Aufgaben:Exercise 5.1Z: Sampling of Harmonic Oscillations - 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"

Javier: Text replacement - "analyse" to "analyze"

2021-09-22T11:47:58Z

Text replacement - "analyse" to "analyze"

Guenter at 14:44, 24 May 2021

2021-05-24T14:44:10Z

Guenter at 14:37, 13 May 2021

2021-05-13T14:37:23Z

Guenter at 14:22, 13 May 2021

2021-05-13T14:22:28Z

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

2021-05-03T15:10:38Z

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

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

2021-03-23T12:39:40Z

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

Noah at 18:00, 20 March 2021

2021-03-20T18:00:29Z

Noah at 12:27, 1 March 2021

2021-03-01T12:27:07Z

← Older revision		Revision as of 09:03, 11 October 2021
Line 149:		Line 149:

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

@@ Line 32: / Line 32: @@
 ''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 10: / Line 10: @@
 The oscillation parameters&nbsp; $f_0$&nbsp; and&nbsp; $A$&nbsp; can be taken from the graph.
-It is assumed that the signals are sampled equidistantly at times&nbsp; $\nu \cdot T_{\rm A}$&nbsp; whereby the parameter values&nbsp; $T_{\rm A} = 80 \ &micro; \text{s}$&nbsp; and&nbsp; $T_{\rm A} = 100 \ &micro; \text{s}$&nbsp; are to be analysed.
+It is assumed that the signals are sampled equidistantly at times&nbsp; $\nu \cdot T_{\rm A}$&nbsp; whereby the parameter values&nbsp; $T_{\rm A} = 80 \ &micro; \text{s}$&nbsp; and&nbsp; $T_{\rm A} = 100 \ &micro; \text{s}$&nbsp; are to be analyzed.
 The signal reconstruction at the receiver takes place through a low-pass filter&nbsp; $H(f)$,&nbsp; which forms the signal&nbsp; $y_{\rm A}(t) = x_{\rm A}(t)$&nbsp; from the sampled signal&nbsp; $y(t)$&nbsp;. It applies:

@@ Line 85: / Line 85: @@
-[[File:P_ID1130__Sig_Z_5_1_c.png|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal - real part and imaginary part.]]
+[[File:P_ID1130__Sig_Z_5_1_c.png|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal &ndash; real part and imaginary part]]
 '''(3)'''&nbsp; The sampling rate is now&nbsp; $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.
 *Only in the special case of the cosine signal the sampling theorem is  satisfied, and it holds:
@@ Line 138: / Line 138: @@
-[[File:P_ID1133__Sig_Z_5_1_e.png|right|frame|Reconstruction of a harmonic oscillation with &nbsp; $60^\circ$ phase]]
+[[File:P_ID1133__Sig_Z_5_1_e.png|right|frame|Reconstruction of a harmonic oscillation with&nbsp; $60^\circ$ phase]]
 '''(5)'''&nbsp; Despite &nbsp; &rArr; &nbsp; $\varphi = 60^\circ$ gilt $\varphi_3 = 0$ &nbsp; &rArr; &nbsp; the reconstructed signal&nbsp; $y_3(t)$&nbsp;  is cosine-shaped, too.&nbsp;
 *The amplitude is equal to

