Signal Representation/Discrete-Time Signal Representation - Revision history

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	{{Display}}		{{Display}}
			[[de:Signaldarstellung/Zeitdiskrete_Signaldarstellung]]

Guenter at 12:51, 22 June 2023

2023-06-22T12:51:21Z

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	*At the output of an ideal transmission system,  the  likewise discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.		*At the output of an ideal transmission system,  the  likewise discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.

	*The question now is how the block  ~~$\text{~~signal reconstruction}$  is to be designed so that  $y(t) = x(t)$  applies.		*The question now is how the block  »'''signal reconstruction'''«  is to be designed so that  $y(t) = x(t)$  applies.

Guenter at 12:48, 22 June 2023

2023-06-22T12:48:21Z

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	<br>		<br>
	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Theorem:}$  Developing the  »Dirac comb«  into a  [[Signal_Representation/Fourier_Series\|»Fourier series&~~raquo~~;]]  and transforming it into the frequency domain using the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|»shifting theorem«]]  gives the following Fourier correspondence:		$\text{Theorem:}$  Developing the  »Dirac comb«  into a  [[Signal_Representation/Fourier_Series\|»Fourier series«]]   and transforming it into the frequency domain using the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|»shifting theorem«]]  gives the following Fourier correspondence:

	:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot		:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot

Guenter at 12:38, 22 June 2023

2023-06-22T12:38:35Z

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	==Sampling theorem==		==Sampling theorem==
	<br>		<br>
	The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the source signal ~~has been chosen correctly~~.		The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  has been chosen correctly corresponding to the bandwidth  $B_{\rm NF}$  of the source signal.

	From the graph in the  [[Signal_Representation/Discrete-Time_Signal_Representation#Signal_reconstruction\|"last section"]] , it can be seen that the following condition must be fulfilled:		From the graph in the  [[Signal_Representation/Discrete-Time_Signal_Representation#Signal_reconstruction\|»last section«]] , it can be seen that the following condition must be fulfilled:

	:$$f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$$		:$$f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$$

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Sampling Theorem:}$  If an analog signal  $x(t)$  has spectral components in the range  $\vert f \vert < B_{\rm NF}$, it only can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  if the sampling rate is sufficiently large:		$\text{Sampling Theorem:}$

			*If an analog signal  $x(t)$  has spectral components in the range  $\vert f \vert < B_{\rm NF}$,  it only can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  if the sampling rate at the transmitter was sufficiently large:
	:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$		:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$
			*Accordingly, the following must apply to the distance between two samples:
	Accordingly, the following must apply to the distance between two samples:

	:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}		:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}


	If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,		If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,  then,  in order to reconstruct the analog signal from its sampled values,  one must use an ideal,  rectangular low-pass filter with cut-off frequency
	*then,  in order to reconstruct the analog signal from its sampled values,		:$$f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A}).$$
	*one must use an ideal, rectangular low-pass filter with cut-off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$~~ .~~


	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 3:}$  The graph ~~above~~ shows the spectrum  $X(f)$  of an analog signal limited to  $\pm\text{ 5 kHz}$.  Below you see the spectrum  $X_{\rm A}(f)$  of the sampled signal with  ~~$T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.~~		$\text{Example 3:}$  The upper graph shows the spectrum  $X(f)$  of an analog signal limited to  $\pm\text{ 5 kHz}$.  Below you see the spectrum  $X_{\rm A}(f)$  of the sampled signal with
	[[File:P_ID1125__Sig_T_5_1_S6_neu.png\|right\|frame\|Sampling theorem in the frequency domain]]		[[File:P_ID1125__Sig_T_5_1_S6_neu.png\|right\|frame\|Sampling theorem in the frequency domain]]
	~~<br>~~Additionally drawn is the frequency response  $H(f)$  of the low-pass filter for signal reconstruction~~, whose~~ cut-off frequency must be ~~ ~~ $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$.		:$$T_{\rm A} =\,\text{ 100 µs}   \ ⇒ \   f_{\rm A}=\,\text{ 10 kHz}.$$
			Additionally drawn is the frequency response  $H(f)$  of the low-pass filter for signal reconstruction.  The cut-off frequency must be
			:$$f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}.$$

			With any other  $f_{\rm G}$ value, the result would be  $Y(f) \neq X(f)$:
			#For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ parts are missing.
			# For  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$  due to convolution operations.


