Aufgaben:Exercise 5.1Z: Sampling of Harmonic Oscillations - 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.~~1Z_Zeitdiskrete_Harmonische~~]]		[[de:Aufgaben:Aufgabe 5.1Z: Abtastung harmonischer Schwingungen]]

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	The signal reconstruction at the receiver takes place through a low-pass filter  $H(f)$,  which forms the signal  $y_{\rm A}(t) = x_{\rm A}(t)$  from the sampled signal  $y(t)$ . It applies:		The signal reconstruction at the receiver takes place through a low-pass filter  $H(f)$,  which forms the signal  $y_{\rm A}(t) = x_{\rm A}(t)$  from the sampled signal  $y(t)$ . It applies:
	:$$H(f) = \left\{ \begin{array}{c} 1 \\ 0.5 \\		:$$H(f) = \left\{ \begin{array}{c} 1 \\ 0.5 \\0 \\ \end{array} \right.\quad\begin{array}{{5}c} {\rm{{\rm{f\ddot{u}r}}}}\\ {\rm{{\rm{f\ddot{u}r}}}} \\ {\rm{{\rm{f\ddot{u}r}}}} \\\end{array}\begin{array}{{5}c}\|f\| < f_{\rm G} \hspace{0.05cm}, \\\|f\| = f_{\rm G} \hspace{0.05cm}, \\\|f\| > f_{\rm G} \hspace{0.05cm}, \\\end{array}$$
	0 \\ \end{array} \right.\quad
	\begin{array}{*{5}c} {\rm{{\rm{f\ddot{u}r}}}}
	\\ {\rm{{\rm{f\ddot{u}r}}}} \\ {\rm{{\rm{f\ddot{u}r}}}} \\
	\end{array}\begin{array}{*{5}c}
	\|f\| < f_{\rm G} \hspace{0.05cm}, \\
	\|f\| = f_{\rm G} \hspace{0.05cm}, \\
	\|f\| > f_{\rm G} \hspace{0.05cm}, \\
	\end{array}$$
	Here  $f_{\rm G}$  indicates the cut-off frequency of the rectangular low-pass filter.  For this shall apply:		Here  $f_{\rm G}$  indicates the cut-off frequency of the rectangular low-pass filter.  For this shall apply:
	:$$f_{\rm G} = \frac{1}{ 2 \cdot T_{\rm A}}\hspace{0.05cm}.$$		:$$f_{\rm G} = \frac{1}{ 2 \cdot T_{\rm A}}\hspace{0.05cm}.$$

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	This result is now to be derived mathematically, whereby a phase  $\varphi$  in the input signal is already taken into account with regard to the remaining subtasks:		This result is now to be derived mathematically, whereby a phase  $\varphi$  in the input signal is already taken into account with regard to the remaining subtasks:
	:$$x(t) = A \cdot \cos (2 \pi \cdot f_0 \cdot t - \varphi)		:$$x(t) = A \cdot \cos (2 \pi \cdot f_0 \cdot t - \varphi)\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	*Then, for the spectral function sketched in the graph above:		*Then, for the spectral function sketched in the graph above:
	:$$X(f) = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm e}^{{\rm j} \hspace{0.05cm}		:$$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(f+ f_{\rm 0} ) + {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \varphi} \cdot \delta(f- f_{\rm 0} )\hspace{0.05cm}.$$
	\cdot \hspace{0.05cm} \varphi} \cdot \delta
	(f+ f_{\rm 0} ) + {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm} {\rm e}^{-{\rm j} \hspace{0.05cm}
	\cdot \hspace{0.05cm} \varphi} \cdot \delta
	(f- f_{\rm 0} )\hspace{0.05cm}.$$
	*With the abbreviations		*With the abbreviations
	:$$R = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm}		:$$R = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm}\cos(\varphi) \hspace{0.5cm}{\rm and} \hspace{0.5cm}I ={A}/{2} \hspace{0.05cm} \cdot\hspace{0.05cm} \sin(\varphi)$$
	\cos(\varphi) \hspace{0.5cm}{\rm and} \hspace{0.5cm}I ={A}/{2} \hspace{0.05cm} \cdot
	\hspace{0.05cm} \sin(\varphi)$$
	:can also be written for this:		:can also be written for this:
	:$$X(f) = (R + {\rm j} \cdot I) \cdot \delta		:$$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}.$$
	(f+ f_{\rm 0} ) + (R - {\rm j} \cdot I) \cdot \delta
	(f- f_{\rm 0} )\hspace{0.05cm}.$$
	*The spectrum of the signal  $x_{\rm A}(t)$  sampled with  $f_{\rm A} = 2f_0$  is thus:		*The spectrum of the signal  $x_{\rm A}(t)$  sampled with  $f_{\rm A} = 2f_0$  is thus:
	:$$X_{\rm A}(f) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}		:$$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}.$$
	)= \sum_{\mu = - \infty }^{+\infty} X (f- 2\mu \cdot f_{\rm 0}
	)\hspace{0.05cm}.$$
	:*The bottom graph shows that  $X_{\rm A}(f)$  consists of Dirac functions at  $\pm f_0$,  $\pm 3f_0$,  $\pm 5f_0$,  and so on.		:*The bottom graph shows that  $X_{\rm A}(f)$  consists of Dirac functions at  $\pm f_0$,  $\pm 3f_0$,  $\pm 5f_0$,  and so on.
	:*All weights are purely real and equal to  $2 \cdot R$.		:*All weights are purely real and equal to  $2 \cdot R$.
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	*If one further takes into account the rectangular low-pass filter whose cut-off frequency lies exactly at  $f_{\rm G} = f_0$,  as well as   $H(f_{\rm G}) = 0.5$, 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  $f_{\rm G} = f_0$,  as well as   $H(f_{\rm G}) = 0.5$, one obtains for the spectrum after signal reconstruction:
	:$$Y(f) = R \cdot \delta		:$$Y(f) = R \cdot \delta(f+ f_{\rm 0} ) + R \cdot \delta(f- f_{\rm 0} )\hspace{0.05cm}, \hspace{0.5cm} R = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm}\cos(\varphi)\hspace{0.05cm}.$$
	(f+ f_{\rm 0} ) + R \cdot \delta
	(f- f_{\rm 0} )\hspace{0.05cm}, \hspace{0.5cm} R = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm}
	\cos(\varphi)\hspace{0.05cm}.$$