@@ Line 87: / Line 87: @@
 [[File:P_ID1130__Sig_Z_5_1_c.png|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal - real part and imaginary part.]]
 '''(3)'''&nbsp; The sampling rate is now&nbsp; $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.
-*Only in the special case of the cosine signal is the sampling theorem now satisfied and it holds:
+*Only in the special case of the cosine signal the sampling theorem is  satisfied, and it holds:
 :$$y_1(t) = x_1(t) &nbsp; &rArr; &nbsp; A_1 \; \underline{=2 \ \text{V}} \text{ und }\varphi_1 \; \underline{= 0}.$$
 This result is now to be derived mathematically, whereby a phase&nbsp; $\varphi$&nbsp; in the input signal is already taken into account with regard to the remaining subtasks:
@@ Line 102: / Line 101: @@
 *With the abbreviations
 :$$R = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm}
-\cos(\varphi) \hspace{0.5cm}{\rm und}  \hspace{0.5cm}I ={A}/{2} \hspace{0.05cm} \cdot
+\cos(\varphi) \hspace{0.5cm}{\rm and}  \hspace{0.5cm}I ={A}/{2} \hspace{0.05cm} \cdot
 \hspace{0.05cm} \sin(\varphi)$$
 :can also be written for this:
@@ Line 108: / Line 107: @@
   (f+ f_{\rm 0} ) + (R - {\rm j} \cdot I) \cdot  \delta
   (f- f_{\rm 0} )\hspace{0.05cm}.$$
-*The spectrum of the signal&nbsp; $f_{\rm A} = 2f_0$&nbsp; sampled with&nbsp; $x_{\rm A}(t)$&nbsp; is thus:
+*The spectrum of the signal&nbsp; $x_{\rm A}(t)$&nbsp; sampled with&nbsp; $f_{\rm A} = 2f_0$&nbsp; is thus:
 :$$X_{\rm A}(f) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
   )= \sum_{\mu = - \infty }^{+\infty} X (f- 2\mu \cdot f_{\rm 0}
   )\hspace{0.05cm}.$$
-:*The bottom graph shows that&nbsp; $X_{\rm A}(f)$&nbsp; consists of Dirac functions at&nbsp; $\pm f_0$,&nbsp; $\pm 3f_0$,&nbsp; $\pm 5f_0$,&nbsp; , and so on.
+:*The bottom graph shows that&nbsp; $X_{\rm A}(f)$&nbsp; consists of Dirac functions at&nbsp; $\pm f_0$,&nbsp; $\pm 3f_0$,&nbsp; $\pm 5f_0$,&nbsp; and so on.
 :*All weights are purely real and equal to&nbsp; $2 \cdot R$.
 :*The imaginary parts of the periodically continued spectrum cancel out.
-*If one further takes into account the rectangular low-pass filter whose cut-off frequency lies exactly at&nbsp; $f_{\rm G} = f_0$&nbsp;, as well as &nbsp; $H(f_{\rm G}) = 0.5$, one obtains for the spectrum after signal reconstruction:
+*If one further takes into account the rectangular low-pass filter whose cut-off frequency lies exactly at&nbsp; $f_{\rm G} = f_0$,&nbsp; as well as &nbsp; $H(f_{\rm G}) = 0.5$, one obtains for the spectrum after signal reconstruction:
 :$$Y(f) = R \cdot  \delta
   (f+ f_{\rm 0} ) + R  \cdot  \delta
@@ Line 122: / Line 121: @@
 \cos(\varphi)\hspace{0.05cm}.$$
-*The Fourier inverse transformation leads to
+*The inverse Fourier transformation leads to
-[[File:P_ID1131__Sig_Z_5_1_d.png|right|frame|Reconstruction of the sampled sinusoidal signal]]
+[[File:P_ID1131__Sig_Z_5_1_d.png|right|frame|Reconstruction of the sampled sine signal]]
 :$$y(t) =  A \cdot \cos (\varphi)\cdot \cos (2 \pi \cdot f_0 \cdot t )
   \hspace{0.05cm}.$$
@@ Line 140: / Line 139: @@
 [[File:P_ID1133__Sig_Z_5_1_e.png|right|frame|Reconstruction of a harmonic oscillation with &nbsp; $60^\circ$ phase]]
-'''(5)'''&nbsp; Despite&nbsp; $\varphi = 60^\circ$ gilt $\varphi_3 = 0$ &nbsp; &rArr; &nbsp; the reconstructed signal&nbsp; $y_3(t)$ ist cosinusförmig. is also cosine-shaped. The amplitude is equal to
+'''(5)'''&nbsp; Despite &nbsp; &rArr; &nbsp; $\varphi = 60^\circ$ gilt $\varphi_3 = 0$ &nbsp; &rArr; &nbsp; the reconstructed signal&nbsp; $y_3(t)$&nbsp;  is cosine-shaped, too.&nbsp;
 :$$A_3 =  A \cdot \cos (60^{\circ})= {A}/{2} \hspace{0.15 cm}\underline{= 1\,{\rm V}}
   \hspace{0.05cm}.$$
-*If you look at the samples drawn in red in the graph, you will admit that as a "signal reconstructor" you would not make any other decision than the low-pass.
+*If you look at the samples drawn in red in the graph, you will admit that you as a "signal reconstructor" would not make any other decision than the low-pass.
-*After all, you do not know the turquoise curve.
+*Because, you don't know the turquoise curve.
 {{ML-Fuß}}

