizzierte Signal&nbsp; $x_{\rm A}(t)$.
+*By sampling with the sampling rate&nbsp; $f_{\rm A}$&nbsp;, the signal&nbsp; $x_{\rm A}(t)$ 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:
 :$$H(f)   = \left\{ \begin{array}{c} 1  \\
@@ Line 18: / Line 18: @@
 \end{array}$$
-Der Bereich zwischen den Frequenzen&nbsp; $f_1$&nbsp; und&nbsp; $f_2 > f_1$&nbsp; ist für die Lösung dieser Aufgabe nicht relevant.
+The range between the frequencies&nbsp; $f_1$&nbsp; and&nbsp; $f_2 > f_1$&nbsp; is not relevant for the solution of this task.
-Die Eckfrequenzen&nbsp; $f_1$&nbsp; und&nbsp; $f_2$&nbsp; sind so zu bestimmen, dass das Ausgangssignal&nbsp; $y(t)$&nbsp; des Tiefpasses mit dem Signal&nbsp; $x(t)$&nbsp; exakt übereinstimmt.
+The corner frequencies&nbsp; $f_1$&nbsp; and&nbsp; $f_2$&nbsp; are to be determined in such a way that the output signal&nbsp; $y(t)$&nbsp; of the low-pass filter exactly matches the signal&nbsp; $x(t)$&nbsp;.
@@ Line 30: / Line 30: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation|Zeitdiskrete Signaldarstellung]].
+*This task belongs to the chapter&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation|Time Discrete Signal Representation]].
-*Zu der hier behandelten Thematik gibt es ein interaktives Applet: &nbsp;[[Applets:Abtastung_periodischer_Signale_und_Signalrekonstruktion_(Applet)|Abtastung periodischer Signale & Signalrekonstruktion]]
+*There is an interactive applet for the topic dealt with here: &nbsp;[[Applets:Abtastung_periodischer_Signale_und_Signalrekonstruktion_(Applet)|Abtastung periodischer Signale & Signalrekonstruktion]]
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Ermitteln Sie aus der Grafik die zugrundeliegende Abtastrate.
+{Determine the underlying sampling rate from the graph.
 |type="{}"}
 $f_{\rm A}\ = \ $  { 10 3% } &nbsp;$\text{kHz}$
-{Bei welchen Frequenzen besitzt die Spektralfunktion&nbsp;  $X_{\rm A}(f)$&nbsp; mit Sicherheit <u>keine Anteile</u>?
+{At which frequencies does the spectral function&nbsp;  $X_{\rm A}(f)$&nbsp; have <u>no components</u> with certainty?
 |type="[]"}
 - $f =  2.5 \ \text{kHz},$
@@ Line 50: / Line 50: @@
 + $f=  34.5 \ \text{kHz}.$
-{Wie groß muss die untere Eckfrequenz&nbsp; $f_1$&nbsp; mindestens sein, damit das Signal perfekt rekonstruiert wird?
+{What is the minimum size of the lower cut-off frequency&nbsp; $f_1$&nbsp; that the signal is perfectly reconstructed?
 |type="{}"}
 $f_{1,\ \text{min}}\ = \ ${ 4 3% } &nbsp;$\text{kHz}$
-{Wie groß darf die obere Eckfrequenz&nbsp; $f_2$&nbsp; höchstens sein, damit das Signal perfekt rekonstruiert wird?
+{What is the maximum size of the upper corner frequency&nbsp; $f_2$&nbsp; for the signal to be perfectly reconstructed?
 |type="{}"}
 $f_{2,\ \text{max}}\ = \ ${ 6 3% } &nbsp;$\text{kHz}$
@@ Line 61: / Line 61: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Der Abstand zweier benachbarter Abtastwerte beträgt&nbsp; $T_{\rm A} = 0.1 \ \text{ms}$. Somit erhält man für die Abtastrate&nbsp; $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$.
+'''(1)'''&nbsp;  The distance between two adjacent samples is&nbsp; $T_{\rm A} = 0.1 \ \text{ms}$. Thus, for the sampling rate&nbsp; $f_{\rm A} = 1/ T_{\rm A} \;\underline {= 10 \ \text{kHz}}$is obtained.