	*With any other  $f_{\rm G}$ value, the result would be  $Y(f) \neq X(f)$.		If the sampling at the transmitter had been done with sampling rate
	*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ parts are missing.		:$$f_{\rm A} < 10\,\text{ kHz}   \ ⇒ \   T_{\rm A} >100 \,{\rm µ s},$$
	* For  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution operations.
	~~<br clear=all>~~
	If the sampling at the transmitter had been done with a sampling rate~~ ~~ $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \,{\rm µ s}$, ~~the analog signal ~~ $~~y(t) = x(t)~~$~~  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}~~

			the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case.

	''Note'':   There is an interactive applet on the topic covered here:   [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction\|"Sampling of Analog Signals and Signal Reconstruction"]]		<u>Note:</u>   There is an interactive applet on the topic covered here:   [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction\|»Sampling of Analog Signals and Signal Reconstruction«]].}}

Guenter at 11:58, 22 June 2023

2023-06-22T11:58:54Z

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	==Frequency domain representation==		==Frequency domain representation==
	<br>		<br>
	The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain\|"Convolution Theorem in the frequency domain"]].  This states that multiplication in the time domain corresponds to convolution in the spectral domain:		The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain\|»Convolution Theorem in the frequency domain«]].  This states that multiplication in the time domain corresponds to convolution in the spectral domain:

	:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}		:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
	X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$		X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

	From the spectrum  $X(f)$  by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$  we get:		*From the spectrum  $X(f)$  by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$  we get:

	:$$X(f) \star \delta		:$$X(f) \star \delta
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	)\hspace{0.05cm}.$$		)\hspace{0.05cm}.$$

	Applying this result to all Dirac delta lines of the Dirac comb, we finally obtain:		*Applying this result to all Dirac delta lines of the Dirac comb,  we finally obtain:

	:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta		:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Conclusion:}$  The sampling of the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  leads in the spectral domain to a  ~~$\text{~~periodic continuation}$  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$. }}		$\text{Conclusion:}$  The sampling of the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  leads in the spectral domain to a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$. }}


	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 2:}$		$\text{Example 2:}$
	The upper graph shows  (schematically!)  the spectrum  $X(f)$  of an analog signal  $x(t)$, which includes frequencies up to  $5 \text{ kHz}$ .		The upper graph shows  $($schematically!$)$  the spectrum  $X(f)$  of an analog signal  $x(t)$,  which includes frequencies up to  $5 \text{ kHz}$ .

	[[File:P_ID1122__Sig_T_5_1_S4_neu.png\|right\|frame\|Spectrum of the sampled signal]]		[[File:P_ID1122__Sig_T_5_1_S4_neu.png\|right\|frame\|Spectrum of the sampled signal]]

	Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e. at the respective distance  $T_{\rm A}\, = {\rm 50 \, µs}$  we obtain the periodic spectrum  $X_{\rm A}(f)$  sketched below.
	*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high-frequency components.		Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$,  i.e. at the respective distance  $T_{\rm A}\, = {\rm 50 \, µs}$  we obtain the periodic spectrum  $X_{\rm A}(f)$  sketched below.
	*Accordingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}
			*Since the Dirac delta functions are infinitely narrow,  the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high-frequency components.

			*Accordingly,  the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}


	==Signal reconstruction==		==Signal reconstruction==
	<br>		<br>
	Signal sampling is not an end in itself in a digital transmission system; it must be reversed at some point.  Consider, for example, the following system:		Signal sampling is not an end in itself in a digital transmission system;  it must be reversed at some point.  Consider,  for example,  the following system:

			[[File:EN_Sig_T_5_1_S2_v2.png\|right\|frame\|Sampling and reconstruction of a signal]]

	[[File:~~EN_Sig_T_5_1_S2_v2~~.png\|~~left~~\|frame\|~~Sampling and reconstruction~~ of a signal]]		[[File:P_ID1124__Sig_T_5_1_S5b_neu.png\|right\|frame\|Frequency domain representation of the signal reconstruction process]]

	*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.		*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
	*At the output of an ideal transmission system, the likewise discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
			*At the output of an ideal transmission system,  the  likewise discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.