@@ Line 10: / Line 10: @@
 The oscillation parameters&nbsp; $f_0$&nbsp; and&nbsp; $A$&nbsp; can be taken from the graph.
-It is assumed that the signals are sampled equidistantly at the times&nbsp; $\nu \cdot T_{\rm A}$&nbsp;, whereby the parameter values&nbsp; $T_{\rm A} = 80 \ &micro; \text{s}$&nbsp; and&nbsp; $T_{\rm A} = 100 \ &micro; \text{s}$&nbsp; are to be analysed.
+It is assumed that the signals are sampled equidistantly at times&nbsp; $\nu \cdot T_{\rm A}$&nbsp; whereby the parameter values&nbsp; $T_{\rm A} = 80 \ &micro; \text{s}$&nbsp; and&nbsp; $T_{\rm A} = 100 \ &micro; \text{s}$&nbsp; are to be analysed.
-The signal reconstruction at the receiver is carried out by a low-pass filter&nbsp; $H(f)$, which forms the signal&nbsp; $y_{\rm A}(t) = x_{\rm A}(t)$&nbsp; from the sampled signal&nbsp; $y(t)$&nbsp;. It applies:
+The signal reconstruction at the receiver takes place through a low-pass filter&nbsp; $H(f)$,&nbsp; which forms the signal&nbsp; $y_{\rm A}(t) = x_{\rm A}(t)$&nbsp; from the sampled signal&nbsp; $y(t)$&nbsp;. It applies:
 :$$H(f)   = \left\{ \begin{array}{c} 1  \\ 0.5 \\
   \\  \end{array} \right.\quad
@@ Line 22: / Line 22: @@
 |f| > f_{\rm G}  \hspace{0.05cm}, \\
 \end{array}$$
-Here&nbsp; $f_{\rm G}$&nbsp; indicates the cut-off frequency of the rectangular low-pass filter. For this shall apply:
+Here&nbsp; $f_{\rm G}$&nbsp; indicates the cut-off frequency of the rectangular low-pass filter.&nbsp; For this shall apply:
 :$$f_{\rm G} = \frac{1}{ 2 \cdot T_{\rm   A}}\hspace{0.05cm}.$$
 The sampling theorem is fulfilled if&nbsp; $y(t) = x(t)$&nbsp; holds.
@@ Line 32: / Line 32: @@
 ''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: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: @@
-{What is the reconstructed signal&nbsp; $y_1(t) = A_1 \cdot \cos (2\pi f_0 t – \varphi_1)$&nbsp; mit dem Abtastabstand&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$? Interpretieren Sie das Ergebnis.
+{What is the reconstructed signal&nbsp; $y_1(t) = A_1 \cdot \cos (2\pi f_0 t – \varphi_1)$&nbsp; with the sampling distance&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$? Interpret the result.
 |type="{}"}
 $A_1\hspace{0.2cm} = \ ${ 2 3% } &nbsp;$\text{V}$
@@ Line 60: / Line 60: @@
-{What is the amplitude&nbsp; $A_2$&nbsp; of the reconstructed signal&nbsp; $y_2(t)$, when the sinusoidal signal&nbsp; $x_2(t)$&nbsp; is present? Let&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$ still apply.
+{What is the amplitude&nbsp; $A_2$&nbsp; of the reconstructed signal&nbsp; $y_2(t)$, when the sinusoidal signal&nbsp; $x_2(t)$&nbsp; is present?&nbsp; Let&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$ still apply.
 |type="{}"}
 $A_2\hspace{0.2cm} = \ $ { 0. } &nbsp;$\text{V}$