	*The inverse Fourier transformation leads to		*The inverse Fourier transformation leads to
	[[File:P_ID1131__Sig_Z_5_1_d.png\|right\|frame\|Reconstruction of the sampled sine signal]]		[[File:P_ID1131__Sig_Z_5_1_d.png\|right\|frame\|Reconstruction of the sampled sine signal]]
	:$$y(t) = A \cdot \cos (\varphi)\cdot \cos (2 \pi \cdot f_0 \cdot t )		:$$y(t) = A \cdot \cos (\varphi)\cdot \cos (2 \pi \cdot f_0 \cdot t )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Thus, a cosine-shaped progression results independent of the input phase  $\varphi$ .		*Thus, a cosine-shaped progression results independent of the input phase  $\varphi$ .
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	'''(5)'''  Despite   ⇒   $\varphi = 60^\circ$ gilt $\varphi_3 = 0$   ⇒   the reconstructed signal  $y_3(t)$  is cosine-shaped, too.		'''(5)'''  Despite   ⇒   $\varphi = 60^\circ$ gilt $\varphi_3 = 0$   ⇒   the reconstructed signal  $y_3(t)$  is cosine-shaped, too.
	*The amplitude is equal to		*The amplitude is equal to
	:$$A_3 = A \cdot \cos (60^{\circ})= {A}/{2} \hspace{0.15 cm}\underline{= 1\,{\rm V}}		:$$A_3 = A \cdot \cos (60^{\circ})= {A}/{2} \hspace{0.15 cm}\underline{= 1\,{\rm V}}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	*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.		*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.
	*Because, you don't know the turquoise curve.		*Because, you don't know the turquoise curve.

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	'''(3)'''  The sampling rate is now  $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.		'''(3)'''  The sampling rate is now  $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.
	*Only in the special case of the cosine signal the sampling theorem is satisfied, and it holds:		*Only in the special case of the cosine signal the sampling theorem is satisfied, and it holds:
	:$$y_1(t) = x_1(t)   ⇒   A_1 \; \underline{=2 \ \text{V}} \text{ und }\varphi_1 \; \underline{= 0}.$$		:$$y_1(t) = x_1(t)   ⇒   A_1 \; \underline{=2 \ \text{V}} \text{und}\varphi_1 \; \underline{= 0}.$$

	This result is now to be derived mathematically, whereby a phase  $\varphi$  in the input signal is already taken into account with regard to the remaining subtasks:		This result is now to be derived mathematically, whereby a phase  $\varphi$  in the input signal is already taken into account with regard to the remaining subtasks:

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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:5.1Z_Zeitdiskrete_Harmonische]]

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

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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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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]]

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

2021-09-22T11:47:58Z

Text replacement - "analyse" to "analyze"

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	The oscillation parameters  $f_0$  and  $A$  can be taken from the graph.		The oscillation parameters  $f_0$  and  $A$  can be taken from the graph.

	It is assumed that the signals are sampled equidistantly at times  $\nu \cdot T_{\rm A}$  whereby the parameter values  $T_{\rm A} = 80 \ µ \text{s}$  and  $T_{\rm A} = 100 \ µ \text{s}$  are to be ~~analysed~~.		It is assumed that the signals are sampled equidistantly at times  $\nu \cdot T_{\rm A}$  whereby the parameter values  $T_{\rm A} = 80 \ µ \text{s}$  and  $T_{\rm A} = 100 \ µ \text{s}$  are to be analyzed.