-[[File:P_ID1127__Sig_A_5_1_b.png|450px|right|frame|Spektrum&nbsp; $X_{\rm A}(f)$&nbsp; des abgetasteten Signals (schematische Darstellung)]]
+[[File:P_ID1127__Sig_A_5_1_b.png|450px|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal (schematic representation)]]
-'''(2)'''&nbsp;   Richtig sind die <u>Lösungsvorschläge 2 und 4</u>:
+'''(2)'''&nbsp;   Proposed <u>solutions 2 and 4</u> are correct:
-*Das Spektrum&nbsp; $X_{\rm A}(f)$&nbsp; des abgetasteten Signals erhält man aus&nbsp; $X(f)$&nbsp; durch periodische Fortsetzung im Abstand&nbsp; $f_{\rm A} =  10 \ \text{kHz}$.
+*The spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal is obtained from&nbsp; $X(f)$&nbsp; by periodic continuation at a distance of&nbsp; $f_{\rm A} =  10 \ \text{kHz}$.
-*Aus der Skizze erkennt man, dass&nbsp; $X_{\rm A}(f)$&nbsp; durchaus Anteile bei&nbsp; $f =  2.5 \ \text{kHz}$&nbsp; und&nbsp;  $f =  6.5 \ \text{kHz}$&nbsp; besitzen kann.
+*From the sketch you can see that&nbsp; $X_{\rm A}(f)$&nbsp; can have parts at&nbsp; $f =  2.5 \ \text{kHz}$&nbsp; and&nbsp;  $f =  6.5 \ \text{kHz}$&nbsp;.
-* Dagegen gibt es  bei&nbsp;  $f =  5.5 \ \text{kHz}$&nbsp; keine Anteile.
+*In contrast, there are no components at&nbsp;  $f =  5.5 \ \text{kHz}$&nbsp;.
-*Auch bei&nbsp;  $f =  34.5 \ \text{kHz}$&nbsp; wird  auf jeden Fall&nbsp; $X_{\rm A}(f) = 0$&nbsp; gelten.
+*Also at&nbsp;  $f =  34.5 \ \text{kHz}$&nbsp; will be valid in any case.&nbsp; $X_{\rm A}(f) = 0$&nbsp;.
 <br clear=all>
-'''(3)'''&nbsp; Es muss sichergestellt sein, dass alle Frequenzen des Analogsignals mit&nbsp; $H(f) = 1$&nbsp; bewertet werden.
+'''(3)'''&nbsp; It must be ensured that all frequencies of the analogue signal are weighted with&nbsp; $H(f) = 1$&nbsp;.
-*Daraus folgt entsprechend der Skizze:
+*From this follows according to the sketch:
 :$$f_{1, \ \text{min}} = B_{\rm NF} \;\underline{= 4 \ \text{kHz}}.$$
-'''(4)'''&nbsp; Ebenso muss garantiert werden, dass alle Spektralanteile von&nbsp; $X_{\rm A}(f)$, die in&nbsp; $X(f)$&nbsp; nicht enthalten sind, durch den Tiefpass entfernt werden.
+'''(4)'''&nbsp; Likewise, it must be guaranteed that all spectral components of&nbsp; $X_{\rm A}(f)$, that are not contained in&nbsp; $X(f)$&nbsp; are removed by the low-pass filter.
-*Entsprechend der Skizze muss also gelten:
+*According to the sketch, the following must therefore apply:
 :$$f_{2, \ \text{max}} = f_{\rm A} – B_{\rm NF} \;\underline{= 6 \ \text{kHz}}.$$

← Older revision	Revision as of 18:47, 29 November 2020
(No difference)

← Older revision		Revision as of 18:46, 29 November 2020
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−	[[Category:Exercises for Signal Representation\|^5. ~~Zeit- und frequenzdiskrete Signaldarstellung~~^]]	+	[[Category:Exercises for Signal Representation\|^5.1 Time Discrete Signal Representation^]]

← Older revision		Revision as of 13:27, 1 September 2020
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−	{{quiz-Header\|Buchseite=~~Signaldarstellung~~/~~Zeitdiskrete Signaldarstellung~~	+	{{quiz-Header\|Buchseite=Signal_Representation/Time_Discrete_Signal_Representation
	}}		}}