← Older revision		Revision as of 12:39, 23 March 2021
Line 149:		Line 149:

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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 41: / Line 41: @@
 <quiz display=simple>
-{Wie groß sind Amplitude und Frequenz der Signale&nbsp; $x_1(t)$,&nbsp; $x_2(t)$&nbsp; und&nbsp; $x_3(t)$?
+{What are the amplitude and frequency of the signals&nbsp; $x_1(t)$,&nbsp; $x_2(t)$&nbsp; and&nbsp; $x_3(t)$?
 |type="{}"}
 $A \hspace{0.25cm} = \ $  { 2 3% } &nbsp;$\text{V}$
@@ Line 47: / Line 47: @@
-{Bei welchen Eingangssignalen ist das Abtasttheorem erfüllt  &nbsp; &rArr;  &nbsp;  $y(t) = x(t)$, wenn&nbsp; $\underline{T_{\rm A} = 80 \ {\rm &micro;} \text{s}}$&nbsp; beträgt?
+{For which input signals is the sampling theorem satisfied  &nbsp; &rArr;  &nbsp;  $y(t) = x(t)$, when&nbsp; $\underline{T_{\rm A} = 80 \ {\rm &micro;} \text{s}}$&nbsp;?
 |type="[]"}
 + $x_1(t)$,
@@ Line 54: / Line 54: @@
-{Wie lautet das rekonstruierte Signal&nbsp; $y_1(t) = A_1 \cdot \cos (2\pi f_0 t – \varphi_1)$&nbsp; mit dem Abtastabstand&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$? Interpretieren Sie das Ergebnis.
+{What is the reconstructed signal&nbsp; $y_1(t) = A_1 \cdot \cos (2\pi f_0 t – \varphi_1)$&nbsp; mit dem Abtastabstand&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$? Interpretieren Sie das Ergebnis.
 |type="{}"}
 $A_1\hspace{0.2cm} = \ ${ 2 3% } &nbsp;$\text{V}$
-$\varphi_1\hspace{0.2cm} = \ $ { 0. } &nbsp;$\text{Grad}$
+$\varphi_1\hspace{0.2cm} = \ $ { 0. } &nbsp;$\text{Deg}$
-{Welche Amplitude&nbsp; $A_2$&nbsp; besitzt das rekonstruierte Signal&nbsp; $y_2(t)$, wenn das Sinussignal&nbsp; $x_2(t)$&nbsp; anliegt? Es gelte weiterhin&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$.
+{What is the amplitude&nbsp; $A_2$&nbsp; of the reconstructed signal&nbsp; $y_2(t)$, when the sinusoidal signal&nbsp; $x_2(t)$&nbsp; is present? Let&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$ still apply.
 |type="{}"}
 $A_2\hspace{0.2cm} = \ $ { 0. } &nbsp;$\text{V}$
-{Welche Amplitude&nbsp; $A_3$&nbsp; besitzt das rekonstruierte Signal&nbsp; $y_3(t)$, wenn das Signal&nbsp; $x_3(t)$&nbsp; anliegt? Es gelte weiterhin&nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$.
+{What is the amplitude&nbsp; $A_3$&nbsp; of the reconstructed signal&nbsp; $y_3(t)$, when the signal&nbsp; $x_3(t)$&nbsp; is present? &nbsp; $\underline{T_{\rm A} = 100 \ {\rm &micro;} \text{s}}$ is still valid.
 |type="{}"}
 $A_3\hspace{0.2cm} = \ $ { 1 3% } &nbsp;$\text{V}$
@@ Line 86: / Line 86: @@
 [[File:P_ID1130__Sig_Z_5_1_c.png|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal - real part and imaginary part.]]