	*The question now is how the block  $\text{signal reconstruction}$  is to be designed so that  $y(t) = x(t)$  applies.		*The question now is how the block  $\text{signal reconstruction}$  is to be designed so that  $y(t) = x(t)$  applies.
	~~<br clear=all>~~
	The solution is simple if one considers the spectral functions:  One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass with  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function\|~~$\text{~~frequency response}$]]  $H(f)$, which
	~~[[File:P_ID1124__Sig_T_5_1_S5b_neu.png\|right\|frame\|Frequency domain representation of the signal reconstruction process]]~~		The solution is simple if one considers the spectral functions:

			One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass with  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function\|»frequency response«]]  $H(f)$, which
	*passes the low frequencies unaltered:		*passes the low frequencies unaltered:
	:$$H(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} \|f\| \le B_{\rm		:$$H(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} \|f\| \le B_{\rm
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	NF}\hspace{0.05cm}.$$		NF}\hspace{0.05cm}.$$

	Further it can be seen from the graph that the frequency response  $H(f)$  can be arbitrarily shaped in the range of  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$,  as long as both of the above conditions are met,		Further it can be seen from the graph that the frequency response  $H(f)$  can be arbitrarily shaped in the range of  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$,  <br>as long as both of the above conditions are met,
	*~~for example~~, linearly sloping (dashed line)		*e.g.,  linearly sloping  $($dashed line$)$
	*or ~~also~~ rectangular-in-frequency.		*or rectangular-in-frequency.

Guenter at 10:26, 22 June 2023

2023-06-22T10:26:49Z

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Guenter at 09:40, 22 June 2023

2023-06-22T09:40:34Z

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	*Let the continuous-time signal be  $x(t)$.  Let the signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.		*Let the continuous-time signal be  $x(t)$.  Let the signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.

	*Let the ~~control~~ variable  $\nu$  of the sample be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers\|$\text{integer}$]]:		*Let the run variable  $\nu$  of the sample be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers\|$\text{integer}$]]:
	:$$\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} .$$		:$$\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} .$$
	*Outside the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) = 0$.		*Outside the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) = 0$.  At the equidistant sampling times,  the result is with the constant  $K$:

	*At the equidistant sampling times with the constant $K$~~, the result is~~:

	:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$		:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$
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	<br>		<br>
	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Definition:}$  ~~Here,~~&~~nbsp~~; ~~$\text{"~~sampling~~"}$~~  shall be understood as the multiplication of the continuous-time signal  $x(t)$  by the  Dirac comb  $p_{\delta}(t)$:		$\text{Definition:}$  »'''sampling'''«  shall be understood as the multiplication of the continuous-time signal  $x(t)$  by the  »Dirac comb«  $p_{\delta}(t)$:

	:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$}}		:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$}}


	It should be noted that other description forms are found in the literature. However, to the authors, the form chosen here appears to be the most appropriate in terms of spectral representation and derivation of the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|$\text{Discrete Fourier Transform}$]]  $\rm (DFT)$.		It should be noted that other description forms are found in the literature.  However,  to the authors,  the form chosen here appears to be the most appropriate in terms of spectral representation and derivation of the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|$\text{Discrete Fourier Transform}$]]  $\rm (DFT)$.

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=

Guenter at 09:27, 22 June 2023

2023-06-22T09:27:13Z

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	A prerequisite for the system-theoretical investigation of digital systems or for their computer simulation is a suitable discrete-time signal description.		A prerequisite for the system-theoretical investigation of digital systems or for their computer simulation is a suitable discrete-time signal description.

	This chapter clarifies the mathematical transition from continuous-time to discrete-time signals,  starting from the  [[Signal_Representation/~~Fourier_Transform_and_Its_Inverse~~\|»Fourier transform theorems«]].		This chapter clarifies the mathematical transition from continuous-time to discrete-time signals,  starting from the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse\|»Fourier transform theorems«]].

	The chapter includes in detail:		The chapter includes in detail:

Guenter at 09:25, 22 June 2023

2023-06-22T09:25:51Z

← Older revision		Revision as of 11:25, 22 June 2023
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	A prerequisite for the system-theoretical investigation of digital systems or for their computer simulation is a suitable discrete-time signal description.		A prerequisite for the system-theoretical investigation of digital systems or for their computer simulation is a suitable discrete-time signal description.