	The signal reconstruction at the receiver takes place through a low-pass filter  $H(f)$,  which forms the signal  $y_{\rm A}(t) = x_{\rm A}(t)$  from the sampled signal  $y(t)$ . It applies:		The signal reconstruction at the receiver takes place through a low-pass filter  $H(f)$,  which forms the signal  $y_{\rm A}(t) = x_{\rm A}(t)$  from the sampled signal  $y(t)$ . It applies:

Guenter at 14:44, 24 May 2021

2021-05-24T14:44:10Z

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

Guenter at 14:37, 13 May 2021

2021-05-13T14:37:23Z

← Older revision		Revision as of 16:37, 13 May 2021
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	[[File:P_ID1130__Sig_Z_5_1_c.png\|right\|frame\|Spectrum  $X_{\rm A}(f)$  of the sampled signal - real part and imaginary part.]]		[[File:P_ID1130__Sig_Z_5_1_c.png\|right\|frame\|Spectrum  $X_{\rm A}(f)$  of the sampled signal - real part and imaginary part.]]
	'''(3)'''  The sampling rate is now  $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.		'''(3)'''  The sampling rate is now  $f_{\rm A} = 2 \cdot f_0 = 10 \ \text{kHz}$.
	*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:
	:$$y_1(t) = x_1(t)   ⇒   A_1 \; \underline{=2 \ \text{V}} \text{ und }\varphi_1 \; \underline{= 0}.$$		:$$y_1(t) = x_1(t)   ⇒   A_1 \; \underline{=2 \ \text{V}} \text{ und }\varphi_1 \; \underline{= 0}.$$


	This result is now to be derived mathematically, whereby a phase  $\varphi$  in the input signal is already taken into account with regard to the remaining subtasks:		This result is now to be derived mathematically, whereby a phase  $\varphi$  in the input signal is already taken into account with regard to the remaining subtasks:
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	*With the abbreviations		*With the abbreviations
	:$$R = {A}/{2} \hspace{0.05cm} \cdot \hspace{0.05cm}		:$$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)$$		\hspace{0.05cm} \sin(\varphi)$$
	:can also be written for this:		:can also be written for this:
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	(f+ f_{\rm 0} ) + (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}.$$		(f- f_{\rm 0} )\hspace{0.05cm}.$$
	*The spectrum of the signal  $f_{\rm A} ~~= 2f_0~~$  sampled with  $x_{\rm A}~~(t)~~$  is thus:		*The spectrum of the signal  $x_{\rm A}(t)$  sampled with  $f_{\rm A} = 2f_0$  is thus:
	:$$X_{\rm A}(f) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}		:$$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}		)= \sum_{\mu = - \infty }^{+\infty} X (f- 2\mu \cdot f_{\rm 0}
	)\hspace{0.05cm}.$$		)\hspace{0.05cm}.$$
	:*The bottom graph shows that  $X_{\rm A}(f)$  consists of Dirac functions at  $\pm f_0$,  $\pm 3f_0$,  $\pm 5f_0$,  , and so on.		:*The bottom graph shows that  $X_{\rm A}(f)$  consists of Dirac functions at  $\pm f_0$,  $\pm 3f_0$,  $\pm 5f_0$,  and so on.
	:*All weights are purely real and equal to  $2 \cdot R$.		:*All weights are purely real and equal to  $2 \cdot R$.
	:*The imaginary parts of the periodically continued spectrum cancel out.		:*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  $f_{\rm G} = f_0$ , as well as   $H(f_{\rm G}) = 0.5$, 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  $f_{\rm G} = f_0$,  as well as   $H(f_{\rm G}) = 0.5$, one obtains for the spectrum after signal reconstruction:
	:$$Y(f) = R \cdot \delta		:$$Y(f) = R \cdot \delta
	(f+ f_{\rm 0} ) + R \cdot \delta		(f+ f_{\rm 0} ) + R \cdot \delta
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	\cos(\varphi)\hspace{0.05cm}.$$		\cos(\varphi)\hspace{0.05cm}.$$

	*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]]
	:$$y(t) = A \cdot \cos (\varphi)\cdot \cos (2 \pi \cdot f_0 \cdot t )		:$$y(t) = A \cdot \cos (\varphi)\cdot \cos (2 \pi \cdot f_0 \cdot t )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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	[[File:P_ID1133__Sig_Z_5_1_e.png\|right\|frame\|Reconstruction of a harmonic oscillation with   $60^\circ$ phase]]		[[File:P_ID1133__Sig_Z_5_1_e.png\|right\|frame\|Reconstruction of a harmonic oscillation with   $60^\circ$ phase]]
	'''(5)'''  Despite  $\varphi = 60^\circ$ gilt $\varphi_3 = 0$   ⇒   the reconstructed signal  $y_3(t)$ ~~ist cosinusförmig.~~ is ~~also~~ cosine-shaped. The amplitude is equal to		'''(5)'''  Despite   ⇒   $\varphi = 60^\circ$ gilt $\varphi_3 = 0$   ⇒   the reconstructed signal  $y_3(t)$  is cosine-shaped, too.
			*The amplitude is equal to
	:$$A_3 = A \cdot \cos (60^{\circ})= {A}/{2} \hspace{0.15 cm}\underline{= 1\,{\rm V}}		:$$A_3 = A \cdot \cos (60^{\circ})= {A}/{2} \hspace{0.15 cm}\underline{= 1\,{\rm V}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*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}}		{{ML-Fuß}}