-'''(3)'''&nbsp; Die Abtastrate beträgt nun&nbsp; $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.
+'''(3)'''&nbsp; The sampling rate is now&nbsp; $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.
 *Only in the special case of the cosine signal is the sampling theorem now satisfied and it holds:
 :$$y_1(t) = x_1(t) &nbsp; &rArr; &nbsp; A_1 \; \underline{=2 \ \text{V}} \text{ und }\varphi_1 \; \underline{= 0}.$$
@@ Line 100: / Line 100: @@
   \cdot \hspace{0.05cm} \varphi} \cdot  \delta
   (f- f_{\rm 0} )\hspace{0.05cm}.$$
-*Mit den Abkürzungen
+*With the abbreviations
 :$$R = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm}
 \cos(\varphi) \hspace{0.5cm}{\rm und}  \hspace{0.5cm}I ={A}/{2} \hspace{0.05cm} \cdot
 \hspace{0.05cm} \sin(\varphi)$$
-:kann hierfür auch geschrieben werden:
+:can also be written for this:
 :$$X(f) = (R + {\rm j} \cdot I) \cdot  \delta
   (f+ f_{\rm 0} ) + (R - {\rm j} \cdot I) \cdot  \delta
   (f- f_{\rm 0} )\hspace{0.05cm}.$$
-*Das Spektrum des mit&nbsp; $f_{\rm A} = 2f_0$&nbsp; abgetasteten Signals&nbsp; $x_{\rm A}(t)$&nbsp; lautet somit:
+*The spectrum of the signal&nbsp; $f_{\rm A} = 2f_0$&nbsp; sampled with&nbsp; $x_{\rm A}(t)$&nbsp; is thus:
 :$$X_{\rm A}(f) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
   )= \sum_{\mu = - \infty }^{+\infty} X (f- 2\mu \cdot f_{\rm 0}
   )\hspace{0.05cm}.$$
-:*Die untere Grafik zeigt, dass&nbsp; $X_{\rm A}(f)$&nbsp; aus Diracfunktionen bei&nbsp; $\pm f_0$,&nbsp; $\pm 3f_0$,&nbsp; $\pm 5f_0$,&nbsp; usw. besteht.
+:*The bottom graph shows that&nbsp; $X_{\rm A}(f)$&nbsp; consists of Dirac functions at&nbsp; $\pm f_0$,&nbsp; $\pm 3f_0$,&nbsp; $\pm 5f_0$,&nbsp; , and so on.
-:*Alle Gewichte sind rein reell und gleich&nbsp; $2 \cdot R$.
+:*All weights are purely real and equal to&nbsp; $2 \cdot R$.
-:*Die Imaginärteile des periodisch fortgesetzten Spektrums heben sich auf.
+:*The imaginary parts of the periodically continued spectrum cancel out.
-*Berücksichtigt man weiter den rechteckförmigen Tiefpass, dessen Grenzfrequenz exakt bei&nbsp; $f_{\rm G} = f_0$&nbsp; liegt, sowie&nbsp; $H(f_{\rm G}) = 0.5$, so erhält man für das Spektrum nach der Signalrekonstruktion:
+*If one further takes into account the rectangular low-pass filter whose cut-off frequency lies exactly at&nbsp; $f_{\rm G} = f_0$&nbsp;, as well as &nbsp; $H(f_{\rm G}) = 0.5$, one obtains for the spectrum after signal reconstruction:
 :$$Y(f) = R \cdot  \delta
   (f+ f_{\rm 0} ) + R  \cdot  \delta
@@ Line 122: / Line 122: @@
 \cos(\varphi)\hspace{0.05cm}.$$
-*Die Fourierrücktransformation führt auf
+*The Fourier inverse transformation leads to
 [[File:P_ID1131__Sig_Z_5_1_d.png|right|frame|Reconstruction of the sampled sinusoidal signal]]