	This chapter clarifies the mathematical transition from continuous-time to discrete-time signals, starting from the  [[Signal_Representation/Fourier_Transform_and_Its_Inverse\|~~$\text{~~Fourier transform theorems}$]].		This chapter clarifies the mathematical transition from continuous-time to discrete-time signals,  starting from the  [[Signal_Representation/Fourier_Transform_and_Its_Inverse\|»Fourier transform theorems«]].

	The chapter includes in detail:		The chapter includes in detail:
	#The  »time and frequency domain representation«  of discrete-time signals,		#The  »time and frequency domain representation«  of discrete-time signals,
	#the  »sampling theorem«, which must be strictly observed in time discretization,		#the  »sampling theorem«,  which must be strictly observed in time discretization,
	#the  »reconstruction of the analog signal«  from the discrete-time representation,		#the  »reconstruction of the analog signal«  from the discrete-time representation,
	#the  »~~Discrete~~ Fourier ~~Transform~~«  $\rm (DFT)$  and its inverse  $\rm (IDFT)$,		#the  »discrete Fourier transform«  $\rm (DFT)$  and its inverse  $\rm (IDFT)$,
	#the  »possibilities for error«  when applying DFT and IDFT,		#the  »possibilities for error«  when applying DFT and IDFT,
	#the application of  »spectral analysis«  to the improvement of metrological procedures, and		#the application of  »spectral analysis«  to the improvement of metrological procedures,  and
	#the  »FFT algorithm«  particularly suitable for computer implementation.		#the  »FFT algorithm«  particularly suitable for computer implementation.

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	==Principle and motivation==		==Principle and motivation==
	<br>		<br>
	Many source signals are analog and thus simultaneously  [[Signal_Representation/Signal_classification#Continuous-time_and_discrete-time_signals\|~~$\text{~~continuous-time}$]]  and  [[Signal_Representation/Signal_classification#Continuous-valued_and_discrete-valued_signals\|~~$\text{~~continuous}$]].  If such an analog signal is to be transmitted by means of a digital system, the following preprocessing steps are required:		Many source signals are analog and thus simultaneously  [[Signal_Representation/Signal_classification#Continuous-time_and_discrete-time_signals\|»continuous-time«]]  and  [[Signal_Representation/Signal_classification#Continuous-valued_and_discrete-valued_signals\|»continuous–valued«]].  If such an analog signal is to be transmitted by means of a digital system,  the following preprocessing steps are required:
	*the  ~~$\text{~~sampling}$  of the source signal  $x(t)$, which is expediently - but not necessarily - performed at equidistant times   ⇒   ~~$\text{~~time discretization~~}$,~~		*the  »'''sampling'''«  of the source signal  $x(t)$,  which is expediently – but not necessarily – performed at equidistant times   ⇒   »time discretization«,
	*the  $\text{quantization}$&~~nbsp~~; ~~of the samples~~, ~~so as to limit the number  $M$  of possible values to a finite value   ⇒   $\text{value discretization}$.~~

			*the  »'''quantization'''«  of the samples,  so as to limit the number  $M$  of possible values to a finite value   ⇒   »value discretization«.

	Quantization is not discussed in detail until the chapter  [[Modulation_Methods/Pulscodemodulation\|"Pulse Code Modulation"]]  of the book  "Modulation Methods".
			Quantization is not discussed in detail until the chapter  [[Modulation_Methods/Pulscodemodulation\|»Pulse Code Modulation«]]  of the book  "Modulation Methods".

	[[File:P_ID1120__Sig_T_5_1_S1_neu.png\|right\|frame\|On time discretization of the continuous-time signal  $x(t)$]]		[[File:P_ID1120__Sig_T_5_1_S1_neu.png\|right\|frame\|On time discretization of the continuous-time signal  $x(t)$]]

	In the following, we use the following nomenclature to describe the sampling:		In the following,  we use the following nomenclature to describe the sampling:
	*Let the continuous-time signal be  $x(t)$.		*Let the continuous-time signal be  $x(t)$.  Let the signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
	*Let the ~~sampled~~ signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
	*Let the ~~run~~ variable  $\nu$  of the sample be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers\|$\text{integer}$]]:		*Let the control variable  $\nu$  of the sample be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers\|$\text{integer}$]]:
	:$$\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} .$$		:$$\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} .$$
	*Outside the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) = 0$.		*Outside the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) = 0$.