 :$$y(t) =  A \cdot \cos (\varphi)\cdot \cos (2 \pi \cdot f_0 \cdot t )
   \hspace{0.05cm}.$$
-*Es ergibt sich also unabhängig von der Eingangsphase&nbsp; $\varphi$&nbsp; ein cosinusförmiger Verlauf.
+*Thus, a cosine-shaped progression results independent of the input phase&nbsp; $\varphi$&nbsp;.
-*Ist&nbsp; $\varphi = 0$&nbsp; wie beim Signal&nbsp; $x_1(t)$, so ist auch die Amplitude des Ausgangssignals gleich&nbsp; $A$.
+*If&nbsp; $\varphi = 0$&nbsp; as with the signal&nbsp; $x_1(t)$, the amplitude of the output signal is also equal to&nbsp; $A$.
-'''(4)'''&nbsp; Das Sinussignal hat die Phase&nbsp; $90^\circ$.
+'''(4)'''&nbsp; The sine signal has the phase&nbsp; $90^\circ$.
-*Daraus folgt direkt&nbsp; $y_2(t) = 0$  &nbsp; &rArr; &nbsp;     Amplitude $\underline{A_2 = 0}$.
+*From this follows directly&nbsp; $y_2(t) = 0$  &nbsp; &rArr; &nbsp;     amplitude $\underline{A_2 = 0}$.
-*Dieses Ergebnis wird verständlich, wenn man sich die Abtastwerte in der Grafik betrachtet.
+*This result becomes understandable if you look at the samples in the graph.
-*Alle Abtastwerte (rote Kreise) sind Null, so dass auch nach dem Filter kein Signal vorhanden sein kann.
+*All samples (red circles) are zero, so there can be no signal even after the filter.
-[[File:P_ID1133__Sig_Z_5_1_e.png|right|frame|Rekonstruktion einer harmonischen Schwingung mit&nbsp; $60^\circ$ Phase]]
+[[File:P_ID1133__Sig_Z_5_1_e.png|right|frame|Reconstruction of a harmonic oscillation with &nbsp; $60^\circ$ phase]]
-'''(5)'''&nbsp; Trotz&nbsp; $\varphi = 60^\circ$ gilt $\varphi_3 = 0$ &nbsp; &rArr; &nbsp; auch das rekonstruierte Signal&nbsp; $y_3(t)$ ist cosinusförmig. Die Amplitude ist gleich
+'''(5)'''&nbsp; Despite&nbsp; $\varphi = 60^\circ$ gilt $\varphi_3 = 0$ &nbsp; &rArr; &nbsp; the reconstructed signal&nbsp; $y_3(t)$ ist cosinusförmig. is also cosine-shaped. The amplitude is equal to
 :$$A_3 =  A \cdot \cos (60^{\circ})= {A}/{2} \hspace{0.15 cm}\underline{= 1\,{\rm V}}
   \hspace{0.05cm}.$$
-*Wenn Sie die rot eingezeichneten Abtastwerte in der Grafik betrachten, so werden Sie zugeben, dass Sie als „Signalrekonstrukteur” keine andere Entscheidung treffen würden als der Tiefpass.
+*If you look at the samples drawn in red in the graph, you will admit that as a "signal reconstructor" you would not make any other decision than the low pass.
-*Sie kennen ja den türkisfarbenen Verlauf nicht.
+*After all, you do not know the turquoise curve.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1129__Sig_Z_5_1.png|right|frame|Drei harmonische Schwingungen gleicher Frequenz&nbsp; $f_0$&nbsp; und gleicher Amplitude&nbsp; $A$]]
+[[File:P_ID1129__Sig_Z_5_1.png|right|frame|Three harmonic oscillations of equal frequency&nbsp; $f_0$&nbsp; and equal amplitude&nbsp; $A$]]
-Wir betrachten drei harmonische Schwingungen mit gleicher Frequenz und gleicher Amplitude:
+We consider three harmonic oscillations with the same frequency and the same amplitude:
 :$$x_1(t)   =   A \cdot \cos (2 \pi \cdot f_0 \cdot t) \hspace{0.05cm}, $$
 :$$ x_2(t)  =   A \cdot \sin (2 \pi \cdot f_0 \cdot t) \hspace{0.05cm}, $$
 :$$ x_3(t)  =   A \cdot \cos (2 \pi \cdot f_0 \cdot t - 60^{\circ}) \hspace{0.05cm}.$$
-Die Schwingungsparameter&nbsp; $f_0$&nbsp; und&nbsp; $A$&nbsp; können Sie der Grafik entnehnen.
+The oscillation parameters&nbsp; $f_0$&nbsp; and&nbsp; $A$&nbsp; can be taken from the graph.
-Angenommen wird, dass die Signale äquidistant zu den Zeitpunkten&nbsp; $\nu \cdot T_{\rm A}$&nbsp; abgetastet werden, wobei die Parameterwerte&nbsp; $T_{\rm A} = 80 \ &micro; \text{s}$&nbsp; und&nbsp; $T_{\rm A} = 100 \ &micro; \text{s}$&nbsp; analysiert werden sollen.
+It is assumed that the signals are sampled equidistantly at the times&nbsp; $\nu \cdot T_{\rm A}$&nbsp;, whereby the parameter values&nbsp; $T_{\rm A} = 80 \ &micro; \text{s}$&nbsp; and&nbsp; $T_{\rm A} = 100 \ &micro; \text{s}$&nbsp; are to be analysed.
-Die Signalrekonstruktion beim Empfänger erfolgt durch einen Tiefpass&nbsp; $H(f)$, der aus dem abgetasteten Signal&nbsp; $y_{\rm A}(t) = x_{\rm A}(t)$&nbsp; das Signal&nbsp; $y(t)$&nbsp; formt. Es gelte:
+The signal reconstruction at the receiver is carried out by a low-pass filter&nbsp; $H(f)$, which forms the signal&nbsp; $y_{\rm A}(t) = x_{\rm A}(t)$&nbsp; from the sampled signal&nbsp; $y(t)$&nbsp;. It applies:
 :$$H(f)   = \left\{ \begin{array}{c} 1  \\ 0.5 \\
   \\  \end{array} \right.\quad
@@ Line 22: / Line 22: @@
 |f| > f_{\rm G}  \hspace{0.05cm}, \\
 \end{array}$$
-Hierbei gibt&nbsp; $f_{\rm G}$&nbsp; die Grenzfrequenz des rechteckförmigen Tiefpassfilters an. Für diese soll gelten:
+Here&nbsp; $f_{\rm G}$&nbsp; indicates the cut-off frequency of the rectangular low-pass filter. For this shall apply:
 :$$f_{\rm G} = \frac{1}{ 2 \cdot T_{\rm   A}}\hspace{0.05cm}.$$
-Das Abtasttheorem ist erfüllt, wenn&nbsp; $y(t) = x(t)$&nbsp; gilt.
+The sampling theorem is fulfilled if&nbsp; $y(t) = x(t)$&nbsp; holds.
@@ Line 31: / Line 31: @@
-''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===
 <quiz display=simple>
@@ Line 72: / Line 73: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;   Aus der Grafik erkennt man die Amplitude&nbsp; $\underline{A = 2\ \text{V}}$&nbsp; sowie die Periodendauer&nbsp; $T_0 = 0.2 \ \text{ms}$.
+'''(1)'''&nbsp;   The graph shows the amplitude&nbsp; $\underline{A = 2\ \text{V}}$&nbsp; and the period&nbsp; $T_0 = 0.2 \ \text{ms}$.
-*Daraus ergibt sich die Signalfrequenz&nbsp; $f_0 = 1/T_0 \; \underline{= 5 \ \text{kHz}}$.
+*This results in the signal frequency&nbsp; $f_0 = 1/T_0 \; \underline{= 5 \ \text{kHz}}$.
-'''(2)'''&nbsp; Richtig sind <u>alle Löungsvorschläge</u>:
+'''(2)'''&nbsp; <u>All proposed solutions</u> are correct:
-*Die Abtastrate beträgt hier&nbsp; $f_{\rm A} = 1/T_{\rm A} = 12.5 \ \text{kHz}$.
+*The sampling rate here is&nbsp; $f_{\rm A} = 1/T_{\rm A} = 12.5 \ \text{kHz}$.
-*Dieser Wert ist größer als&nbsp; $2 \cdot f_0 = 10 \ \text{kHz}$.
+*This value is greater than&nbsp; $2 \cdot f_0 = 10 \ \text{kHz}$.
-*Damit ist das Abtasttheorem unabhängig von der Phase erfüllt, und es gilt stets&nbsp; $y(t) = x(t)$.
+*Thus the sampling theorem is fulfilled independently of the phase and&nbsp; $y(t) = x(t)$ always applies.
-[[File:P_ID1130__Sig_Z_5_1_c.png|right|frame|Spektrum&nbsp; $X_{\rm A}(f)$&nbsp; des abgetasteten Signals &ndash; Realteil und Imaginärteil]]
+[[File:P_ID1130__Sig_Z_5_1_c.png|right|frame|Spectrum&nbsp; $X_{\rm A}(f)$&nbsp; of the sampled signal - real part and imaginary part.]]
 '''(3)'''&nbsp; Die Abtastrate beträgt nun&nbsp; $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.
-*Nur im Sonderfall des Cosinussignals ist jetzt das Abtasttheorem erfüllt und es gilt:
+*Only in the special case of the cosine signal is the sampling theorem now satisfied and it holds:
 :$$y_1(t) = x_1(t) &nbsp; &rArr; &nbsp; A_1 \; \underline{=2 \ \text{V}} \text{ und }\varphi_1 \; \underline{= 0}.$$
-Dieses Ergebnis soll nun noch mathematisch hergeleitet werden, wobei im Hinblick auf die noch anstehenden Teilaufgaben bereits auch eine Phase&nbsp; $\varphi$&nbsp; im Eingangssignal berücksichtigt wird:
+This result is now to be derived mathematically, whereby a phase&nbsp; $\varphi$&nbsp; in the input signal is already taken into account with regard to the remaining subtasks:
 :$$x(t) =  A \cdot \cos (2 \pi \cdot f_0 \cdot t - \varphi)
   \hspace{0.05cm}.$$
-*Dann gilt für die Spektralfunktion, die in der oberen Grafik skizziert ist:
+*Then, for the spectral function sketched in the graph above:
 :$$X(f) = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm e}^{{\rm j} \hspace{0.05cm}
   \cdot \hspace{0.05cm} \varphi} \cdot  \delta
@@ Line 122: / Line 123: @@
 *Die Fourierrücktransformation führt auf
-[[File:P_ID1131__Sig_Z_5_1_d.png|right|frame|Rekonstruktion des abgetasteten Sinussignals]]
+[[File:P_ID1131__Sig_Z_5_1_d.png|right|frame|Reconstruction of the sampled sinusoidal signal]]
 :$$y(t) =  A \cdot \cos (\varphi)\cdot \cos (2 \pi \cdot f_0 \cdot t )
   \hspace{0.05cm}.$$